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Assignment Subject: Chemical Engineering Thermodynamics-I



Topic: Numerical Problems related to     



Flow through Nozzles, Turbines, Compressors Throttling Gas & Vapor Power Cycles Refrigeration & Air Conditioning Liquefaction



Submitted To: Prof.Dr.Mahmood Saleem



Submitted By: Hamza Shafiq



Roll #: CE-M10-41



Date of Issue: 10th Week (Wednesday)



Date of Submission:



Nozzles, Throttling, Expanders/Turbine, Compressors/Pumps Numerical 1: Steam expands adiabatically in a nozzle from inlet conditions of 130(psia), 420 ℉ and a velocity of 230(ft)( s−1 ¿ to a discharge pressure of 35(psia) where its velocity is 2,000(ft)( s−1 ¿. What is the state of steam at the nozzle exit and what is SG for the process? Solution: u1=230 ft s−1 u2=2,000 ft s−1 P1=130 psia P2=35 psia T 1=420 ℉ From Steam Table at initial condition



H 1=1233.6 Btu/ lb S1=1.6310



Btu lb . R



As we know that



∆ H=



∆ u2 2



Therefore,



∆ H =(u¿¿ 22 −u21)/2 ¿ Thus,



∆ H =78.8



Btu lb



As we know that



∆ H =H 2−H 1 Therefore,



H 2=∆ H + H 1 H 2=1154.8



Btu lb



Again from the Steam Table at final condition



H liq =228.03



Btu lb



H vap =1167.1



Btu lb



Sliq =0.3809



Btu lb . R



Svap =1.6872



Btu lb . R



As we know that Quality = Fraction Vapour = x Also



x=



H 2−H vap H vap−H liq



x=0.987 Similarly for the two or multiphase system, we know that



S2=S liq + x (S vap −S liq ) S2=0.3809+ 0.987 ( 1.9872−0.3809 ) S2=1.67



Btu lb . R



For S g,



S g=∆ S=S 2−S1 S g=1.67−1.631 S g=0.039



Btu lb . R



Numerical 2: A gas at upstream conditions (T 1 , P1 ¿ is throttled to a downstream pressure of 1.2 bar. Using the Redlich/Kwong equation to estimate the downstream temperature and ∆ S of the gas for one of the following a) b) c) d)



Carbondioxide, with T 1=350K and P1= 80 bar Ethylene, with T 1=350K and P1=60 bar Nitrogen, with T 1=250K and P1=60 bar Propane, with T 1=400K and P1=20 bar



Solution: a) P2=1.2 ¯¿ T 1=350 K P1=80 ¯¿ T c =304.2 K Pc =73.83 ¯ ¿ ω=0.224 A=5.457 B=1.045 x 10−3 C=0 D=−1.157 x 105 T r=1.151



Pr =1.084 Ω=0.08664 ψ=0.42748 As we know that



Pr Tr



β=Ω.



β=0.08159 q=(



Ψ ) Ω. T 1.5 r



q=3.9956 Guess z=1;



z=1+ β−qβ



z−β z (z + β )



z=0.8047 For i=1



I i=ln (



z+β ) z



I =2.3853 As we know that



H r =R .T 1 [ ( z−1 ) −1.5 qI ] H r =−42169.92 kJ /mol Also



Sr =R [ ln ( z ) −0.5 qI ] Sr =4.



9826 kJ mol . K



The most convenient way will be by guess the final downstream temperature initially and then calculating it. Assume, T 2=280 K For this value of this temperature, the value of I will be



Z=0.721 Thus from the above mentioned formula



H r =−2.681 kJ /mol And



Sr =



−5.177 kJ mol . K



From the appendix we can deduce the value of C p=4.467 As we know that



T 2=



Hr +T 1 Cp



T 2=349.39 K And



Δ S=C p ln (T ¿ ¿ 2/T 1)−Rln Therefore



Δ S=31.549 kJ /mol



b) P2=1.2 ¯¿ T 1=350 K P1=60 ¯¿ T c =282.3 K



P2 −S r ¿ P1



Pc =50.40 ¯¿ ω=0.087 A=1.424 B=14.394 x 10−3 C=−4.392 x 10−3 D=0 T r=1.24 Pr =1.19 Ω=0.08664 ψ=0.42748 As we know that



β=Ω.



Pr Tr



β=0.083 q=(



Ψ ) Ω. T 1.5 r



q=42.06 Guess z=1;



z=1+ β−qβ



z−β z (z + β )



z=−1.872 For i=1



I i=ln (



z+β ) z



I =3.070 As we know that



H r =R .T 1 [ ( z−1 ) −1.5 qI ] H r =46134.7 kJ /mol Also



Sr =R [ ln ( z ) −0.5 qI ] Sr =−531.57



kJ mol . K



The most convenient way will be by guess the final downstream temperature initially and then calculating it. Assume, T 2=302 K For this value of this temperature, the value of I will be



Z=0.773 Thus from the above mentioned formula



H r =−2.253 kJ /mol And



Sr =−4.346



kJ mol . K



From the appendix we can deduce the value of C p=5.325 As we know that



T 2=



Hr +T 1 Cp



T 2=302 K And



Δ S=C p ln (T ¿ ¿ 2/T 1)−Rln Therefore



P2 −S r ¿ P1



Δ S=30 kJ /mol



c) P2=1.2 ¯¿ T 1=250 K P1=60 ¯¿ T c =162.2 K Pc =34 ¯¿ ω=0.038 A=3.820 B=0.593 x 10−3 C=0 D=0.040 x 105 T r=1.981 Pr =1.765 Ω=0.08664 ψ=0.42748 As we know that



β=Ω.



Pr Tr



β=0.077 q=(



Ψ ) Ω. T 1.5 r



q=1.7695 Guess z=1;



z=1+ β−qβ



z−β z (z + β )



z=0.960 For i=1



I i=ln (



z+β ) z



I =0.077 As we know that



H r =R .T 1 [ ( z−1 ) −1.5 qI ] H r =−507.95 kJ /mol Also



Sr =R [ ln ( z )−0.5 qI ] Sr =−0.2255 kJ /molK



The most convenient way will be by guess the final downstream temperature initially and then calculating it. Assume, T 2=232 K For this value of this temperature, the value of I will be



Z=0.956 Thus from the above mentioned formula



H r =−0.521 kJ /mol And



Sr =−1.396



kJ mol . K



From the appendix we can deduce the value of C p=3.502 As we know that



T 2=



Hr +T 1 Cp



T 2=232.0 K And



Δ S=C p ln (T ¿ ¿ 2/T 1)−Rln Therefore



Δ S=32 kJ /mol



d) P2=1.2 ¯¿ T 1=400 K P1=20 ¯¿ T c =369.8 K Pc =42.28 ¯¿ ω=0.152 A=1.213 B=28.785 x 10−3 C=−8.824 x 10−6 D=0 T r=1.081 Pr =0.471 Ω=0.08664 ψ=0.42748 As we know that



P2 −S r ¿ P1



β=Ω.



Pr Tr



β=0.0377 q=(



Ψ ) Ω. T 1.5 r



q=4.3899 Guess z=1;



z=1+ β−qβ



z−β z (z + β )



z=0.88422 For i=1



I i=ln (



z+β ) z



I =3.96 As we know that



H r =R .T 1 [ ( z−1 ) −1.5 qI ] H r =871.03 kJ /mol Also



Sr =R [ ln ( z ) −0.5 qI ] Sr =−8.859



kJ mol . K



The most convenient way will be by guess the final downstream temperature initially and then calculating it. Assume, T 2=380 K For this value of this temperature, the value of I will be



Z=0.86 Thus from the above mentioned formula



H r =−1.39 kJ /mol And



Sr =−2.33



kJ mol . K



From the appendix we can deduce the value of C p=9.01 As we know that



T 2=



Hr +T 1 Cp



T 2=385.1 K And



Δ S=C p ln (T ¿ ¿ 2/T 1)−Rln



P2 −S r ¿ P1



Therefore



Δ S=22.163 kJ /mol



Numerical 3: For a pressure-explicit equation of state, prove that the Joule/Thompson inversion curve is the locus of states for which: T(



∂Z ∂Z ) = p( ) ∂T p ∂P T



Apply this equation to (a) the van der Waals equation; (b) the Redlich/Kwong equation. Discuss the results.



Solution: We know that the Joule Thompson inversion curve is locus of states for which ( On applying general equation of differential calculus



(∂ x /∂ y) z=(



∂x ∂ x ∂w ) +( ) ( ) ∂ y w ∂w y ∂ y z



For this it become



(



∂Z ∂Z ∂Z ∂ρ ) =( ) +( ) ( ) ∂ T P ∂T ρ ∂ ρ T ∂ T P



Therefore it become



(



∂Z ∂Z ∂Z ∂ρ ) =( ) +( ) ( ) ∂ T ρ ∂T P ∂ ρ T ∂ T P



We know that



PV =ZRT P=ZRT / V P= ρZRT Because ρ=1/V Thus



ρ=



P ZRT



On taking differential on both sides



(



(



∂ρ R −1 ∂Z ) = { Z +T 2 ∂ T P T ( ZT ) ∂T



( ( ))



}



P



∂ρ R ) = ¿ ∂T P T



Taking (



∂Z ) =0 , the preceding equations became ∂T ρ



∂Z ) =0 . ∂T ρ



(



∂Z ∂Z ∂ρ ) =−( ) ( ) ∂T ρ ∂ ρ T ∂T P



And



(



∂ρ −P − ρ ) = = ∂ T P ZR T 2 T



On comparing these two equations



(



∂Z −ρ ∂ Z )= ( ) ∂T ρ T ∂ ρ T



Thus



T(



∂Z ∂Z ) =−ρ( ) ∂T ρ ∂ρ T a)



As we know the van der Waal equation



P=



V a − V −b V 2



Multiplying both sides with V / RT



PV V a = − RT V −b VRT Substituting Z=PV /RT ; V =1 / ρ



Z=



1 aρ − 1−bρ RT



Also we know that q=a /bRT , ξ=bρ Then



Z=



1 −qξ 1−ξ



On differentiating it



(



∂Z ∂Z −ξdq ) =( )= ∂ T ρ ∂T ξ dT



For van der Waal equation we know that



α =1, q=



Ψ ΩT r



Therefore on taking differential of q on both sides −1 dT r 2 dq Ψ T r = . dT Ω dT



( )



dq −Ψ = dT Ω T 2r T c dq −Ψ −q = = dT Ω TT r T Therefore



(



∂Z qξ )= ∂T ρ T



Also



(



∂Z ) =b ¿ ∂ρ T b)



For Redlich/Kwong equation we know that



P=



RT a − V −b V (V +b)



Proceeding exactly as above with same definitions



Z=



1 qξ − 1−ξ 1+ξ



For Redlich/Kwong, we know that



α =T −0.5 r Also q=



Ψ ΩT 1.5 r



This leads to



dq −1.5 q = dT T And



(



∂Z 1.5 qξ )= ∂ T ρ T ( 1+ ξ)



Moreover



(



∂Z b bq ) = − 2 ∂ ρ T (1−ξ ) ( 1+ξ )2



Substituting these two equations lead to



q=



1+ ξ 2 1 ( ) 1−ξ 2.5+1.5 ξ



( )



Numerical 4: Isobutene expands adiabatically in a turbine from 5,000kPa and 250℃ to 500kPa at the rate of 0.7 kgmol s−1 . If the turbine efficiency is 0.80, what is the power output if the turbine and what is the temperature of the isobutane leaving the turbine? Solution: P1=5000 kPa T 1=250 ℃=523.13 K T C =408.1 K Pc =36.48 ¯¿ ηo =0.7 kgmol s−1 P2=500 kPa For turbine



∆ S=0 A=1.677 B=37.853 x 10−3



C=−11.94 x 10−6 D=0 T r 1=1.282 Pr 1=1.3706 Pr 2=0.137 First we will have to calculate the final temperature assuming that the iso-butane is behaving ideal As we know that



∆ S=C p ln



T2 P2 −Rln ( ) T1 P1



( )



Assuming that ∆ S=21.06 We get



T 2=445.17 K Therefore



T r 2=1.092 For the power output T2



∆ H ig =∫ C p dT T1



T2



∆ H ig =R ∫ T1



Cp dT R



T2



∆ H ig =R ∫ (A +BT +C T 2) dT T1



Thus



∆ H ig =



−11.07 kJ mol



In actual case



∆ H ' =∆ H ig + ∆ H R



For ∆ H R



∆ H R=H R 1−H R 2 H R1 =Pr 1 ¿ RTC Where



Bo=0.083−



0.422 T 1.6 r1



B1=0.139−



0.172 T 4.2 r1



d B o 0.675 = d T r 1 T 2.6 r1 d B 1 0.7 .22 = 5.2 d T r1 T r1 Also



H R2 =Pr 2 ¿ RTC Where



Bo=0.083−



0.422 T 1.6 r2



B1=0.139−



0.172 T 4.2 r2



d B o 0.675 = d T r 2 T 2.6 r2 d B 1 0.7 .22 = 5.2 d T r2 T r2 Thus on substituting the value in above mentioned equations for the determination of ∆ H R, we get



∆ H R=−8320.32



kJ mol



Therefore we get



∆ H ' =−8331.4



kJ mol



For η=0.8



Δ H =η Δ H ' Δ H =−6665



kj mol



For η=0.7 We get



W =η . ΔH W =−4666.1 kW For the actual final temperature, we will use the following equation



∆ H =C p ln



T2 −∆ H R T1



( )



Thus, we get



T 2=457.9 K



Numerical 5: The steam rate to a turbine for variable output is controlled by a throttle valve in the inlet line. Steam is supplied to the throttle valve at 1,700kPa and 225℃. During a test run, the pressure at the turbine inlet is 1,000kPa, the exhaust steam at 10kPa has a quality of 0.95, the steam flow rate is 0.5kg s−1 , and the power output of the turbine is 180kW a) What are the heat losses from the turbine? b) What would be the power output if the steam supplied to the throttle valve were expanded isentropically to the final pressure? Solution: P1=1700 kPa T 1=225 ℃=498.13 K



H 1=2851



kj kg



S1=6.513



kJ kg



At 10 kPa



x 2=0.95 Sliq =0.6993



kJ kg



H liq =191.832 kJ /kg H vap =2584.4 kJ /kg Svap =8.511 kJ /kgK m . =0.5



kg s



Thus we can calculate W ., as



W . =−180 kW As we know that



H 2=H liq + x2 ( H vap −H liq ) H 2=2465



kJ kg



Thus, as



Δ H =H 2−H 1 Δ H =−385.85



kJ kg



As we know that



Q=m. . Δ H −W . Q=−12.9



kJ sec



For isentropic expansion



'



x 2=



S1 −S liq S vap−S liq



x '2=0.78 H '2=H liq + x'2 ( H vap −H liq ) H '2=2063



kJ kg



As



W . =m . Δ H =m . (H ¿¿ 2' −H 1 )¿ W . =−392.4 kW



Numerical 6: Turbine can be used to recover energy from high-pressure liquid streams. However they are not used when the high-pressure stream is saturated liquid. Why? Illustrate by determining the downstream state for isentropic expansion of saturated liquid water at 5 bar to a final pressure of 1 bar. Solution: When a saturated liquid is expanded in a turbine some of the liquid vaporizes. A turbine properly designed for expansion of liquids cannot handle the much larger volumes resulting from the formation of vapor. Thus the expansion will be irreversible and the fraction of liquid vaporize will be greater



Numerical 7: Saturated steam at 125kPa is compressed adiabatically in a centrifugal compressor to 700kPa at the rate of 2.5 kg s−1 . The compressor efficiency is 78%. What is the power requirement of the compressor and what are the enthalpy and entropy of the steam in its final state? Solution: P1=125 kPa P2=700 kPa m . =2.5 kg/ s



From steam table at 125 kPa



H 1=2685.2 S1=7.284



kj kg



kJ kgK



At 700 kPa



H '2=30513.3 S2=7.458



kJ kg



kJ kgK



η=78 % As we know that



Δ H =η ¿ H '2 −H 1 ΔH= η Δ H =469.36



kJ kg



Therefore



H 2= Δ H+ H 1 H 2=3154.7



kj kg



Therefore using the final enthalpy value we can deduce the value of entropy Thus



S2=7.46



kJ molK



Also



W =m ΔH Thus



W =1173.46 kW



Numerical 8: Propylene is compressed adiabatically form 11.5 bar and 30℃ to 18 bars at the rate of 1kgmol s−1 . If the compressor efficiency is 0.8, what is the power requirement if the compressor and what is the discharge temperature of the propylene? Solution: P1=11.5 ¯¿ T 1=30 ℃ T C =365.6 K Pc =46.65 ¯¿ η=0.7 kgmo l s−1 P2=18 ¯¿ For turbine



∆ S=0 A=1.67 B=22.706 x 10−3 C=−6.905 x 10−6 D=0 T r 1=0.8292 Pr 1=0.2465 Pr 2=0.386 First we will have to calculate the final temperature assuming that the propane is behaving ideal As we know that



∆ S=C p ln



T2 P2 −Rln ( ) T1 P1



( )



Assuming that ∆ S=−3.11 We get



T 2=324.127 K Therefore



T r 2=0.887 For the power output T2



∆ H ig =∫ C p dT T1



T2



∆ H ig =R ∫ T1



Cp dT R



T2



∆ H ig =R ∫ (A +BT +C T 2) dT T1



Thus



∆ H ig =



1.409 kJ mol



In actual case



∆ H ' =∆ H ig + ∆ H R For ∆ H R



∆ H R=H R 1−H R 2 H R1 =Pr 1 ¿ RTC Where



Bo=0.083−



0.422 T 1.6 r1



B1=0.139−



0.172 T 4.2 r1



d B o 0.675 = d T r 1 T 2.6 r1 d B 1 0.7 .22 = 5.2 d T r1 T r1 Also



H R2 =Pr 2 ¿ RTC Where



Bo=0.083−



0.422 T 1.6 r2



B1=0.139−



0.172 T 4.2 r2



d B o 0.67 5 = 2.6 d T r2 T r2 d B 1 0.7 .22 = 5.2 d T r2 T r2 Thus on substituting the value in above mentioned equations for the determination of ∆ H R, we get



∆ H R=−0.4449



kJ mol



Therefore we get



∆ H ' =1.0241



kJ mol



For η=0.8



Δ H =η Δ H ' Δ H =0.81928



kj mol



Also m=1 kgmol . s−1we get



W =m. ΔH



W =0.82 kW For the actual final temperature, we will use the following equation



∆ H =C p ln



T2 −∆ H R T1



( )



Thus, we get



T 2=327.1 K



Numerical 9: Steam enters a nozzle at 800kPa and 280℃ at negligible velocity and discharges at a pressure of 525kPa. Assuming isentropic expansion of the steam in the nozzle, what is the exit velocity and what is the cross-sectional area at the nozzle exit for a flow rate of 0.75kg s−1 ? Solution: P1=800 kPa T 1=280 ℃ From steam table we get



H 1=3014.9 kJ /kg S1=7.195 kJ /kgK Similarly at



P2=52 kPa S2=7.1595 kJ /kgK From steam table we get



H 2=2885.2 kJ /kg V 2=531.2 c m3 / g With heat, work potential energy and initial velocity equals to zero Form the 1st law of conservation of thermodynamics, we get



u22 ∆ H + =0 2 u2= √−2( H 2−H 1) u2=565.2 m s−1 Also we know that



A2=



m. V 2 u2



We get



A2=7.05 c m



2



Numerical 10: Air at 1 atm and 35 ℃ is compressed in a staged reciprocating compressor (with intercooling) to a final pressure of 50 atm. For each stage, the inlet gas temperature is 35 ℃ and the maximum allowable outlet temperature is 200℃. Mechanical power is the same for all stages and isentropic efficiency is 65% for each stag. The volumetric flowrate of air is 0.5m 3 s−1 at the inlet to the first stage a) b) c) d)



How many stages are required? What is the mechanical power requirement per stage? What is the heat duty for each intercooler? What is the coolant for each intercooler? It enters at 25℃ and leaves at 45℃. What is the cooling water rate per intercooler?



( 72 ) R.



Assume air is an ideal gas withC p= Solution: P1=1 atm T 1=35 ℃=308.13 K P2=5 ¯¿ T 2=200 ℃=473.13 K



η=65 % V =0.5



m3 s



C p=3.5 R a) As we know that



T 2=T 1 +



T '2 −T 1 η



T '2=[ ( T 2−T 1) η+T 1 ] T '2=415.46 K As for one step



T 2=T 1 ¿ For n step



T 2=T 1 ¿ P2 ) P1 R n= Cp T ln ⁡( 2 ) T1 ln ⁡(



n=3.74=4≈. b) For the power requirement per stage we know that



W=



W ideal η C p T1[



W=



P2 P1



( )



R n cp



−1]



η



On substituting the value we get



W =444.62 kW The work obtained is for the whole process Thus the work done in per step will be



W . =111.15 kW c) Since the working gas is ideal and is entering and leaving the compressor and intercooler respectively at the same temperature, therefore there is no change in the enthalpy. Thus 1st law of thermodynamic yield



Q . =W . Therefore



Q . =−111.15 kW d) From steam table for the saturated liquid



Δ HW =



( 188.4−104.8 ) kJ kg



Δ H W =83.6



kJ kg



m . =Q . / Δ H W m . =1.329



kg s



Numerical 11: Liquid (identified below) at 25℃ are completely vaporized at 1 atm in a counter current heat exchanger. Saturated steam is the heating medium, available at four pressures: 4.5,9,17 and 33 bars. Which variety of steam is most appropriate for each case? Assume a minimum approach ∆ T of 10℃ for heat exxhange. a) Benzene; b) n-Decane;



c) Ethylene glycol; d) o-Xylene Solution: For this we required the liquid with lowest saturated temperature to satisfy ΔT =0 For this benzene is the most appropriate for each case.



Numerical 12: For isentropic expansion in a converging/diverging nozzle with negligible entrance velocity, sketch graphs of mass flow rate m, ˙ velocity u, and area ratio A/ A1 vs the pressure ratio P/ P1. Here, A is the cross-sectional area of the nozzle at the point in the nozzle where the pressure is P, and the subscript 1 denotes the nozzle entrance. Solution: The mass-flow rate m . is of course constant throughout the nozzle from entrance to exit. The velocity u rises monotonically from nozzle entrance to nozzle exit as the pressure ratio decrease. The area ratio decreases at the nozzle entrance to a minimum value at the throat and thereafter increases to the nozzle exit.



Power cycles Numerical 13: A power plant operating on heat recovered from the exhaust gases of internal combustion engine uses isobutane as the working medium in a modified Rankine Cycle in which the upper pressure level is above the critical pressure of isobutane. Thus the isobutane doesn’t undergo a change of phase as it absorbs heat prior to its entry into turbine. Isobutane vapor is heated at 4,800kPa to 260℃, and enters the turbine as a supercritical fluid at these conditions: Isentropic expansion in the turbine produces a superheated vapor at 450kPa, which is cooled and condensed at constant pressure. The resulting saturated liquid enters the pump for return to the heater. If the power output of the modified Rankine cycle is 1,000kW, what is the isobutnae flow rate, the heat-transfer rates in the heater and condenser, and the thermal efficiency of thee cycle? The vapor pressure of isobutnae is given in Table B.2 of App. B.



Solution: P1=4800 kPa T 1=533.13 K T C =408.1 K Pc =36.48 ¯¿ P2=450 kPa For turbine



∆ S=0 A=1.677 B=37.853 x 10−3 C=−11.94 x 10−6 D=0 T r 1=1.3064 Pr 1=1.3158 Pr 2=0.123 Assume T 2=455 K , for the sake of ease of solving the problem Thus



T2



∆ H ig =∫ C p dT T1



T2



∆ H ig =R ∫ T1



Cp dT R



T2



∆ H ig =R ∫ (A +BT +C T 2) dT T1



Thus



∆ H ig =



−1.14 kJ mol



In actual case



∆ H ' =∆ H ig + ∆ H R For ∆ H R



∆ H R=H R 1−H R 2 H R1 =Pr 1 ¿ RTC Where



Bo=0.083−



0.422 T 1.6 r1



B1=0.139−



0.172 T 4.2 r1



d B o 0.675 = d T r 1 T 2.6 r1 d B 1 0.7 .22 = 5.2 d T r1 T r1 Also



H R2 =Pr 2 ¿ RTC Where



Bo=0.083−



0.422 T 1.6 r2



B1=0.139−



0.172 T 4.2 r2



d B o 0.675 = d T r 2 T 2.6 r2



d B 1 0.7 .22 = 5.2 d T r2 T r2 Thus on substituting the value in above mentioned equations for the determination of ∆ H R, we get



∆ H R=−8848.4



kJ mol



Therefore



∆ H turbine =Δ H=−8850 kj/mol For calculating the work of pump we will have to find the volume of iso-butane as a saturated liquid 450 kPa. For this we will first have to calculate the saturation temperature, which can be calculated by the following Antoine equation As



ln Psat =A−



B T +C



Therefore



T sat =



B −C A−ln P sat



T sat =34 ℃ =307.15 K For the table we know that



V c =262. c m3 /mol Z c =0.282 T rsat =0.753 As we know that 1−T rsat )



V liq =V c Z (c



V liq =112.36



2/7



c m3 mol



W pump =V liq ( P1 −P 2 ) W pump =489 kW



For the mass rate



m. =



1000 W turbine+ W pump



m. =0.12



kg s



For the heat-transfer rates we will first have to calculate the enthalpy changes in the cooler/condenser with the saturated temperature as the initial temperature. T2



∆ H ig =∫ C p dT T sat



T2



∆ H ig =R ∫



T sat



Cp dT R



T2



∆ H ig =R ∫ ( A +BT +C T 2)dT T sat



Thus



∆ H ig =



−1.746 kJ mol



Therefore



∆ H a =∆ H ig +∆ H R Implies that



∆ H a =−1.



808 kJ mol



Also



T n=261.4 K T rn=0.641 We know that



∆ Hn=



1.092 R T n ( ln Pc −1.013 ) 0.93−T rn



∆ H n =21.18



kJ mol



For the ∆ H b



1−T rsat ∆ H b =−∆ H n 1−T rn



(



0.38



)



Therefore



∆ H b =−18.37



kJ mol



Since we know that



Q out =m. ( ∆ H a +∆ H b ) Qout =−4360 kW Also



Q¿ =(W turbine+ W pump )m. +Qout Q ¿ =5360 kW As we know that



η=



1000 Q¿



η=0.18 %η=18 %



Numerical 14: For comparison of Diesel and Otto-engine cycles: a) Show that the thermal efficiency of the air-standard Diesel cycle can be expressed as 1 η=1− r



()



γ−1



r cγ−1 γ ( r c −1 )



where r is the compression ration and r c is the cutoff ratio, defined as r c =V A /V D . (See fig.8.10)



b) Show that for the same compression ratio the thermal efficiency of the air-standard Otto engine is greater than the thermal efficiency of the air standard Diesel cycle. 1 Hint: show that the fraction which multiplies r



γ −1



()



in the equation for η is greater



than unity by expanding r γc in a Taylor series with the remainder taken to the first derivative c) If γ =1.4, how does the thermal efficiency of an air-standard Otto cycle with a compression ratio 0f 8 compare with the thermal efficiency of an air-standard Diesel cycle with the same compression ratio and cutoff ratio of 27 how the compression changed if the cutoff ratio is 3? Solution: a) As we know that



1 γ 1 γ − r 1 re η=1− [ ] γ 1 1 − re r



( ) ()



Where



r e =expansionratio=V B /V A r =compression ratio=V c /V D r c =cutoff ratio=V A /V D Since V A =V B



r =(V ¿ ¿ c /V D) /(V ¿ ¿ c /V A )=r c ¿ ¿ re Or



1 rc = re r Therefore



rc γ 1 γ − r 1 r η=1− [ ] γ rc 1 − r r



( ) ()



1 r



()



η=1−



γ−1



r γc −1 γ (r c −1)



b) We wish to show that



r γc −1 >1 γ (r c −1) Taylor series with remainder, taken to the first derivative can be written as γ −2 1 r γc =1+ γ ( r c −1 ) + γ ( γ −1 ) [ 1+θ ( r c −1 ) ] . ( r c −1 )2 2



For a> 1 and r c >1 also



γ −2 1 γ ( γ −1 ) [ 1+θ ( r c −1 ) ] . ( r c −1 ) 2> 0 2



Therefore



r γc >1+γ ( r c −1 ) r γc −1> γ ( r c −1 ) On dividing both side with γ ( r c −1 ), we will get



r γc −1 >1 γ ( r c −1 ) Hence proved that with the same compression ratio the thermal efficiency of Otto engine is



greater than that of Diesel cycle.



c) If γ =1.4 and r =8 Then the thermal efficiency of the air standard Otto engine



1 r



()



η=1−



γ−1



ηotto =0.564 Similarly the thermal efficiency of the diesel engine can be given as



1 η=1− r



()



γ−1



r γc −1 γ (r c −1)



Therefore If r c =2



ηdiesel =0.49 And for r c =3



ηdiesel =0.43



Numerical 15: An air-standard Diesel cycle absorbs 1,500Jmo l−1 of heat (step DA of fig.8.10, which simulates combustion). The pressure and temperature at the beginning of the compression step are 1 bar and 20℃, and the pressure at the end of the compression step is 4 bar. Assuming air to



( 72 ) R and C =( 52 ) R, what are the compression ratio and the



be an ideal gas for which C p= expansion ratio of the cycle? Solution:



v



7 C p= R 2 Pc =1 ¯¿ T C =293.13 K P D=5 ¯¿ γ =1.4 As we know that



Pc V γc =PD V γD Vc P =r= D VD Pc



1γ



( )



r =3.1 Also we know that



Q DA=1,500 Jmol−1 Q DA=C p ( T A −T D ) T A=



Q DA +T D CP



T D =515.8 K re=



VB VA



As we know that



V B =V c Therefore



RT C V B Vc PC re= = = V A V A RT A PA



Since



P A =PD Therefore



r e =2.8



Numerical 16: Consider an air-standard cycle for the turbojet power plant shown in the figure 8.13. The temperature and pressure of the air entering the compressor are 1 bar and 30℃. The pressure ratio in the compressor is 6.5, and the temperature at the turbine inlet is 1,100℃. If the expansion in the nozzle is isentropic and if the nozzle exhausts at 1 bar, what is the pressure at the nozzle inlet(turbine exhaust) and what is the velocity if the air leaving the nozzle? Solution:



From this given figure



T A=303.15 K T C =1373.15 K



7 CP= R 2 P A =PE =1 ¯¿ Pc =P B=6.5 ¯¿ Molecular Wt. of air = 29



g mol



For the work, we know that



P W AB =C P T A [ B PA



( )



R CP



−1]



2



W AB =C P T A [r comp7 −1] Also



W CD =C P T C [



PD PC



( )



R CP



−1]



2 7



W CD =C P T C [r exp −1] Both work terms are equal but have opposite sign Therefore



[



2



]



2



T C r exp 7 −1 =−T A [r comp 7 −1] As we know that



r comp =6.5 Therefore



r exp =0.55 Also we know



r exp = And



PD PC



r comp =



PB PA



Therefore



PD =r r P E exp comp As we know that



u2E−u 2D =



2 γ PD V D P [1− E γ −1 PD



( )



γ −1 γ



]



On making following substitution



u D=0 γ −1 2 = γ 7 P D V D =nR T D We get



√



u E= 2



2 7 RTD 1 7 [1− ] 2n r exp r comp



(



)



u E=843.33m s−1 Also



PD =r r P E exp comp P D=P E r exp r comp Thus



P D=3.56 ¯¿



Numerical 17: Liquefied natural gas (LNG) is transported in very large tankers, stored as liquid in equilibrium with its vapor at approximately atmospheric pressure. If LNG is essentially pure methane, the



storage temperature then is about 111.4K, the normal boiling point of methane. The enormous amount of cold liquid can in principle serve as a heat sink for an on-board heat engine. Energy discarded to the LNG serves for its vaporization. If the heat source is ambient air at 300K, and if the efficiency of a heat engine is 60% of its Carnot value, estimate the vaporization rate in moles vaporized per kJ of power output. For methane, ∆ H lvn =8.206 kjmo l −1. Solution: As



T C =111.4 K T H =300 K Δ H lvn =8.206



kJ mol



As we know that



ηcar not =1−



TC TH



ηcarnot =0.629 Also given



ηengine =0.6 ηcarnot ηengine =0.377 Basis W =1 kJ ηengine = Q heat =



W Q heat



W ηengine



Q heat =2.65 kJ Also



Qheat ( 1−ηengine )=Qcold



Q cold =1.65 kJ Also as



Q cold Δ H lvn mol =0.20 W kJ



Numerical 18: Air-standard power cycles are conventionally displayed on PV diagrams. An alternative is the PT diagram. Sketch air-standard cycles on PT diagram for the following: a) b) c) d)



Carnot cycle; Otto-cycle; Diesel cycle; Brayton cycle.



Why would a PT diagram not be helpful for depicting power cycles involving liquid/vapor phase changes? Solution: a) Carnot Cycle



b) Otto-Cycle



c) Diesel Cycle



d) Brayton Cycle



Numerical 19: Steam enters the turbine of a power plant operating on the Rankine cycle (fig.8.3) at 3,300kPa and exhausts at 50kPa. To show the effect of superheating on the performance of the cycle, calculate the thermal efficiency of the cycle and the quality of the exhaust steam form the turbine for turbine-inlet steam temperatures of 450, 550 and 650℃ . Solution:



At 50 kPa saturated liquid P1=50 kPa P4 =3300 kPa V 4 =1.030 c m3 / g H 4 =340.564 kJ /kg



Sliq =1.09



kJ molK



At 50 kPa saturated vapor H vap =2646 kJ /kg Svap =7.59



kJ kgK



As we know that W pump =V 4 ( P 4−P1 ) W pump =3.34 kW Δ H =W pump W pump =H 1 + H 4 H 1=W pump + H 4 H 1=343.91



kJ kg



For the given values of temperatures 450 T = 550 ℃ 650



{



3340.6 H 2= 3565.3 kJ /kg 3792.9



{ {



7.037 3 S2= 7.3282 kJ / kgK 7.5891 S'3=S 2 As S '3−S liq x= S vap−S liq ' 3



Thus 0.914 x '3= 0.959 0.999



{



H '3=H liq + x'3 (H vap−H liq ) 2447.73 H '3= 2551.477 kJ /kg 2643.69



{



W turbine =H '3−H 2 −892.87 W turbine = −789.13 kW −696.91



{



Q H =H 2−H 1 2996.68 Q H = 3221.389 kJ 3448.98



{



Also η=



W turbine +W pump QH



0.297 η= 0.314 0.332



{



Numerical 20: Steam enters the turbine of a power plant operating on the Rankine cycle (fig.8.3) at 600 ℃ and exhaust at 30kPa. To show the effect of boiler pressure on the performance of the cycle, calculate the thermal efficiency of the cycle and the quality if the exhaust steam from the turbine for boiler pressures of 5,000, 7,500 and 10,000kPa. Solution:



At30 kPa saturated liquid



V 4 =1.022 c m3 H 4 =89.302 kJ /kg P1=30 kPa H liq =289.302 kJ /kg Sliq =7.7695 kJ /kg At 30 kPa saturated vapor



H vap =265.4 kJ /kg Svap =0.9441 kJ /kg For the given values of pressure



5000 P4 = 7500 kPa 10000



{ { {



3664.5 H 2= 3643.7 kJ /kg 3622.7 7.2578 S2= 7.0526 kJ /kgK 6.9013



As we know that



W pump =V 4 ( P 4−P1 ) 5079.34 W pump = 7634.34 kW 10189.34



{



Also



W pump =H 1−H 4 H 1=H 4 + W pump 294.381 H 1= 296.936 kJ /kg 299.491



{



S'3=S 2 As S '3−S liq x= S vap−S liq ' 3



Thus 0.925 x '3= 0.895 0.873



{



H '3=H liq + x'3 (H vap−H liq ) 2450.192 H '3= 2380.109 kJ /kg 2328.715



{



W turbine =H '3−H 2 −1214.308 W turbine = −1263.591 kW −1293.985



{



Q H =H 2−H 1



3370.119 Q H = 3346.764 kJ 3323.209



{



Also η=



W turbine+ W pump QH



0.359 η= 0.375 0.386



{



Numerical 21: A steam power plant operating on a regenerative cycle, illustrated in fig. 8.5 includes just one feedwater heater. Steam enters the turbine at 650 psia and 900℉ and exhaust at 1 psia. Steam for the feedwater heater is extracted from the turbine at 50 psia, and in condensing raises the temperature at 50 psia. If the turbine and pump efficiencies are both 0.78, what is the thermal efficiency of the cycle and what fraction of the steam entering the turbine is extracted for the feedwater heater? Solution:



At 4500 kPa∧500 ℃ from steam table



H 2=3439.3 kJ /kg S2=7.3011 kJ /kg S'3=S 2 At 350 kPa & this entropy



H '3=2770.6 kJ /kg η=0.78 As we know that



W 1=η ( H '3−H 2 ) W 1=−521.586 kW Also



W 1= Δ H=H 3 −H 2 or



H 3=H 2 +W 1 Thus



H 3=2.98 x 10 3 kJ /kg At 20 kPa isentropic expansion



S'4 =S2 H liq =251.453 kJ /kg H vap =2609.9 kJ /kg Sliq =0.8321



kJ kgK



Svap =7.9094 kJ /kjK Also as we know that



x '4 =



S '4−S liq S vap−Sliq



x '4 =0.876 Also



H '4 =H liq + x '4 (H vap−H liq ) H '4 =2.317 x 103 kJ /kg Also



H 4 =H 2+ η ( H '4 −H 2 ) H 4 =2.564 x 103 kJ /kg We know that



H 5=H liq V 5=1.017 m3 / g P5=20 kPa P6=4500 kPa W pump =



V 5 ( P 6−P5 ) η



W pump =5.841 kW Also



W pump = Δ H=H 6−H 5 or



H 6=H 5 +W pump Thus



H 6=257.294 kJ /kg At 350 kPa for saturated liquid



H 7=584.270 kJ /kg T 7=138.86 ℃ T 1=138.87−6=132.87 ℃ At this temperature at steam table



H satliq =558.2 kJ /kg Psat =294.26 kPa V satliq =1.073 c m3 /g By definition of the volume expansivity



β=



1 V satliq



(



1083−1.063 ) 20



β=9.32 x 10−4 P1=P6 As we know that



H 1=H satliq +V satliq ( 1−β T 1 ) (P ¿ ¿ 1−P sat ) ¿ H 1=561.305 kJ /kg Also by the energy balance on feed water heater, we get



m=



H 1−H 6 H 3−H 7



m=0.132 kg Also



W 2 =( 1−m ) (H 4 −H 3) W 2 =−307.567 kW Thus overall work will be



W net =( W 1+W pump ) +W 2 W net =−823.3 kW Also



Q H =H 2−H 1 Q H =2878 kJ Thus the efficiency will be



η=



W net QH



η=0.286



Refrigeration & Air Conditioning Numerical 22: A vapor-compression refrigeration system operates on the cycle of fig.9.1. The refrigerant is water. Given that the evaporation t=4℃, the condensation t= 14℃, η(compressor) =0.76, and the refrigeration rate =1,200 kJ s−1 , determine the circulation rate of the refrigerant, the heat-transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle, and the coefficient of performance of the cycle, and the coefficient of performance of a Carnot refrigeration cycle operating between the same temperature levels. Solution:



From the



T 2=277.13 K T 4=307.13 K η=0.76 Q C =1200 kJ / s H 2=2508.9 kJ /kg



steam table



S2=9.0529



kJ kgK



H 4 =142.4 kJ / kg As it is an isentropic expansion



S'2=S 2 Via interpolation



H '3=2814.7 kJ /kg H '3−H 2 Δ H 23 = η Therefore



Δ H 23 =402.386 kJ /kg Also we know



Δ H 23 =H 3 −H 2 H 3= Δ H 23+ H 2 Therefore



H 3=2.911 x 103 kJ /kg Thus



m=QC /Δ H 12 m=Q C / H 2−H 1 m=



0.507 kg s



Also



Q H =m ( H 4−H 3 ) Q H =−1404 Similarly



kJ s



W =m Δ H 23 W =204 kW ω=



Qc W



ω=5.881 Also



ω carnot =



T2 T 4 −T 2



Thus



ω carnot =9.238



Numerical 23: A vapor-compression refrigeration system is conventional except that a countercurrent heat exchanger is installed to subcool the liquid from the condenser by heat exchange with the vapor stream from the evaporator. The minimum temperature difference for heat transfer is 10℉. Tetrafluroenthane is the refrigerant (Table 9.1, fig.G.2), evaporating at 20℉ and condensing at 80℉. The heat load on the evaporator is 2,000 Btu s−1 . If the compressor efficiency is 75%, what is the power requirement? How does this result compare with the power required by the compressor if the system operates without the heat exchanger? How do the refrigerant circulation rates compare the two cases? Solution: At 20 ℉∧33.110 psia



H 2=105.907 Btu/lb S2=0.22325 Btu/lbR At point 2A same pressure and at 70 ℉



H 2 A =116 Btu /lb S2 A =0.2435 Btu /lbR At 80 ℉



H 4 =37.978 Btu /lb S4 =0.07892 Btu /lbR According to energy balance, heat exchanger



H 1=H 4 −H 2 A + H 2 H 1=27.885 Btu/lb As



Q c =2000



Btu s



Therefore



m=



Qc H 2−H 1



m=25.634



lb s



At entropy of 0.2434 and pressure of 101.37 psia



H '3=127



Btu lb



η=0.75 H '3 −H 2 A Δ H comp= η Therefore



Δ H comp=14.667



Btu lb



Thus



W =m Δ H comp Therefore



W =390.66 kW If heat exchanger is limited



H 1=H 4 As



m=



QC H 2−H 4



So



m=29.443



lb s



As we know



H '3=116 Btu /lb H '3 −H 2 Δ H comp= η Therefore



Δ H comp=13.457



Btu lb



Thus



W =m Δ H comp Therefore



W =418.032 kW



Numerical 24: A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a low temperature heat source; during the summer, it acts as a high-temperature heat sink. The heat transfer rate through the walls and inside and outside of the house, summer and winter. The heat-pump motor is rated at 1.5kW. Determine the minimum outside temperature for which the house can be maintained at 20℃ during the winter and the maximum outside temperature for which the house can be maintained at 25 ℃ during the summer. Solution:



Winter



T H =293.13 K W =1.5 kW As



Q H =−0.75(T H −T C ) Guess T C =250 K Therefore



Q H =−32.3475 kJ As



W /Q H =



T H −T c TH



Therefore



T c =268.94 K Summer



T C =293.13 K Guess T H =300 K Therefore



Q C =0.75(T H −T C ) Thus



Q C =1.5975 kJ As



T −T c W = H 0.75(T H −T c ) TH Therefore



T H =322.5 K



Numerical 25: The contents of the freezer in a home refrigerator are maintained at -20℃. The kitchen temperature is 20℃. If the heat leaks amount to 125,000kJ per day, and if the electricity costs $0.08 per kWhr, estimate the yearly cost of running the refrigerator. Assume a coefficient of performance equal to 60% of the Carnot value. Solution: T H =20 ℃ =293.13 K T C =−20 ℃=253.13 K QC =125000 kJ /day As we know that



ω carnot =



Tc T H −T c



Thus



ω carnot =6.328 As



ω=0.6 ω carnot Therefore



ω=3.979 As



W=



QC ω



Thus



W =0.381 kW For the cost We know that



cost=



0.08 W kWhr



Therefore



co st=267.18



dolllars yr



Numerical 26: The condenser of a home refrigerator is commonly underneath the appliance; thus the condensing the refrigerant exchanges heat with household air, which has an average temperature of about 70℉. It is proposed to reconfigure a refrigerator so that the condenser is outside the home, where the average yearly temperature is about 50℉. Discuss the pros and cons of this proposal. Assume a freezer temperature of 0℉ and an actual coefficient of performance 60% that of a Carnot refrigerator. Solution: On average, the coefficient of performance will increase, thus providing savings on electric casts. On the other hand, installation casts would be higher. The proposed arrangement would result in cooling of the kitchen, as the refrigerator would act as an air conditioner. This would be detrimental in the winter, but beneficial in the summer, at least in temperate climates.



Numerical 27: Which is the more efficient way to increase the coefficient of performance of a Carnot refrigerator to increase T c with T H constant, or to decrease T H with T c constant? For a real refrigerator, does either of these strategies make sense? Solution: As we know that



ω=



Tc T H −T c



On differentiating it



(



TH ∂ω ) = ∂ T C T ( T H −T c )2 H



And



(



Tc ∂ω ) = ∂ T H T ( T H −T c )2 c



Because T H >T c, the more effective procedure is to increase T c. For a real refrigeration system, increasing T cis hardly an option if refrigeration is required at a particular value of T c. Decreasing T H is no more realistic, because for all practical purposes, T H is fixed by environmental conditions, and not subject to control



Numerical 28: A refrigeration system requires 1.5kW for a refrigeration rate of 4kJ s−1 a) What is the coefficient of performance? b) How much heat is rejected in the condenser? c) If heat rejection is at 40℃, what is the lowest temperature the system can possibly maintain? Solution: a)



Q c =4 kJ /s W =1.5 kW As we know that



ω=Qc /W Therefore



ω=2.66 b) As



W =QH −Qc



Or



Q H =W +Qc Thus



Q H =5.5



kJ s



c)



T H =40 ℃+ 273.13=313.13 K As



ω=



Tc T H −T c



Therefore



T c =T H



ω ( ω+1 )



Thus



T c =227.2 K