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Psychrometric chart Manoj PJ Associate professor(MECH) 1

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Page 1: 2. psychrometric chart

Psychrometric chart

Manoj PJ Associate professor(MECH)1

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Manoj PJ Associate professor(MECH)2

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Psychrometric chart is prepared to represent

graphically all the necessary moist air properties

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It gives

Specific humidity.

RH

Specific volume of the air-vapour mixture.

Enthalpy of air-vapour mixture (with datum 0

degree C)

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Adiabatic saturation process.

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Problem

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The sling psychrometer reads 40dgree C DBT

and 28 degree C WBT. Calculate.

Sp humidity , Relative humidity, Dew point

temperature, enthalpy , specific volume / kg of

dry air

Assume atmospheric pressure to be 1.03 bar.

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Table 2.1 page2.1

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Partial pressure of water vapour

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ω

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Due point temperature is the saturation

temperature of the water vapour at the existing

pressure of water vapour.

From the steam table, the saturation temperature

at 0.03038 bar is 25 degree C

DPT= 25 degree C (table 2.1 page 2.1)

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Note

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Using chart

RH= 42%

Specific humidity= 0.019 kg/ kg of dry air

.h= 90 KJ/kg of dry air

Specific volume= 0.9 cu.m/kg

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A room contains humid air Tdb = 25 0C and wet

bulb temperature 19 0C . Calculate (a) the relative

humidity, (b) specific humidity, and (c) dew point

Assume standard atmospheric pressure.

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Using chart

RH=56%

Specific humidity= 0.012 kg/kg of dry air.

.h= 55 kJ/kg of dry air

Sp vol= 0.85 m3/kg.

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Sensible heating and Sensible

cooling

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Heating or cooling of air without addition or

subtraction of moisture is termed as sensible

heating or cooling.

Heating the air by a electric heater.

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Latent heating and latent

cooling

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Heating or cooling of air due to addition or

subtraction of moisture is termed as latent

heating or cooling.

Ex. Steam emitted from a hot food.

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Sensible heating

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Sensible cooling

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By pass factor.

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B*Cp*T1 +(1 – B)Cp*T2 = 1*Cp* T3

B= (T3-T2)/(T1-T2)

B= (h3-h2)/(h1-h2)

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Also for cooling

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B= (h3-h2)/(h1-h2)

B= (T3-T2)/(T1-T2)

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Chemical Dehumidification

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Example –silica gel

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Humidification by steam injection.

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Heating and Humidification (winter

air conditioning)

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Summer air conditioning (Cooling

and dehumidification).

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Mixing of Air Streams

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The makeup air at rate of 100 m3/min from the

environment having tdb = 40°C and twb = 27°C is

mixed with 600 m3/min of return air from the

conditioned space having state tdb = 23°C and

relative humidity 50%. Compute dry and wet-bulb

temperatures and specific humidity of the mixture.

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At state 1

Specific vol= 0.913 cu m / kg

Sp humidity= 0.017 kg/ kg

.h1= 86 KJ/Kg-K

At state 2

Specific vol= 0.852 cu m / kg

Sp humidity= 0.009 kg/ kg

.h2= 67 KJ/Kg-K

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.m1*h1+m2*h2=(m1+m2)h3

109*86+705*46=(109+705)*h3

.h3=51 kJ/Kg-K

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Locate the point 3

DBT at point 3= 26 degree C,

Specific humidity at 3=0.0102 Kg/ Kg of dry air

WBT=18 degree C

(point 3 can also be located by dividing the line 1-

2 in m1/m2 ratio)

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July 2014(10 marks)

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One kg. of air at 40°C DBT and 50 % R.H. is

mixed with two kg. of air at 20°C DBT and 12°C

dew-point temperature. Calculate the temperature

and specific humidity of the mixture.

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2013

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A mixture of dry air and water vapour is at a

temperature of 21°C. The dew point temperature

is 15°C.Determine:(a) Partial pressure of water.

(b) Relative humidity. (c) Specific humidity.

Assume atmospheric pressure as 1.03 bar

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Ref table 2.1 page 2.1

Partial pressure of water vapour= saturation

pressure at DPT , that is 150C = 0.017 Bar

RH = Pv/ Pvs -------(1)

But Pvs= Saturation pressure at DBT , That is at

210C

= 0.0249 Bar

From (1) RH= 0.017/0.0249 = 68%

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2012

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The atmospheric air at 25°C DBT and 12°C WBT

is flowing at the rate of 100 cubic m/minute

through the duct. Dry saturated steam at 100°C is

injected into air stream at the rate of 72 kg/hr.

Calculate specific humidity and enthalpy of

leaving air. Also determine dry bulb temperature,

wet bulb temperature and relative humidity of

leaving air.

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2012

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800 cubic m /minute of re-circulated air at 22°C

DBT and l0°C dew point temperature is to be

mixed with 300 cubic m/minute of fresh air at

30°C DBT and 50% RH. Determine enthalpy,

specific volume, humidity ratio and dew point

temperature of the mixture

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Different heat sources of a room

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Sensible heat load of the room

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I. Heat flows through the exterior walls, ceilings,

floors, windows and doors due to the temperature

difference between their two sides.

2. Load due to solar radiation (sun load) is

divided into two forms.

(a) Heat transmitted directly by radiation through

glass of windows and ventilators.

b) Heat from sun will be absorbed by the walls

and roof and later on transferred to room by

conduction.

3. Heat received from the occupants.

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4. Heat received from different equipments which

are commonly used in the air-conditioned

building.

5. Heat received from the infiltrated air from

outside through cracks in doors, windows and

ventilators and through their frequent openings.

6. Miscellaneous heat sources which include the

followings

(a) Heat gain by the ducts carrying the

conditioned air and passing through

unconditioned space.

(b) Heat transferred through interior partition of

rooms in the same building which are not air-

conditioned.

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The latent heat load of the room.

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1. The latent heat load from the air entering into

the air-conditioned space by infiltration.

2. The latent heat load from the occupants.

3. The latent heat load from cooking foods and

from stored materials.

4. Moisture passing directly into the air-

conditioned space through permeable walls

where the water vapour pressure is higher.

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Sensible heat factor(SHF) or

sensible heat ratio

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RSHF (Room Sensible Heat

Factor)

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Reference point or circle or

alignment circle

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Tdb = 26°C and ф = 50%

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Grand(gross) Sensible Heat Factor

(GSHF)

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Grand total heat load = room heating load +

outdoor load on the air conditioning unit due to

mixing of fresh air

Line joining mixture condition(after mixing with

fresh air) to ADP.

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GSHF line indicates the condition of air as it

moves through the cooling coil.

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apparatus dew point (coil ADP)

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If the GSHF line is extended, it strikes the

saturation curve known as apparatus dew point

(ADP)

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Infiltration load

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The load on the air conditioning unit due to air

leak through doors ,windows etc.

Infiltration load is considered as room heating

load.

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Effective sensible heat factor(ESHF)

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It is the line connecting room desired condition to

ADP(coil ADP)

Effective sensible heat= room sensible heat +

portion of the out door air sensible heat which is

considered as being bypassed through the

conditioning coil.

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Effective sensible heat factor(ESHF)

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ESHF

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Approximate method.

To relate BPF and ADP.

Simplify the calculation.

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Air Conditioning Processes

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An office for seating 30 occupants is to be maintained at 22°C DBTand55% RH. The outdoor conditions are 36°C DBT and 27°C WBT.

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The various loads In the office are:

Solar heat gain 8500W,

Sensible heat gain per occupant 83W,

Latent heat gain per occupant 100W,

Lighting load 2500W,

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Sensible heat load from other sources 12000W,

infiltration load 15 cubic meter/minute .

Assuming 40% fresh air and 60% of re-circulated air passing through the evapourator coil and

ADP of the coil is 8 0C.

Find capacity of the plant and

mass flow rate of air

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The flow diagram for the given air-conditioning

system is shown in Fig.

• Locate point I at the intersection of 36°C DBT

and 27°C WBT lines.

• Locale point 2 at the intersection of 22°C DBT

line and 55% RH curve.

• Locate point A by drawing vertical and horizontal

lines through points I and 2 respectively.

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Since bypass factor is 0.15, and ADP is8 degree

Divide line 1-2 in 4: 6

Mark point 3 near to 2

Since ADP is 8 0Cdegree c, draw line 3-6 . Find intersection 4.

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problem

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An air conditioned space is maintained at 260 C

DBT and 50% RH. When out side air conditions

are 350C DBT and 280C WBT.

(a) if the space has a sensible heat gain of 17.6

kW and air is supplied to the room at a condition

of 80C saturated, calculate

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1. the mass and volume flow rate of air supplied

to the room.

2. the latent heat gain of the space

3. the cooling load of the refrigeration plant if 25%

of total weight of the air supplied to the space is

fresh air and the reminder is recalculated air

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Problem

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An air-conditioned plant is to be designed for a

small office room for winter conditions.

Out-door conditions = 10°C DBT and 8°C WBT.

Required indoor-conditions = 20°C DBT and 60%

R.H.

Amount of free air circulation = 0.3

m3/min/person.

Seating capacity of the office = 50.

The required condition is achieved first by heating

and then by adiabatic humidifying. Find the

followings:

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(a) Heating capacity of the coil in kW and the

surface temperature required if the bypass factor

of the coil is 0.32.

(b) The moisture added per kg of dry air in the

humidifer.

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Locate the points ‘a’ and ‘c’ on the psychrometric

chart as their conditions are known and then draw

a constant enthalpy line through ‘c’ and constant

specific humidity line through a. The point b is

located as an intersection of the above two

mentioned lines.

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Effective temperature.

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is a measure of feeling warmth or cold to the

human body in response to the -air temperature,

moisture content and-air motion.

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Effective temperature.

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It is the dry bulb temperature of a sample of

saturated air which will give a particular feeling of

comfort to the same percentage of people as any

other combination of dry bulb temperature and

relative humidity.

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Effective temperature is effected by clothing, age

, sex, and degree of work.

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Comfort chart.

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The chart which gives different percentages of

people ,feeling comfort at different - effective

temperatures is known as comfort chart.

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Factor Governing Optimum

Effective Temperature.

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1. Climatic and Seasonal

Differences.

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The people living in colder climates feel

comfortable at lower effective than the people

living in warmer regions.

The comfort chart shows that the optimum

effective temperature in winter is 19°C is shifted

to 22°C optimum effective temperature in

summer.

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2. Clothing.

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Light clothing requires less optimum effective

temperature compared with heavy clothing.

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3.Age and Sex.

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The metabolic rate of women is less than man by

nature itself. So the women require greater

effective temperature (1°C) than the man.

Similar, case exists for young and old people

also,

The children require higher effective temperature

compared with adults.

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4. Activity.

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The dancing people require lower effective

temperature whereas the visitors seating in the

dancing room require higher effective

temperature than the dancers.

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5. Duration of stay.

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For longer duration use indoor effective

temperature.

Short duration –use - out-door effective

temperature .

The thermal shock

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6. Air velocity.

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Higher air velocities less effective temperature.

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Industrial application.

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manufacturing chemicals,

petroleum refinery - crystallize wax and separate it out, fractional distillation of the lighter hydrocarbons .

Rubber industries

paper and pulp industries,

where one of the main purposes is to It is also needed. It has also applications in many heat treatment.

Ice plants.

food preservation.

Transport refrigeration.

Marine application