The gas turbine in Figure is to be simulated. Compressor and turbine characteristics are given in...

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The gas turbine in Figure is to be simulated. Compressor and turbine characteristics are given in Figures (b) and (c) respectively. Use air as the working fluid. T is 100 F, and 210 gal/hr of JP-4 (heat content 130000 Btu/gal) fuel is used. Assuming p = p3 and neglecting the effects of my on the mass flow rate, complete the following in order: a) Generate curve fits for p = p2(m) and E(m) b) Generate curve fits for m = m(p3, T3) and E = E(p3, T3) c) Develop the system of equations required to model the system. d) Simulate the system. e) List the steady state value of all variables. f) What is the turbine exit temperature? g) If the unit is an industrial gas turbine, examine the potential for energy recovery from the exhaust gases. An oil supply of 0.31 kg/s must be heated from 15C during 7200 hours per year of operation. The exhaust gases are available for preheating, if worthwhile. The energy gained by preheating will save 1 cent/MJ, but the heat exchanger will cost $250/m yr to own and operate. should the preheater be installed? If so, how large should it be, what would it cost, and what would be the savings per year? Take cp = 2015 J/kg K for the oil. Compressor T, P T21 P2 T3, P4 Combustor P4, T5 E Figure 1: System schematic Turbine Es Flow rate (bm/s) 30 20 10 70 0 Discharge pressure (psi) 20 20 30 40 50 60 8 10- 0 10 20 30 40 Compressor power (BTU/s) 2000 1600 1200 800 20 Flow rate (Ibm/s) 30 400 20 30 40 50 60 70 Discharge pressure (psi) Figure (b) Compressor characteristics T = 1,500F T-1,700F T-1,900F 50 60 70 Inlet pressure (psil Turbine power (BTU/s) 4,000 3,000 2,000 1,000 20 30 40 FIGURE P5-6(c) Turbine characteristics Figure (c) Turbine characteristics T-1,900F T-1,700F T-1,500F 50 60 70 Inlet pressure (psil Least-Square Curve Fit T P=[69 55 26] X=regress (m,P2,2) m[11 20 30] T X 13 3 2 70.544 0.638 -0.071] a=170.544 0.638 -0.071] j:=0..2 ap =Xj P (m)=a+am+amm 2 70- 60+ 50 P (ma) P2 40 30 20 10 0- 10 m=0,0.1..30 ma ma 30 Least-Square Curve Fit3 T E[1000 1210 1600]" T m=[11 20 30] X=regress (m,E,2) X=[3 3 2 924.737 -2.228 0.825] G:=0..2 =X =x+j a'= [924.737 -2.228 0.825] 2 e(m)=a+am+a+m 2000 1600 e (ma) E 1200- 800- 400 10 m=0,0.1..30 ma ma Parametric Curve Fit T3 [1500 1700 1900] 21 Pg== 40 60 T 12 12 12 m21 18 17 28 26 23 P30 -1 (0) moPs, (Ps) = @m0= aml am2 T 1 P 22 -0.611 0.667 -0.003] [7.182 0.184 0.002] [7.267 0.206 9.447.10] 1 P 1 P3 1 P 3g P30 1 P mr (P3)=a+am, Ps+m, P3 2 mz (P3)=1+0mlP3+m12 P3 m3 (P3)=0m+0m2, P3+mP3 .my(2) 1 T 30 bo 1 Ts, (T3) 1 T -1 amla -1 1 T 30 b=1 Ts 1 T 30 30 b 1 T, (T3) 1 T am2 1 T am2, 32 bo [-3 -304.771 0.347-9.636.10] b =[20.372 -0.023 6.307-10 b =[-0.252 2.885-10-8.182.10] mr(P3,T3) = (bo+bo T3+bo-T3)+(b0+bT3+bT) P3+ (b+b T3+bT ) P 10 P3 = 0,0.1..70 i==0..2 20 30 P3 P3 40 60 500 40 P3 P3, Ps Ps mr (P3, 1500) mr (P3, 1700) mr (Ps, 1900) mr. Parametric Curve Fit T T[1500 1700 1900]" 21 P 40 60 900 500 500 E2300 2000 1900 4200 3700 3400 -1 E (40) 2 P E (2) ------ P P 1 2 P 0-188.259 40.344 0.547] T q=[-1.028.10 69.48 0.155] 012 2-[-1.019.103 71.626 0.034] E(P)+ P3+P3 2 E2 (P3)=0+ P3+P3 Es (P3)=0+ P3+P3 -1 -1 00 1 Ts 2 1 T32 jawa dela 0120 0,2 1 T 1 T (T bo [3.313-10-38.107 0.011] b [-1.039.10 1.225 -3.374-10] e. (P3,T3)=(bo+bo, T3+ba, T) + ( +1 T3P3 to " 2 10 ==0..2 0122 b [12.085 -0.013 3.374-10"] 2 5000 4000 3000 2000 1000 0- 20 P=0,0.1..70 30 P3 P3 P3 P3 Psi P3, 60 70 e, (P3, 1900) e (Ps, 1700) e, (P3, 1500) E E2 Solver Contralues Solve T P=169 55 26] T m[11 20 30]" E=1000 1210 1600]" T T C[0.240 0.240 0.240] T T T2 [1500 1700 1900] T T=[100 100 100] q=[210 210 210] mt=10001 T=1500 1700 1900] T m20 E=1200 E=1500 2 P=a+ap, m+a.m E=a+P+ P E=m.C.(T2-T) q=(m+m)-C(T3-T) P3=P mram (T3)+ami ( (T3) P3+2 (T3) P32 E= (T3)+(T3) P3+2 (T3) P3 E =E+E, + ma == Find (P,ma, ET2, T3, P3,m,E,E,,q,m) 2 45.722 P= 45.722 45.722 2.547.103 E=2.547-103 2.547-10 [248.78 T=248.78 248.78 [45.722 P3= 45.722 45.722 2.381-103 T3= 3.145-101 -1.808.107 mr=1.331.107 E =-1.787-10 E=-1.787.10 -1.608.1 3.101 -34.275 -2.353.105 1.364-10 m -34.163 -34.302 The gas turbine in Figure is to be simulated. Compressor and turbine characteristics are given in Figures (b) and (c) respectively. Use air as the working fluid. T is 100 F, and 210 gal/hr of JP-4 (heat content 130000 Btu/gal) fuel is used. Assuming p = p3 and neglecting the effects of my on the mass flow rate, complete the following in order: a) Generate curve fits for p = p2(m) and E(m) b) Generate curve fits for m = m(p3, T3) and E = E(p3, T3) c) Develop the system of equations required to model the system. d) Simulate the system. e) List the steady state value of all variables. f) What is the turbine exit temperature? g) If the unit is an industrial gas turbine, examine the potential for energy recovery from the exhaust gases. An oil supply of 0.31 kg/s must be heated from 15C during 7200 hours per year of operation. The exhaust gases are available for preheating, if worthwhile. The energy gained by preheating will save 1 cent/MJ, but the heat exchanger will cost $250/m yr to own and operate. should the preheater be installed? If so, how large should it be, what would it cost, and what would be the savings per year? Take cp = 2015 J/kg K for the oil. Compressor T, P T21 P2 T3, P4 Combustor P4, T5 E Figure 1: System schematic Turbine Es Flow rate (bm/s) 30 20 10 70 0 Discharge pressure (psi) 20 20 30 40 50 60 8 10- 0 10 20 30 40 Compressor power (BTU/s) 2000 1600 1200 800 20 Flow rate (Ibm/s) 30 400 20 30 40 50 60 70 Discharge pressure (psi) Figure (b) Compressor characteristics T = 1,500F T-1,700F T-1,900F 50 60 70 Inlet pressure (psil Turbine power (BTU/s) 4,000 3,000 2,000 1,000 20 30 40 FIGURE P5-6(c) Turbine characteristics Figure (c) Turbine characteristics T-1,900F T-1,700F T-1,500F 50 60 70 Inlet pressure (psil Least-Square Curve Fit T P=[69 55 26] X=regress (m,P2,2) m[11 20 30] T X 13 3 2 70.544 0.638 -0.071] a=170.544 0.638 -0.071] j:=0..2 ap =Xj P (m)=a+am+amm 2 70- 60+ 50 P (ma) P2 40 30 20 10 0- 10 m=0,0.1..30 ma ma 30 Least-Square Curve Fit3 T E[1000 1210 1600]" T m=[11 20 30] X=regress (m,E,2) X=[3 3 2 924.737 -2.228 0.825] G:=0..2 =X =x+j a'= [924.737 -2.228 0.825] 2 e(m)=a+am+a+m 2000 1600 e (ma) E 1200- 800- 400 10 m=0,0.1..30 ma ma Parametric Curve Fit T3 [1500 1700 1900] 21 Pg== 40 60 T 12 12 12 m21 18 17 28 26 23 P30 -1 (0) moPs, (Ps) = @m0= aml am2 T 1 P 22 -0.611 0.667 -0.003] [7.182 0.184 0.002] [7.267 0.206 9.447.10] 1 P 1 P3 1 P 3g P30 1 P mr (P3)=a+am, Ps+m, P3 2 mz (P3)=1+0mlP3+m12 P3 m3 (P3)=0m+0m2, P3+mP3 .my(2) 1 T 30 bo 1 Ts, (T3) 1 T -1 amla -1 1 T 30 b=1 Ts 1 T 30 30 b 1 T, (T3) 1 T am2 1 T am2, 32 bo [-3 -304.771 0.347-9.636.10] b =[20.372 -0.023 6.307-10 b =[-0.252 2.885-10-8.182.10] mr(P3,T3) = (bo+bo T3+bo-T3)+(b0+bT3+bT) P3+ (b+b T3+bT ) P 10 P3 = 0,0.1..70 i==0..2 20 30 P3 P3 40 60 500 40 P3 P3, Ps Ps mr (P3, 1500) mr (P3, 1700) mr (Ps, 1900) mr. Parametric Curve Fit T T[1500 1700 1900]" 21 P 40 60 900 500 500 E2300 2000 1900 4200 3700 3400 -1 E (40) 2 P E (2) ------ P P 1 2 P 0-188.259 40.344 0.547] T q=[-1.028.10 69.48 0.155] 012 2-[-1.019.103 71.626 0.034] E(P)+ P3+P3 2 E2 (P3)=0+ P3+P3 Es (P3)=0+ P3+P3 -1 -1 00 1 Ts 2 1 T32 jawa dela 0120 0,2 1 T 1 T (T bo [3.313-10-38.107 0.011] b [-1.039.10 1.225 -3.374-10] e. (P3,T3)=(bo+bo, T3+ba, T) + ( +1 T3P3 to " 2 10 ==0..2 0122 b [12.085 -0.013 3.374-10"] 2 5000 4000 3000 2000 1000 0- 20 P=0,0.1..70 30 P3 P3 P3 P3 Psi P3, 60 70 e, (P3, 1900) e (Ps, 1700) e, (P3, 1500) E E2 Solver Contralues Solve T P=169 55 26] T m[11 20 30]" E=1000 1210 1600]" T T C[0.240 0.240 0.240] T T T2 [1500 1700 1900] T T=[100 100 100] q=[210 210 210] mt=10001 T=1500 1700 1900] T m20 E=1200 E=1500 2 P=a+ap, m+a.m E=a+P+ P E=m.C.(T2-T) q=(m+m)-C(T3-T) P3=P mram (T3)+ami ( (T3) P3+2 (T3) P32 E= (T3)+(T3) P3+2 (T3) P3 E =E+E, + ma == Find (P,ma, ET2, T3, P3,m,E,E,,q,m) 2 45.722 P= 45.722 45.722 2.547.103 E=2.547-103 2.547-10 [248.78 T=248.78 248.78 [45.722 P3= 45.722 45.722 2.381-103 T3= 3.145-101 -1.808.107 mr=1.331.107 E =-1.787-10 E=-1.787.10 -1.608.1 3.101 -34.275 -2.353.105 1.364-10 m -34.163 -34.302