14 Psychrometric Chart

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SKF 2123 - Energy Balance

Instructor: Assoc. Prof. Dr Azeman Mustafa

Balances on Nonreactive Processes© JKG- FKKKSA Page 8 - 1

Psychrometric Chart (or Humidity Chart)

• A psychrometric chart is a graphical representation of thethermodynamic properties of air-water system.

• There are a number of different charts. They differ withrespect to the barometric pressure, range of temperature, andthe choice of coordinates.

• The most popular charts have absolute humidity plotted againstdry-bulb temperature.

• Most published charts are for standard barometric pressure (1atm).








The psychrometric chart has seven lines.

A Absolute

Humidity Axis

B Specific VolumeC

Dry Bulb Axis

D

Relative

Humidity

D

Dew Point E

Enthalpy

F

Wet Bulb &

Saturation Curve

G






Description of the psychrometric chart

• The horizontal axis forms the dry-bulb temperature scale and vertical

lines are lines of constant dry-bulb temperature.• The dry-bulb temperature is the air temperature as measured by a

thermometer, thermocouple or other conventional temperature-measuringdevice

Dry Bulb Temperature (oC or oF)

constant dry-bulbTemperature (T db)


• The vertical axis is the absolute humidity(or humidity ratio, moisture content)scale and horizontal lines are lines ofconstant humidity.

AbsoluteHumidity

ormoisturecontent

[kg H2O(v)/kgDry Air (DA)]

constanthumidity






• Curves lines are used to represent relative humidity (RH or hT ).

Relative humidity, ( )100%x

Tp

ph

* OH

OH

T

2

2=


• The line of saturation (100% RH) is used to provide the dew-pointtemperature (T dp) and the wet-bulb temperature (T wb) scales.






• Lines of constant dew-point are horizontal lines.


• Constant wet-bulb temperatures form diagonal lines on the chart.






• A different set of diagonal lines represent lines of constant specificvolume.


• Lines of constant enthalpy at saturation are closely aligned to the lines ofconstant wet-bulb many charts use a single set of lines to represent constantwet-bulb temperature and constant enthalpy . An enthalpy scale is typicallyplaced above the saturation line.

• Reference states for Enthalpy :- liquid water at 1 atm & 0oC, dry air at 1 atm &0oC






• Enthalpy deviation – to correct the enthalpy of humid air that is not saturated(see psychometric chart)

Enthalpydeviation


Example 13 - How to read the psychometricchart?

The dry bulb reading is 78°F and the wet bulb is 65°F. Using the chart determinethe following values. What is the:

Relative Humidity? 50 percent

Dew Point? 57.8 degrees

Absolute Humidity? 0.0102 lbs. water / lb. dry air

Specific Volume?

13.78 cu. ft. / lb. dry air Enthalphy?

30 BTU / lbs. dry air The amount of water in 1 ft3 of air?

7.4 x 10-4 lbs. water





Balances on Nonreactive Processes© JKG- FKKKSA - 15

Working session 11 – Problem 8.69(a)

The latest weather report states that the temperature is 24°C and therelative humidity is 50%.

a. A thermometer is mounted at the back porch of your house. Whattemperature would it read

b. Using the chart determine the following values.

i. Wet bulb temperature (oC)ii. Dew Point (oC)iii. Absolute Humidity (kg H2O(v)/kg DA)iv. Specific Volume (m3/kg DA)v. Specific Enthalphy (kJ/kg DA)vi. The amount of water (kg) in 1 m3 of humid air


Working session 11 – Problem 8.69(a)

i. Wet bulb temperature = 17 oC

ii. Dew Point = 13 oC

iii. Absolute Humidity (kg H2O(v)/kg DA) = 0.0092 kg H2O(v)/kg DA

iv. Specific Volume (m3/kg DA) = 0.855 m3/kg DA

v. Specific Enthalphy (kJ/kg DA) = (48 – 0.2) kJ/kg DA = 47.8 kJ/kg DA

vi. The amount of water (kg) in 1 m3 of humid air






What happens during heating and cooling atconstant pressure?

a. There is no change in the absolute humidity of the air-vapormixture, as long as no condensation occurs.

b. Cooling occurs from right to left… if superheated humid is cooledthe system follows a horizontal path to the left until thesaturation curve (dew pt temp & 100% RH); thereafter the gasphase follows the saturation curve

c. Heating occurs from left to right.

d. There is a change in the sensible heat of

the air-vapor mixture.

e. Heat must be added or subtracted

to cause the temperature change.

Dry Bulb Temperature

Heating

Cooling


Example 14 – Problem 8.69(a, modified)

The latest weather report states that the temperature is 24°C and the relative humidity is50%. A sample of outside air is cooled at constant pressure.

a. At what temperature would condensation begin?

T dp = 13 oC

b. Calculate the rate at which heat (kJ/s) must be removed for every 1 m3/s of humid airbefore condensation occurs. (draw a process path on the psychometric chart)

T 1 = T db=24oC

T 2=13oCRH =50%

1 m3 /s HAmDA kg/s dry air24°CRH= 50%

mDA kg/s dry air13 °CRH=100%

Q= ? KJ






Example 14 – Problem 8.69(a, modified)

Simplified energy balance, ….. Ĥi (kJ/kg DA)

Mass of dry air,

Rate of heat removed,

12DA HHmQ −= &&

DAs

kg 1.1696

HAm 0.855

DA kg 1.00 x(HA)AirHumid

s

m 1.00m

3

3

DA

=

=

( ) ( )

s

kJ 13.22-

DAkg

kJ47.8-36.5DA

s

kg 1.1696HHmQ 12DA

=

=−=


Working session 12 – Recall Example 14

c. If the air is cooled to 10 °C at constant pressure, calculate the fraction of water thatcondenses and the rate at which heat (KJ/s) must be removed for every 1 m3/s of humidair. (draw a process path on the psychometric chart)

Psychometric chart

Mass balance,

1 m3/s HAmDA kg/s dry airm1 kg/s water24°CRH= 50%

mDA kg/s dry airm2= kg/s water10 °CRH=100%

10°Cm3= kg/s water condensed

Q= ? KJ







From figure 8.4.1

The fraction of water that condenses,

OHs

kg 0.0019m

OHs

kg 0.0089

DAkg

OHkg 0.0076DA

s

kg 1.1696m

OHs

kg 0.0108

DAkg

OHkg 0.0092DA

s

kg 1.1696m

mmm

23

22

2

22

1

321

=∴

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =

+=

&

&

&

&&&

DAkg

OHkg 0.0076h

DAkg

OHkg 0.0092h

2a,2

2a,1

=

=

0.176fedOHkg0.0108

condensedOHkg0.0019

m

m

2

2

1

3

==∴



Simplified energy balance,

Reference states : liquid water at 1 atm & 0oC, dry air at 1 atm & 0oC

The values of mDA (kg DA) & m3 (kg condensed water) are as previously calculated The values of Ĥ1 & Ĥ2 in kJ/kg DA are obtained from figure 8.4.2 The value of Ĥ3 in kJ/kg is relative to the specific enthalpy of liquid water at 1 atm &

0oC

∑∑ −=in

ii

out

ii HmHmQ &&&

Ĥ3 kJ/kgm3 kg--H2O (l)

Ĥ2 kJ/kg DAmDA kg DAĤ1 kJ/kg DAmDA kg DAHumid Air

Ĥout(kJ/kg)

@10oC

m out (kg)Ĥin(kJ/kg)@24oC

m in (kg)Substance






Working session 13

Calculate the amount of sensible heat that must be added to 10 kg of airat 13°c dry-bulb and 40% relative humidity to raise the temperature ofthe air to 32°C dry bulb. What is the corresponding relative humidity?

13 32

40%

Enthalpy

H2

H1

oC


Working session 13

a. Locate the 13°C dry bulb and 40% RH point.

b. Follow the enthalpies line to Ĥ1.

c. Move horizontally to intercept the vertical line of 32°C dry bulb

d. Follow enthalpies line Ĥ2.

e. Ĥ2 = 42-0.45 = 41.55 kJ/kg DA.

f. Ĥ1 = 22.75-0.075 = 22.675 kJ/kg DA.

40%

Enthalpy

Ĥ2

Ĥ1

13 32oC

kJ188.75 QDAkg

kJ22.675)-(41.55kg10Q

)H-H(mQ 12

=

=

=

The relative humidity is now ~ 12%.






What is humidification and dehumidification?

A. Humidification adds moisture to the air which increase the absolutehumidity.

B. Dehumidificaiton remove moisture from the air which decrease theabsolute humidity.

Dehumidification

Humidification


How is humidity increased?

A. Water is added in vapor form.

B. Water is converted from liquid to gas.

C. There is an increase in the energy level.

Humidification






How is humidity decreased?

A. There is a change from gas to liquid.

B. There is a decrease in the energy level.

C. With the loss of energy, condensation occurs.

Dehumidification


Example 15

How much moisture is needed to raise the relative humidity of an air-vapor mixture from 12% to 50% and the temperature stays at 75°F.

75°

12%

50%

0.0094

0.0022

lbs water/

lb dry airWater needed,

= final absolute humidity - initial absolute humidity

= 0.0094 - 0.0022 = 0.0072 lbs water / lb dry air






Working session 14

How much heat is needed to evaporate the water that’s needed to increasethe relative humidity from 12% to 50% and the temperature stays at 75°.

75°db

12%

50%

0.0094

0.0022

lbs water/

lb dry air


Working session 14 :- The amount of heat ….

BTU needed = final enthalpy - initial enthalpy

BTU needed = 28.0 BTU/lb dry air - 21.0 BTU’s/lb dry air

BTU needed = 7.0 BTU per lb dry air

75°db

12%

50%H1

H2






Adiabatic Cooling

• Adiabatic cooling is said to have taken place when a warm gas isbrought into contact with a cold liquid, causing the gas to cool andsome of the liquid to evaporate and with an assumption that no heatis lost to its surrounding during the gas-to-liquid heat transferprocess.

• Examples of common adiabatic cooling process

• Spray cooling , spray humidification• Spray dehumidification• Drying• Spray drying


Adiabatic Cooling Process

• Air undergoing adiabatic cooling through contact with liquid watermoves along a constant wet-bulb temperature line on thepsychrometric chart from its initial condition towards the 100%relative humidity curve






Adiabatic Saturation Temperature (T as)

Assuming;

1. (Cp)air, (Cp)H2O, and (∆Hvap)H2O, are independent of temperature at theprevailing process conditions.

2. The enthalpy changes undergone by the unevaporated liquid water and thesolid (if there is one) in going fromT

2 to T

4 are negligible compared to the

changes undergone by the entering wet air and the evaporated water.

3. The heat required to raise liquid water from T 2 to T 3 is negligible comparedto the heat of vaporization of water.


and the process is adiabatic (Q=0), then the simplified energyequation becomes;

If the outlet temperature T 3 is low enough, the air leaves

saturated with water (i.e. RH = 100%) then the temperaturecorresponding to this condition is called the adiabaticsaturation temperature (T as) & happens to coincide with the wetbulb temperature (i.e. T as = T wb)






What is evaporative cooling?

Through evaporation, moisture in theair accumulates and the airtemperature decreases.

Sensible heat from the air vaporizeswater from its liquid to gaseous phase.

There is no loss or gain of heat withinthe system because the amount ofsensible heat removed equals latentheat added to the water.

The process follows a constantenthalpy line and hence the processis also known as adiabatic cooling

The maximum temperature reductionis the difference between startingdry bulb temperature and its wet bulbtemperature.

Lowering of dry bulb temp

Addition ofmoisture

Evaporative cooling lowers the dry bulbtemperature. It is used where a lower

temperature is desired.


Example 16

A chicken producer is considering installing evaporative cooling ina breeding herd building. The outside air has a 32°C dry bulbtemperature and 35 percent relative humidity,

a. What is the lowest temperature that can theoretically beobtained from the air coming off the cooling pads. (~ 21oC )

b. However, due to the inefficiency of evaporative coolers, thetemperature of the air coming off the cooling pads will be 2oCabove the lowest attainable temperature in part (a). Estimate therelative humidity under these conditions. (~ 23oC , RH~85%)






Working session 15

A chicken producer is considering installing an evaporativecooling system in a breeding herd building. Air enters thecooler at 35°C dry bulb and 30% relative humidity andleaves at 27°C

a. Determine the amount of water (kg/hr) needed for anevaporative cooler when 1000 m3/hr humid air passesthrough a livestock building per hour.

b. Calculate also the flowrate (m3/hr) of air entering thecooler and the corresponding relative humidity of theair leaving the cooler.


Working session 15

0.0143

0.0106

kg water/

kg dry air

35°27°

30%

Dry Bulb Temp.

Vhumid ~ 0.8694 m3/kg DA 0.888 m3/kg DA






a. Read absolute humidity value for 35°C and 30% relative humidity y1=0.0106 kg water/kg DA

b. Extend enthalphy line toward saturation curve.c. Read absolute humidity value for 27°C

y3=0.0143 kg water/kg DA.

d. Water added/kg dry air:0.0143 - 0.0106 = 0.0037 kg water/kg DA

e. Kg of dry air required= (1000 m3 air /hr) x (kg DA /0.8694 m3 air) = 1150kg DA /hr

f. Total water added= 0.0037 kg water/kg DA x 1150 kg DA/hr = 4.256 kg/hr water

g. Air flowrate entering a cooler= 1150 kg DA /hr x 0.888 m3/kg DA = 1021 m3/hr

h. Relative humidity of air leaving a cooler

= ~ 65%

Working session 15


Question #2 – Final Exam 2005/06

Analysis of an air stream gives a dry bulb and wet bulb temperature of25°C and 18°C, respectively. Based on this information, answer thefollowing questions. You are also required to show your work on the chartprovided.

a. What is the relative humidity? (~51%)b. If the stream is passed through a 100% efficient evaporative cooler, it

will be fully saturated. In this condition, what will be the temperature?(18°C)

c. If the air stream enters a storage room with a temperature of 0°C and arelative humidity of 95%, will it increase or decrease the moisture

content. Show your calculations. (0.01 kg/kg DA 0.0038 kg/kg DA,moisture content decreases)d. If the air that leaves a wet-coil evaporator at 0°C and 100% relative

humidity is heated to 2°C, what is its relative humidity? (~80%).

14 Psychrometric Chart

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