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˙3−m˙4=m˙a(ω2−ω1)
Here, mass flow rate of water at inlet and exit is m˙w,i and m˙w,e respectively, specific humidity at state 1 and 2 is ω1 and ω2 respectively; mass flow rate at state 2, 3 and 4 is m˙2, m˙3 and m˙4 respectively.
Express the energy balance.
E˙in−E˙out=ΔE˙systemE˙in−E˙out=0E˙in=E˙out∑m˙ihi=∑m˙ehe
0=∑m˙ehe−∑m˙ihi0=m˙a2h2+m˙4h4−m˙a1h1−m˙3h3m˙a=m˙3(h3−h4)(h2−h1)−(ω2−ω1)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˙i and m˙e respectively, enthalpy at inlet and exit is hi and he respectively and enthalpy at state 1, 2, 3 and 4 is h1,h2,h3 and h4 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−(ω2−ω1) (III)
Determine the change in entropy of water steam.
Δswater=m4s4−m3s3 (IV)
Determine the change in entropy of water vapour in the air stream.
Δsvapour=ω2sg,2−ω1sg,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=P1−Pvap,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=P2−Pvap,2 (IX)
Here, the pressure at the state 1 is P1.
Determine the entropy change of dry air.
Δsa=s2−s1=cp×lnT2T1−R×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=21.8 kJ/kg dry airω1=0.00264 kg H2O/kg dry airν1=0.820 m3/kg dry air
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=49.3 kJ/kg dry airω2=0.0123 kg H2O/kg dry air
Refer Table A-4, “saturated water-temperature table”, and write the enthalpy and entropy at state 3 at temperature of 30°C.
h3=hf=125.74 kJ/kg H2O
s3=sf=8.7803 kJ/kg⋅K
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=(x2−x1)(y3−y1)(x3−x1)+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°C and 25°C for x1, x2 and x3 respectively, 83.915 kJ/kg for y1 and 104.83 kJ/kg for y3 in Equation (XIV).
y2=(22°C−20°C)(104.83 kJ/kg−83.915 kJ/kg)(25°C−20°C)+83.915 kJ/kg=92.28 kJ/kg=hf@22°C
Substitute 92.28 kJ/kg for hf@22°C in Equation (XIII).
h4=92.28 kJ/kg H2O
Repeat the Equation (XIV), obtain the value of entropy for water streams at 22°C of temperature as 0.3249 kJ/kg⋅K.
Substitute 5 kg/s for m˙3, 125.74 kJ/kg H2O for h3, 92.28 kJ/kg H2O for h4, 49.3 kJ/kg dry air for h2, 21.8 kJ/kg dry air for h1, 0.00264 kg H2O/kg dry air for ω1 and 0.0123 kg H2O/kg dry air for ω2 in Equation (I).
m˙a=(5 kg/s)(125.74−92.28) kJ/kg H2O(49.3−21.8) kJ/kg−(0.0123−0.00264)(92.28 kJ/kg H2O)=6.29 kg/s
Substitute 5 kgwater/s for m˙3 and 6.29 kg dry air/s for m˙a in Equation (II).
m3=5 kgwater/s6.29 kg dry air/s=0.7949 kg water/kg dry air
Substitute 0.7949 kg water/kg dry air for m3, 0.00264 kg H2O/kg dry air for ω1 and 0.0123 kg H2O/kg dry air for ω2 in Equation (III).
m4=(0.7949 kg water/kg dry air)−(0.0123−0.00264)kg H2O/kg dry air=(0.7949 kg water/kg dry air)−(0.00966 kg H2O/kg dry air)=0.7852 kg water/kg dry air
Substitute 0.7852 kg water/kg dry air for m4, 0.7949 kg water/kg dry air for m3, 0.3249 kJ/kg⋅K for s4, and 8.7803 kJ/kg⋅K for s3 in Equation (IV).
Δswater=[(0.7852 kg water/kg dry air)×(0.3249 kJ/kg⋅K)−(0.7949 kg water/kg dry air)×(8.7803 kJ/kg⋅K)]=(0.255111 kJ/kg⋅K dry air)−(0.347212 kJ/kg⋅K dry air)=−0.0921 kJ/kg⋅K dry air
Refer Table A-4, “saturated water-temperature table”, and write the entropy at state 1 at temperature of 15°C.
sg1=sg@15°C=8.7803 kJ/kg⋅K
Repeat the Equation (XIV), obtain the value of entropy for water vapour at 18°C of temperature as 8.7112 kJ/kg⋅K.
Substitute 8.7803 kJ/kg⋅K for sg1, 8.7112 kJ/kg⋅K for sg2, 0.00264 kg H2O/kg dry air for ω1 and 0.0123 kg H2O/kg dry air for ω2 in Equation (V).
Δsvapour=[(0.0123 kg H2O/kg dry air)×(8.7112 kJ/kg⋅K)−(0.00264 kg H2O/kg dry air)×(8.7803 kJ/kg⋅K)]=[(0.107148 kJ/K⋅kg dry air)−(0.02318 kJ/K⋅kg dry air)]=0.083968 kJ/K⋅kg dry air
Substitute 0.25 for ϕ1 and 1.7057 kPa for Psat@15°C in Equation (VI).
Pvap,1=(0.25)×(1.7057 kPa)=0.4264 kPa
Substitute 1 atm for P1 and 0.4264 kPa for Pvap,1 in Equation (VII).
Pa1=1 atm−(0.4264 kPa)=1 atm×((101.325 kPa)1 atm)−(0.4264 kPa)=101.325 kPa−0.4264 kPa=100.8986 kPa
Substitute 0.95 for ϕ2 and 2.065 kPa for Psat@18°C in Equation (VIII).
Pvap,2=(0.95)×(2.065 kPa)=1.962 kPa
Substitute 1 atm for P2 and 1.962 kPa for Pvap,2 in Equation (IX).
Pa2=1 atm−(1.962 kPa)=1 atm×((101.325 kPa)1 atm)−(1.962 kPa)=101.325 kPa−1.962 kPa=99.36 kPa
Substitute 1.005 kJ/kg⋅K for cp, 15°C for T1, 18°C for T2, 0.287 kJ/kg⋅K for R, 99.36 kPa for Pa2, and 100.90 kPa for Pa1 in Equation (X).
Δsa=(1.005 kJ/kg⋅K)×ln(18°C)(15°C)−(0.287 kJ/kg⋅K)×ln(99.36 kPa)(100.90 kPa)=(1.005 kJ/kg⋅K)×ln(18°C+273)(15°C+273)−(0.287 kJ/kg⋅K)×ln(99.36 kPa)(100.90 kPa)=(1.005 kJ/kg⋅K)×ln(1.010417 K)−(0.287 kJ/kg⋅K)×ln(0.984737 kPa)=0.01483 kJ/kg dry air
Substitute −0.09210 kJ/kg dry air for Δswater, 0.08397 kJ/kg⋅K dry air for Δsvapour and 0.01483 kJ/kg dry air for Δsa in Equation (XI)
sgen=[(−0.09210 kJ/kg dry air)+(0.08397 kJ/kg⋅K dry air)+(0.01483 kJ/kg dry air)]=0.00670 kJ/kg dry air
Substitute 0.00670 kJ/kg dry air for sgen and 15°C for T0 in Equation (XII).
xdest=(15°C)×(0.00670 kJ/kg dry air)=(15°C+273)×(0.00670 kJ/kg dry air)=(288 K)×(0.00670 kJ/kg dry air)=1.9296 kJ/kg dry air
≅1.93 kJ/kg dry air
Thus, the exergy lost in the cooling tower is 1.93 kJ/kg dry air.