EGR 334 ThermodynamicsChapter 12: Sections 5-7

Lecture 39: Humidity andPsychrometric Applications Quiz Today?

Today’s main concepts:• Demonstrate understanding of psychrometric

terminology, including humidity ratio, relative humidity, mixture enthalpy, and dew point temperature.

• Apply mass, energy, and entropy balances to analyze air-conditioning processes.

Reading Assignment:

Homework Assignment:

Read Chapter 13, Sections 1-5

Problems from Chap 12: 46,51, 55, 67

3Sec 12.5 : Psychrometric applications

Greek: psuchra = coldMetron = measure

Psychrometric: Study of systems containing “dry air” and water vapor

May also include condensed water.

Humidity: a measure of the amount of water in the air or “moist air”

• Absolute humidity• % humidity• Wet bulb temperature• Relative humidity• Dew Point

Temperature• Humidity Ratio• Mixture Enthalpy

Terms to understand:

for Moist Air1. The overall mixture and each component, dry air and water vapor, obey the ideal gas equation of state.

2. Dry air and water vapor within the mixture are considered as if they each exist alone in volume V at the mixture temperature T while each exerts part of the mixture pressure.

3. The partial pressures pa and pv of dry air and water vapor are, respectively

where ya and yv are the mole fractions of the dry air and water vapor.

aa a

n RTp y p

V

VV V

n RTp y p

V

and

Humidity Ratio:

4. Humidity Ratio is the ratio of the mass of the vapor to the mass of the “dry air”.

0.622V V V V

a a a V

m M p p

m M p p p

Mixture pressure, p

,T

for Moist Air

5. The mixture pressure is the sum of the partial pressures of the dry air and the water vapor:

6. A typical state of water vapor in moist air is fixed using partial pressure pv and the mixture temperature T. The water vapor is superheated at this state.

Typical state of the water vapor

in moist air

a vp p p

7. When pv corresponds to pg at

temperature T, the mixture is said to be saturated.

8. The ratio of pv and pg is called the relative humidity, f:

v

, g, ,

V

V sat T p T p

y p

y p

Relative humidity:

6Sec 12.5 : Psychrometric applications

Humidity relates to temperature

7Sec 12.5 : Psychrometric applications

Humidity can be measured using a hygrometer:

Paper-disk hygrometer

Capacitive hygrometer

Whirling hygrometer(sling psychrometer)

Hair/Fiber hygrometer

8Sec 12.5.4 : Evaluating the Dew Point Temperature

Humidity can be measured using a “wet bulb”

The temperature of the “wet bulb” is lower than the dry thermometer.

The evaporation of water is an endothermic process (requires heat) which is obtained from the environment (bulk water).

The “wet bulb” temperature = dew point temperature.

This is the lowest temperature that can hold the current water vapor content of the air.

Mixture pressure, p

,T

Typical state of the water vapor

in moist air

a vp p p

Summarize:

Relative humidity:

Relative Humidity:

Dew point

v

, g, ,

V

V sat T P T P

y p

y p

Temperature at which pv=pg:

_ int (@ )dew po V gT p p

Humidity Ratio:

0.622V V V V

a a a V

m M p p

m M p p p

10Sec 12.5.3 : Evaluating U, H, and S

For mixtures, recall the internal energy, enthalpy, and entropy are equal to the parts added together

waterwaterDryAirDryAirwaterDryAir umumUUU

waterwaterDryAirDryAirwaterDryAir hmhmHHH

va v a v

a a

mHh h h h

m m

Using the definition of the humidity ratio, :

DryAir water DryAir DryAir water waterS S S m s m s

va v a v

a a

mUu u u u

m m

v

a

mm

and

11Sec 12.5.3 : Evaluating U, H, and S

To evaluate enthalpy:

For air: a) use Table A-22 or b) use Δh=cp_air (Δ T)

2

1

lnPc T dT ps R

T p

2

1

0 ln ln lnv

g

pps R R R

p p

ln, RTspTs gv Note: hg and sg are taken from the steam table.

For vapor: use hv hg(T)

a va

Hh h

m

notice how h ≈ constant for low pressure superheated vapor on Molier diagram.

To evaluate entropy:

Here is therelative humidity

12

Example (12.47): A lecture hall having a volume of 106 ft3 contains air at 80 oF, 1 atm, and a humidity ratio of 0.01 lbm of water per lbm of dry air. Determinea) relative humidityb) the dew point temperature in degrees F.c) the mass of water vapor contained in the room

13

Example (12.47): A lecture hall having a volume of 106 ft3 contains air at 80 oF, 1 atm, and a humidity ratio of 0.01 lbm of water per lbm of dry air. Determinea) relative humidityb) the dew point temperature in degrees F.c) the mass of water vapor contained in the room

0.622 V

V

p

p p

0.622

1V

pp

10.01582

0.6221

0.01

V

atmp atm

0.5073 0.0345gp psi atm

At T = 80 F: Look up pg using Table A2E

v

g ,

0.015820.459 45.9%

0.0345T p

p

p

14

Example (12.47): A lecture hall having a volume of 106 ft3 contains air at 80 oF, 1 atm, and a humidity ratio of 0.01 lbm of water per lbm of dry air. Determinea) relative humidityb) the dew point temperature in degrees F.c) the mass of water vapor contained in the room

_ int (@ )dew po V gT T p p

0.5073Vp psi

To find the Dew Point:

_ int 79.46odew poT FFrom Table A3E: at pv=pg

v v vp V m RTTo find the mass, use the ideal gas law:2 6 3 2

2

(0.5073 / )(10 ) 1441577.7

(0.11021 / )(540 ) 778fv

v mv m f

lb in ftp V in Btum lb

RT Btu lb R R ft lb ft

pV(T)pg(T)

Tdp

15

Example (12.54): Wet grain at 20°C containing 40% moisture by mass enters a dryer operating at steady state. Dry air enters the dryer at 90°C, 1 atm at a rate of 15 kgair/kgwet_grain entering. Moist air exits the dryer at 38°C, 1 atm, 53% relative humidity. For the grain exiting the dryer, determine the percent moisture by mass.

16

Example (12.54): Wet grain at 20°C containing 40% moisture by mass enters a dryer operating at steady state. Dry air enters the dryer at 90°C, 1 atm at a rate of 15 kgair/kgwet_grain entering. Moist air exits the dryer at 38°C, 1 atm, 53% relative humidity. For the grain exiting the dryer, determine the percent moisture by mass.

Dry air 90°C,1 atm

Wet grain, 20°C40% moisture

Dry grain ?% moisture

Wet air 38°C,1 atm

Mass balance on water:

,1 ,4 ,2 ,3v v v vm m m m

1 2

4 3

,1 ,2 ,30v v vm m m

2

2,2,%

m

mvv

3 53%

,4am

1 ,1 ,1g vm m m 3 ,3 ,3a vm m m

2 ,2 ,2g vm m m

,4

1 _

15a air

wet grain

m kg

m kg

Mass balance on air:

,1 ,4 ,2 ,3a a a am m m m

,4 ,30 0a am m

Mass balance of grain:

,1 ,2g gm m

17

Example (12.54):

40% of wet grain is water:

Dry air 90°C,1 atm15 kgair/kggrain

Wet grain20°C40% moisture

Dry grain?% moisture

Wet air 38°C,1 atm

1 2

4 3

,4 ,3a am m

3 53%

33 (@38 )

V

g

p

p C

,4 1(15 / )a air grainm kg kg m

Mass balance on water:

,1 ,2 ,3v v vm m m

,1 10.40vm m

mass balance of air: 15 kgair / 1 kg wet grain comes in:

Relative Humidity of State 3:

3

10.53(0.06632 ) 0.03469

1.01325V

atmp bar atm

bar

(from Table A2 at T=38C, pg=0.06632 bar)

18

Example (12.54): Dry air 90°C,1 atm15 kgair/kggrain

Wet grain20°C40% moisture

Dry grain?% moisture

Wet air 38°C,1 atm

1 2

4 3Humidity ratio:3 53%

,3 33

,3 3 3

0.622v V

a V

m p

m p p

,3 ,30.02235v am m

Unknowns so far: (10 unknowns) 1 ,1 ,1 2 ,2 ,2 3 ,3 ,3 ,4g v g v a v am m m m m m m m m m

Equations so far: (9 equations)

(0.03469 )0.622

(1 0.03469)

atm

atm

0.02235

1 ,1 ,1g vm m m

3 ,3 ,3a vm m m 2 ,2 ,2g vm m m

,3 ,30.02235v am m

,1 ,2 ,3v v vm m m

,4 ,3a am m,1 10.40vm m

,4 1(15 / )a air grainm kg kg m

,1 ,2g gm m

19

Example (12.54): Dry air 90°C,1 atm15 kgair/kggrain

Wet grain20°C40% moisture

Dry grain?% moisture

Wet air 38°C,1 atm

1 2

4 3But wait:3 53%

Recall the answer is asked for as% moisture per grain out

Solving using IT for the other mass flow rates:

which means that can be set as our basis or let =1 kg/s

,2

2

vm

m

2m 2m

Therefore:

2

2,2,%

m

mvv

0.097419.74%

1

20

Procedure for analysis of air conditioning systems:

1) Identify State Properties of as many individual mixture componentsUse ideal gas law: or Table A20

and A22Steam tables: Tables A2, A3, A4, etc.Humidity definitions:Constant process data: isobaric, isothermal, isentropic, polytropic, etc.

1 1 1, ,pV mRT pv RT pV n RT

/V gp p )/ 0.622 /(V a V Vm m p p p

2) Apply mass balance to each individual component of the mixture.

3) Apply energy balance to each separate stream of the mixture.

in outair airm m 2 2in outH O H Om m

2 2 2 20 0 0 00in in in in out out out outair air H H air air H HQ W m h m h m h m h

4) Solve equations

21

Example (12.67): Moist air at 90°F, 1 atm, 60% relative humidity and a volumetric flow rate of 2000 ft3/min enters a control volume at steady state and flows along a surface maintained at 40°F through which heat transfer occurs. Saturated moist air and condensate, each at 54°F, exit the control volume For the control volume W = KE = PE = 0. Determine

Moist Air at 90°F and 1 atm RH = 60%, (AV)1=2000 ft3/min

Condensate at 54°F

QSaturated air at 54°F

1 2

3

(a) The rate of heat transfer, in kW(b) The rate of enthalpy change, in KW/K.

1 1 1a vm m m 2 2 2a vm m m

3 3vm m

22

Example (12.67): Set property states: State 1: T1=90 °F, p1 = 1 atm ,

RH = 60%, (AV)1=2000 ft3/min

Q

1 23

1 1 1a vm m m 2 2 2a vm m m

3 3vm m

State 2: saturated air at T2= 54 °F

State 3: condensate at T3=54 °F refers to saturated fluid at 54 deg. F.

Using V

g

p

p

V gp p from Table A2E: pg at 90 F = 0.6988 psi

1 0.60(0.6988 ) 0.4193 0.0285Vp psi psi atm then

1 1 1 10.622 /( ) 0.622(0.0285) /(1 0.0285) 0.01825V Vp p p 1

11

v

a

m

m

1 10.01825v am m

for saturated air: pv = pg From Table A2E: pg at 54 F = 0.2064 psi = 0.01404 atm

2 2 2 20.622 /( ) 0.622(0.0140) /(1 0.0140) 0.00883V Vp p p

2 20.00883v am m22

2

v

a

m

m

23

Example (12.67): Set property states: State 1: T1=90 °F, p1 = 1 atm ,

RH = 60%, (AV)1=2000 ft3/min

Q

1 23

1 1 1a vm m m 2 2 2a vm m m

3 3vm m

State 2:

State 3:

air: water:1 1 90VT T F

1 10.01825v am m

2 20.00883v am m

1 1 1 0.9715a vp p p atm 1 0.0285Vp atm1 1 90aT T F

air: water:2 2 54aT T F 2 2 54VT T F 2 2 2 0.9866a vp p p atm 2 0.01404Vp atm

1 131.46 /a mh Btu lb 1 1 1100.7 /v g mh h Btu lb

_ 1 0.240 /p a mc Btu lb R

2 122.83 /a mh Btu lb 2 2 1085.1 /v g mh h Btu lb

_ 2 0.240 /p a mc Btu lb R

(from Table A22E) (from Table A2E)

(from Table A22E) (from Table A2E)

3 54T F

3 3 22.07 /v f mh h Btu lb (from Table A2E)

24

Example (12.67): Determine(a) The rate of heat transfer, in kW(b) The rate of enthalpy change, in KW/K.

Use ideal gas law to establish mass flow rate

Q

1 23

1 1 1a vm m m 2 2 2a vm m m

3 3vm m( V)pV p A mRT

23 21

1 21

14.7 /(0.9715 )(2000 / min) 144

(0.06855 / )(550 ) 1 1 778fa

aa m f

lb inp V atm ft in Btum

R T Btu lb R R atm ft lb ft

140.2 / minmlb

For air in: (ideal gas)

For water vapor in:

2 20.00883 0.00883(140.2) 1.238 / minv a mm m lb

1 10.01825 2.5585 / minv a mm m lb

Air Mass Balance:1 2a am m 2 1 140.2 / mina a mm m lb

For water/vapor out:

25

Example (12.67): Determine(a) The rate of heat transfer, in kW(b) The rate of enthalpy change, in KW/K.

Set up Mass balances:

Q

1 23

1 1 1a vm m m 2 2 2a vm m m

3 3vm m

1 2 140.2 / mina a a mm m m lb For air:

For water/vapor:1 2 3v v vm m m

Set up Energy balance:

1 1 2 2 3 30 CVQ W m h m h m h

1 1 2 2 1 1 2 2 3 30 CV a a a a v v v v v vQ m h m h m h m h m h

2 1 2 2 3 3 1 1( )CV a a a v v v v v vQ m h h m h m h m h

3 1 2 2.559 1.238 1.321 / minv v v mm m m lb

140.2(122.83 131.46) 1.238(1085.1) 1.321(22.07) 2.559(1100.7)CVQ

2654 /minBtu

26

end of slides for lecture 39