Free Essay

Power Plant Design Project 1

In: Science

Submitted By luisescobar76
Words 4855
Pages 20
Introduciton:

In this experiment, three differnet thermodynamic cycle scenaro’s were given. Each was to be determined via tables and graphs: cold air-standard Brayton cycle, an ideal steam Rankine cycle, an an ideal vapor-compression refrigeration cycle. In each question, their would be a constant set to perform different calculations. This was done the desired amount. In the cold air-standard Brayton cycle, the thermal efficiency and net specific work output was needed with constant compressor pressure ratios of 6, 9, 12, and 15; while the turbine inlet temperatures were 800ºC, 1000ºC, 1200ºC, and 1400ºC. Then, the turbine inlet temperatures remained constant at 800ºC, 1000ºC, 1200ºC, and 1400ºC; while the compressor pressure ratios was 6, 9, 12, and 15. After all data was collected, four graphs were to be created by plotting the turbine inlet temperatures and the compressor pressure ratio versus the cycle thermal efficiency and net specific work output. For the ideal steam Rankine cycle, needed was to plot the cycle thermal efficiency by first setting the turbine exhaust pressures at a constant of 10, 50, and 100 kPa; and going through the turbine inlet pressures at 1, 5, and 8 MPa. Secondaly, it was needed to set the turbine exhaust pressures constant and going through a turbine inlet temperature of 300ºC, 500ºC, 700ºC. After all data was collected, two graphs were to be created. The first was plotted: the cycle thermal efficiency versus the turbine inlet temperatures as the exhaust pressures were and turbine inlet pressures were constant. The second was plotted: the cycle thermal efficiency versus the turbine inlet pressure as each exhaust pressure and turbine inlet temperature were constant. Lastly, for the ideal vapor-compression refrigeration cycle, it was required to calculate and plot the COP as the condenser and evaporator pressures effected the COP. In order to do this, it was needed to set the condenser pressures constant and going through the different evaporator pressures. Then a graph was calculated by the condenser pressure versus the COP. Next, remained constant was the evaporator pressures and went through each of the condenser pressures. Also, a graph was calculated by evaporator pressure versus the COP.

Problem 1

Data (Part 1):

INPUTS k= 1.4 cp= 1.005
P1= 100 [KPa]
T1= 300 [K] PR = 6 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] T3 [K] nth
500.6 201.6 575.3 643.1 432.1 230.5 1073 0.4007
500.6 201.6 776.3 763.0 512.6 311.0 1273 0.4007
500.6 201.6 977.3 882.8 593.1 391.6 1473 0.4007
500.6 201.6 1178.3 1002.7 673.7 472.1 1673 0.4007 PR = 9 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] T3 [K] nth
562.0 263.3 513.5 572.7 502.8 239.4 1073 0.4662
562.0 263.3 714.5 679.5 596.5 333.1 1273 0.4662
562.0 263.3 915.5 786.3 690.2 426.8 1473 0.4662
562.0 263.3 1116.5 893.0 783.9 520.5 1673 0.4662 PR = 12 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] T3 [K] nth
610.2 311.7 465.1 527.5 548.2 236.4 1073 0.5083
610.2 311.7 666.1 625.9 650.4 338.6 1273 0.5083
610.2 311.7 867.1 724.2 752.5 440.8 1473 0.5083
610.2 311.7 1068.1 822.5 854.7 543.0 1673 0.5083 PR = 15 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] T3 [K] nth
650.4 352.1 424.8 495.0 580.9 228.8 1073 0.5387
650.4 352.1 625.8 587.2 689.2 337.1 1273 0.5387
650.4 352.1 826.8 679.5 797.5 445.4 1473 0.5387
650.4 352.1 1027.8 771.7 905.8 553.7 1673 0.5387

Graphs (Part 1):

Data (Part 2):

INPUTS OUTPUTS k= 1.4 cp= 1.005
P1= 100 [KPa]
T1= 300 [K]

T3 [K]

1073 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] PR = nth
500.6 201.6 575.3 643.1 432.1 230.5 6 0.4007
562.0 263.3 513.5 572.7 502.8 239.4 9 0.4662
610.2 311.7 465.1 527.5 548.2 236.4 12 0.5083
650.4 352.1 424.8 495.0 580.9 228.8 15 0.5387

T3 [K] 1273 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] PR = nth
500.6 201.6 776.3 763.0 512.6 311.0 6 0.4007
562.0 263.3 714.5 679.5 596.5 333.1 9 0.4662
610.2 311.7 666.1 625.9 650.4 338.6 12 0.5083
650.4 352.1 625.8 587.2 689.2 337.1 15 0.5387 T3 [K] 1473 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] PR = nth
500.6 201.6 977.3 882.8 593.1 391.6 6 0.4007
562.0 263.3 915.5 786.3 690.2 426.8 9 0.4662
610.2 311.7 867.1 724.2 752.5 440.8 12 0.5083
650.4 352.1 826.8 679.5 797.5 445.4 15 0.5387 T3 [K] 1673 T2 [K] wc,in [KJ/kg] qin [KJ/kg] T4 [K] wt,out [KJ/kg] w,net [KJ/kg] PR = nth
500.6 201.6 1178.3 1002.7 673.7 472.1 6 0.4007
562.0 263.3 1116.5 893.0 783.9 520.5 9 0.4662
610.2 311.7 1068.1 822.5 854.7 543.0 12 0.5083
650.4 352.1 1027.8 771.7 905.8 553.7 15 0.5387

Graphs (Part 2):

Sample Calculations:

Given rp= P_2/P_1 = P_3/P_4 =6
Specific heat (k) =1.005,
P1 = 100 KPa, T1= 300 K, T3=1073 K

T_2= T_1 (〖P_2/P_1 )〗^((k-1)/k) T_2= 100 K(〖6)〗^((1.4-1)/1.4)=500.6 K w_(c,in)=C_p (T_2- T_1) w_(c,in)=1.005(500.6- 300)=201.6 KJ/Kg q_in=C_p (T_3- T_2) q_in=1.005(1073- 500.6)=573.5 KJ/Kg
T_4s= T_3 (〖P_3/P_4 )〗^((k-1)/k) T_4= 1073(〖1/6)〗^((1.4-1)/1.4)=643.1 K w_(T,out)=C_p (T_3- T_4) w_(T,out)=1.005(1073- 643.1)=432.1 KJ/Kg
W_(net,out)= W_(T,out)- W_(c,in) W_(net,out)=432.1- 201.6=230.5 KJ/Kg
〖 η〗_th= W_(net,out)/q_in 〖 η〗_th= 230.5/573.5=.4007

Problem 2

Data (Part 1):
P1 (kpa) 10
T3 (°C) 300

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
1.00 192.81 2858.7901 0.86 2257.23 2065.42 793.37 0.28
5.04 196.85 2728.8501 0.74 1965.85 1774.04 954.81 0.35
8.07 199.88 2586.6201 0.69 1832.72 1640.91 945.71 0.37
P1 (kpa) 50
T3 (°C) 300

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
1.00 192.81 3286.2901 0.95 2461.24 2269.43 1016.86 0.31
5.04 196.85 3237.8501 0.84 2210.50 2018.69 1219.16 0.38
8.07 199.88 3199.6201 0.81 2130.28 1938.47 1261.15 0.39
P1 (kpa) 100
T3 (°C) 300

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
1.00 192.81 3731.2901 1.02 2624.32 2432.51 1298.78 0.35
5.04 196.85 3703.4501 0.92 2381.30 2189.49 1513.96 0.41
8.07 199.88 3682.3201 0.88 2307.50 2115.69 1566.63 0.43 P1 (kpa) 10
T3 (°C) 500 Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
0.98 341.52 2710.0815 0.93 2479.17 2138.63 571.45 0.21
5.10 345.64 2580.0615 0.79 2155.37 1814.83 765.23 0.30
8.19 348.73 2437.7715 0.72 2007.41 1666.87 770.90 0.32
P1 (kpa) 50
T3 (°C) 500

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
0.98 341.52 3137.5815 1.03 2705.89 2365.35 772.23 0.25
5.10 345.64 3089.0615 0.91 2427.24 2086.70 1002.36 0.32
8.19 348.73 3050.7715 0.87 2338.10 1997.56 1053.22 0.35
P1 (kpa) 100
T3 (°C) 500

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
0.98 341.52 3582.5815 1.10 2887.13 2546.59 1035.99 0.29
5.10 345.64 3554.6615 0.99 2617.06 2276.52 1278.14 0.36
8.19 348.73 3533.4715 0.95 2535.04 2194.50 1338.97 0.38
P1 (kpa) 10
T3 (°C) 700

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
0.94 418.45 3060.6513 1.07 2826.05 2408.54 652.11 0.21
5.11 422.62 3012.0793 0.94 2533.03 2115.52 896.56 0.30
8.24 425.75 2973.7503 0.90 2439.28 2021.77 951.98 0.32

P1 (kpa)

50
T3 (°C) 700

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
0.94 418.45 2633.1513 0.96 2587.64 2170.13 463.03 0.18
5.11 422.62 2503.0793 0.81 2247.12 1829.61 673.47 0.27
8.24 425.75 2360.7503 0.74 2091.53 1674.02 686.73 0.29
P1 (kpa) 100
T3 (°C) 700

Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
0.94 418.45 3505.6513 1.15 3016.64 2599.13 906.52 0.26
5.11 422.62 3477.6793 1.03 2732.64 2315.13 1162.55 0.33
8.24 425.75 3456.4503 0.99 2646.38 2228.87 1227.58 0.36

Table Values
S 3=S4(KJ/Kg k) P1(Kpa) Sf Ssg Hf Hfg
7.1246 10 0.6492 7.4996 191.81 2392.1
7.7642 50 1.0912 6.5019 340.54 2304.7
8.2755 100 1.3028 6.0562 417.51 2257.5
6.2111
6.9781
7.5136
5.7937
6.7266
7.2822

Graph (Part 1):

Data (Part 2):

P1 (kpa) 10
P3 (kpa) 1000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 191.81 0.00101 1.00 192.81 2858.7901 0.86 2257.23 2065.42 793.37 0.28
500 191.81 0.00101 5.04 196.85 2728.8501 0.74 1965.85 1774.04 954.81 0.35
700 191.81 0.00101 8.07 199.88 2586.6201 0.69 1832.72 1640.91 945.71 0.37
P1 (kpa) 10
P3 (kpa) 5000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 191.81 0.00101 1.00 192.81 3286.2901 0.95 2461.24 2269.43 1016.86 0.31
500 191.81 0.00101 5.04 196.85 3237.8501 0.84 2210.50 2018.69 1219.16 0.38
700 191.81 0.00101 8.07 199.88 3199.6201 0.81 2130.28 1938.47 1261.15 0.39
P1 (kpa) 10
P3 (kpa) 8000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 191.81 0.00101 1.00 192.81 3731.2901 1.02 2624.32 2432.51 1298.78 0.35
500 191.81 0.00101 5.04 196.85 3703.4501 0.92 2381.30 2189.49 1513.96 0.41
700 191.81 0.00101 8.07 199.88 3682.3201 0.88 2307.50 2115.69 1566.63 0.43

P1 (kpa)
50
P3 (kpa) 1000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 340.54 0.00103 0.98 341.52 2710.0815 0.93 2479.17 2138.63 571.45 0.21
500 340.54 0.00103 5.10 345.64 2580.0615 0.79 2155.37 1814.83 765.23 0.30
700 340.54 0.00103 8.19 348.73 2437.7715 0.72 2007.41 1666.87 770.90 0.32
P1 (kpa) 50
P3 (kpa) 5000
P3 (Kpa) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 340.54 0.00103 0.98 341.52 3137.5815 1.03 2705.89 2365.35 772.23 0.25
500 340.54 0.00103 5.10 345.64 3089.0615 0.91 2427.24 2086.70 1002.36 0.32
700 340.54 0.00103 8.19 348.73 3050.7715 0.87 2338.10 1997.56 1053.22 0.35
P1 (kpa) 50
P3 (kpa) 8000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 340.54 0.00103 0.98 341.52 3582.5815 1.10 2887.13 2546.59 1035.99 0.29
500 340.54 0.00103 5.10 345.64 3554.6615 0.99 2617.06 2276.52 1278.14 0.36
700 340.54 0.00103 8.19 348.73 3533.4715 0.95 2535.04 2194.50 1338.97 0.38
P1 (kpa) 100
P3 (kpa) 1000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 417.51 0.001043 0.94 418.45 2633.1513 0.96 2587.64 2170.13 463.03 0.18
500 417.51 0.001043 5.11 422.62 2503.0793 0.81 2247.12 1829.61 673.47 0.27
700 417.51 0.001043 8.24 425.75 2360.7503 0.74 2091.53 1674.02 686.73 0.29
P1 (kpa) 100
P3 (kpa) 5000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 417.51 0.001043 0.94 418.45 3060.6513 1.07 2826.05 2408.54 652.11 0.21
500 417.51 0.001043 5.11 422.62 3012.0793 0.94 2533.03 2115.52 896.56 0.30
700 417.51 0.001043 8.24 425.75 2973.7503 0.90 2439.28 2021.77 951.98 0.32
P1 (kpa) 100
P3 (kpa) 8000
T3 (°C) H1 (KJ/Kg) V1 (KJ/Kg K) Wp,in (KJ/Kg) H2(KJ/Kg) qin(KJ/Kg) X4 (KJ/Kg) H4(KJ/Kg) qout(KJ/Kg) Wnet nth
300 417.51 0.001043 0.94 418.45 3505.6513 1.15 3016.64 2599.13 906.52 0.26
500 417.51 0.001043 5.11 422.62 3477.6793 1.03 2732.64 2315.13 1162.55 0.33
700 417.51 0.001043 8.24 425.75 3456.4503 0.99 2646.38 2228.87 1227.58 0.36

Values
P (KPa) T (°C) H3 (KJ/Kg) S 3=S4(KJ/Kg k) P1(Kpa) Sf Ssg Hf Hfg
1000 300 3051.6 7.1246 10 0.6492 7.4996 191.81 2392.1 500 3479.1 7.7642 50 1.0912 6.5019 340.54 2304.7 700 3924.1 8.2755 100 1.3028 6.0562 417.51 2257.5
5000 300 2925.7 6.2111 500 3434.7 6.9781 700 3900.3 7.5136
8000 300 2786.5 5.7937 500 3399.5 6.7266 700 3882.2 7.2822

Graph (Part 2):

Sample Calculations:
Turbine Exhaust Pressures of 10, 50, and 100 kPa
Turbine Inlet Pressures of 1, 5, and 8 MPa
Turbine Inlet Temperatures 300C, 500C, 700C
Process 1 - 2 State 1 P1 = 10 kPa => h1 = 191.81 KJ/Kg v1 = 0.00101 m3/Kg wp,in = V1 (P2 – P1) = (0.00101 m3/Kg)*(1000-10) = .9999 KJ/Kg h2 = h1 + wp,in = 191.81 + .9999 = 192.8099 KJ/Kg

Process 2 – 3 State 3 P3 = 1000 kPa, T3 = 300C => h3 = 3051.6 KJ/Kg s3 = 7.1246 KJ/Kg . K
Energy Balance:
2q3 + h2 – h3 = 0 => 2q3 = h3 – h2 => qin = h3 – h2 = 0 = (3051.6 – 192.8099) = 2858.7901 KJ/Kg

Process 3 – 4 State 4 P4 = 10 kPa, s3 = s4 => h4 = hf – x4hfg = 191.81 + (.8634327164)*(2392.1) = 2257.227401 KJ/Kg

Process 4 – 1
Energy Balance:
4q1 + h1 – h4 = 0 => 4q1 = h4 – h1 => qout = h4 – h1 = 0 = (2257.227401 – 191.81) = 2065.417401 KJ/Kg wnet = qin – qout = (2858.7901 - 2065.417401) = 793.372699 KJ/Kg th = = 

Problem 3

Data (Part 1): Pc = 400 Pe (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
100 234.44 0.95183 262.71 28.27 63.94 198.77 7.03
180 242.86 0.93965 259.21 16.35 63.94 195.27 11.94
240 247.28 0.93458 257.75 10.47 63.94 193.81 18.51
360 253.81 0.92836 255.94 2.13 63.94 192.00 90.22 Pc = 600 Pe (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
100 234.44 0.95183 271.40 36.96 81.51 189.89 5.14
180 242.86 0.93965 267.74 24.88 81.51 186.23 7.48
240 247.28 0.93458 266.22 18.94 81.51 184.71 9.75
360 253.81 0.92836 264.36 10.55 81.51 182.85 17.33 Pc = 800 Pe (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
100 234.44 0.95183 277.67 43.23 95.47 182.20 4.21
180 242.86 0.93965 273.87 31.01 95.47 178.40 5.75
240 247.28 0.93458 272.31 25.03 95.47 176.84 7.06
360 253.81 0.92836 270.39 16.58 95.47 174.92 10.55 Pc = 1000 Pe (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
100 234.44 0.95183 282.53 48.09 107.32 175.21 3.64
180 242.86 0.93965 278.64 35.78 107.32 171.32 4.79
240 247.28 0.93458 277.03 29.75 107.32 169.71 5.70
360 253.81 0.92836 275.04 21.23 107.32 167.72 7.90 Pc = 1200 Pe (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
100 234.44 0.95183 286.50 52.06 117.77 168.73 3.24
180 242.86 0.93965 282.51 39.65 117.77 164.74 4.15
240 247.28 0.93458 280.85 33.57 117.77 163.08 4.86
360 253.81 0.92836 278.81 25.00 117.77 161.04 6.44 TABLE VALUES NEEDED
P (KPa) T (°C) H (KJ/Kg) S (KJ/Kg k) P (KPa) T (°C) H (KJ/Kg) S (KJ/Kg k)
400 sat 255.5 0.9269 1000 40 271.71 0.9179 10 256.58 0.9305 50 282.74 0.9525 20 265.86 0.9628 1200 50 278.27 0.9267
600 sat 262.4 0.9218 60 289.64 0.9614 30 270.81 0.9499 40 280.58 0.9816
800 sat 267.29 0.9183 40 276.45 0.948 50 286.69 0.9802

Grpah (Part 1):

Data (Part 2):

Pe 100 (KPa) Pc (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
400 234.44 0.95183 262.71 28.27 63.94 198.77 7.03
600 234.44 0.95183 271.40 36.96 81.51 189.89 5.14
800 234.44 0.95183 277.67 43.23 95.47 182.20 4.21
1000 234.44 0.95183 282.53 48.09 107.32 175.21 3.64
1200 234.44 0.95183 286.50 52.06 117.77 168.73 3.24 Pe 180 (KPa) Pc (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
400 242.86 0.93965 259.21 16.35 63.94 195.27 11.94
600 242.86 0.93965 267.74 24.88 81.51 186.23 7.48
800 242.86 0.93965 273.87 31.01 95.47 178.40 5.75
1000 242.86 0.93965 278.64 35.78 107.32 171.32 4.79
1200 242.86 0.93965 282.51 39.65 117.77 164.74 4.15 Pe 240 (KPa) Pc (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
400 247.28 0.93458 257.75 10.47 63.94 193.81 18.51
600 247.28 0.93458 266.22 18.94 81.51 184.71 9.75
800 247.28 0.93458 272.31 25.03 95.47 176.84 7.06
1000 247.28 0.93458 277.03 29.75 107.32 169.71 5.70
1200 247.28 0.93458 280.85 33.57 117.77 163.08 4.86 Pe 360 (KPa) Pc (Kpa) H1 (KJ/Kg) S1 (KJ/Kg K) H2 (KJ/Kg) Win (KJ/Kg) H3 (KJ/Kg) QH (KJ/Kg) COP
400 253.81 0.92836 255.94 2.13 63.94 192.00 90.22
600 253.81 0.92836 264.36 10.55 81.51 182.85 17.33
800 253.81 0.92836 270.39 16.58 95.47 174.92 10.55
1000 253.81 0.92836 275.04 21.23 107.32 167.72 7.90
1200 253.81 0.92836 278.81 25.00 117.77 161.04 6.44

Table Values Needed
P (KPa) T (°C) H (KJ/Kg) S (KJ/Kg k) P (KPa) T (°C) H (KJ/Kg) S (KJ/Kg k)
400 sat 255.5 0.9269 600 sat 262.4 0.9218 10 256.58 0.9305 30 270.81 0.9499 20 265.86 0.9628 40 280.58 0.9816
800 sat 267.29 0.9183 1000 40 271.71 0.9179 40 276.45 0.948 50 282.74 0.9525 50 286.69 0.9802
1200 50 278.27 0.9267 60 289.64 0.9614

Graph (Part 2):

Sample Calculations:

Process 1 - 2 State 1 Pe = 100 kPa sat vapor => h1 = 234.44 KJ/Kg S1 = 0.95183 KJ/Kg k State 2 Pc = 400 kPa S1 = S2 = 0.95183 KJ/Kg k h_2=256.58+(.95183- .9305) ((256.86-256.58))/((9628- .9305))=262.71 KJ/Kg Energy Balance: Q ̇-W ̇+ ∑▒〖(m_in ) ̇(θ ̇_in+∑▒(m_out ) ̇ ((θ_out ) ̇= 0) ̇ ) ̇ 〗 -1w2=h_2- h_1=28.27 KJ/Kg win = 28.27 KJ/Kg
Process 2 – 3 State 2 h2 = 262.71 KJ/Kg State 3 Pc = 400 kPa sat liq. => h3 = 64.94 KJ/Kg Energy Balance: Q ̇-W ̇+ ∑▒〖(m_in ) ̇(θ ̇_in+∑▒(m_out ) ̇ ((θ_out ) ̇= 0) ̇ ) ̇ 〗 2q3=h_3- h_2=-198.77 KJ/Kg Qh = 198.77 KJ/Kg COP = q_h/w_in =7.03

Observations:

For the first cycle, cold air-standard Brayton, the thermal efficiency and net specific work output was needed with constant compressor pressure ratios of; while the turbine inlet temperatures cycled. Then the turbine inlet temperatures remained constant, while the compressor pressure ratios was cycled. After all data was collected, four graphs were created by plotting the turbine inlet temperatures and the compressor pressure ratio versus the cycle thermal efficiency and net specific work output. Overlooking the data, as the compressor pressure ratio increased from 6 to 15, the work output also increased. This also occured as the turbine inlet temperatures increased. Also, as the compressor pressure ratio increased, the work of the compressor increased. However, for a pressure ratio of 6 and temperature of 1073 K, the cycle thermal efficiency remained the same. This occured for each of the pressure ratios cycling through the turbine inlet temperatures.
The second cycle was the ideal steam Rankine cycle. The cycle thermal efficiency was needed and two plots were to be determined by the the data. This was done by first setting the turbine exhaust pressures constant and going through the turbine inlet pressures, and secondaly setting the turbine exhaust pressures constant and going through the turbine inlet temperatures. The first was plotted: the cycle thermal efficiency versus the turbine inlet temperatures as the exhaust pressures were and turbine inlet pressures were constant. The second was plotted: the cycle thermal efficiency versus the turbine inlet pressure as each exhaust pressure and turbine inlet temperature were constant. Observing the table, whenever the turbine exhaust pressure would increase, 10 to 100 kPa, the cycle thermal efficiency would decrease. However, as the turbine inlet temperatures increased while the pressures of the turbine at the inlet and exit remained constant, the cycle thermal efficiency increased. Also Qin decreased for every inlet and exhaust pressure; while going through different temperatures.
Last was the ideal vapor-compression refrigeration cycle. It was required to calculate and plot the COP. This was done by setting the condenser pressure constant and going through the different evaporator pressures. A graph was calculated by the condenser pressure versus the COP. Then the evaporator pressure was constant as the condenser pressure was cycled through. Also, a graph was calculated by evaporator pressure versus the COP. Noting from the table, the COP increased as the condenser pressure increased while going through the evaporator pressures. However, the COP decreased for a constant evaporator pressure, while going through the condenser pressure.

Conclusion:

In this experiment, three differnet thermodynamic cycle scenaro’s were given. The first was the ideal, cold air-standard Brayton cycle. It was noted that in order to increase the the work output, the compressor pressure ratios and the turbine inlet temperatures were to be increased. Also, the cycle thermal efficiency would reamin the same as the pressure ratio and temperature were constant. Each of the pressure ratios cycling through the turbine inlet temperatures showed this happening. For the Rankine cycle, the cycle thermal efficiency would decrease whenever the turbine exhaust pressure would increase. Also, in order to increase the cycle thermal efficiency, the turbine inlet temperatures needed to be increased. While cycling through different temperatures, Qin would decreased for every inlet and exhaust pressure. The last cycle was the refrigeration cycle. As the condenser pressure increased while going through the evaporator pressures, the COP would also increase. But, while going through the condenser pressure the COP decreased.…...

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