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Lect 7 thermo

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Aniruddha Das

The maximum work that can be obtained from a system initially at a higher temperature T than the environment at T0 is equal to the initial available energy minus the available energy at the final state. For an irreversible process, the actual work is always less than the maximum reversible work, with the difference equal to T0 times the entropy generation. The maximum work from a flow process is equal to the change in availability plus any heat input times 1 minus the ratio of environment to heat source temperatures.

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Maximum work or available energy from a system Initially a body of mass m, is at temp T, what is the maximum work that can be obtained out of it if the environment is at ð‘‡0 which is lower than the body temperature T. ð‘‡0 ð‘„â„Ž ð‘„ð‘™ w ð‘šð‘ 𑣠𑇠ð‘„â„Ž = ð‘šð‘ 𑣠𑇠− ð‘‡0 ; ∆𑠠ð‘ = ð‘šð‘ ð‘£ ln ð‘‡0 𑇠∆ð‘ ð‘ ð‘¢ð‘Ÿ = ð‘„ð‘™/ð‘‡0 For max work output, ∆ð‘ ð‘ ð‘¢ð‘Ÿ + ∆𑠠ð‘ = 0 ð‘„ð‘™ = −ð‘‡0 ð‘š ð‘ ð‘£ ln ð‘‡0 𑇠; 𑤠= ð‘„â„Ž − ð‘„ð‘™ = ð‘šð‘ 𑣠𑇠− ð‘‡0 − ð‘‡0 ð‘šð‘ ð‘£ ln 𑇠ð‘‡0 𑤠= ð‘šð‘ 𑣠𑇠− ð‘‡0 − ð‘‡0 ln 𑇠ð‘‡0 = ð‘ˆ1 − ð‘ˆ0 − ð‘‡0(ð‘†1 − ð‘†0) = ð‘ˆ1 − ð‘‡0 ð‘†1 − ð‘ˆ0 − ð‘‡0 ð‘†0 = ð¹1 − ð¹0; ð‘¤â„Žð‘’ð‘Ÿð‘’ ð¹ = 𑈠− ð‘‡0 ð‘†, ð¹ ð‘–ð‘  ð‘ð‘Žð‘™ð‘™ð‘’ð‘‘ ð‘¡â„Žð‘’ ð‘Žð‘£ð‘Žð‘–ð‘™ð‘Žð‘ð‘–ð‘™ð‘–ð‘¡ð‘¦ ð‘œð‘Ÿ ð»ð‘’ð‘™ð‘šð‘œð‘™ð‘¡ð‘§ ð‘“ð‘¢ð‘›ð‘ð‘¡ð‘–ð‘œð‘› If the system boundary expands against the atmospheric pressure p0 then F becomes: 𜑠= 𑈠− ð‘‡0 𑆠+ ð‘0 ð‘‰, ðœ‘1 − ðœ‘2 = ð‘Žð‘£ð‘Žð‘–ð‘™ð‘Žð‘ð‘™ð‘’ ð‘¤ð‘œð‘Ÿð‘˜ = 𑤠ð‘šð‘Žð‘¥
cm ð‘‡0 ð‘„12 ð‘„0 w12 ð‘‡â„Ž Control mass process when the system receives heat from a source Maximum work or available energy ð‘ˆ1 − ð‘ˆ2 ð‘†1 − ð‘†2 The system goes from 1-2, does work w12 and rejects heat ð‘„0 Lets us consider two situations: the body rejects heat either reversibly or irreversibly, but it receives heat reversibly. ð‘„12 − ð‘„0 ð‘–ð‘Ÿð‘Ÿ = ð‘ˆ2 − ð‘ˆ1 + ð‘¤12 ð‘Žð‘ð‘¡ ð‘  ð‘”ð‘’ð‘› = ∆ð‘ ð‘ð‘š + ∆ð‘ ð‘ ð‘¢ð‘Ÿ + ∆ð‘ ð‘Ÿð‘’ð‘  > 0, for irrev process ð‘ ð‘”ð‘’ð‘› ð‘Ÿð‘’ð‘£ = ð‘†2 − ð‘†1 + ð‘„0 ð‘Ÿð‘’ð‘£ ð‘‡0 − ð‘„12 ð‘‡â„Ž = 0 for rev process ð‘„0 ð‘Ÿð‘’ð‘£ = −ð‘‡0 ð‘†2 − ð‘†1 + ð‘‡0 ð‘‡â„Ž Q12, From 1st law for the CM we get: ð‘„12 − ð‘„0 ð‘Ÿð‘’ð‘£ = ð‘¤12 ð‘Ÿð‘’ð‘£ + ð‘ˆ2 − ð‘ˆ1 ð‘¤12 ð‘Ÿð‘’ð‘£ = ð‘„12 − ð‘„0 ð‘Ÿð‘’𑣠− ð‘ˆ2 − ð‘ˆ1 ð‘¤12 ð‘Ÿð‘’ð‘£ = ð‘„12 + ð‘‡0 ð‘†2 − ð‘†1 − ð‘„12 ð‘‡0 ð‘‡â„Ž − ð‘ˆ2 − ð‘ˆ1 ð‘¤12 ð‘Ÿð‘’ð‘£ = ð‘„12 1 − ð‘‡0 ð‘‡â„Ž + ð‘ˆ1 − ð‘‡0 ð‘†1 − ð‘ˆ2 − ð‘‡0 ð‘†2 ð‘¤12 ð‘Ÿð‘’ð‘£ = 𑤠ð‘šð‘Žð‘¥ = ð‘„12 1 − ð‘‡0 ð‘‡â„Ž + ð¹1 − ð¹2 sur res ð‘¤12 ð‘Žð‘ð‘¡ = ð‘„12 − ð‘„0 ð‘–ð‘Ÿð‘Ÿ − ð‘ˆ2 − ð‘ˆ1 ð¼ = ð¼ð‘Ÿð‘Ÿ = ð‘¤12 ð‘Ÿð‘’𑣠− ð‘¤12 ð‘Žð‘ð‘¡ = ð‘‡0 ð‘†2 − ð‘†1 − ð‘„12 ð‘‡0 ð‘‡â„Ž + ð‘„0 ð‘–ð‘Ÿð‘Ÿ = ð‘‡0 ð‘†2 − ð‘†1 − ð‘„12 ð‘‡â„Ž + ð‘„0 ð‘–ð‘Ÿð‘Ÿ ð‘‡0 = ð‘‡0(∆ð‘ ð‘ð‘š + ∆ð‘ ð‘Ÿð‘’ð‘  + ∆ð‘ ð‘ ð‘¢ð‘Ÿ) = ð‘‡0∆𑠠ð‘¢ð‘›ð‘–ð‘£ = ð‘‡0 ð‘ ð‘”ð‘’ð‘› Measure of irreversibility
NH3 100C Example problem: 1kg of NH3 is contained in a spring loaded piston/cylinder as saturated liquid at -20C. Heat is added from a reservoir at 100C until a final state of 800kpa, 70C is reached. Find the work, heat transfer, and entropy generation assuming the process to be internally reversible. p1=pressure(r717,t=-20,x=0) v1=volume(r717,t=-20,x=0) u1=intenergy(r717,t=-20,x=0) s1=entropy(r717,t=-20,x=0) v2=volume(r717,p=800,t=70) u2=intenergy(r717,t=70,p=800) s2=entropy(r717,t=70,p=800) w12=(800+p1)*.5*(v2-v1) q12=w12+u2-u1 s_gen+q12/373=s2-s1
Maximum work in a flow process 1 2 w ð‘š ð‘š q ð‘‡â„Ž ð‘‡0ð‘ž0, Objective: to find the maximum work for the Case shown in picture 1st Law for the CV: ð‘š â„Ž1 + ð‘£1 2 2 + ð‘”ð‘§1 + 𑞠− ð‘ž0 = ð‘š â„Ž2 + ð‘£2 2 2 + gz2 + w ð‘šð‘ 1 + ð‘ž ð‘‡â„Ž − ð‘ž0 ð‘‡0 + ð‘ ð‘”ð‘’ð‘› = ð‘šð‘ 2, ð‘ ð‘”ð‘’ð‘› = 0, ð‘“ð‘œð‘Ÿ 𑤠ð‘¡ð‘œ ð‘ð‘’ ð‘šð‘Žð‘¥2nd Law for the CV ð‘š ð‘ 1 − ð‘ 2 + ð‘ž ð‘‡â„Ž = ð‘ž0 ð‘‡0  ð‘ž0 = ð‘š ð‘ 1 − ð‘ 2 ð‘‡0 + ð‘žð‘‡0 ð‘‡â„Ž (1) (2) Now use eqn (3) in (1) to obtain max work as: (3) 𑤠ð‘šð‘Žð‘¥ = ð‘š{ â„Ž1 − ð‘‡0 ð‘ 1 − â„Ž2 − ð‘‡0 ð‘ 2 + ð‘£1 2 − ð‘£2 2 2 + ð‘” ð‘§1 − ð‘§2 } + ð‘ž 1 − ð‘‡0 ð‘‡â„Ž ï¹1 = h − T0s1 + v1 2 2 + gz1; Keenan function 𑤠ð‘šð‘Žð‘¥ = ð‘š ï¹1 − ï¹2 + ð‘ž 1 − ð‘‡0 ð‘‡â„Ž Second law efficiency: 𑤠ð‘Žð‘ð‘¡/𑤠ð‘šð‘Žð‘¥ = 𑤠ð‘Žð‘ð‘¡/(ï¹ð‘– − ï¹ ð‘’)
T S 1 2 2s P=p1 P=p2 (â„Ž1−ℎ2)/(â„Ž1 − â„Ž2ð‘ ) = ï¨_ð‘–ð‘ ð‘’ð‘›, ð‘¡ð‘¢ð‘Ÿ For the case of a turbine expansion P=p2 1 2s 2 T S P=p1 For the case of a compressor (â„Ž1−ℎ2ð‘ )/(â„Ž1 − â„Ž2) = ï¨_ð‘–ð‘ ð‘’ð‘›, ð‘ð‘œð‘šð‘ An insulated steam turbine, receives 30kg/s of steam at 3 Mpa, 350C . In the turbine where the pressure is 500kpa, steam is beld off at the rate of 5kg/s, for processing equipment. The temperature of this steam is 200C. The balance of the steam leaves the turbine at 15kpa, 90% quality. Find the availability per kg of steam at the entry, the isentropic efficiency and the second law efficiency of the turbine. What is the actual work output of the turbine Ans: 1110 kj/kg, eta_isen=.7975, eta_2nd=.8176, w_act=20144 kw Example problem
m1=30 [kg/s]; m2= 5[kg/s]; m3=25 [kg/s] p1=3000; p2=500; p3=15; x3=.9; p0=100 t1= 350; t2=200; t0=25; t0k=298 h1=enthalpy(steam,p=p1,t=t1) s1=entropy(steam,p=p1,t=t1) h0=enthalpy(water,p=p0,t=t0) s0=entropy(water,p=p0,t=t0) h2=enthalpy(steam,p=p2,t=t2) s2=entropy(steam,p=p2,t=t2) h3=enthalpy(steam,p=p3,x=x3) s3=entropy(steam,p=p3,x=x3) a1=h1-t0k*s1-(h0-t0k*s0) { availability at entry} a2=h2-t0k*s2-(h0-t0k*s0) a3=h3-t0k*s3-(h0-t0k*s0) m1*h1=m2*h2+m3*h3+w3 { w3 = actual work output} m1*s1+s_gen=m2*s2+m3*s3 w_rev=w3+t0k*s_gen eta_2nd=w3/w_rev h2s=enthalpy(steam,p=p2,s=s1) h3s=enthalpy(steam,p=p3,s=s1) m1*h1=m2*h2s+m3*h3s+w3_s {w3s = isentropic work of turbine} eta_isen=w3/w3_s 30kg/s 3MPa, 350C 5kg/s .5MPa, 200C 25kg/s 15kPa, 90%quality 1 2 3 ð‘¤cv
Lect 7 thermo

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Lect 7 thermo

  • 1. Maximum work or available energy from a system Initially a body of mass m, is at temp T, what is the maximum work that can be obtained out of it if the environment is at ð‘‡0 which is lower than the body temperature T. ð‘‡0 ð‘„â„Ž ð‘„ð‘™ w ð‘šð‘ 𑣠𑇠ð‘„â„Ž = ð‘šð‘ 𑣠𑇠− ð‘‡0 ; ∆𑠠ð‘ = ð‘šð‘ ð‘£ ln ð‘‡0 𑇠∆ð‘ ð‘ ð‘¢ð‘Ÿ = ð‘„ð‘™/ð‘‡0 For max work output, ∆ð‘ ð‘ ð‘¢ð‘Ÿ + ∆𑠠ð‘ = 0 ð‘„ð‘™ = −ð‘‡0 ð‘š ð‘ ð‘£ ln ð‘‡0 𑇠; 𑤠= ð‘„â„Ž − ð‘„ð‘™ = ð‘šð‘ 𑣠𑇠− ð‘‡0 − ð‘‡0 ð‘šð‘ ð‘£ ln 𑇠ð‘‡0 𑤠= ð‘šð‘ 𑣠𑇠− ð‘‡0 − ð‘‡0 ln 𑇠ð‘‡0 = ð‘ˆ1 − ð‘ˆ0 − ð‘‡0(ð‘†1 − ð‘†0) = ð‘ˆ1 − ð‘‡0 ð‘†1 − ð‘ˆ0 − ð‘‡0 ð‘†0 = ð¹1 − ð¹0; ð‘¤â„Žð‘’ð‘Ÿð‘’ ð¹ = 𑈠− ð‘‡0 ð‘†, ð¹ ð‘–ð‘  ð‘ð‘Žð‘™ð‘™ð‘’ð‘‘ ð‘¡â„Žð‘’ ð‘Žð‘£ð‘Žð‘–ð‘™ð‘Žð‘ð‘–ð‘™ð‘–ð‘¡ð‘¦ ð‘œð‘Ÿ ð»ð‘’ð‘™ð‘šð‘œð‘™ð‘¡ð‘§ ð‘“ð‘¢ð‘›ð‘ð‘¡ð‘–ð‘œð‘› If the system boundary expands against the atmospheric pressure p0 then F becomes: 𜑠= 𑈠− ð‘‡0 𑆠+ ð‘0 ð‘‰, ðœ‘1 − ðœ‘2 = ð‘Žð‘£ð‘Žð‘–ð‘™ð‘Žð‘ð‘™ð‘’ ð‘¤ð‘œð‘Ÿð‘˜ = 𑤠ð‘šð‘Žð‘¥
  • 2. cm ð‘‡0 ð‘„12 ð‘„0 w12 ð‘‡â„Ž Control mass process when the system receives heat from a source Maximum work or available energy ð‘ˆ1 − ð‘ˆ2 ð‘†1 − ð‘†2 The system goes from 1-2, does work w12 and rejects heat ð‘„0 Lets us consider two situations: the body rejects heat either reversibly or irreversibly, but it receives heat reversibly. ð‘„12 − ð‘„0 ð‘–ð‘Ÿð‘Ÿ = ð‘ˆ2 − ð‘ˆ1 + ð‘¤12 ð‘Žð‘ð‘¡ ð‘  ð‘”ð‘’ð‘› = ∆ð‘ ð‘ð‘š + ∆ð‘ ð‘ ð‘¢ð‘Ÿ + ∆ð‘ ð‘Ÿð‘’ð‘  > 0, for irrev process ð‘ ð‘”ð‘’ð‘› ð‘Ÿð‘’ð‘£ = ð‘†2 − ð‘†1 + ð‘„0 ð‘Ÿð‘’ð‘£ ð‘‡0 − ð‘„12 ð‘‡â„Ž = 0 for rev process ð‘„0 ð‘Ÿð‘’ð‘£ = −ð‘‡0 ð‘†2 − ð‘†1 + ð‘‡0 ð‘‡â„Ž Q12, From 1st law for the CM we get: ð‘„12 − ð‘„0 ð‘Ÿð‘’ð‘£ = ð‘¤12 ð‘Ÿð‘’ð‘£ + ð‘ˆ2 − ð‘ˆ1 ð‘¤12 ð‘Ÿð‘’ð‘£ = ð‘„12 − ð‘„0 ð‘Ÿð‘’𑣠− ð‘ˆ2 − ð‘ˆ1 ð‘¤12 ð‘Ÿð‘’ð‘£ = ð‘„12 + ð‘‡0 ð‘†2 − ð‘†1 − ð‘„12 ð‘‡0 ð‘‡â„Ž − ð‘ˆ2 − ð‘ˆ1 ð‘¤12 ð‘Ÿð‘’ð‘£ = ð‘„12 1 − ð‘‡0 ð‘‡â„Ž + ð‘ˆ1 − ð‘‡0 ð‘†1 − ð‘ˆ2 − ð‘‡0 ð‘†2 ð‘¤12 ð‘Ÿð‘’ð‘£ = 𑤠ð‘šð‘Žð‘¥ = ð‘„12 1 − ð‘‡0 ð‘‡â„Ž + ð¹1 − ð¹2 sur res ð‘¤12 ð‘Žð‘ð‘¡ = ð‘„12 − ð‘„0 ð‘–ð‘Ÿð‘Ÿ − ð‘ˆ2 − ð‘ˆ1 ð¼ = ð¼ð‘Ÿð‘Ÿ = ð‘¤12 ð‘Ÿð‘’𑣠− ð‘¤12 ð‘Žð‘ð‘¡ = ð‘‡0 ð‘†2 − ð‘†1 − ð‘„12 ð‘‡0 ð‘‡â„Ž + ð‘„0 ð‘–ð‘Ÿð‘Ÿ = ð‘‡0 ð‘†2 − ð‘†1 − ð‘„12 ð‘‡â„Ž + ð‘„0 ð‘–ð‘Ÿð‘Ÿ ð‘‡0 = ð‘‡0(∆ð‘ ð‘ð‘š + ∆ð‘ ð‘Ÿð‘’ð‘  + ∆ð‘ ð‘ ð‘¢ð‘Ÿ) = ð‘‡0∆𑠠ð‘¢ð‘›ð‘–ð‘£ = ð‘‡0 ð‘ ð‘”ð‘’ð‘› Measure of irreversibility
  • 3. NH3 100C Example problem: 1kg of NH3 is contained in a spring loaded piston/cylinder as saturated liquid at -20C. Heat is added from a reservoir at 100C until a final state of 800kpa, 70C is reached. Find the work, heat transfer, and entropy generation assuming the process to be internally reversible. p1=pressure(r717,t=-20,x=0) v1=volume(r717,t=-20,x=0) u1=intenergy(r717,t=-20,x=0) s1=entropy(r717,t=-20,x=0) v2=volume(r717,p=800,t=70) u2=intenergy(r717,t=70,p=800) s2=entropy(r717,t=70,p=800) w12=(800+p1)*.5*(v2-v1) q12=w12+u2-u1 s_gen+q12/373=s2-s1
  • 4. Maximum work in a flow process 1 2 w ð‘š ð‘š q ð‘‡â„Ž ð‘‡0ð‘ž0, Objective: to find the maximum work for the Case shown in picture 1st Law for the CV: ð‘š â„Ž1 + ð‘£1 2 2 + ð‘”ð‘§1 + 𑞠− ð‘ž0 = ð‘š â„Ž2 + ð‘£2 2 2 + gz2 + w ð‘šð‘ 1 + ð‘ž ð‘‡â„Ž − ð‘ž0 ð‘‡0 + ð‘ ð‘”ð‘’ð‘› = ð‘šð‘ 2, ð‘ ð‘”ð‘’ð‘› = 0, ð‘“ð‘œð‘Ÿ 𑤠ð‘¡ð‘œ ð‘ð‘’ ð‘šð‘Žð‘¥2nd Law for the CV ð‘š ð‘ 1 − ð‘ 2 + ð‘ž ð‘‡â„Ž = ð‘ž0 ð‘‡0  ð‘ž0 = ð‘š ð‘ 1 − ð‘ 2 ð‘‡0 + ð‘žð‘‡0 ð‘‡â„Ž (1) (2) Now use eqn (3) in (1) to obtain max work as: (3) 𑤠ð‘šð‘Žð‘¥ = ð‘š{ â„Ž1 − ð‘‡0 ð‘ 1 − â„Ž2 − ð‘‡0 ð‘ 2 + ð‘£1 2 − ð‘£2 2 2 + ð‘” ð‘§1 − ð‘§2 } + ð‘ž 1 − ð‘‡0 ð‘‡â„Ž ï¹1 = h − T0s1 + v1 2 2 + gz1; Keenan function 𑤠ð‘šð‘Žð‘¥ = ð‘š ï¹1 − ï¹2 + ð‘ž 1 − ð‘‡0 ð‘‡â„Ž Second law efficiency: 𑤠ð‘Žð‘ð‘¡/𑤠ð‘šð‘Žð‘¥ = 𑤠ð‘Žð‘ð‘¡/(ï¹ð‘– − ï¹ ð‘’)
  • 5. T S 1 2 2s P=p1 P=p2 (â„Ž1−ℎ2)/(â„Ž1 − â„Ž2ð‘ ) = ï¨_ð‘–ð‘ ð‘’ð‘›, ð‘¡ð‘¢ð‘Ÿ For the case of a turbine expansion P=p2 1 2s 2 T S P=p1 For the case of a compressor (â„Ž1−ℎ2ð‘ )/(â„Ž1 − â„Ž2) = ï¨_ð‘–ð‘ ð‘’ð‘›, ð‘ð‘œð‘šð‘ An insulated steam turbine, receives 30kg/s of steam at 3 Mpa, 350C . In the turbine where the pressure is 500kpa, steam is beld off at the rate of 5kg/s, for processing equipment. The temperature of this steam is 200C. The balance of the steam leaves the turbine at 15kpa, 90% quality. Find the availability per kg of steam at the entry, the isentropic efficiency and the second law efficiency of the turbine. What is the actual work output of the turbine Ans: 1110 kj/kg, eta_isen=.7975, eta_2nd=.8176, w_act=20144 kw Example problem
  • 6. m1=30 [kg/s]; m2= 5[kg/s]; m3=25 [kg/s] p1=3000; p2=500; p3=15; x3=.9; p0=100 t1= 350; t2=200; t0=25; t0k=298 h1=enthalpy(steam,p=p1,t=t1) s1=entropy(steam,p=p1,t=t1) h0=enthalpy(water,p=p0,t=t0) s0=entropy(water,p=p0,t=t0) h2=enthalpy(steam,p=p2,t=t2) s2=entropy(steam,p=p2,t=t2) h3=enthalpy(steam,p=p3,x=x3) s3=entropy(steam,p=p3,x=x3) a1=h1-t0k*s1-(h0-t0k*s0) { availability at entry} a2=h2-t0k*s2-(h0-t0k*s0) a3=h3-t0k*s3-(h0-t0k*s0) m1*h1=m2*h2+m3*h3+w3 { w3 = actual work output} m1*s1+s_gen=m2*s2+m3*s3 w_rev=w3+t0k*s_gen eta_2nd=w3/w_rev h2s=enthalpy(steam,p=p2,s=s1) h3s=enthalpy(steam,p=p3,s=s1) m1*h1=m2*h2s+m3*h3s+w3_s {w3s = isentropic work of turbine} eta_isen=w3/w3_s 30kg/s 3MPa, 350C 5kg/s .5MPa, 200C 25kg/s 15kPa, 90%quality 1 2 3 ð‘¤cv