Contents
Team KAB's Analysis of Lab 5
Programmer: Aaron Klapheck
% Lab #5 The Gas Turbine Engine 10-Oct-08 clear, clc, home fprintf('The date and time: %s \n', datestr(now))
The date and time: 14-Oct-2008 12:35:34
Data
% P1 remains constant throughout this experament: P1 = 14.69; % P1 is atmospheric pressure in psi. % The following data changes with time. % There are a total of 9 variables being measured. The 9 variable are % being measured by the computer. The computer takes multiple measurements % at each "steady state" so % the values being measured are then averaged at each of the four % steady-states. The data can be distinguished from each other according to % the different engine loads. % % Units: % Voltage in Volts (V) % Current in Amps (A) % Temperature in degrees Fahrenheit (F) % Pressure in pounds per square inch gauge pressure (psig) % % T's and P's: % T1 is the Temperature of the air entering the compressor % T2 is the Temperature of the air exiting the compressor % T3 is the Temperature of the air entering the turbine % T4 is the Temperature of the air exiting the turbine % T5 is the Temperature of the air entering the cooling cabin % T6 is the Temperature of the air exiting the cooling cabin % P2 is the Pressure of the air exiting the compressor Load_30 = [ 84.3712 76.1905 107.4481 1040.8212 1487.1918 854.3895 36.5574 270.1465 43.3455 84.1270 76.0684 107.8144 1039.7559 1485.061 856.5202 36.4353 271.2454 43.5897 84.3712 76.1905 107.3260 1042.9518 1487.1918 856.5202 36.6184 270.8791 43.4676 84.3712 76.1905 107.8144 1039.7559 1482.9304 858.6508 36.4963 270.5128 43.1013 84.1270 76.3126 107.8144 1040.8212 1480.7998 857.5855 36.4353 271.0623 42.9792 84.1270 76.4347 107.9365 1044.0171 1482.9304 856.5202 36.4353 270.3297 43.2234 83.7607 76.4347 107.9365 1041.8865 1488.2571 857.5855 36.2521 269.7802 43.4676 83.6386 76.3126 107.8144 1039.7559 1489.3224 855.4549 36.5574 270.8791 43.2234 83.7607 76.1905 107.9365 1044.0171 1483.9958 854.3895 36.4353 270.1465 43.5897]; Load_40 = [ 86.5690 77.2894 113.3089 1083.4341 1485.061 903.3944 37.0458 269.7802 56.8987 86.5690 77.7778 113.5531 1085.5647 1482.9304 901.2638 37.0458 271.0623 57.2650 86.3248 77.2894 113.7973 1088.7607 1485.061 902.3290 36.8626 269.7802 57.6313 86.2027 77.7778 113.5531 1085.5647 1481.8651 905.5251 36.8016 270.6960 57.3871 86.4469 77.7778 113.5531 1086.6301 1486.1265 900.1984 36.9847 269.7802 57.5092 86.8132 77.4115 113.5531 1084.4994 1481.8651 902.3290 36.8626 269.9634 57.5092 86.3248 77.4115 113.4310 1088.7607 1483.9958 904.4597 36.6795 270.5128 57.3871 86.3248 77.7778 113.3089 1087.6953 1488.2571 902.3290 36.8016 269.9634 57.6313 86.6911 77.6557 113.7973 1085.5647 1482.9304 900.1984 36.9847 270.6960 57.6313]; Load_50 = [ 89.8657 78.3883 119.4139 1150.5494 1519.1514 963.0526 37.4121 270.3297 71.5507 89.4994 78.6325 119.1697 1148.4188 1523.4127 957.7259 37.0458 269.9634 71.5507 89.1331 78.5104 119.5360 1152.6802 1523.4127 960.9219 37.0458 270.5128 71.1844 89.2552 78.8767 119.2918 1150.5494 1522.3474 964.1178 37.1679 270.1465 71.4286 89.1331 78.8767 119.2918 1153.7455 1512.7595 959.8565 37.1679 270.6960 71.3065 88.7668 78.6325 119.6581 1150.5494 1514.8901 957.7259 37.3510 270.3297 71.5507 88.8889 78.6325 119.2918 1151.6149 1518.0862 963.0526 37.2900 269.9634 71.5507]; Load_60 = [ 90.8425 78.0220 122.2222 1230.4489 1564.9603 1023.7759 37.5342 270.3297 85.4701 90.3541 78.0220 122.2222 1234.7101 1556.4377 1017.3840 37.4121 270.5128 85.4701 90.4762 78.0220 122.4664 1233.6448 1554.3071 1021.6453 37.5952 270.1465 84.9817 90.4762 78.3883 122.1001 1232.5795 1560.6991 1023.7759 37.5342 270.5128 85.7143 90.2320 77.8999 122.5885 1232.5795 1550.0459 1023.7759 37.4731 271.0623 85.9585]; T1 = [mean(Load_30(:,1)); mean(Load_40(:,1)); mean(Load_50(:,1)); mean(Load_60(:,1))]; T5 = [mean(Load_30(:,2)); mean(Load_40(:,2)); mean(Load_50(:,2)); mean(Load_60(:,2))]; T6 = [mean(Load_30(:,3)); mean(Load_40(:,3)); mean(Load_50(:,3)); mean(Load_60(:,3))]; T2 = [mean(Load_30(:,4)); mean(Load_40(:,4)); mean(Load_50(:,4)); mean(Load_60(:,4))]; T3 = [mean(Load_30(:,5)); mean(Load_40(:,5)); mean(Load_50(:,5)); mean(Load_60(:,5))]; T4 = [mean(Load_30(:,6)); mean(Load_40(:,6)); mean(Load_50(:,6)); mean(Load_60(:,6))]; P2 = [mean(Load_30(:,7)); mean(Load_40(:,7)); mean(Load_50(:,7)); mean(Load_60(:,7))]; V = [mean(Load_30(:,8)); mean(Load_40(:,8)); mean(Load_50(:,8)); mean(Load_60(:,8))]; I = [mean(Load_30(:,9)); mean(Load_40(:,9)); mean(Load_50(:,9)); mean(Load_60(:,9))]; % Convert all units to metric % Kelvin = (Fahrenheit - 32)5/9 + 273.15 Load = [30; 40; 50; 60]; % kW T1 = (T1 - 32)./1.8 + 273.15; % K T2a = (T2 - 32)./1.8 + 273.15 % K T3 = (T3 - 32)./1.8 + 273.15 % K T4a = (T4 - 32)./1.8 + 273.15; % K T5 = (T5 - 32)./1.8 + 273.15; % K T6 = (T6 - 32)./1.8 + 273.15; % K P2 = (P2 + P1).*6.89; % kPa P1 = 101; % kPa - atmopheric pressure is 101 kPa
T2a =
834.0008
858.8583
894.9046
940.2570
T3 =
1.0e+003 *
1.0805
1.0799
1.0993
1.1205
Graph pressure ratios as a function of load
P_ratio = P2/P1; fprintf('\nPressure ratios for an increasing load: %g %g %g %g \n \n', P_ratio) FitToLine = polyfit(Load, P_ratio, 1); fprintf('Notice the rate of pressure ratio increase is very small: %g \n \n', FitToLine(1,1)) x = [29:0.01:61]; FitToLine = polyval(FitToLine, x); plot(Load, P_ratio, 'o', x, FitToLine), ... xlabel('Load (kW)'), title('Pressure Ratio vs. Load'), ... ylabel('Pressure ratios (P2/P1 unitless)'), ... legend('Data', 'Fit to Line', 'Location', 'NorthWest')
Pressure ratios for an increasing load: 3.48997 3.51912 3.54061 3.56095 Notice the rate of pressure ratio increase is very small: 0.0023444
Calculations
% Assume specific heat is constant(k = c_p/c_v). k = 1.4; % assuming T = 300K is constant (which it is not). Cp = 1.005; % kJ/(kg*K) Assuming T = 300K is approximately constant. m_dot = 0.844; % kg/s. This value is assumed. % 1st. Calculate the isentropic efficiency of the turbine and the % compresser using the 4 equations below. % % For a compressor with constant specific heats: % T2s/T1 = (P2/P1)^((k-1)/k) ... (eqn 1). % Deffinition of isentropic efficiancey of a compressor (n_c): % n_c = ((T2s - T1)/(T2a - T1)) ... (eqn 2). % For a turbine with constant specific heats: % T4s/T3 = (P1/P2)^((k-1)/k) ... (eqn 3). % Deffinition of isentropic efficiancey of a turbine (n_t): % n_t = ((T3 - T4a)/(T3 - T4s)) ... (eqn 4). fprintf('%%%%%% Get n_c using equations 1 and 2 %%%%%% \n') T2s = T1.*(P2./P1).^((k-1)/k); n_c = ((T2s - T1)./(T2a - T1)); [num2str(round(100000.*n_c)./1000), ['%';'%';'%';'%']] fprintf('%%%%%% Get n_t using equations 3 and 4 %%%%%% \n') T4s = T3.*(P1./P2).^((k-1)/k); n_t = ((T3 - T4a)./(T3 - T4s)); [num2str(round(100000.*n_t)./1000), ['%';'%';'%';'%']] % 2nd. Calculate the Thermal efficiency, Back work ratio, and Generator % efficiency using the 4 equations below. Because n_c and n_t were both % clearly wrong we make the assumption that n_t and n_c are 75%. n_c = 0.75; n_t = n_c; % % Rearange (eqn 2): T2a = (T2s-T1)/n_c + T1 ... (eqn 5) % Rearange (eqn 4): T3 = (T4a-n_t*T4s)/(1-n_t) ... (eqn 6) % Plug (eqn 6) into (eqn 3) and solve for T4s: % T4s = 1./(((1-n_t)/((P1/P2)^((k-1)/k)))+n_t/T4a) ... (eqn 7) % The thermal efficiency of an ideal Brayton cycle: % n_th,Brayton = 1 - 1/((P2/P1)^(1-1/k)) ... (eqn 8) fprintf('%%%%%% Get T2a using equation 5 %%%%%% \n') T2a = (T2s-T1)./n_c + T1 fprintf('%%%%%% Get T3 using equations 6 and 7 %%%%%% \n') T4s = 1./((((1-n_t)./((P1./P2).^((k-1)/k)))+n_t)./T4a); T3 = (T4a-n_t.*T4s)./(1-n_t) fprintf('%%%%%% The work done by the turbine %%%%%% \n') w_t = Cp.*(T3-T4a) fprintf('%%%%%% The work done compressing the air in compresser %%%%%% \n') w_c = Cp.*(T2a-T1) fprintf('%%%%%% The heat generated in the burner %%%%%% \n') q_in = Cp.*(T3-T2a) fprintf('%%%%%% The net work %%%%%% \n') w_net = w_t - w_c fprintf('%%%%%% The ideal thermal efficiency using equation 8 %%%%%% \n') n_th_ideal = 1-1./((P2./P1).^(1-1/k)); [num2str(round(100000.*n_th_ideal)./1000), ['%';'%';'%';'%']] fprintf('%%%%%% The actual thermal efficiency %%%%%% \n') n_th = w_net./q_in; [num2str(round(100000.*n_th)./1000), ['%';'%';'%';'%']] fprintf('%%%%%% The back work ratio %%%%%% \n') r_bw = w_c./w_t fprintf('%%%%%% The generator efficiency %%%%%% \n') n_gen = Load./(m_dot.*w_net); [num2str(round(100000.*n_gen)./1000), ['%';'%';'%';'%']]
%%% Get n_c using equations 1 and 2 %%%
ans =
24.374%
23.631%
22.489%
21.067%
%%% Get n_t using equations 3 and 4 %%%
ans =
107.672%
99.113%
93.048%
87.197%
%%% Get T2a using equation 5 %%%
T2a =
474.9477
478.4210
481.8406
483.9013
%%% Get T3 using equations 6 and 7 %%%
T3 =
1.0e+003 *
0.9437
0.9783
1.0215
1.0666
%%% The work done by the turbine %%%
w_t =
213.6119
222.6664
233.4326
244.6669
%%% The work done compressing the air in compresser %%%
w_c =
173.7328
175.8827
177.7861
179.1559
%%% The heat generated in the burner %%%
q_in =
471.0962
502.3689
542.3466
585.6516
%%% The net work %%%
w_net =
39.8791
46.7837
55.6465
65.5110
%%% The ideal thermal efficiency using equation 8 %%%
ans =
30.031%
30.197%
30.318%
30.432%
%%% The actual thermal efficiency %%%
ans =
8.465%
9.313%
10.26%
11.186%
%%% The back work ratio %%%
r_bw =
0.8133
0.7899
0.7616
0.7322
%%% The generator efficiency %%%
ans =
89.132%
101.303%
106.461%
108.516%