Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Question
Chapter 14.7, Problem 113P
To determine

The exergy lost in the cooling tower.

Expert Solution & Answer
Check Mark

Answer to Problem 113P

The exergy lost in the cooling tower is 1.93kJ/kgdryair.

Explanation of Solution

As the process is a steady flow and thus the mass flow rate of dry air remains constant during the entire process.

m˙a1=m˙a2=m˙a

Here, the mass flow rate of air at inlet is m˙a1, mass flow rate of dry air at exit is m˙a2 and mass flow rate of dry air is m˙a.

Express the water mass balance:

m˙w,i=m˙w,em˙3+m˙a1ω1=m˙4+m˙2ω2m˙3m˙4=m˙a(ω2ω1)

Here, mass flow rate of water at inlet and exit is m˙w,iandm˙w,e respectively, specific humidity at state 1 and 2 is ω1andω2 respectively; mass flow rate at state 2, 3 and 4 is m˙2,m˙3andm˙4 respectively.

Express the energy balance.

E˙inE˙out=ΔE˙systemE˙inE˙out=0E˙in=E˙outm˙ihi=m˙ehe

0=m˙ehem˙ihi0=m˙a2h2+m˙4h4m˙a1h1m˙3h3m˙a=m˙3(h3h4)(h2h1)(ω2ω1)h4 (I)

Here, the rate of total energy entering the system is E˙in, the rate of total energy leaving the system is E˙out, the rate of change in the total energy of the system is ΔE˙system, initial and exit mass flow rate is m˙iandm˙e respectively, enthalpy at inlet and exit is hiandhe respectively and enthalpy at state 1, 2, 3 and 4 is h1,h2,h3andh4 respectively.

Determine the mass flow rate of steam at state 3 per unit mass of dry air.

m3=m˙3m˙a (II)

Determine the mass flow rate of steam at state 4 per unit mass of dry air.

m4=m3(ω2ω1) (III)

Determine the change in entropy of water steam.

Δswater=m4s4m3s3 (IV)

Determine the change in entropy of water vapour in the air stream.

Δsvapour=ω2sg,2ω1sg,1 (V)

Determine the partial pressure of water vapour at state 1 for air steam.

Pvap,1=ϕ1×Pg1=ϕ1×Psat@15°C (VI)

Determine the partial pressure of dry air at state 1 for air steam.

Pa1=P1Pvap,1 (VII)

Here, the pressure at the state 1 is P1.

Determine the partial pressure of water vapour at state 2 for air steam.

Pvap,2=ϕ2×Pg2=ϕ2×Psat@18°C (VIII)

Determine the partial pressure of dry air at state 2 for air steam.

Pa2=P2Pvap,2 (IX)

Here, the pressure at the state 1 is P1.

Determine the entropy change of dry air.

Δsa=s2s1=cp×lnT2T1R×lnPa,2Pa,1 (X)

Here, the temperature at the state 1 is T1, the temperature at the state 2 is T2, and the universal gas constant is R.

Determine the entropy generation in the cooling tower is the total entropy change.

sgen=Δswater+Δsvapour+Δsa (XI)

Determine the exergy destruction per unit mass of dry air.

xdest=T0sgen (XII)

Conclusion:

Refer Figure A-31, “psychometric chart at 1 atm total pressure”, and write the properties corresponding to dry bulb temperature of 15°C and relative humidity of 25%.

h1=21.8kJ/kgdryairω1=0.00264kgH2O/kgdryairν1=0.820m3/kgdryair

Refer Figure A-31, “psychometric chart at 1 atm total pressure”, and write the properties corresponding to dry bulb temperature of 18°C and relative humidity of 95%.

h2=49.3kJ/kgdryairω2=0.0123kgH2O/kgdryair

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy and entropy at state 3 at temperature of 30°C.

h3=hf=125.74kJ/kgH2O

s3=sf=8.7803kJ/kgK

Here, enthalpy of saturation liquid is hf.

Refer Table A-4, “saturated water-temperature table”, and write the enthalpy at state 4 at temperature of 22°C using an interpolation method.

h4=hf@22°C (XIII)

Here, enthalpy of saturation liquid at temperature of 22°C is hf@22°C.

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+y1 (XIV)

Here, the variables denote by x and y is temperature and specific enthalpy at state 4 respectively.

Show the specific enthalpy at state 4 corresponding to temperature as in Table (1).

Temperature

T(°C)

Specific enthalpy at state 4

h4@hf(kJ/kg)

20 (x1)83.915 (y1)
22 (x2)(y2=?)
25 (x3)104.83 (y3)

Substitute 20°C,22°Cand25°C for x1,x2andx3 respectively, 83.915kJ/kg for y1 and 104.83kJ/kg for y3 in Equation (XIV).

y2=(22°C20°C)(104.83kJ/kg83.915kJ/kg)(25°C20°C)+83.915kJ/kg=92.28kJ/kg=hf@22°C

Substitute 92.28kJ/kg for hf@22°C in Equation (XIII).

h4=92.28kJ/kgH2O

Repeat the Equation (XIV), obtain the value of entropy for water streams at 22°C of temperature as 0.3249kJ/kgK.

Substitute 5kg/s for m˙3, 125.74kJ/kgH2O for h3, 92.28kJ/kgH2O for h4, 49.3kJ/kgdryair for h2, 21.8kJ/kgdryair for h1, 0.00264kgH2O/kgdryair for ω1 and 0.0123kgH2O/kgdryair for ω2 in Equation (I).

m˙a=(5kg/s)(125.7492.28)kJ/kgH2O(49.321.8)kJ/kg(0.01230.00264)(92.28kJ/kgH2O)=6.29kg/s

Substitute 5kgwater/s for m˙3 and 6.29kgdryair/s for m˙a in Equation (II).

m3=5kgwater/s6.29kgdryair/s=0.7949kgwater/kgdryair

Substitute 0.7949kgwater/kgdryair for m3, 0.00264kgH2O/kgdryair for ω1 and 0.0123kgH2O/kgdryair for ω2 in Equation (III).

m4=(0.7949kgwater/kgdryair)(0.01230.00264)kgH2O/kgdryair=(0.7949kgwater/kgdryair)(0.00966kgH2O/kgdryair)=0.7852kgwater/kgdryair

Substitute 0.7852kgwater/kgdryair for m4, 0.7949kgwater/kgdryair for m3, 0.3249kJ/kgK for s4, and 8.7803kJ/kgK for s3 in Equation (IV).

Δswater=[(0.7852kgwater/kgdryair)×(0.3249kJ/kgK)(0.7949kgwater/kgdryair)×(8.7803kJ/kgK)]=(0.255111kJ/kgKdryair)(0.347212kJ/kgKdryair)=0.0921kJ/kgKdryair

Refer Table A-4, “saturated water-temperature table”, and write the entropy at state 1 at temperature of 15°C.

sg1=sg@15°C=8.7803kJ/kgK

Repeat the Equation (XIV), obtain the value of entropy for water vapour at 18°C of temperature as 8.7112kJ/kgK.

Substitute 8.7803kJ/kgK for sg1, 8.7112kJ/kgK for sg2, 0.00264kgH2O/kgdryair for ω1 and 0.0123kgH2O/kgdryair for ω2 in Equation (V).

Δsvapour=[(0.0123kgH2O/kgdryair)×(8.7112kJ/kgK)(0.00264kgH2O/kgdryair)×(8.7803kJ/kgK)]=[(0.107148kJ/Kkgdryair)(0.02318kJ/Kkgdryair)]=0.083968kJ/Kkgdryair

Substitute 0.25 for ϕ1 and 1.7057kPa for Psat@15°C in Equation (VI).

Pvap,1=(0.25)×(1.7057kPa)=0.4264kPa

Substitute 1 atm for P1 and 0.4264kPa for Pvap,1 in Equation (VII).

Pa1=1atm(0.4264kPa)=1atm×((101.325kPa)1atm)(0.4264kPa)=101.325kPa0.4264kPa=100.8986kPa

Substitute 0.95 for ϕ2 and 2.065kPa for Psat@18°C in Equation (VIII).

Pvap,2=(0.95)×(2.065kPa)=1.962kPa

Substitute 1 atm for P2 and 1.962kPa for Pvap,2 in Equation (IX).

Pa2=1atm(1.962kPa)=1atm×((101.325kPa)1atm)(1.962kPa)=101.325kPa1.962kPa=99.36kPa

Substitute 1.005kJ/kgK for cp, 15°C for T1, 18°C for T2, 0.287kJ/kgK for R, 99.36 kPa for Pa2, and 100.90kPa for Pa1 in Equation (X).

Δsa=(1.005kJ/kgK)×ln(18°C)(15°C)(0.287kJ/kgK)×ln(99.36kPa)(100.90kPa)=(1.005kJ/kgK)×ln(18°C+273)(15°C+273)(0.287kJ/kgK)×ln(99.36kPa)(100.90kPa)=(1.005kJ/kgK)×ln(1.010417K)(0.287kJ/kgK)×ln(0.984737kPa)=0.01483kJ/kgdryair

Substitute 0.09210kJ/kgdryair for Δswater, 0.08397kJ/kgKdryair for Δsvapour and 0.01483kJ/kgdryair for Δsa in Equation (XI)

sgen=[(0.09210kJ/kgdryair)+(0.08397kJ/kgKdryair)+(0.01483kJ/kgdryair)]=0.00670kJ/kgdryair

Substitute 0.00670kJ/kgdryair for sgen and 15°C for T0 in Equation (XII).

xdest=(15°C)×(0.00670kJ/kgdryair)=(15°C+273)×(0.00670kJ/kgdryair)=(288K)×(0.00670kJ/kgdryair)=1.9296kJ/kgdryair

      1.93kJ/kgdryair

Thus, the exergy lost in the cooling tower is 1.93kJ/kgdryair.

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Chapter 14 Solutions

Thermodynamics: An Engineering Approach

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