Tut1 Solutions

Presented on 17/01/2012 and 19/01/2012 at as a part of ME306 Tutorial © Heat Pump Laboratory IITB 1 of 40

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APPLIED THERMODYNAMICS

Psychrometry Tutorial Solutions

Presentation by

Prof M V Rane

Prof Sreedhara Sheshadri

Department of Mechanical Engineering

Indian Institute of Technology Bombay

Powai, Mumbai 400 076 INDIA

Presented by Yogesh S Padiya on 22/02/2008 at IIT Bombay, as a part of HPL activity

Saved as D:\Student\PhD\YSP\other\slide format file last updated on 21/02/200


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ASSIGNMENT PROBLEM(Moran & Shapiro (2006) P 12.34 1 of 2

Exploring Psychrometric Principles


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SOLUTIONSlide 1 of 2

Problem 12.34

• Known

Operating data at steady state as shown in diagram

• Find

The Volumetric flow rate of the entering dry air

The rate through dried particles exit

• Assumptions

Ideal gas principles apply to the dry air and moist air streams

• Analysis

The volumetric flow rate of the dry air can be found by equating the mass flow rate of dry air at locations 2 and 3: (i = 2,3) (AV) = Volume flow rate, ma = mass flow rate of dry air, pa = partial pressure of dry air, va =

specific volume of dry air

ai

iai

v

AVm

)(

ai

aipM

TRv

.

.


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Problem 12.34

Pa2 = p2

Use steam tables and RH values to calculate pa3

The volumetric flow rate of the dry air (AV)2 = 342.96 m3/min

The rate at which the particles exit at 4 must equal the rate at which the particles enter at 1.

mv3 = mass flow rate of water vapor at 3

mv3 equals the mass flow rate of moisture at 1

the rate at which dried particles enter at 1 is proportional to mass flow rate at 1

mass flow rate of dried particles = 9.77 kg/min

3

2

2

332 ..)()(

T

T

p

pAVAV

a

a

3

33

)(

v

vv

AVm

3

3.

.

v

vpM

TRv


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Problem 12.48

• Known

Operating data at steady state as shown in diagram

• Find

The energy removed by heat transfer in the dehumidifier section in tons

The energy added by heat transfer in the heating section in kW

• Assumptions

For control volume Qcv = 0 , Wcv = 0, kinetic potential energy effects negligible

RH = 100% at section 2

The condensate is saturated at T2

The moist air is modeled as an ideal gas mixture

• Analysis

Since heating section does not add or remove moisture ω2 = ω3

Find ω1 and ω3 from the psychrometric charts ω1= 0.017; ω2 = ω3 = 0.0105; Pg2 = 0.0168 bar ; T2 = 15 oC


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Problem 12.48

Calculate mass flow rate of dry air = 0.054 kg/s

Apply energy balance to dehumidifier control volume

Apply mass balance for water to obtain m4

Qcv = -14.78 kJ/s = -4.2 tons

Apply energy balance to heating control volume

Qcv = 6.05 kW

Values of hg and hf are obtained from steam tables

44222111 ][][0 fgaagaacv hmhhmhhmQ

)( 214 amm

)]()[( 2323 ggaaacv hhhhmQ

Presented on 17/01/2012 and 19/01/2012 at as a part of ME306 Tutorial

ASSIGNMENT PROBLEMMoran & Shapiro (2006) P 12.50 1 of 2


© Heat Pump Laboratory IITB 11 of 40

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ASSIGNMENT PROBLEMMoran & Shapiro (2006) P 12.50 1 of 2



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Problem 12.50


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Problem 12.50

• Known

Steady state operating conditions as shown in figure

• Find

The temperature of the air exiting the dehumidifier

The volumetric flow rate at the air conditioner inlet

The rate of water condensation

The rate of heat transfer to the air in the heating unit

• Analysis

Carry out mass rate balances for air and water

The mass flow rate of dry air is constant = ma

m4 = mv1 – mv3


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Problem 12.50

Humidity ratio ω2 = ω3

Use ω2 to find saturated vapour pressure at exit 2 and hence find T2

ω2 = ω3 = 0.007516; T2 = 9.6 oC

ω1 = 0.01691

Energy balance at dehumidifier control volume

Use hf value for hw

Use hg values for hv1 and hv2

Use cpaT for ha1 and ha2; cpa = 1.005 kJ/kg.K

Obtain ma = 54.81 kg/s and (AV)1 = 48.63 m3/min

Energy balance at heating section

Qcv = 804.1 kJ/min


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wwvvaavvaacv hmhmhmhmhmQ ][][0 222111

)]()[( 23323 ggaaacv hhhhmQ


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Problem 12.51

• Known

Air conditioning system operating at steady state as shown in diagram

Moist air at 32 ºC, ϕ = 77 % enters the system and passes through the spray section, exiting saturated

The moist air is then heated at fixed moisture content to 25 ºC, ϕ = 45 %

• Find

The temperature of the moist air leaving the spray section

The change in amount of water vapor contained in the moist air passing through the system


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Problem 12.51

• Assumptions

Two control volumes at steady state are considered: one enclosing the overall system and one enclosing the heating section

P = 1 atm throughout

At 2 moist air is saturated

Ideal gas principles apply to the moist air

• Analysis

As no moisture is added and removed from the heating section

2 3

3 ,3,3

3

,3 3 ,3

0.622 0.622gv

v g

PP

P P P P


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Problem 12.51

Pg,3 can be obtained from steam table at T3

Humidity ratio at 3 are obtained from the above equation

are used to calculate Pg,2

T2 =12.3 ºC corresponding to Pg,2 from steam table (by interpolation)

Considering the overall system, change in the amount of water vapor contained in the moist air is given by

Now change in the amount of water vapor contained in the moist air per unit mass of dry air is given by

Humidity ratio at 1 from are obtained from similar relation that are given for humidity ratio at 3.

The change in the amount of water vapor contained in the moist air per unit mass of dry air is -0.0144 kg/kg (d.a)

3 1

3 1 3 1( )v v am m m


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Problem 12.52

• Known

Air is removed from classroom at 27 ºC and of 50 % relative humidity

Moisture is added to air at a rate of 4.5 kg/h in saturated condition at 33 ºC

Heat transfer into the occupied space at a rate of 34000 kJ/h

P =1 atm throughout

• Find

For a supply air volumetric flow rate of 40m3/min, the supply air temperature and relative humidity


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Problem 12.52

Temperature and relative humidity for air volumetric flow rate ranging from 35 to 90 m3/min

• Assumptions

Control volume at steady state as shown in diagram

Kinetic and potential energy are neglected,

Ideal gas principles apply to the moist air

• Analysis

At steady state mass rate balance give

1 2v w vm m m

2 1

wa

mm

1 2a a am m m

(1)

0cvW


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Problem 12.52

Subscript a1 and a2 are for dry air at section 1 and 2 respectively

Subscript v1 and v2 are for water vapor at section 1 and 2 respectively

Subscript w is for moisture added by the occupants

Pg,2 can be obtained from steam table at T2

Humidity ratio at 2 is 0.0111

By energy rate balance at steady state

hg2 and hg3 can be calculated from steam table at the corresponding temperature

Enthalpy of air ha1 and ha2 are given by the relation cpa(T1-T2) , cpa =1.005 kJ/kg-K

In the equation (2) unknowns are T1 humidity ratio at 1 and mass flow rate of air.

(2) 1 2 1 1 2 2 30 cv a a a g g w gQ m h h h h m h

1 1 1 3 2 2 20 cv cv a a v v w a a v vQ W m h m h m h m h m h

2 222

2 2 2

0.622 0.622gv

v g

PP

P P P P


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Problem 12.52

The volume flow rate of air is given by

Solve equations (1) ,(2),(3) and (4) simultaneously using an iterative procedure

Supply temperature T1 = 15.1 ºC, humidity ratio at 1 = 0.009546 , Pv1= 0.01532 bar, relative humidity = 0.869

1 1

112 1 1 2 1 1

/ /a w awa a

a v

R M T m R M TmAV m v

P P P

11

1

0.622 v

v

P

P P

(3)

(4)


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Problem 12.56

• Known

Steady state operating data are provided for adiabatic mixing of two moist air streams

( AV )1 = 35 m3 / min, T1 = 14 ˚C, φ1 = 80 %

( AV )2 = 80 m3 / min, T1 = 40 ˚C, φ1 = 40 %

• Find

Relative humidity of exit stream

Temperature of exit stream

• Assumptions

Steady state flow

Pressure remains constant at 1 atm


CONFIDENTIAL

kg

m

v

AVm

a

a

3

1

11 48.42

kg

m

v

AVm

a

a

3

2

22 62.87

221133 aaa mmm

221133 mamama hmhmhm

gam hhh

213 aaa mmm


Problem 12.56


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31

23

2

1

mm

mm

a

a

hh

hh

m

m

31

23

2

1

mm

mm

a

a

m

m

31

23

31

23

2

1

mm

mm

mm

mm

a

a

hh

hh

m

m


Problem 12.56


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CT o32

%52

3

3

givesAfigvaluesthesengusiakgkJhakg

vkgm 9.),(/5.70,

)(

)(0153.0 33


Problem 12.56


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• Known

Moist air at mixes adiabatically with saturated air at

to produce single mixed stream at

• Find

Relative humidity and temperature of exit stream

Energy destruction rate

• Assumptions

Steady state flow

• Analysis

At staedy state, mass flow rate balancing gives, ,

and . Also for each stream

min/5,1,5 222 kgmbarPCT o min/3%,50,1,30 1111 kgmbarPCT o

barP 13

213 mmm 213 aaa mmm

213 vvv mmm )1( ava mmmm


Problem 12.58


CONFIDENTIAL

accordingly, 221133 aaa mmm

2

2

1

1

2

2

21

1

1

21

22113

11

.1

.1

2

mm

mm

mm

mm

aa

aa


Problem 12.58


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To fin and

Then

1 2

02123.004246.05.0. 111 TpP gv

00872.0. 222 TpP gv

)(

)(01349.0

02123.01

02123.0622.01

akg

vkg

)(

)(00547.0

00872.01

00872.0622.02

akg

vkg


Problem 12.58


CONFIDENTIAL

Accordingly,

)(

)(008463.0

00547.1

5

01349.1

3

)5(00547.1

00547.0)3(

01349.1

01349.0

3akg

vkg

01342.063046.0

)008463.0)(1(

622.0

.

3

333

pPv

)(

01342.0

)( 33

33

TpTp

p

gg

v


Problem 12.58


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To find

333322221111 ......0 vvaavvaavvaacvcv hmhmhmhmhmhmQ

21

22221111

333

...

aa

gaagaa

gamm

hhmhhmhh

min/)(96.21/ 111 akgmma

min/)(9728.41/ 222 akgmma

givesbalancingenergyT3


Problem 12.58


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Energy destruction rate is obtained from

9728.496.2

)6.2510)(00547.0(1.2789728.4)3.2556)(01349.0(21.303)96.2(. 333

ga hh

)(/95.308 akgkJ

%)8.81(818.001641.0

01342.03

cvgvaaagvaaavvvaaa

j

jTsmPTsmTsmPTsmPTsmPTsm

T

Q

)(),()(),(),(),(.0 3333322222111111

CT o4.143

givesstatesteadyatbalancingentropy

productionentropyofratetheiswhereTE cvcvod ,


Problem 12.58


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)(),(),(),(),(.),( 22222111111333333 TsPTsmPTsPTsmPTsPTsm gaaavvaaavvaaacv

)(),(

),(),(),(),(),(

2221111

33332233211331

TsmPTsm

PTsmPTsPTsmPTsPTsm

gavva

vvaaaaaaaaaaa


Problem 12.58


CONFIDENTIAL

From previous calculations

Evaluating

2

3

2

32

1

3

1

21 lnlnlnln

a

a

a

paa

a

a

a

paacvP

P

M

R

T

TCm

P

P

M

R

T

TCm

11113333 ln)(ln)(

v

ga

v

gaM

RTsm

M

RTsm

)( 222 Tsm ga

barPandbarPbarP aaa 98658.001342.0199128.0,97877.0 321

Kkg

kJCwith pacv

.005.1


Problem 12.58


CONFIDENTIAL

kWEd 048.0)009828.0)(293(

99128.0

98658.0ln

97.28

314.8

278

4.287ln005.19728.4

97877.0

98658.0ln

97.28

314.8

303

4.287ln005.196.2cv

5.0ln

02.18

314.84533.8)01349.0)(96.2(818.0ln

02.18

314.879544.8)008463.0)(9323.7(

K

kJ

.min009828.0)0257.9)(00547.0)(9728.4(

The energy destruction rate is


Problem 12.58

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