2. psychrometric chart

Psychrometric chart
Manoj PJ Associate professor(MECH)1

 Psychrometric chart is prepared to represent
graphically all the necessary moist air properties

 It gives
 Specific humidity.
 RH
 Specific volume of the air-vapour mixture.
 Enthalpy of air-vapour mixture (with datum 0
degree C)

Adiabatic saturation process.

Problem
 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.

Table 2.1 page2.1
 Partial pressure of water vapour

 ω

 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)

Note

 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

 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.

 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.

Sensible heating and Sensible
cooling
 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.

Latent heating and latent
cooling
 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.

Sensible heating

Sensible cooling

By pass factor.

 B*Cp*T1 +(1 – B)Cp*T2 = 1*Cp* T3
 B= (T3-T2)/(T1-T2)
 B= (h3-h2)/(h1-h2)

Also for cooling
 B= (h3-h2)/(h1-h2)
 B= (T3-T2)/(T1-T2)

Chemical Dehumidification
 Example –silica gel

Humidification by steam injection.

Heating and Humidification (winter
air conditioning)

Summer air conditioning (Cooling
and dehumidification).

Mixing of Air Streams

 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.

 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

 .m1*h1+m2*h2=(m1+m2)h3
 109*86+705*46=(109+705)*h3
 .h3=51 kJ/Kg-K

 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)

July 2014(10 marks)
 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.

2013
 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

 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%

2012
 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.

2012
 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

Different heat sources of a room

Sensible heat load of the room
 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.

 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

The latent heat load of the room.
 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.

Sensible heat factor(SHF) or
sensible heat ratio

RSHF (Room Sensible Heat
Factor)

Reference point or circle or
alignment circle
 Tdb = 26°C and ф = 50%

Grand(gross) Sensible Heat Factor
(GSHF)
 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.

 GSHF line indicates the condition of air as it
moves through the cooling coil.

apparatus dew point (coil ADP)
 If the GSHF line is extended, it strikes the
saturation curve known as apparatus dew point
(ADP)

Infiltration load
 The load on the air conditioning unit due to air
leak through doors ,windows etc.
 Infiltration load is considered as room heating
load.

Effective sensible heat factor(ESHF)
 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.

Effective sensible heat factor(ESHF)

ESHF
 Approximate method.
 To relate BPF and ADP.
 Simplify the calculation.

Air Conditioning Processes

 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.

 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,

 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

 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.

 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.

problem
 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

 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

Problem
 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:

 (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.

 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.

Effective temperature.
 is a measure of feeling warmth or cold to the
human body in response to the -air temperature,
moisture content and-air motion.

Effective temperature.
 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.

 Effective temperature is effected by clothing, age
, sex, and degree of work.

Comfort chart.
 The chart which gives different percentages of
people ,feeling comfort at different - effective
temperatures is known as comfort chart.

Factor Governing Optimum
Effective Temperature.

1. Climatic and Seasonal
Differences.
 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.

2. Clothing.
 Light clothing requires less optimum effective
temperature compared with heavy clothing.

3.Age and Sex.
 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.

4. Activity.
 The dancing people require lower effective
temperature whereas the visitors seating in the
dancing room require higher effective
temperature than the dancers.

5. Duration of stay.
 For longer duration use indoor effective
temperature.
 Short duration –use - out-door effective
temperature .
 The thermal shock

6. Air velocity.
 Higher air velocities less effective temperature.

Industrial application.
 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

2. psychrometric chart

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