Chapter 6. Space Heating Load

• Heating Load:

1- Heat loss through walls, windows,…etc

2- Heat to warm outside air (1) desired (2) infiltration

• The ideal heating system would provide just enough heat to match heat

loss from structure.

• Load is not a constant

• Outdoor temp changes as well as wind velocity, sunlight (solar), internal

heat gains, appliances, people.

• Heating system must handle the coldest “outside” average temperature of

the year.

• Most equipment operate at partial load most of the time

• Unheated Rooms:

– To< Tu< Ti – Assume steady state heat transfer

)( oi TTUAq −=&

The following Factors should be considered before

performing heat load calculations:

1. Is the heat capacity of the structure high or low ?

2. Is the structure insulated ?

3. Is the structure exposed to high wind ?

4. Is the ventilation or infiltration load high ?

5. Is there more glass area than normal in the structure ?

6. During what part of the day will the structure be used ?

7. What is the nature of occupancy ?

8. Will there be long periods of operation at reduced indoor temperature ?

9. What is the amplitude between maximum and minimum daily temperature in the locality ?

10. Are there local conditions that cause significant variation fromtemperatures listed ?

11. What auxiliary heating devices will be in the building ?

12. What is the expected cost of fuel or energy ?

Chapter 4 – Design Temp for inside. Low end (70ºF at Ø= 30%) of comfort zone

Representative inside air design temperature

Air change method, ACH Air Change/HR

0.1 as low as becan and 0.25.0

hourper changeair ofnumber ACH

))((

Re →=

=

sidential

T

ACH

C

VACHQ&

•Experience shows nominal values for air changes per hour

•Experience shows nominal values for air changes per hour

Crack Method

•Q=AC ∆∆∆∆Pn

•A= effective leakage area of the cracks

•C= flow coefficients, which depends on the type of crack and nature

of flow in the crack

•∆P=outside - inside pressure difference, Po-Pi

•n=exponent that depends on the nature of the flow crack 0.4<n<1.0

• Crack method is complex but in general more accurate.

• We need to be able to calculate ∆ P to put in the equation

essureBuildingSw PPPP Pr∆+∆+∆=∆

∆Pw=Pressure difference due to wind

Cp= function of angle of wind, Height and length to width ratio of building

∆Ps=Pressure difference stack effect

Cd range from 1 for buildings with no doors in the stairways to about 0.65-0.85

for modern office buildings

∆Pp=Pressure difference due to building pressurization

c

pwwind

g

VwCPP

2

2ρ=∆=∆

)11

(

iToTcgaR

ghoPd

C

sP −=∆

• Figure 6.1 Windows and door infiltration Characteristic

• Figure 6.2 Variation of wall averaged pressure coefficients for low rise buildings

• Figure 6.3 Variation of wall averaged pressure coefficients for tall buildings

• Figure 6.4 Average roof pressure coefficients for a tall building

• Figure 6.5 Pressure difference due to stack effect.

• Figure 6.6 Curtain wall infiltration for one room or one room floor

• Figure 6.7 Infiltration through cracks around a closed swinging door

• Figure 6.8 Swinging door infiltration characteristics with traffic

• Figure 6.9 Flow coefficient dependence on traffic rate

• Supply Air for Space Heating

• Source Media for space heating

∆

Figure 6.1 Windows and door

infiltration Characteristic

Fig. 6.2 Variation of wall averaged pressure

coefficients for low rise buildings

Fig. 6.3 Variation of wall averaged pressure

coefficients for tall buildings

Fig. 6. 4 Average roof pressure

coefficients for a tall building

Fig. 6. 5 Pressure difference due to

stack effect.

Fig.6.6 Curtain wall infiltration for

one room or one room floor

Fig. 6.7 Infiltration through cracks

around a closed swinging door

Fig. 6.8 Swinging door infiltration

characteristics with trafficFigure 6.9 Flow coefficient

dependence on traffic rate

Procedure for calculating heat losses from a structure:

1. Select the design outdoor weather conditions: temperature, humidity, wind

direction, and speed.

2. Select the design indoor air temperature to be maintained in each heated

space when outside design conditions exist.

3. Estimate temperatures in adjacent un-heated spaces.

4. Select or compute the heat transmission coefficient (U factors) for all

building components through which heat losses are to be calculated.

5. Determine all surface areas through which the heat is lost.

6. Compute heat transmission losses for each kind of wall, glass, floor, ceiling,

and roof in the building by multiplying the heat transmission coefficient in

each case by the area of the surface and the temperature difference between

indoor and outdoor air, or adjacent unheated space.

7. Compute heat losses from basement or grade-level slab floors by methods

presented in Chapter 5.

8. Select infiltration air quantities and compute the heat load caused by infiltration around doors and windows.

9. When positive ventilation using outdoor air is provided by an air-heating or air-conditioning unit, the energy required to warm the outdoor air to room temperature must be provided by the unit. The principle for calculation of this load component is identical to that for infiltration. If mechanical exhaust from the room is provided in an amount equal to the out-door air drawn in by the unit, the unit must also provide for the natural infiltration losses. If no mechanical exhaust is used, and the outdoor ventilation air sup-plied equals or exceeds the natural infiltration that would occur without ventilation, natural infiltration may be neglected.

10. The sum of the transmission losses or heat transmitted through the confining walls, floors, ceiling, glass, and other surfaces, plus the energy associated with the cold entering air by infiltration or required to replace mechanical exhaust, represents the total heating load.

11. In buildings that have a reasonably steady internal heat release of appreciable magnitude from sources other than the heating system, a computation of this heat release under design conditions should be made and deducted from the total of the heat losses computed above.

12. Additional heating capacity may be required for intermittently heated buildings to bring the temperature of the air, confining surfaces, and building contents to the design indoor temperature within a specified time.

Source Media for space heating

“Hot water as heating media”

q = mw cp (t2 – t1)

where:

q = heating required, Btu/hr or W

mw = mass flow rate of hot water, lbm/hr or kg/s

cp = specific heat of water, Btu/lbm or kJ/(kg-C)

t2 = water temperature leaving coil, F or C

t1 = water temperature entering coil, F or C

Assuming that cp is constant

q = 500 Qgpm(t2 – t1) English Units

q = 4.2 QL/s(t2 – t1) SI units

Source Media for space heating

“Steam as heating fluid”

q = mv (i2 – i1)

where:

q = heating required, Btu/hr or W

mv = mass flow rate of the vapor, lbm/hr or kg/s

i2 = enthalpy of the vapor leaving the coil, Btu/lbm or kJ/kg

i1 = enthalpy of the vapor entering the coil, Btu/lbm or kJ/kg

If saturated vapor is the heating medium,

i2 – i1 = the enthalpy of vaporization, i fg.

Source Media for space heating

“Furnace Heating”

In the case of a furnace where combustion gases heat the

air directly, the heating value of the fuel and a furnace

efficiency must be known. A general relation from which

m f or Q f can be found is

q f= mf (HV)ηηηη = Qf (HV)ηηηηwhere:

q f = heating required, Btu/hr or W

mf = rate at which fuel is used, lbm/hr or kg/s

Qf = Volume flow rate ft3/min or m3/s

HV = heating value of the fuel, Btu/lbm or kJ/kg

η = furnace efficiency