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1 | Page COOLING LOAD CALCULATIONS FOR HOSTEL BUILDING UET LAHORE NAROWAL CAMPUS A Project Submitted to fulfil the requirement of B.Sc. Mechanical Engineering degree at Department of Mechanical Engineering University of Engineering & Technology Lahore-Narowal Campus Under supervision and kind guidance of Engr. Aqib Hussain By JABIR ALI SIDDIQUE 2012-ME-519 RIZWAN MINHAS 2012-ME-508 MUHAMMAD QAMAR NAEEM 2012-ME-524

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COOLING LOAD CALCULATIONS FOR

HOSTEL BUILDING UET LAHORE

NAROWAL CAMPUS

A Project Submitted to fulfil the requirement of B.Sc. Mechanical Engineering degree at

Department of Mechanical Engineering

University of Engineering & Technology Lahore-Narowal Campus

Under supervision and kind guidance of

Engr. Aqib Hussain

By

JABIR ALI SIDDIQUE 2012-ME-519 RIZWAN MINHAS 2012-ME-508

MUHAMMAD QAMAR NAEEM 2012-ME-524

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Internal Examiner

NAME: ______________________________

SIGNATURE: _________________________

DATED: __________________________________

External Examiner

NAME: ______________________________

SIGNATURE: _________________________

DATED: __________________________________

PROJECT SUPERVISOR SIGNATURE: ______________________

Department of Mechanical Engineering- Narowal Campus

University of Engineering & Technology Lahore-PAKISTAN

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Abstract: This document represents the Final Year Project report for the “Cooling load Calculations of Hostel Building”, located in Narowal Campus of University of engineering & Technology Lahore. A well designed and adequate HVAC system is essential to maintaining the comfortable, productive and health living environment. The system is being designed to meet the minimum ASHRAE and building code standards. Narowal campus was inaugurated in 2012 and the campus building is under construction consisting of total area of 200 acres. The hostel building will host more than 450 students and is being built with total area of about 112,222 sq. feet. The building is to be built on concrete slab with masonry walls. The building comprises of three floors, named as Ground Floor, First Floor and Second Floor.

The baseline load calculations were manipulated for:

Outdoor/Indoor design conditions Building Components Ductwork conditions Ventilation/Infiltration conditions Worst Case Scenario (Combining All the safety Factors)

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ACKNOWLEDGMETS: We would like to express our sincere gratitude to Engr. Aqib Hussain for his solid efforts in initiation of this project and his extended approach to give us familiarity with basics of HVAC. Perhaps this motivation was really necessary for starting this Project. We would also like to extend our gratitude to Engr. Ijaz-Ul-Haq for his continuous support in the completion of our B.Sc. Mechanical Engineering degree. Perhaps it was not possible to successfully complete this task without his kind advice that was available at every moment.

Finally, we would like to honor our parents for their continuous support throughout our academic career. Perhaps they are a true source of inspiration for us.

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Table of Contents Abstract: ................................................................................................................................................. 3

ACKNOWLEDGMETS: .............................................................................................................................. 4

List of Figures: ....................................................................................................................................... 10

List of Tables: ........................................................................................................................................ 10

1.1 Introduction: ................................................................................................................................... 11

1.2 Overview:........................................................................................................................................ 11

1.2.1 Individual systems: ................................................................................................................... 11

1.2.2 District networks: ..................................................................................................................... 12

1.3 Heating: .......................................................................................................................................... 12

1.3.1 Generation: .............................................................................................................................. 12

1.3.2 Distribution: ............................................................................................................................. 12

1.3.2(a) Water/steam: .................................................................................................................... 12

1.3.2(b) Air: .................................................................................................................................... 12

1.4 Ventilation: ..................................................................................................................................... 13

1.4.1 Mechanical or forced ventilation: ............................................................................................ 13

1.4.2 Natural ventilation: .................................................................................................................. 13

1.5 Air conditioning: ............................................................................................................................. 14

1.5.1 Refrigeration cycle: .................................................................................................................. 15

1.5.2 Free cooling: ............................................................................................................................. 15

1.5.3 Central vs. split system:............................................................................................................ 15

1.5.4 Dehumidification: .................................................................................................................... 16

2 Purpose: ............................................................................................................................................. 16

2.1 Problem Definition: ..................................................................................................................... 16

2.2 Scope: ......................................................................................................................................... 16

3 Assumptions: ..................................................................................................................................... 17

4 Zoning the Office: .............................................................................................................................. 17

4.1 Introduction to Zoning: ............................................................................................................... 17

4.2 Need of Zoning: ........................................................................................................................... 17

4.3 Final Zoning Choice: .................................................................................................................... 18

5 Space heat gain: ................................................................................................................................. 18

5.1 The method of how the heat enters the space:........................................................................... 18

5.2 Sensible heat: .............................................................................................................................. 18

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5.3 The factors involve the sensible heat load: ................................................................................. 18

5.4 Latent Heat Loads: ...................................................................................................................... 19

5.4.1 The factors involve the sensible heat load: ............................................................................. 19

5.5 Space heat gain and cooling load (heat storage effect): .............................................................. 19

5.6 Space cooling and cooling load (Coil): ......................................................................................... 20

5.7 Components of cooling load:....................................................................................................... 20

6 CLTD/SCL/CLF METHOD OF LOAD CALCULATION: .............................................................................. 22

6.1 External Cooling Load: ................................................................................................................. 22

6.1.1 Roof: ..................................................................................................................................... 22

6.1.2 Walls: .................................................................................................................................... 23

6.1.3 Solar load through glass: ........................................................................................................ 23

6.1.4 Partition, ceilings and floors: .................................................................................................. 25

6.2 Internal Cooling Loads: ................................................................................................................ 25

6.2.1 People: .................................................................................................................................. 26

6.2.2 Lights: ................................................................................................................................... 27

6.2.3 Appliances: ............................................................................................................................ 28

6.2.4 Infiltration Air: ....................................................................................................................... 28

Excel Calculations ................................................................................................................................. 29

Zone 1: Residential Rooms .................................................................................................................... 29

Zone 2: Toilets ...................................................................................................................................... 33

Zone 3: Main Dining Hall & Kitchen ...................................................................................................... 35

Zone 4: TV Lounge ................................................................................................................................ 35

Manual Calculations ............................................................................................................................. 36

7 Zone1(Residential Rooms): ................................................................................................................ 36

7.1 Specifications: ............................................................................................................................. 36

7.2 Conduction through exterior surfaces: ........................................................................................ 36

7.3 Conduction through Walls:.......................................................................................................... 36

Room 1 & 27 ......................................................................................................................................... 37

7.4 Solar radiation through Windows Glass: ..................................................................................... 37

7.5 Conduction through interior surfaces: ........................................................................................ 38

7.6 Conduction through Lightings: .................................................................................................... 38

7.7 Conduction through Task Lightings and Bracket Fans: ................................................................ 38

7.8 Conduction through electrical equipment: .................................................................................. 38

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7.9 Conduction through People: ....................................................................................................... 39

7.10 Infiltration Load: ....................................................................................................................... 39

Total Cooling load of Each Room 1 & 27: .............................................................................................. 39

Room 2, 3, 17, 18, 24, 25 & 26 .............................................................................................................. 40

Total Cooling load of each Room 2, 3, 17, 18, 24, 25 & 26: .................................................................... 42

Room 4 ................................................................................................................................................. 42

Total Cooling load of Room 4: ............................................................................................................... 45

Room 5 ................................................................................................................................................. 45

Total Cooling load of Room 5: ............................................................................................................... 47

Room 6, 7, 37, 38 & 39 .......................................................................................................................... 47

Total Cooling load of Room 6, 7, 37, 38 & 39: ....................................................................................... 50

Room 8 & 36 ......................................................................................................................................... 50

Total Cooling load of Room 8 & 36:....................................................................................................... 52

Room 9 & 35 ......................................................................................................................................... 53

Total Cooling load of each Room 9 & 35: .............................................................................................. 55

Room 10, 11, 12, 33 & 34 ...................................................................................................................... 55

Total Cooling load of each Room 10, 11, 12, 33 & 34: ........................................................................... 58

Room 13, 14, 15, 21, 22, 23,29 & 30 ...................................................................................................... 58

Total Cooling load of each Room 13, 14, 15, 21, 22, 23,29 & 30: ........................................................... 60

Room 16, 28 & 20(Electric Room) ......................................................................................................... 60

Total Cooling load of each Room 16 28 & 20:........................................................................................ 63

Room 19 (Warden Room) ..................................................................................................................... 63

Total Cooling load of Room 19: ............................................................................................................. 65

Room 31 ............................................................................................................................................... 66

Total Cooling load of Room 31: ............................................................................................................. 68

Room 32 ............................................................................................................................................... 68

Total Cooling load of Room 32: ............................................................................................................. 71

Room 41, 42, 56, 57, 63, 64 & 65........................................................................................................... 71

Total Cooling load of each Room 41, 42, 56, 57, 63, 64 & 65 ................................................................. 73

Room 58 (Warden Room) ..................................................................................................................... 73

Total Cooling load of Room 58: ............................................................................................................. 76

8 Zone2(Toilets): ................................................................................................................................... 77

8.1 Specifications: ............................................................................................................................. 77

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8.2 Conduction through exterior surfaces: ........................................................................................ 77

8.3 Conduction through Walls:.......................................................................................................... 77

Toilet 1, 2, 9, 10, 13, 14, 22, 23, 30, 31, 34 & 35 .................................................................................... 78

8.4 Solar radiation through Windows Glass: ..................................................................................... 78

8.5 Conduction through interior surfaces: ........................................................................................ 79

8.6 Conduction through Lightings: .................................................................................................... 79

8.7 Conduction through Task Lightings and Bracket Fans: ................................................................ 79

8.8 Conduction through electrical equipment: .................................................................................. 79

8.9 Conduction through People: ....................................................................................................... 79

Total Cooling load of each Toilet 1, 2, 9, 10, 13, 14, 22, 23, 30, 31, 34 & 35: ......................................... 79

Toilet 3, 4, 15, 16, 24, 25, 36 & 37 ......................................................................................................... 80

Total Cooling load of each Toilet 3, 4, 15, 16, 24, 25, 36 & 37: .............................................................. 81

Toilet 5, 6, 17, 18, 26, 27, 38 & 39 ......................................................................................................... 82

Total Cooling load of each Toilet 5, 6, 17, 18, 26, 27, 38 & 39: .............................................................. 83

Toilet 7, 8, 11, 12, 19, 20, 28, 29, 32, 33, 40 & 41 .................................................................................. 84

Total Cooling load of each Toilet 7, 8, 11, 12, 19, 20, 28, 29, 32, 33, 40 & 41: ........................................ 85

9 Zone3(Main Dining Hall & Kitchen) .................................................................................................... 86

9.1 Specifications: ............................................................................................................................. 86

9.2 Conduction through exterior surfaces: ........................................................................................ 86

9.3 Conduction through Walls:.......................................................................................................... 86

9.4 Solar radiation through Windows Glass: ..................................................................................... 87

9.5 Conduction through interior surfaces: ........................................................................................ 88

9.6 Conduction through Lightings: .................................................................................................... 88

9.7 Conduction through Task Lightings and Bracket Fans: ................................................................ 88

9.8 Conduction through electrical equipment: .................................................................................. 88

9.9 Conduction through People: ....................................................................................................... 88

Total Cooling load of Main Kitchen: ...................................................................................................... 89

Main Dining Hall ................................................................................................................................... 89

Total Cooling load of Main Dining Hall: ................................................................................................. 92

10 Zone4(TV Lounge) ............................................................................................................................ 93

10.1 Specifications: ........................................................................................................................... 93

10.2 Conduction through exterior surfaces: ...................................................................................... 93

10.3 Conduction through Walls: ........................................................................................................ 93

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10.4 Solar radiation through Windows Glass: ................................................................................... 94

10.5 Conduction through interior surfaces: ...................................................................................... 95

10.6 Conduction through Lightings: .................................................................................................. 95

10.7 Conduction through Task Lightings and Bracket Fans: .............................................................. 95

10.8 Conduction through electrical equipment: ................................................................................ 95

10.9 Conduction through People: ..................................................................................................... 96

Total Cooling load of TV Lounge: .......................................................................................................... 96

Total Cooling load of Hostel Building .................................................................................................... 97

Conclusion: ........................................................................................................................................... 98

References: ........................................................................................................................................... 99

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List of Figures: Figure 1: Residential HVAC Design Process……………………………………………………………………………………10

Figure 2: Difference b/w space heat and space cooling load………………………………………………………….15

Figure 3: Solar Heat Gain for Lights……………………………………………………………………………………………….16

Figure 4: External Loads and internal Loads…………………………………………………………………………..….….16

Figure 5: Heat Gain Locations………………………………………………………………………………………………….…...17

Figure 6: Winter and Summer Comfort Zone………………………………………………………………………………….18

List of Tables:

Table 1: Cooling Load Temperature Difference for Roof and External Walls (Dark)……………………….19

Table 2: Cooling Load Factor for Window Glass with Indoor Shading Devices………………………………..20

Table 3: Maximum Solar Heat Gain Factor for Sunlit Glass on Average Cloudiness Days………………..21

Table 4: Heat Gain from Occupants at Various Activities……………………………………………………………….22

Table 5: Cooling Load Factors (CLF) for Lights………………………………………………………………………………..23

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1.1 Introduction: It was natural that the first HVAC controllers would be pneumatic since engineers understood fluid control. Thus, mechanical engineers could use their experience with the properties of steam and air to control the flow of heated or cooled air. After the control of air flow and temperature was standardized, the use of electromechanical relays in ladder logic to switch dampers became standardized. Eventually, the relays became electronic switches, as transistors eventually could handle greater current loads [1]. By 1985, pneumatic controls could no longer compete with this new technology although pneumatic control systems (sometimes decades old) are still common in many older buildings. By the year 2000, computerized controllers were common. Today, some of these controllers can even be accessed by web browsers, which need no longer be in the same building as the HVAC equipment. This allows some economies of scale, as a single operations center can easily monitor multiple buildings. [2]

HVAC (heating, ventilating, and air conditioning) is the technology of indoor and vehicular environmental comfort. Its goal is to provide thermal comfort and acceptable indoor air quality. HVAC system design is a sub discipline of mechanical engineering, based on the principles of thermodynamics, fluid mechanics, and heat transfer. HVAC is important in the design of medium to a large industrial and office buildings such as skyscrapers, onboard vessels, and in marine environments such as aquariums, where safe and healthy building conditions are regulated with respect to temperature and humidity, using fresh air from outdoors. Ventilating or ventilation is the process of exchanging or replacing air in any space to provide high indoor air quality which involves temperature control, oxygen replenishment, and removal of moisture, odors, smoke, heat, dust, airborne bacteria, and carbon dioxide. Ventilation removes unpleasant smells and excessive moisture, introduces outside air, keeps interior building air circulating, and prevents stagnation of the interior air. Ventilation includes both the exchange of air to the outside as well as circulation of air within the building. It is one of the most important factors for maintaining acceptable indoor air quality in buildings. Methods for ventilating a building may be divided into mechanical/forced and natural types. [3]

1.2 Overview: The three central functions of heating, ventilation, and air-conditioning are interrelated, especially with the need to provide thermal comfort and acceptable indoor air quality within reasonable installation, operation, and maintenance costs. HVAC systems can provide ventilation, reduce air infiltration, and maintain pressure relationships between spaces. The means of air delivery and removal from spaces is known as room air distribution. [4]

1.2.1 Individual systems: In modern buildings the design, installation, and control systems of these functions are integrated into one or more HVAC systems. For very small buildings, contractors normally estimate the capacity and select HVAC systems and equipment. For larger buildings, building service designers, mechanical engineers, or building services engineers analyze, design, and specify the HVAC systems. Specialty mechanical contractors then fabricate and commission the systems. Building permits and code-compliance inspections of the installations are normally required for all sizes of building.

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1.2.2 District networks: Although HVAC is executed in individual buildings or other enclosed spaces (like NORAD's underground headquarters), the equipment involved is in some cases an extension of a larger district heating (DH) or district cooling (DC) network, or a combined DHC network. In such cases, the operating and maintenance aspects are simplified and metering becomes necessary to bill for the energy that is consumed, and in some cases energy that is returned to the larger system. For example, at a given time one building may be utilizing chilled water for air conditioning and the warm water it returns may be used in another building for heating, or for the overall heating-portion of the DHC network (likely with energy added to boost the temperature). [5]

1.3 Heating: Heaters are appliances whose purpose is to generate heat for the building. This can be done via central heating. Such a system contains a boiler, furnace or heat pump to heat water, steam or air in a central location such as a furnace room in a home or a mechanical room in a large building. The heat can be transferred by convection, conduction or radiation.

1.3.1 Generation: Heaters exist for various types of fuel, including solid fuels, liquids, and gases. Another type of heat source is electricity, typically heating ribbons made of high resistance wire. This principle is also used for baseboard heaters and portable heaters. Electrical heaters are often used as backup or supplemental heat for heat pump systems.

The heat pump gained popularity in the 1950s in the US and Japan [6]. Heat pumps can extract heat from various sources, such as environmental air, exhaust air from a building, or from the ground. Initially, heat pump HVAC systems were used in moderate climates, but with improvements in low temperature operation and reduced loads due to more efficient homes, they are increasing in popularity in cooler climates.

1.3.2 Distribution:

1.3.2(a) Water/steam:

In the case of heated water or steam, piping is used to transport the heat to the rooms. Most modern hot water boiler heating systems have a circulator, which is a pump, to move hot water through the distribution system (as opposed to older gravity-fed systems). The heat can be transferred to the surrounding air using radiators, hot water coils (hydro-air), or other heat exchangers. The radiators may be mounted on walls or installed within the floor to give floor heat. The use of water as the heat transfer medium is known as hydronic. The heated water can also supply an auxiliary heat exchanger to supply hot water for bathing and washing. [7]

1.3.2(b) Air:

Warm air systems distribute heated air through duct work systems of supply and return air through metal or fiberglass ducts. Many systems use the same ducts to distribute air cooled by an evaporator

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coil for air conditioning. The air supply is typically filtered through air cleaners to remove dust and pollen particles. [8]

1.4 Ventilation: Ventilation is the process of changing or replacing air in any space to control temperature or remove any combination of moisture, odors, smoke, heat, dust, airborne bacteria, or carbon dioxide, and to replenish oxygen. Ventilation includes both the exchange of air with the outside as well as circulation of air within the building. It is one of the most important factors for maintaining acceptable indoor air quality in buildings. Methods for ventilating a building may be divided into mechanical/forced and natural types.

1.4.1 Mechanical or forced ventilation: Mechanical or forced ventilation is provided by an air handler and used to control indoor air quality. Excess humidity, odors, and contaminants can often be controlled via dilution or replacement with outside air. However, in humid climates much energy is required to remove excess moisture from ventilation air. Kitchens and bathrooms typically have mechanical exhausts to control odors and sometimes humidity. Factors in the design of such systems include the flow rate (which is a function of the fan speed and exhaust vent size) and noise level. Direct drive fans are available for many applications, and can reduce maintenance needs. Ceiling fans and table/floor fans circulate air within a room for the purpose of reducing the perceived temperature by increasing evaporation of perspiration on the skin of the occupants. Because hot air rises, ceiling fans may be used to keep a room warmer in the winter by circulating the warm stratified air from the ceiling to the floor. [9]

1.4.2 Natural ventilation: Natural ventilation is the ventilation of a building with outside air without using fans or other mechanical systems. It can be via operable windows, louvers, or trickle vents when spaces are small and the architecture permits. In more complex schemes, warm air is allowed to rise and flow out high building openings to the outside (stack effect), causing cool outside air to be drawn into low building openings. Natural ventilation schemes can use very little energy, but care must be taken to ensure comfort. In warm or humid climates, maintaining thermal comfort solely via natural ventilation may not be possible. Air conditioning systems are used, either as backups or supplements. Air-side economizers also use outside air to condition spaces, but do so using fans, ducts, dampers, and control systems to introduce and distribute cool outdoor air when appropriate.

An important component of natural ventilation is air change rate or air changes per hour: the hourly rate of ventilation divided by the volume of the space. For example, six air changes per hour means an amount of new air, equal to the volume of the space, is added every ten minutes. For human comfort, a minimum of four air changes per hour is typical, though warehouses might have only two. Too high of an air change rate may be uncomfortable, akin to a wind tunnel which have thousands of changes per hour. The highest air change rates are for crowded spaces, bars, night clubs, commercial kitchens at around 30 to 50 air changes per hour. Room pressure can be either positive or negative with respect to outside the room. Positive pressure occurs when there is more air being supplied than exhausted, and is common to reduce the infiltration of outside contaminants. [10]

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[11]

1.5 Air conditioning: An air conditioning system, or a standalone air conditioner, provides cooling and humidity control for all or part of a building. Air conditioned buildings often have sealed windows, because open windows would work against the system intended to maintain constant indoor air conditions. Outside, fresh air is generally drawn into the system by a vent into the indoor heat exchanger section, creating positive air pressure. The percentage of return air made up of fresh air can usually be manipulated by adjusting the opening of this vent. Typical fresh air intake is about 10%. Air conditioning and refrigeration are provided through the removal of heat. Heat can be removed through radiation, convection, or conduction. Refrigeration conduction media such as water, air, ice, and chemicals are referred to as refrigerants. A refrigerant is employed either in a heat pump system in which a compressor is used to drive thermodynamic refrigeration cycle, or in a free cooling system which uses pumps to circulate a cool refrigerant (typically water or a glycol mix). [12]

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1.5.1 Refrigeration cycle: The refrigeration cycle uses four essential elements to cool.

The system refrigerant starts its cycle in a gaseous state. The compressor pumps the refrigerant gas up to a high pressure and temperature.

From there it enters a heat exchanger (sometimes called a condensing coil or condenser) where it loses energy (heat) to the outside, cools, and condenses into its liquid phase.

An expansion valve (also called metering device) regulates the refrigerant liquid to flow at the proper rate.

The liquid refrigerant is returned to another heat exchanger where it is allowed to evaporate, hence the heat exchanger is often called an evaporating coil or evaporator. As the liquid refrigerant evaporates it absorbs energy (heat) from the inside air, returns to the compressor, and repeats the cycle. In the process, heat is absorbed from indoors and transferred outdoors, resulting in cooling of the building.

In variable climates, the system may include a reversing valve that switches from heating in winter to cooling in summer. By reversing the flow of refrigerant, the heat pump refrigeration cycle is changed from cooling to heating or vice versa. This allows a facility to be heated and cooled by a single piece of equipment by the same means, and with the same hardware. [13]

1.5.2 Free cooling: Free cooling systems can have very high efficiencies, and are sometimes combined with seasonal thermal energy storage so the cold of winter can be used for summer air conditioning. Common storage mediums are deep aquifers or a natural underground rock mass accessed via a cluster of small-diameter, heat exchanger equipped boreholes. Some systems with small storages are hybrids, using free cooling early in the cooling season, and later employing a heat pump to chill the circulation coming from the storage. The heat pump is added-in because the storage acts as a heat sink when the system is in cooling (as opposed to charging) mode, causing the temperature to gradually increase during the cooling season. [14]

Some systems include an "economizer mode", which is sometimes called a "free cooling mode". When economizing, the control system will open (fully or partially) the outside air damper and close (fully or partially) the return air damper. This will cause fresh, outside air to be supplied to the system. When the outside air is cooler than the demanded cool air, this will allow the demand to be met without using the mechanical supply of cooling (typically chilled water or a direct expansion "DX" unit), thus saving energy. The control system can compare the temperature of the outside air vs. return air, or it can compare the enthalpy of the air, as is frequently done in climates where humidity is more of an issue. In both cases, the outside air must be less energetic than the return air for the system to enter the economizer mode.

1.5.3 Central vs. split system: Central air conditioning systems (or package systems) with a combined outdoor condenser/evaporator unit are often installed in modern residences, offices, and public buildings, but are difficult to retrofit (install in a building that was not designed to receive it) because of the bulky air ducts required [15]. An alternative to central systems is the use of separate indoor and

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outdoor coils in split systems. These systems, although most often seen in residential applications, are gaining popularity in small commercial buildings. The evaporator coil is connected to a remote condenser unit using refrigerant piping between an indoor and outdoor unit instead of ducting air directly from the outdoor unit. Indoor units with directional vents mount onto walls, suspended from ceilings, or fit into the ceiling. Other indoor units mount inside the ceiling cavity, so that short lengths of duct handle air from the indoor unit to vents or diffusers around the rooms.

1.5.4 Dehumidification: Dehumidification (air drying) in an air conditioning system is provided by the evaporator. Since the evaporator operates at a temperature below the dew point, moisture in the air condenses on the evaporator coil tubes. This moisture is collected at the bottom of the evaporator in a pan and removed by piping to a central drain or onto the ground outside.

A dehumidifier is an air-conditioner-like device that controls the humidity of a room or building. It is often employed in basements which have a higher relative humidity because of their lower temperature (and propensity for damp floors and walls). In food retailing establishments, large open chiller cabinets are highly effective at dehumidifying the internal air. Conversely, a humidifier increases the humidity of a building. [16]

2 Purpose: The aim of the project is to find the cooling and heating load of under construction Boys Hostel of UET Lahore’s Narowal campus. A well designed and adequate HVAC system is essential to maintaining the comfortable, productive and health living environment. The system is being designed to meet the minimum ASHRAE and building code standards.

2.1 Problem Definition: University of Engineering and Technology Lahore is pioneer engineering institution in Pakistan providing engineering serving the cause of education since 1923. The university is currently managing 4 sub campuses in addition to main campus. Narowal campus was inaugurated in 2012 and the campus building is under construction consisting of total area of 200 acres.

The hostel building will host more than 450 students and is being built with total area of about 112,222 Sft. The building will be oriented so that the front entrance will be pointing south. The building is to be built on concrete slab with masonry walls.

2.2 Scope: The scope of the design will involve the following considerations.

The zones of the complete hostel building The building insulation, doors and windows for consideration of heat transfer The internal heat generation of the building from equipment and lighting

Heating and cooling load Ventilation requirements for each zone

Humidification requirements Dehumidification requirements

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Ducting layout and specifications Air distribution means

Annual cooling cost Overall energy uses

3 Assumptions: Refer to ASHRAE Standard 62 for ventilation requirements.

The typical values below:

Auditoriums, theaters - 15 cfm/person

Sleeping rooms - 15 cfm/person Bedroom - 30 cfm/room

Classroom - 15 cfm/person Communication centers - 20 cfm/person Conference rooms - 20 cfm/person

Corridors - 0.1 cfm/sq ft Dining - 20 cfm/person

Lobbies - 15 cfm/person Locker, dressing rooms - 0.5 cfm/sq ft

Lounges, bars - 30 cfm/person Offices - 20 cfm/person Toilet, bath (private) - 35 cfm/room

Toilet (public) - 50 cfm/water closet or urinal [17]

4 Zoning the Office: 4.1 Introduction to Zoning: A temperature zoning system is a system that allows to:

Support a consistent temperature in a given part of a house, regardless of external conditions, such as sun, rain, snow, clouds, wind or phase of moon.

Support different temperatures in different parts of the house at different times as desired by the owner. In an essence, each zone will have its own climate. It is possible, for example, to have some rooms be cooler or warmer than others (individual preferences), and it is also possible to keep the same rooms at different temperature depending on, for example, whether the rooms are [supposed to be] occupied or not.

When using a zone control system, it is important that realistic demands are made on the system such as keeping all the zones desired temperatures to within 5 to 8c of other zone temperatures or of the master controlling thermostat. [18]

4.2 Need of Zoning:

We need Zoned Temperature Control in our Building if one or more of these conditions exist:

Family lifestyles dictate different temperatures in different areas of the home.

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Heating and cooling temperature patterns vary at different times of the day.

Our Building has: More than one level.

Large, open areas such as vaulted ceilings or lofts, an atrium or a solarium. A room off the back or over the garage.

Finished rooms in the basement or attic. A room or rooms with expansive glass areas.

A portion built over a concrete slab floor. A rambling floor plan or wings extending off the main living area. [19]

4.3 Final Zoning Choice: There are total 4 zones in the hostel building.

Zone1: Residential rooms (129) Zone2: Toilets (64)

Zone3: Main Dining Hall & Main Kitchen Zone4: TV Lounge

5 Space heat gain: It's the rate of heat gained when heat enters the space or heat generated within a space.

5.1 The method of how the heat enters the space: Solar radiation through the window or any transparent surfaces. Heat conduction through walls, roof and windows of the class. Heat conduction through interior partitions, ceilings and floors. The generated heat by the occupants such as lights, appliances, equipment and processes. The loads that are results of ventilation and infiltration of outdoor air. Other miscellaneous heat gains.

5.2 Sensible heat: It's about heat at which a substance absorbs. During rising the temperature of the substance, the substance doesn't change state. Sensible heat gain is directly added to the conditioned space by conduction, convection, and radiation. Note that the sensible heat gain entering the conditioned space does not equal the sensible cooling load during the same time interval because of the stored heat in the building envelope. Only the convective heat becomes cooling load instantaneously. [20]

5.3 The factors involve the sensible heat load: Heat transmitted through floors, ceilings, walls. Occupant's body heat. Appliance & Light heat. Solar Heat gain through glass. Infiltration of outside air. Air introduced by Ventilation.

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5.4 Latent Heat Loads: Latent heat gain occurs when moisture is added to the space either from internal sources (e.g. vapor emitted by occupants and equipment) or from outdoor air as a result of infiltration or ventilation to maintain proper indoor air quality.

5.4.1 The factors involve the sensible heat load:

Moisture-laden outside air form Infiltration & Ventilation. Occupant Respiration & Activities. Moisture from Equipment & Appliances.

5.5 Space heat gain and cooling load (heat storage effect): The heat that is collected from the heat sources (conduction, convection, solar radiation, lightning, people, equipment, etc...) doesn't go directly to heating the room. However, only some part of the heat sources that is absorbed air in the conditioned space (class), leading to a quick change in its temperature. Most of the radiation heat especially from sun, lighting, people is first absorbed by the internal surfaces, which include ceiling, floor, internal walls, furniture etc. Due to the large but finite thermal capacity of the roof, floor, walls etc., their temperature increases slowly due to absorption of radiant heat. The radiant portion introduces a time lag and also a decrement factor depending upon the dynamic characteristics of the surfaces. Due to the time lag, the effect of radiation will be felt even when the source of radiation, in this case the sun is removed. [21]

[22]

The relation between heat gain and cooling load and the effect of the mass of the structure (light, medium & heavy) is shown below. From figure it is evident that, there is a delay in the peak heat, especially for heavy construction.

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[23]

5.6 Space cooling and cooling load (Coil): Space cooling is the rate at which heat must be removed from the spaces to maintain air temperature at a constant value. Cooling load, on the other hand, is the rate at which energy is removed at the cooling coil that serves one or more conditioned spaces in any central air conditioning system. [24]

5.7 Components of cooling load:

[25]

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The total residential room cooling load consists of heat transferred through the room envelope (walls, roof, floor, windows, doors etc.) and heat generated by occupants, equipment, and lights. The load due to heat transfer through the envelope is called as external load, while all other loads are The total cooling load on any building consists of both sensible as well as latent load components. The sensible load affects the dry bulb temperature, while the latent load affects the moisture content of the conditioned space. [26]

Residential room may be classified as externally loaded and internally loaded as you can see from figure. In externally loaded room, the cooling load on the room is mainly due to heat transfer between the surroundings and the internal conditioned space. Since the surrounding conditions are highly variable in any given day, the cooling load of an externally loaded building varies widely.

[27]

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6 CLTD/SCL/CLF METHOD OF LOAD CALCULATION: CLTD is a theoretical temperature difference that accounts for the combined effects of inside and outside air temp difference, daily temp range, solar radiation and heat storage in the construction assembly/building mass. It is affected by orientation, tilt, month, day, hour, latitude, etc. CLTD factors are used for adjustment to conductive heat gains from walls, roof, floor and glass. CLF accounts for the fact that all the radiant energy that enters the conditioned space at a particular time does not become a part of the cooling load instantly. The CLF values for various surfaces have been calculated as functions of solar time and orientation and are available in the form of tables in ASHRAE Handbooks. CLF factors are used for adjustment to heat gains from internal loads such as lights, occupancy, power appliances. [28]

6.1 External Cooling Load:

6.1.1 Roof:

If the roof is exposed directly to the sun, it absorbs maximum heat. If there is other room above the air-conditioned room, then the amount of heat gained by the roof reduces. The heat gained by the partitions of the room depends upon the type of partition. Roof calculation formula is given below:

Q = U * A * (CLTD)

Q = cooling load. U = Coefficient of heat transfer roof or wall or glass. A = area of roof. CLTD = cooling load temperature difference.

Since the ASHRAE tables provide hourly CLTD values for one typical set of conditions i.e. outdoor maximum temperature of 95°F with mean temperature of 85°C and daily range of 21°F, the equation is further adjusted to apply correction factors for conditions other than the mentioned base case. Thus,

Q Roof = U * A * CLTD Roof Corrected

[29]

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6.1.2 Walls:

The walls of the room gain heat from the sun by way of conduction. The amount of heat depends on the wall material and its alignment with respect to sun. If the wall of the room is exposed to the west direction, it will gain maximum heat between 2 to 5 pm. The southern wall will gain maximum heat in the mid-day between 12 to 2 pm. The heat gained by the wall facing north direction is the least. The heat gained by the walls in day-time gets stored in them, and it is released into the rooms at the night time thus causing excessive heating of the room. If the walls of the room are insulated the amount of heat gained by them reduces drastically.

The cooling load from walls is treated in a similar way as roof:

Q Wall = U * A * CLTD Wall Corrected

Where

Q Wall = Load through the walls. U = Thermal Transmittance for walls. A = area of walls. CLTD = Cooling Load Temperature Difference for walls.

Table 1 Cooling Load Temperature Difference for Roof and External Walls (Dark). [30]

Solar time, hour 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Roof 14 12 10 8 7 5 4 4 6 8 11 15 18 22 25 28 29 30 29 27 24 21 19 16

External wall

North

North-east

East

South-east

South

South-west

West

North-west

8

9

11

11

11

15

17

14

7

8

10

10

10

14

15

12

7

7

8

9

8

12

13

11

6

6

7

7

7

10

12

9

5

5

6

6

6

9

10

8

4

5

5

5

5

8

9

7

3

4

5

5

4

6

7

6

3

4

5

5

4

5

6

5

3

6

7

5

3

5

5

4

3

8

10

7

3

4

5

4

4

10

13

10

4

4

5

4

4

19

15.5

12

20.5

29

35

41

5

12

17

14

7

5

6

5

6

13

18

16

9

7

6

6

6

13

18

17

11

9

8

7

7

13

18

18

13

12

10

8

8

14

18

18

15

15

12

10

9

14

18

18

16

18

17

12

10

14

17

17

16

20

10

15

11

13

17

17

16

21

11

17

11

13

16

16

15

21

12

18

10

12

15

15

14

20

11

17

10

11

13

14

13

19

11

16

9

10

12

12

12

17

19

15

6.1.3 Solar load through glass:

Solar load through glass has two components: 1) Conductive and 2) Solar Transmission The absorbed and then conductive portion of the radiation through the windows is treated like the roof & walls where CLTD values for standard glazing are tabulated in ASHARE fundamentals handbook.

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For solar transmission, the cooling load is calculated by the cooling load SCL factor and shading coefficient (SC).

The cooling load equations for glass are:

Conductive Q Glass Conductive = U * A * CLTD Glass Corrected

Solar Transmission Q Glass Solar = A * SC * SCL

Where

Q Conductive = Conductive load through the glass. Q Solar = Solar transmission load through the glass. U = Thermal Transmittance for glass. A = area of glass. CLTD = Cooling Load Temperature Difference for glass. SC = Shading coefficient. SCL = Solar Cooling Load Factor.

Table 2 Cooling Load Factor for Window Glass with Indoor Shading Devices. [31]

Solar time,

hour

1

2

3

4

5

6

7

8

9

10

11

12

Orientation:

North

North-east

East

South-east

South

South-west

West

North-west

Horizontal

0.08

0.03

0.03

0.03

0.04

0.05

0.05

0.05

0.06

0.07

0.02

0.02

0.03

0.04

0.05

0.05

0.04

0.05

0.06

0.02

0.02

0.02

0.03

0.04

0.04

0.04

0.04

0.06

0.02

0.02

0.02

0.03

0.04

0.04

0.03

0.04

0.07

0.02

0.02

0.02

0.03

0.03

0.03

0.03

0.03

0.73

0.56

0.47

0.30

0.09

0.07

0.06

0.07

0.12

0.66

0.76

0.72

0.57

0.16

0.11

0.09

0.11

0.27

0.65

0.74

0.80

0.74

0.23

0.14

0.11

0.14

0.44

0.73

0.58

0.76

0.81

0.38

0.16

0.13

0.17

0.59

0.80

0.37

0.62

0.79

0.58

0.19

0.15

0.19

0.72

0.86

0.29

0.41

0.68

0.75

0.22

0.16

0.20

0.81

0.91

0.86

0.81

0.82

0.83

0.81

0.82

0.865

0.85

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Table 3 Maximum Solar Heat Gain Factor for Sunlit Glass on Average Cloudiness Days [32]

Month Maximum solar heat gain factor for 22-degree north latitude, W/m2

North North-east /

north-west

East / west South-east /

south-west-

South Horizontal

January.

February.

March.

April

May

June

July

August

September

October

November

December

88

97

107

119

142

180

44

123

112

100

88

84

140

265

404

513

572

589

565

502

388

262

142

101

617

704

743

719

687

666

149

694

705

676

606

579

789

759

663

516

404

355

391

496

639

735

786

790

696

578

398

210

139

134

171

223

392

563

686

730

704

808

882

899

892

880

877

879

854

792

699

657

6.1.4 Partition, ceilings and floors:

The various internal loads consist of sensible and latent heat transfers due to occupants, products, processes appliances and lighting. The lighting load is only sensible. The conversion of sensible heat gains (from lighting, people, appliances, etc.) to space cooling load is affected by the thermal storage characteristics of that space and is thus subject to appropriate cooling load factors (CLF) to account for the time lag of the cooling load caused by the building mass. The weighting factors equation determines the CLF factors.

CLF = Q cooling load / Q internal gains [33]

6.2 Internal Cooling Loads: The various internal loads consist of sensible and latent heat transfers due to occupants, products, processes appliances and lighting. The lighting load is only sensible. The conversion of sensible heat gains (from lighting, people, appliances, etc.) to space cooling load is affected by the thermal storage characteristics of that space and is thus subject to appropriate cooling load factors (CLF) to account for the time lag of the cooling load caused by the building mass. The weighting factors equation determines the CLF factors.

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CLF = Q cooling load / Q internal gains

Note that the latent heat gains are considered instantaneous.

6.2.1 People:

Human beings release both sensible heat and latent heat to the conditioned space when they stay in it as you can see the figures in the table (11). You might have noticed that when a small room is filled with people, it tends to become warmer. People emit heat primarily through breathing and perspiration, and, to a lesser extent, through radiation. This heat translates into an increased cooling load on your cooling systems. The heat gain by the occupants in the building is separated into sensible and latent heat. The number of people, the type of activity they are performing, and the CLF determines sensible and latent heat. The CLF is determined by the time the occupants come into the building and for how long they stay in the building. [34]

The heat gain from the occupancy or people is given be equation:

Q sensible = N (QS) (CLF)

Q latent = N (QL)

N = number of students in space (class room). QS, QL = Sensible and Latent heat gain from occupancy is given in Table 3. CLF = Cooling Load Factor, by hour of occupancy in Table 37.

Table 4 Heat Gain from Occupants at Various Activities [35]

Activity Total heat, W Sensible heat, W Latent heat, W

Adult, male Adjusted

Seated at rest

Seated, very light work, writing

Seated, eating

Seated, light work, typing,

Standing, light work or walking slowly,

Light bench work

Light machine work

Heavy work

Moderate dancing

Athletics

115

140

150

185

235

255

305

470

400

585

100

120

170b

150

185

230

305

470

375

525

60

65

75

250

90

100

100

165

120

185

40

55

95

200

95

130

205

305

255

340

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6.2.2 Lights:

The primary source of heat from lighting comes from light-emitting elements. Table (13), indicate and explain more about the lights effects on the heat gain when it's off or on. Calculation of this load component is not straightforward; the rate of heat gain at any given moment can be quite different from the heat equivalent of power supplied instantaneously to those lights. Only part of the energy from lights is in the form of convective heat, which is picked up instantaneously by the air-conditioning apparatus. The remaining portion is in the form of radiation, which affects the conditioned space only after having been absorbed and re-released by walls, floors, furniture, etc. This absorbed energy contributes to space cooling load only after a time lag, with some part of such\ energy still present and reradiating after the lights have been switched off. [36]

Generally, the instantaneous rate of heat gain from electric lighting may be calculated from:

Q = 3.41 x W x FUT x FSA

Cooling load factors are used to convert instantaneous heat gain from lighting to the sensible cooling

load; thus the equation is modified to:

Q = 3.41 x W x FUT x FSA x (CLF).

Where: W = Watts input from electrical lighting plan or lighting load data. FUT = Lighting use factor, as appropriate. FSA = special ballast allowance factor, as appropriate. CLF = Cooling Load Factor, by hour of occupancy, Table 38.

Table 5 Cooling Load Factors (CLF) for Lights: [37] Lights Number of Hours the lights are turned ON

For 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

8 0.85 0.92 0.95 0.96 0.97 0.97 0.97 0.98 0.13 0.06 0.04 0.03 0.02 0.02 0.02 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01

10 0.85 0.93 0.95 0.97 0.97 0.97 0.98 0.98 0.98 0.98 0.14 0.07 0.04 0.03 0.02 0.02 0.02 0.02 0.02 0.02 0.01 0.01 0.01 0.01

12 0.86 0.93 0.96 0.97 0.97 0.98 0.98 0.98 0.98 0.98 0.98 1.0 0.14 0.07 0.04 0.03 0.03 0.02 0.02 0.02 0.02 0.02 0.02 0.02

14 0.86 0.93 0.96 0.97 0.98 0.98 0.98 0.98 0.98 0.98 0.99 0.99 0.99 0.99 0.15 0.07 0.05 0.03 0.03 0.03 0.02 0.02 0.02 0.02

16 0.87 0.94 0.96 0.97 0.98 0.98 0.98 0.99 0.99 0.99 0.99 0.99 0.99 0.99 0.99 0.99 0.15 0.08 0.05 0.04 0.03 0.03 0.03 0.02

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6.2.3 Appliances:

In a cooling load estimate, heat gain from all appliances-electrical, gas, or steam-should be taken into account. Because of the variety of appliances, applications, schedules, use, and installations, estimates can be very subjective. Often, the only information available about heat gain from equipment is that on its name-plate. [38] Q Sensible = Qin x Fu x Fr x (CLF) Q Latent = Qin x Fu Where

Qin = rated energy input from appliances. See Table 5 through 9 or use manufacturer's data. For computers, monitors, printers and miscellaneous office equipment, see 2001 ASHRAE Fundamentals, Chapter 29, Tables 8, 9, & 10.

Fu = Usage factor. See 1997 ASHRAE Fundamentals, Chapter 28, Table 6 and 7. Fr = Radiation factor. See 1997 ASHRAE Fundamentals, Chapter 28, Table 6 and 7. CLF = Cooling Load Factor, by hour of occupancy. See 1997 ASHRAE Fundamentals,

Chapter 28, Table 37 and 39.

6.2.4 Infiltration Air:

Q sensible = 1.08 x CFM x (To - Ti)

Q latent = 4840 x CFM x (Wo - Wi)

Q total = 4.5 x CFM x (ho - hi)

Where

CFM = Infiltration air flow rate. See 1997 ASHRAE Fundamentals, Chapter 25, for determining infiltration

To, Ti = Outside/Inside dry bulb temperature. Wo, Wi = Outside/Inside humidity ratio.

ho, hi = Outside/Inside air enthalpy.

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Excel Calculations

Zone 1: Residential Rooms Serial No.

Q Walls Q Glass Q Lights QLaptop QInf QPeople Q Roof QTotal

R1 2990.49 4700.72 204.72 887.12 2995.64 1800 0 13578.69

R2 1197.69 4700.72 204.72 887.12 2995.64 1800 0 11785.89

R3 1197.69 4700.72 204.72 887.12 2995.64 1800 0 11785.89

R4 2422.77 4700.72 204.72 887.12 2995.64 1800 0 13010.97

R5 3209.85 2975.41 204.72 887.12 2995.64 1800 0 12072.74

R6 1118.259 2975.41 204.72 887.12 2995.64 1800 0 9981.149

R7 1118.259 2975.41 204.72 887.12 2995.64 1800 0 9981.149

R8 2343.34 2975.41 204.72 887.12 2995.64 1800 0 11206.23

R9 2315.949 1528.95 204.72 887.12 2995.64 1800 0 9732.379

R10 1090.869 1528.95 204.72 887.12 2995.64 1800 0 8507.299

R11 1090.869 1528.95 204.72 887.12 2995.64 1800 0 8507.299

R12 1090.869 1528.95 204.72 887.12 2995.64 1800 0 8507.299

R13 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R14 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R15 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R16 3261.9 3327.77 204.72 887.12 2995.64 1800 0 12477.15

R17 1197.69 2939.13 204.72 887.12 2995.64 1800 0 10024.3

R18 1197.69 2939.13 204.72 887.12 2995.64 1800 0 10024.3

R19 WR 1197.69 2939.13 204.72 221.78 2995.64 450 0 8008.96

R20 ER 3261.9 3327.77 204.72 0 2995.64 450 0 10240.03

R21 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R22 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R23 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R24 1197.69 2939.13 204.72 887.12 2995.64 1800 0 10024.3

R25 1197.69 2939.13 204.72 887.12 2995.64 1800 0 10024.3

R26 1197.69 2939.13 204.72 887.12 2995.64 1800 0 10024.3

R27 2990.49 2939.13 204.72 887.12 2995.64 1800 0 11817.1

R28 3261.9 3327.77 204.72 887.12 2995.64 1800 0 12477.15

R29 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R30 1170.3 3327.77 204.72 887.12 2995.64 1800 0 10385.55

R31 2096.58 3327.77 204.72 887.12 2995.64 1800 0 11311.83

R32 2893.629 3308.54 204.72 887.12 2995.64 1800 0 12089.649

R33 1100.829 3308.54 204.72 887.12 2995.64 1800 0 10296.849

R34 1100.829 3308.54 204.72 887.12 2995.64 1800 0 10296.849

R35 2325.909 3308.54 204.72 887.12 2995.64 1800 0 11521.929

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R36 2044.539 4672.23 204.72 887.12 2995.64 1800 0 12604.249

R37 1118.259 4672.23 204.72 887.12 2995.64 1800 0 11677.969

R38 1118.259 4672.23 204.72 887.12 2995.64 1800 0 11677.969

R39 1118.259 4672.23 204.72 887.12 2995.64 1800 0 11677.969

R40 2990.5 2837.77 204.72 887.12 2995.64 1800 0 11715.75

R41 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R42 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R43 2422.77 2837.77 204.72 887.12 2995.64 1800 0 11148.02

R44 3209.85 4511.124 204.72 887.12 2995.64 1800 0 13608.454

R45 1118.26 4511.124 204.72 887.12 2995.64 1800 0 11516.864

R46 1118.26 4511.124 204.72 887.12 2995.64 1800 0 11516.864

R47 2343.34 4511.124 204.72 887.12 2995.64 1800 0 12741.944

R48 2315.95 3194.45 204.72 887.12 2995.64 1800 0 11397.88

R49 1090.869 3194.45 204.72 887.12 2995.64 1800 0 10172.799

R50 1090.869 3194.45 204.72 887.12 2995.64 1800 0 10172.799

R51 1090.869 3194.45 204.72 887.12 2995.64 1800 0 10172.799

R52 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R53 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R54 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R55 3261.9 1484.81 204.72 887.12 2995.64 1800 0 10634.19

R56 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R57 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R58 WR 1197.69 2837.77 204.72 221.78 2995.64 450 0 7907.6

R59 ER 3261.9 1484.81 204.72 0 2995.64 450 0 8397.07

R60 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R61 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R62 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R63 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R64 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R65 1197.69 2837.77 204.72 887.12 2995.64 1800 0 9922.94

R66 2990.49 2837.77 204.72 887.12 2995.64 1800 0 11715.74

R67 3261.9 1484.81 204.72 887.12 2995.64 1800 0 10634.19

R68 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R69 1170.3 1484.81 204.72 887.12 2995.64 1800 0 8542.59

R70 2096.58 1484.81 204.72 887.12 2995.64 1800 0 9468.87

R71 2893.63 3194.45 204.72 887.12 2995.64 1800 0 11975.56

R72 1100.83 3194.45 204.72 887.12 2995.64 1800 0 10182.76

R73 1100.83 3194.45 204.72 887.12 2995.64 1800 0 10182.76

R74 2325.91 3194.45 204.72 887.12 2995.64 1800 0 11407.84

R75 2044.54 4511.124 204.72 887.12 2995.64 1800 0 12443.144

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R76 1118.26 4511.124 204.72 887.12 2995.64 1800 0 11516.864

R77 1118.26 4511.124 204.72 887.12 2995.64 1800 0 11516.864

R78 1118.26 4511.124 204.72 887.12 2995.64 1800 0 11516.864

R79 1975.068 3194.453 307.08 887.12 2995.64 1800 0 11159.361

R80 1975.068 3194.453 307.08 887.12 2995.64 1800 0 11159.361

R81 4188.38 3194.453 307.08 887.12 2995.64 1800 0 13372.673

R82 2363.508 4511.124 204.72 887.12 2995.64 1800 0 12762.112

R83 2071.348 4511.124 307.08 887.12 2995.64 1800 0 12572.312

R84 2071.348 4511.124 307.08 887.12 2995.64 1800 0 12572.312

R85 3036.97 2718.286 204.72 887.12 2995.64 1800 1539.99 13182.726

R86 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R87 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R88 2469.25 2718.286 204.72 887.12 2995.64 1800 1539.99 12615.006

R89 3237.083 4242.61 204.72 887.12 2995.64 1800 1539.99 14907.163

R90 1145.483 4242.61 204.72 887.12 2995.64 1800 1539.99 12815.563

R91 1145.483 4242.61 204.72 887.12 2995.64 1800 1539.99 12815.563

R92 2071.763 4242.61 204.72 887.12 2995.64 1800 1539.99 13741.843

R93 2336.533 3004.3 204.72 887.12 2995.64 1800 1539.99 12768.303

R94 1111.453 3004.3 204.72 887.12 2995.64 1800 1539.99 11543.223

R95 1111.453 3004.3 204.72 887.12 2995.64 1800 1539.99 11543.223

R96 1111.453 3004.3 204.72 887.12 2995.64 1800 1539.99 11543.223

R97 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R98 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R99 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R100 3301.74 1396.42 204.72 887.12 2995.64 1800 1539.99 12125.63

R101 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R102 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R103 WR 1244.17 2718.286 204.72 221.78 2995.64 450 1539.99 9374.586

R104 ER 3301.74 1396.42 204.72 0 2995.64 450 1539.99 9888.51

R105 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R106 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R107 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R108 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R109 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R110 1244.17 2718.286 204.72 887.12 2995.64 1800 1539.99 11389.926

R111 3036.97 2718.286 204.72 887.12 2995.64 1800 1539.99 13182.726

R112 3301.74 1396.42 204.72 887.12 2995.64 1800 1539.99 12125.63

R113 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R114 1210.14 1396.42 204.72 887.12 2995.64 1800 1539.99 10034.03

R115 2136.42 1396.42 204.72 887.12 2995.64 1800 1539.99 10960.31

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R116 2904.253 3004.3 204.72 887.12 2995.64 1800 1539.99 13336.023

R117 1111.453 3004.3 204.72 887.12 2995.64 1800 1539.99 11543.223

R118 1111.453 3004.3 204.72 887.12 2995.64 1800 1539.99 11543.223

R119 2336.533 3004.3 204.72 887.12 2995.64 1800 1539.99 12768.303

R120 2071.763 4242.61 204.72 887.12 2995.64 1800 1539.99 13741.843

R121 1145.483 4242.61 204.72 887.12 2995.64 1800 1539.99 12815.563

R122 1145.483 4242.61 204.72 887.12 2995.64 1800 1539.99 12815.563

R123 1145.483 4242.61 204.72 887.12 2995.64 1800 1539.99 12815.563

R124 1987.933 3004.3 204.72 887.12 2995.64 1800 4277.76 15157.473

R125 1987.933 3004.3 204.72 887.12 2995.64 1800 4277.76 15157.473

R126 4394.914 3004.3 204.72 887.12 2995.64 1800 4990 18276.694

R127 2380.523 4242.61 204.72 887.12 2995.64 1800 2053.324 14563.937

R128 2088.263 4242.61 204.72 887.12 2995.64 1800 4277.76 16496.113

R129 2088.263 4242.61 204.72 887.12 2995.64 1800 4277.76 16496.113

Total 226891.94 389419.26 26920.68 109781.1 386437.56 224100 84213.97 1447764.513

Unit of Heat Gain Q is BTU/hr

Total Heat Gain for Zone 1 is Q =1447764.513 BTU/hr

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Zone 2: Toilets Serial No. Q Walls Q Glass Q Lights Q Inf Q People Q Roof Q Total

T1 679.77 1512.87 102.36 1397.32 0 0 3692.32

T2 679.77 1512.87 102.36 1397.32 0 0 3692.32

T3 398.151 957.6 102.36 1397.32 0 0 2855.431

T4 398.151 957.6 102.36 1397.32 0 0 2855.431

T5 301.04 492.07 102.36 1397.32 0 0 2292.79

T6 301.04 492.07 102.36 1397.32 0 0 2292.79

T7 582.66 1071 102.36 1397.32 0 0 3153.34

T8 582.66 1071 102.36 1397.32 0 0 3153.34

T9 679.77 1512.87 102.36 1397.32 0 0 3692.32

T10 679.77 1512.87 102.36 1397.32 0 0 3692.32

T11 582.66 1071 102.36 1397.32 0 0 3153.34

T12 582.66 1071 102.36 1397.32 0 0 3153.34

T13 679.77 1512.87 102.36 1397.32 0 0 3692.32

T14 679.77 1512.87 102.36 1397.32 0 0 3692.32

T15 398.151 957.6 102.36 1397.32 0 0 2855.431

T16 398.151 957.6 102.36 1397.32 0 0 2855.431

T17 301.04 492.07 102.36 1397.32 0 0 2292.79

T18 301.04 492.07 102.36 1397.32 0 0 2292.79

T19 582.66 1071 102.36 1397.32 0 0 3153.34

T20 582.66 1071 102.36 1397.32 0 0 3153.34

GT21 1111.536 2281.752 204.72 2497.32 1350 0 7445.328

T22 679.77 1512.87 102.36 1397.32 0 0 3692.32

T23 679.77 1512.87 102.36 1397.32 0 0 3692.32

T24 398.151 957.6 102.36 1397.32 0 0 2855.431

T25 398.151 957.6 102.36 1397.32 0 0 2855.431

T26 301.04 492.07 102.36 1397.32 0 0 2292.79

T27 301.04 492.07 102.36 1397.32 0 0 2292.79

T28 582.66 1071 102.36 1397.32 0 0 3153.34

T29 582.66 1071 102.36 1397.32 0 0 3153.34

T30 679.77 1512.87 102.36 1397.32 0 0 3692.32

T31 679.77 1512.87 102.36 1397.32 0 0 3692.32

T32 582.66 1071 102.36 1397.32 0 0 3153.34

T33 582.66 1071 102.36 1397.32 0 0 3153.34

T34 679.77 1512.87 102.36 1397.32 0 0 3692.32

T35 679.77 1512.87 102.36 1397.32 0 0 3692.32

T36 398.151 957.6 102.36 1397.32 0 0 2855.431

T37 398.151 957.6 102.36 1397.32 0 0 2855.431

T38 301.04 492.07 102.36 1397.32 0 0 2292.79

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T39 301.04 492.07 102.36 1397.32 0 0 2292.79

T40 582.66 1071 102.36 1397.32 0 0 3153.34

T41 582.66 1071 102.36 1397.32 0 0 3153.34

GT42 1111.536 2281.752 204.72 2497.32 1350 0 7445.328

T43 722.4154 710.88 102.36 1397.32 0 646.12 3579.0954

T44 722.4154 710.88 102.36 1397.32 0 646.12 3579.0954

T45 423.12 1109.52 102.36 1397.32 0 646.12 3678.44

T46 423.12 1109.52 102.36 1397.32 0 646.12 3678.44

T47 319.92 785.68 102.36 1397.32 0 646.12 3251.4

T48 319.92 785.68 102.36 1397.32 0 646.12 3251.4

T49 619.213 358.85 102.36 1397.32 0 646.12 3123.863

T50 619.213 358.85 102.36 1397.32 0 646.12 3123.863

T51 722.4154 710.88 102.36 1397.32 0 646.12 3579.0954

T52 722.4154 710.88 102.36 1397.32 0 646.12 3579.0954

T53 619.213 358.85 102.36 1397.32 0 646.12 3123.863

T54 619.213 358.85 102.36 1397.32 0 646.12 3123.863

T55 722.4154 710.88 102.36 1397.32 0 646.12 3579.0954

T56 722.4154 710.88 102.36 1397.32 0 646.12 3579.0954

T57 619.213 358.85 102.36 1397.32 0 646.12 3123.863

T58 619.213 358.85 102.36 1397.32 0 646.12 3123.863

T59 319.92 785.68 102.36 1397.32 0 646.12 3251.4

T60 319.92 785.68 102.36 1397.32 0 646.12 3251.4

T61 423.12 1109.52 102.36 1397.32 0 646.12 3678.44

T62 423.12 1109.52 102.36 1397.32 0 646.12 3678.44

GT63 547.883 4268.47 204.72 2497.32 1350 1625.54 10493.933

GT64 414.253 3004.3068 204.72 2497.32 1350 1625.54 9096.1398

Total 34949.8264 68439.2608 6960.48 93828.48 5400 16173.48 225751.5272

Unit of Heat Gain Q is BTU/hr

Total Heat Gain for Zone 2 is Q =225751.5272 BTU/hr

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Zone 3: Main Dining Hall & Kitchen

Serial No. Q Walls Q Glass Q Lights Q Inf Q People Q Total

MDH 2081.55 4586.84 818.88 27000 4491.235 38978.505

Mkitchen 2132.3198 4104.01 511.8 4500 4491.235 15739.3648

Total 4213.8698 8690.85 1330.68 31500 8982.47 54717.8698

Unit of Heat Gain Q is BTU/hr

Total Heat Gain for Zone 3 is Q =54717.8698 BTU/hr

Zone 4: TV Lounge

Serial No. Q Walls Q Glass Q Lights Q Inf Q People Q Total

TV Lounge 2632.84 5950.82 511.8 9000 5188.94 23284.4

Unit of Heat Gain Q is BTU/hr

Total Heat Gain for Zone 4 is Q = 23284.4 BTU/hr

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Manual Calculations

7 Zone1(Residential Rooms):

7.1 Specifications:

Each room direction is 18x12 Ft2

Total number of residential rooms are 129 Ceiling height is taken to be 10 ft. Total occupancy of each room is 4 people Nature of working is regular office work.

7.2 Conduction through exterior surfaces: The cooling load caused by the conduction heat gains through the exterior roofs, walls, windows and glass are each found by the formula

Q = U X A X CLTD

Where Q = Cooling load for roofs, glass or walls in BTU/hr

U = Overall heat co efficient for glass, roofs or walls in BTU/hr-ft2 F

A = Area of roofs, walls or glass in ft2

CLTD = Cooling load temperature difference in F

7.3 Conduction through Walls: Conduction through walls can be found out using the same formula

Qw = Uv X Aw X CLTDVt.

The wall is made up of the following materials having specific thermal resistances.

So overall heat transfer coefficient is

Layer, Inside to Outside Thickness R

Inside surface resistance 0 0.685

4.5 in face brick 4.5 in 0.495

Air space 1 in 1

4.5 in face brick 4.5 in 0.495

Outside surface resistance 0 0.333

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Room 1 & 27 Now conduction through,

South-West Wall:

Area of south-west wall = A = 76.5 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr

East-South Wall:

At ground floor, there is another wall of room is joined with this wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

Area of north-west wall = A = 180 ft2

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 180 X 30 = 1792.8 BTU/hr

East-North Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 308.76 BTU/hr

Total heat gains through walls

Q w = 308.76 + 1792.8+ 888.93+0= 2990.49 BTU/hr

7.4 Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass

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SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr

7.5 Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. In designing the air conditioning of hostel building, every portion of the building has to be conditioned including rooms, toilets, Kitchen, Dining Hall and TV Lounge as well. Therefore, there is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

7.6 Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

7.7 Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

7.8 Conduction through electrical equipment: The heat gain from the electrical equipment plays a major role in the cooling load increment of the space. In hostel building, the laptops are widely used and heat rejected from them must be taken

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into account for the calculation of tonnage of air conditioning. The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr

7.9 Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room1 is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

7.10 Infiltration Load: Q sensible = 1.08 x CFM x (To - Ti)

Q latent = 4840 x CFM x (Wo - Wi)

Q total = 4.5 x CFM x (ho - hi)

Q = 2995.64

Total Cooling load of Each Room 1 & 27: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above.

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 2990.49+4700.72+0+204.72+2995.64+887.12+1800 = 13578.69 BTU/hr

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Room 2, 3, 17, 18, 24, 25 & 26 Now conduction through,

South-West Wall:

Area of south-west wall = A = 76.5 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr

East-South Wall:

At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-North Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 308.76 BTU/hr

Total heat gains through walls

Q w = 308.76 + 0+ 888.93+0= 1197.69 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

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Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0 Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in each Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr

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Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of each Room 2, 3, 17, 18, 24, 25 & 26: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above.

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1197.69+4700.72+0+204.72+2995.64+887.12+1800 = 11785.89 BTU/hr

Room 4 Now conduction through,

South-West Wall:

Area of south-west wall = A = 76.5 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr

East-South Wall:

Area of east-south wall = A = 180 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through south-west wall is

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Q = 0.332 X 180 X 20.5 = 1225.08 BTU/hr

North-West Wall:

At ground floor, there is another wall of room joined with this wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-North Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 308.76 BTU/hr

Total heat gains through walls

Q w = 308.76 + 1225.08+ 888.93+0= 2422.77 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

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Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0 Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

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Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of Room 4: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 2422.77+4700.72+0+204.72+2995.64+887.12+1800 = 13010.33 BTU/hr

Room 5 Now conduction through,

South-West Wall:

Area of south-west wall = A = 180 ft2

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 180 X 35 = 2091.6 BTU/hr

East-South Wall:

Area of east-south wall = A = 76.5 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 20.5 = 520.659 BTU/hr

North-West Wall:

Area of north-west wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 30 = 1195.2 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 597.6 BTU/hr

East-North Wall:

Q=0

Total heat gains through walls

Q w = 2091.6+520.659+597.6= 3209.85 BTU/hr

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Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-south direction = SHGF = 124.5 Cooling load factor for east-south direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 43.5 = 2975.41 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF

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There is no bracket fan used in the building so, QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room5 are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of Room 5: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 3209.85+2975.41+0+204.72+2995.64+887.12+1800 = 12072.1 BTU/hr

Room 6, 7, 37, 38 & 39 Now conduction through,

South-West Wall:

Q = 0 BTU/hr

East-South Wall:

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Area of east-south wall = A = 76.5 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 20.5 = 520.659 BTU/hr

North-West Wall:

Area of north-west wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 30 = 1195.2 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 597.6 BTU/hr

East-North Wall:

Q=0

Total heat gains through walls

Q w = 0+520.659+597.6= 1118.259 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for window

Solar heat gain factor for east-south direction = SHGF = 124.5 Cooling load factor for east-south direction = CLF = 0.825

Heat gain for window is Q = 124.5 X 0.67 X 0.825 X 43.5 = 2975.41 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

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Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

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Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of Room 6, 7, 37, 38 & 39: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1118.259+2975.41+0+204.72+0+887.12+1800 = 6985.509 BTU/hr

Room 8 & 36 Now conduction through,

South-West Wall:

Q = 0 BTU/hr

East-South Wall:

Area of east-south wall = A = 76.5 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 20.5 = 520.659 BTU/hr

North-West Wall:

Area of north-west wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 30 = 1195.2 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 597.6 BTU/hr

East-North Wall:

Area of east-south wall = A = 180 ft2

Considering the wall of type D, for east-north direction CLTDmax =20.5

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So heat gain through east-north wall is

Q = 0.332 X 180 X 15.5 = 926.28 BTU/hr

Total heat gains through walls

Q w = 926.28+520.659+597.6= 2343.34 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-south direction = SHGF = 124.5 Cooling load factor for east-south direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 43.5 = 2975.41 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

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Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of Room 8 & 36: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp Q = 2343.34+2975.41+0+204.72+0+887.12+1800 = 8210.59 BTU/hr

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Room 9 & 35 Now conduction through,

South-West Wall:

Area of wall = A = 120 ft2

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through east-north wall is

Q = 0.332 X 120 X 35 = 1394.4 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 697.2 BTU/hr

East-South Wall:

Area of east-south wall = A = 180 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through south-west wall is

Q = 0.332 X 180 X 20.5 = 1225.08 BTU/hr

North-West Wall:

Q=0

East-North Wall:

Area of east-south wall = A = 76.5 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 76.5 X 15.5 = 393.669 BTU/hr

Total heat gains through walls

Q w = 697.2+0+1225.08+393.669= 2315.949 BTU/hr

Solar radiation through Windows Glass:

Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

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Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-north direction = SHGF = 61 Cooling load factor for east-north direction = CLF = 0.825

Heat gain for window is Q = 61 X 0.67 X 0.86 X 43.5 = 1528.95 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People:

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The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of each Room 9 & 35: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp Q = 2315.949+1528.95+0+204.72+0+887.12+1800 = 6736.739 BTU/hr

Room 10, 11, 12, 33 & 34 Now conduction through,

East-North Wall:

Area of south-west wall = A = 76.6 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =15.5

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 15.5 = 393.669 BTU/hr

East-South Wall:

At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

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At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

South-West Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 120 X 35 = 1394.4 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 697.2 BTU/hr

Total heat gains through walls

Q w = 697.2 + 0+ 393.669+0= 1090.869 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-north direction = SHGF = 61 Cooling load factor for east-north direction = CLF = 0.86

Heat gain for window is Q = 61 X 0.67 X 0.86 X 43.5 = 1528.95 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

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QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in each Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room1 are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room1 is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room1 is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

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Total Cooling load of each Room 10, 11, 12, 33 & 34: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1090.869+1528.95+0+204.72+0+887.12+1800 = 5511.659 BTU/hr

Room 13, 14, 15, 21, 22, 23,29 & 30 Now conduction through,

North-West Wall:

Area of north-west wall = A = 76.6 ft2 (Excluding window)

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 76.5 X 30 = 888.93 BTU/hr

North-East Wall:

At ground floor, there is another wall of room joined from this wall. Therefore, there is no direct exposure of sun light on this wall and heat gain from that wall is considered to be zero i.e. Q =0

South-West Wall:

At ground floor, there is another wall of room joined from this wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-South Wall:

Area of east-south wall = A = 120 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through east-south wall is

Q = 0.332 X 120 X 20.5 = 816.72 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 408.36 BTU/hr

Total heat gains through walls

Q w = 408.36 + 0+ 816.72+0= 1225.08 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF

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Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for window

Solar heat gain factor for north-west direction = SHGF = 132 Cooling load factor for north-west direction = CLF = 0.825

Heat gain for window is Q = 132 X 0.67 X 0.825 X 43.5 = 1528.95 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is

Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF

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Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in each Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of each Room 13, 14, 15, 21, 22, 23,29 & 30: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1090.869+1528.95+0+204.72+0+887.12+1800 = 5511.659 BTU/hr

Room 16, 28 & 20(Electric Room) Now conduction through,

North-West Wall:

Area of north-west wall = A = 76.6 ft2 (Excluding window)

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 76.5 X 30 = 716.94 BTU/hr

North-East Wall:

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At ground floor, there is another wall of room joined from this wall. Therefore, there is no direct exposure of sun light on this wall and heat gain from that wall is considered to be zero i.e. Q =0

South-West Wall:

Area of south-west wall = A = 180 ft2

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 180 X 35 = 2091.6 BTU/hr

East-South Wall:

Area of east-south wall = A = 120 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through east-south wall is

Q = 0.332 X 120 X 20.5 = 816.72 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 408.36 BTU/hr

Total heat gains through walls

Q w = 408.36 + 0+ 2091.6+716.94= 3216.9 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for window

Solar heat gain factor for north-west direction = SHGF = 132 Cooling load factor for north-west direction = CLF = 0.825

Heat gain for window is Q = 132 X 0.67 X 0.825 X 43.5 = 3327.77 BTU/hr

Conduction through interior surfaces:

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The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in each Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room1 are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

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Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room1 is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room1 is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of each Room 16 28 & 20: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 3261.9+3327.77+0+204.72+0+887.12+1800 = 9481.51 BTU/hr

Room 19 (Warden Room) Now conduction through,

South-West Wall:

Area of south-west wall = A = 76.5 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr

East-South Wall:

At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-North Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr

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There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 308.76 BTU/hr

Total heat gains through walls

Q w = 308.76 + 0+ 888.93+0= 1197.69 BTU/hr

7.4 Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 43.5 = 4700.72 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

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Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in this Room is 1 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 1 X [3.412 X 65 X 1] = 221.78 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 1 X 1 = 250 BTU/hr Latent heat gain for Room is

Ql; = 200 X 1 = 200 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 250+200 = 450 BTU/hr

Total Cooling load of Room 19: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1197.69+4700.72+0+204.72+0+221.78+450 = 5013.32 BTU/hr

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Room 31 Now conduction through,

North-West Wall:

Area of north-west wall = A = 76.6 ft2 (Excluding window)

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 76.5 X 30 = 761.94 BTU/hr

North-East Wall:

Area of north-east wall = A = 180 ft2

Considering the wall of type D, for north-east direction CLTDmax =15.5

So heat gain through north-east wall is

Q = 0.332 X 180 X 15.5 = 926.28 BTU/hr

South-West Wall:

At ground floor, there is another wall of room joined from this wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-South Wall:

Area of east-south wall = A = 120 ft2

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through east-south wall is

Q = 0.332 X 120 X 20.5 = 816.72 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 408.36 BTU/hr

Total heat gains through walls

Q w = 408.36 + 761.94+ 926.28+0= 2096.58 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass

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SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for window

Solar heat gain factor for north-west direction = SHGF = 132 Cooling load factor for north-west direction = CLF = 0.865

Heat gain for window is Q = 132 X 0.67 X 0.825 X 43.5 = 3327.77 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in each Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

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Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of Room 31: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 2096.58+3327.77+0+204.72+0+887.12+1800 = 8316.19 BTU/hr

Room 32 Now conduction through,

South-West Wall:

Area of wall = A = 120 ft2

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 120 X 35 = 1394.4 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 697.2 BTU/hr

East-South Wall:

Q = 0 BTU/hr

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North-West Wall:

Area of wall = A = 180 ft2

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 30 X 180 = 1792.8 BTU/hr

East-North Wall:

Area of east-south wall = A = 76.5 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 76.5 X 15.5 = 393.669 BTU/hr

Total heat gains through walls

Q w = 697.2+0+1792.8+393.669= 2893.629 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 7.25 = 43.5 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-north direction = SHGF = 61 Cooling load factor for east-north direction = CLF = 0.825

Heat gain for window is Q = 61 X 0.67 X 0.86 X 43.5 = 1528.95 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

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Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in the Room1 are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

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Qp = Qs + Ql = 1000+800 = 1800 BTU/hr

Total Cooling load of Room 32: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp Q = 2315.949+1528.95+0+204.72+0+887.12+1800 = 6736.739 BTU/hr

Room 41, 42, 56, 57, 63, 64 & 65 Now conduction through,

South-West Wall:

Area of south-west wall = A = 76.5 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr

East-South Wall:

At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-North Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 308.76 BTU/hr

Total heat gains through walls

Q w = 308.76 + 0+ 888.93+0= 1197.69 BTU/hr

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Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 5 = 30 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 30 = 2837.77 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

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QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in each Room are 4 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 4 X [3.412 X 65 X 1] = 887.12 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in each Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for each Room is

Qs = 250 X 4 X 1 = 1000 BTU/hr Latent heat gain for each Room is

Ql; = 200 X 4 = 800 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 1000+800 = 1800 BTU/hr Total Cooling load of each Room 41, 42, 56, 57, 63, 64 & 65: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1197.69+2837.77+0+204.72+0+887.12+1800 = 6927.3 BTU/hr

Room 58 (Warden Room) Now conduction through,

South-West Wall:

Area of south-west wall = A = 76.5 ft2 (Excluding window)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 76.5 X 35 = 888.93 BTU/hr

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East-South Wall:

At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-North Wall:

Area of east-north wall = A = 120 ft2

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 120 X 15.5 = 617.52 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 308.76 BTU/hr

Total heat gains through walls

Q w = 308.76 + 0+ 888.93+0= 1197.69 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula,

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 6 X 5 = 30 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 30 = 2837.77 BTU/hr

Conduction through interior surfaces:

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The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in Room are 2 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 2 X [3.412 X 24 X 1.25 X 1] = 204.72 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of laptops in this Room is 1 The wattage of one laptop doing word processing and mass storage is 65 W So cooling load due to heat gain from laptops is

Qlaptop = 1 X [3.412 X 65 X 1] = 221.78 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Room are 4 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

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Latent heat gain per person is = 200 BTU/hr Sensible heat gain for a Room is

Qs = 250 X 1 X 1 = 250 BTU/hr Latent heat gain for Room is

Ql; = 200 X 1 = 200 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 250+200 = 450 BTU/hr

Total Cooling load of Room 58: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 1197.69+2837.77+0+204.72+0+221.78+450 = 4911.96 BTU/hr

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8 Zone2(Toilets):

8.1 Specifications:

Each toilet direction is 12.5x7 Ft2

Total number of toilets are 64. Ceiling height is taken to be 10 ft.

8.2 Conduction through exterior surfaces: The cooling load caused by the conduction heat gains through the exterior roofs, walls, windows and glass are each found by the formula

Q = U X A X CLTD

Where Q = Cooling load for roofs, glass or walls in BTU/hr

U = Overall heat co efficient for glass, roofs or walls in BTU/hr-ft2 F

A = Area of roofs, walls or glass in ft2

CLTD = Cooling load temperature difference in F

8.3 Conduction through Walls: Conduction through walls can be found out using the same formula

Qw = Uv X Aw X CLTDVt.

The wall is made up of the following materials having specific thermal resistances.

So overall heat transfer coefficient is

Layer, Inside to Outside Thickness R

Inside surface resistance 0 0.685

4.5 in face brick 4.5 in 0.495

Air space 1 in 1

4.5 in face brick 4.5 in 0.495

Outside surface resistance 0 0.333

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Toilet 1, 2, 9, 10, 13, 14, 22, 23, 30, 31, 34 & 35 Now conduction through,

South-West Wall:

Area of south-west wall = A = 56 ft2 (Excluding windows)

Considering the wall of type D, for south-west direction CLTDmax =35

So heat gain through south-west wall is

Q = 0.332 X 56 X 35 = 650.72 BTU/hr

East-South Wall:

At ground floor, there is another wall of room joined from the east-south wall. Therefore, there is no direct exposure of sun light on the east-south wall and heat gain from that wall is considered to be zero i.e. Q =0

North-West Wall:

At ground floor, there is another wall of room joined from the north-west wall. Therefore, there is no direct exposure of sun light on that wall and heat gain from that wall is considered to be zero i.e. Q =0

East-North Wall:

Q=0

Q w = 0 + 0+ 650.72+0= 650.72 BTU/hr

8.4 Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (3.5 X 2) = 14 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67

Now for window Solar heat gain factor for south-west direction = SHGF = 195.5 Cooling load factor for south-west direction = CLF = 0.825

Heat gain for window is Q = 195.5 X 0.67 X 0.825 X 14 = 1512.87 BTU/hr

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8.5 Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

8.6 Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each toilet is 1 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 1 X [3.412 X 24 X 1.25 X 1] = 102.36 BTU/hr

8.7 Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

8.8 Conduction through electrical equipment:

Q= 0 BTU/hr

8.9 Conduction through People: Qp = 0 BTU/hr

Total Cooling load of each Toilet 1, 2, 9, 10, 13, 14, 22, 23, 30, 31, 34 & 35: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 679.77+1512.87+0+102.36+0+1397.32+0 = 3692.32 BTU/hr

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Toilet 3, 4, 15, 16, 24, 25, 36 & 37 Now conduction through,

South-West Wall:

Q = 0 BTU/hr

East-South Wall:

Area of east-south wall = A = 56 ft2 (Excluding windows)

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through east-south wall is

Q = 0.332 X 56 X 20.5 = 398.151 BTU/hr

North-West Wall:

Q =0

East-North Wall:

Q=0

Q w = 0 + 0+ 398.151+0= 398.151 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (3.5 X 2) = 14 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-south direction = SHGF = 124.5 Cooling load factor for east-south direction = CLF = 0.82

Heat gain for window is Q = 124.5 X 0.67 X 0.82 X 14 = 957.60 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be

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QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each toilet is 1 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 1 X [3.412 X 24 X 1.25 X 1] = 102.36 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment:

Q= 0 BTU/hr Conduction through People:

Qp = 0 BTU/hr

Total Cooling load of each Toilet 3, 4, 15, 16, 24, 25, 36 & 37: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 398.151+957.60+0+102.36+0+1397.32+0 = 2855.43 BTU/hr

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Toilet 5, 6, 17, 18, 26, 27, 38 & 39 Now conduction through,

South-West Wall:

Q = 0 BTU/hr

East-South Wall:

Q = 0 BTU/hr

North-West Wall:

Q = 0 BTU/hr

East-North Wall:

Area of east-north wall = A = 56 ft2 (Excluding windows)

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 56 X 15.5 = 301.04 BTU/hr

Q w = 0 + 0+ 301.04+0= 301.04 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (3.5 X 2) = 14 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for east-north direction = SHGF = 61 Cooling load factor for east-north direction = CLF = 0.86

Heat gain for window is Q = 61 X 0.67 X 0.86 X 14 = 492.07 BTU/hr

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Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each toilet is 1 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 1 X [3.412 X 24 X 1.25 X 1] = 102.36 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment:

Q= 0 BTU/hr Conduction through People:

Qp = 0 BTU/hr

Total Cooling load of each Toilet 5, 6, 17, 18, 26, 27, 38 & 39: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 301.04+492.07+0+102.36+0+1397.32+0 = 2292.79 BTU/hr

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Toilet 7, 8, 11, 12, 19, 20, 28, 29, 32, 33, 40 & 41 Now conduction through,

South-West Wall:

Q = 0 BTU/hr

East-South Wall:

Q = 0 BTU/hr

North-West Wall:

Area of north-west wall = A = 56 ft2 (Excluding windows)

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 56 X 30 = 582.66 BTU/hr

East-North Wall:

Q = 0 BTU/hr

Q w = 0 + 0+ 582.66+0= 582.66 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (3.5 X 2) = 14 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for North window

Solar heat gain factor for north-west direction = SHGF = 132 Cooling load factor for north-west direction = CLF = 0.865

Heat gain for window is Q = 132 X 0.67 X 0.865 X 14 = 1071.00 BTU/hr

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Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in each toilet is 1 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 1 X [3.412 X 24 X 1.25 X 1] = 102.36 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

Conduction through electrical equipment:

Q= 0 BTU/hr Conduction through People:

Qp = 0 BTU/hr

Total Cooling load of each Toilet 7, 8, 11, 12, 19, 20, 28, 29, 32, 33, 40 & 41: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 582.66+1071.00+0+102.36+0+1397.32+0 = 3153.34 BTU/hr

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9 Zone3(Main Dining Hall & Kitchen) Main Kitchen:

9.1 Specifications:

Main Kitchen direction is 29.33x37.33 Ft2

Total number of Main Kitchen is 1 Ceiling height is taken to be 10 ft.

9.2 Conduction through exterior surfaces: The cooling load caused by the conduction heat gains through the exterior roofs, walls, windows and glass are each found by the formula

Q = U X A X CLTD

Where Q = Cooling load for roofs, glass or walls in BTU/hr

U = Overall heat co efficient for glass, roofs or walls in BTU/hr-ft2 F

A = Area of roofs, walls or glass in ft2

CLTD = Cooling load temperature difference in F

9.3 Conduction through Walls: Conduction through walls can be found out using the same formula

Qw = Uv X Aw X CLTDVt.

The wall is made up of the following materials having specific thermal resistances.

So overall heat transfer coefficient is

Layer, Inside to Outside Thickness R

Inside surface resistance 0 0.685

4.5 in face brick 4.5 in 0.495

Air space 1 in 1

4.5 in face brick 4.5 in 0.495

Outside surface resistance 0 0.333

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Now conduction through,

North-West Wall:

Q = 0 BTU/hr

East-South Wall:

Area of east-south wall = A = 313.3 ft2 (Excluding windows)

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through east-south wall is

Q = 0.332 X 313.3 X 20.5 = 2132.31 BTU/hr

South-West Wall:

Q = 0 BTU/hr

East-North Wall:

Q = 0 BTU/hr

Q w = 0 + 0+ 0+2132.31= 2132.31 BTU/hr

9.4 Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (7.25 X 6) = 87 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for window

Solar heat gain factor for east-south direction = SHGF = 124.5 Cooling load factor for east-south direction = CLF = 0.82

Heat gain for window is Q = 124.5 X 0.67 X 0.82 X 60 = 4104.01 BTU/hr

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9.5 Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

9.6 Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in Main Kitchen are 5 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 5 X [3.412 X 24 X 1.25 X 1] = 511.8 BTU/hr

9.7 Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

9.8 Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of TV is 0 The wattage of one TV doing word processing and mass storage is 65 W So cooling load due to heat gain from TV is

Qlaptop =0 X [3.412 X 65 X 1] = 0 BTU/hr

9.9 Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

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Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Main Kitchen are 10 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for Main Kitchen is

Qs = 250 X 10 X 1 = 2500 BTU/hr Latent heat gain for Main Kitchen is

Ql; = 200 X 10= 2000 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 2000+2500 = 4500 BTU/hr

Total Cooling load of Main Kitchen: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 2132.31+4104.01+0+511.8+4500+4491.235+0 = 15739.36 BTU/hr

Main Dining Hall Specifications:

Main Dining Hall direction is 53.5x42.08 Ft2

Total number of Main Dining Hall is 1 Ceiling height is taken to be 10 ft.

Conduction through exterior surfaces:

The cooling load caused by the conduction heat gains through the exterior roofs, walls, windows and glass are each found by the formula

Q = U X A X CLTD

Where Q = Cooling load for roofs, glass or walls in BTU/hr

U = Overall heat co efficient for glass, roofs or walls in BTU/hr-ft2 F

A = Area of roofs, walls or glass in ft2

CLTD = Cooling load temperature difference in F

Conduction through Walls:

Conduction through walls can be found out using the same formula

Qw = Uv X Aw X CLTDVt.

The wall is made up of the following materials having specific thermal resistances.

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So overall heat transfer coefficient is

Now conduction through,

North-West Wall:

Q = 0 BTU/hr

East-South Wall:

Q = 0 BTU/hr

South-West Wall:

Q = 0 BTU/hr

East-North Wall:

Area of east-north wall = A = 404.5 ft2 (Excluding windows)

Considering the wall of type D, for east-north direction CLTDmax =15.5

So heat gain through east-north wall is

Q = 0.332 X 404.5 X 15.5 = 2081.55 BTU/hr

Q w = 0 + 0+ 0+2081.55= 2081.55 BTU/hr

Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF

Layer, Inside to Outside Thickness R

Inside surface resistance 0 0.685

4.5 in face brick 4.5 in 0.495

Air space 1 in 1

4.5 in face brick 4.5 in 0.495

Outside surface resistance 0 0.333

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Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (7.25 X 6) = 87 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67 Now for window

Solar heat gain factor for east-north direction = SHGF = 61 Cooling load factor for east-north direction = CLF = 0.86

Heat gain for window is Q = 61 X 0.67 X 0.86 X 130.5 = 4586.84 BTU/hr

Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in Main Dining Hall are 8 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 8 X [3.412 X 24 X 1.25 X 1] = 818.88 BTU/hr

Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

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Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of TV is 0 The wattage of one TV doing word processing and mass storage is 65 W So cooling load due to heat gain from TV is

Qlaptop =0 X [3.412 X 65 X 1] = 0 BTU/hr Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in Main Dining Hall are 60 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for Main Dining Hall is

Qs = 250 X 60 X 1 = 15000 BTU/hr Latent heat gain for Main Dining Hall is

Ql; = 200 X 60= 12000 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 12000+15000 = 27000 BTU/hr

Total Cooling load of Main Dining Hall: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 2081.55+4586.84+0+818.88+27000+4491.235+0 = 38978.505 BTU/hr

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10 Zone4(TV Lounge)

10.1 Specifications:

TV Lounge direction is 42.08x32.75 Ft2

Total number of TV Lounge is 1 Ceiling height is taken to be 10 ft.

10.2 Conduction through exterior surfaces: The cooling load caused by the conduction heat gains through the exterior roofs, walls, windows and glass are each found by the formula

Q = U X A X CLTD

Where Q = Cooling load for roofs, glass or walls in BTU/hr

U = Overall heat co efficient for glass, roofs or walls in BTU/hr-ft2 F

A = Area of roofs, walls or glass in ft2

CLTD = Cooling load temperature difference in F

10.3 Conduction through Walls: Conduction through walls can be found out using the same formula

Qw = Uv X Aw X CLTDVt.

The wall is made up of the following materials having specific thermal resistances.

So overall heat transfer coefficient is

Layer, Inside to Outside Thickness R

Inside surface resistance 0 0.685

4.5 in face brick 4.5 in 0.495

Air space 1 in 1

4.5 in face brick 4.5 in 0.495

Outside surface resistance 0 0.333

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Now conduction through,

North-West Wall:

Area of north-west wall = A = 200 ft2

Considering the wall of type D, for north-west direction CLTDmax =30

So heat gain through north-west wall is

Q = 0.332 X 200 X 30 = 1992 BTU/hr

There is no direct exposure of sun light at east-north wall. there is corridor in front of this wall.

So, Q = 996 BTU/hr

East-South Wall:

Area of east-south wall = A = 240.5 ft2 (Excluding windows)

Considering the wall of type D, for east-south direction CLTDmax =20.5

So heat gain through east-south wall is

Q = 0.332 X 240.5 X 20.5 = 1638.84 BTU/hr

South-West Wall:

Q = 0 BTU/hr

East-North Wall:

Q = 0 BTU/hr

Q w = 0 + 0+ 996+1638.84= 2632.84 BTU/hr

10.4 Solar radiation through Windows Glass: Radiant energy through sun passes through the transparent material such as glass and became a heat gain to a room. The solar cooling load can be found by using the following formula

QG = SHGF X A G X S C X CLF Where SHGF = Maximum solar heat gain factor in BTU/hr-ft2

AG = Area of the glass SC = Shading coefficient CLF = Cooling load factor for a glass.

Area of the window is AG = 2 (7.25 X 6) = 87 ft2

Shading coefficient for the internal shading of Light color Venetian blinds is SC = 0.67

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Now for window Solar heat gain factor for east-south direction = SHGF = 124.5 Cooling load factor for east-south direction = CLF = 0.82

Heat gain for window is Q = 124.5 X 0.67 X 0.82 X 87 = 5950.82 BTU/hr

10.5 Conduction through interior surfaces: The heat that flows through the interior unconditioned spaces to the conditioned spaces through partition, floors and ceilings also increase the cooling load. There is no portion in a floor that remains unconditioned.

Therefore, conduction through interior surfaces is taken to be QINT = 0 BTU/hr

10.6 Conduction through Lightings: The equation for determining cooling load due to heat gain from lighting is

QL = 3.412 X W X BF X CLF Where W = Light capacity in watts BF = Ballast factor CLF = Cooling load factor for lightings

The lights used in Hostel building are recessed and vented. The lights used are fluorescent lights therefore ballast factor i.e. BF = 1.25

Cooling load factor for the light is taken to be 1 in case of operational. Total number of vented lights in TV Lounge are 5 The wattage of one light is 24 W

So cooling load due to heat gain from lighting is Q L = 5 X [3.412 X 24 X 1.25 X 1] = 511.8 BTU/hr

10.7 Conduction through Task Lightings and Bracket Fans: The heat addition through task objects e.g. table lamps, bracket fans and tube lights etc. also increase the cooling load of a building. The heat addition through bracket fans can be calculated as

QBF = 3.412 X W X CLF There is no bracket fan used in the building so,

QBF = 0

10.8 Conduction through electrical equipment: The heat addition through laptops can be calculated as

Qlaptop = 3.412 X W X CLF Cooling load factor for the laptop is taken to be 1 in case of operational. Total number of TV is 1 The wattage of one TV doing word processing and mass storage is 65 W So cooling load due to heat gain from TV is

Qlaptop = 1 X [3.412 X 65 X 1] = 221.78 BTU/hr

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10.9 Conduction through People: The heat gained from people is composed of two parts, sensible heat and latent heat resulting from perspiration. Some of the sensible heat is absorbed by the heat storage effect, but not the latent heat. The equations from sensible and latent heat gains from people are

Qs = q 3 X N X CLF Ql = q l X N

Where qs, ql = Sensible and latent heat gains per person N = Number of people CLF = Cooling load factor for people

Total number of people in TV Lounge are 20 For regular work, Sensible heat gain per person is qs = 250 BTU/hr

Latent heat gain per person is = 200 BTU/hr Sensible heat gain for TV Lounge is

Qs = 250 X 20 X 1 = 5000 BTU/hr Latent heat gain for TV Lounge is

Ql; = 200 X 20= 4000 BTU/hr Total cooling load required for the working of people is

Qp = Qs + Ql = 5000+4000 = 9000 BTU/hr

Total Cooling load of TV Lounge: Total cooling load of the room is simply the combined cooling load of individual components and can be calculated by adding all the cooling loads calculated above

Q = Qw + QG + QlNT + Q L + QBF+ Qlaptop + Qp

Q = 2632.84+5950.82+0+511.8+221.78+9000+5188.94 = 23284.4 BTU/hr

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Total Cooling load of Hostel Building Total Load = Q(Zone1) + Q(Zone2) + Q(Zone3) + Q(Zone4)

Q = 1447764.513+ 225751.5272 + 54717.8698 + 23284.4

Q = 1751518.31 BTU/hr

Q = 1751518.31 TR = 145.959 = 146 TR 12000

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Conclusion: The Manual and Software calculations shall differ slightly because of the assumptions that we make while calculating the values of various constants. The manual and software calculations are tabulated below:

Cooling Load of the building by HAP V5.00 157.399 TR

Cooling load of the building by manual calculations 145.959 TR

Difference 11.44

Percentage Error 7.26 %

The difference between the manual and software calculations arose due to the following reasons:

The Walls of the building are not straight; all the walls are approximately at angle of 45 degrees. Therefore, we have taken the average values between two sides.

The following are the important factors used and their standard values:

1. Cooling Load Temperature Difference(CLTD) North = 19 South = 29 East = 41 West= 12

2. Solar Heat Gain Factor(SHGF) North = 44 South = 171 East = 78 West= 220

3. Cooling Load Factor(CLF)

North = 0.91 South = 0.83 East = 0.81 West= 0.82

The material of which the walls are made is speculated as there was no data available whatsoever regarding the wall material.

The building was approximated to be medium high load. The lobbies were not considered as a part of the space that is to be cooled. The slightly curved walls were taken as straight entities as HAP does not have the option

to process the curved walls.

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References: [1] Ventilation and Infiltration chapter, Fundamentals volume of the ASHRAE Handbook, ASHRAE, Inc., Atlanta, GA, 2005

[2] Swenson, S. Don (1995). HVAC: heating, ventilating, and air conditioning. Homewood, Illinois: American Technical

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[4] Designer's Guide to Ceiling-Based Air Diffusion, Rock and Zhu, ASHRAE, Inc., New York, 2002

[5] Holm L. (2012). Long Term Experiences with Solar District Heating in Denmark. European Sustainable Energy Week,

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[11] Figure 1 Residential HVAC design Process, K; Wang, Xia (2005). "Desiccant cooling air conditioning: a review".

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[31] Table 2 Cooling Load Factors (CLF) for windows Glass. Reproduced with permission from the 1989 ASHRAE Handbook-

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[32] Table 3 Maximum solar heat gain factor for Sunlit glass. Reproduced with permission from the 1989 ASHRAE Handbook-

Fundamentals.

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