Contents
- Team KAB's Analysis of Lab 4
- Data
- Calculations
- Graph of 3 Polytropic processes from i to d for the low pressure pump
- Graph of Polytropic processes from i to d for the high pressure pump
- Find the work done in a single, double-acting cylinder
- Find the total work and power done in a 2-stage double-acting compressor
- Compare the supplied power consumed to the calculated one.
Team KAB's Analysis of Lab 4
Programmer: Aaron Klapheck
% Lab #4 The Reciprocating Air Compressor 26-Sep-08 clear, clc, home fprintf('The date and time: %s \n', datestr(now))
The date and time: 30-Sep-2008 13:29:37
Data
% This section consists of data either recorded by hand or by the computer % used in this experiment. % Recorded data came from the excel file Data.xls. % Computer generated data came from the excel file KAB_COMP_LAB.xls. % The following data is constant throughout the experiment (from Data.xls). P_atm = 30; % Atmospheric Pressure in inches of Hg (inHg). T_room = 23; % Room temperature in degrees Celcius (C). N = 150; % Angular acceleration in revolutions per minute (rpm). d_low = 7; % Diameter of low pressure piston in inches (in). d_high = 3.125; % Diameter of high pressure piston in inches (in). L = 5; % Distance the piston travels in one direction measured % in inches (in) aka stroke. % The following data changes with time. % There are a total of 13 variables being measured. Of those 13, 3 are % being measured by hand (Voltage, Current, and Power), the 10 others are % being measured by the computer. For the variables being measured by hand % there is only one measurement being taken for each "steady-state" reached. % Whereas the computer takes multiple measurements at each steady state so % the values being measured are then averaged at each of the five % steady-states. The data can be distinguished from each other according to % the different receiver pressure values which are P5. In other words each % P5 value corresponds to one steady state. % % Units: % Power in kiloWatts (kW) % Voltage in Volts (V) % Current in Amps (A) % Temperature in degrees Celcius (C) % Pressure in pounds per square inch (psi) % Ch15 refers to inches of water (in H2O) % % T's and P's: % T1 is the Temperature out of the low pressure Cylinder % T4 is the Temperature out of the high pressure Cylinder % T7 is the Temperature into the low pressure Cylinder % T9 is the Temperature into the high pressure Cylinder % P1 is the Pressure out of the low pressure Cylinder % P2 is the Pressure into the high pressure Cylinder % P3 is the Pressure out of the high pressure Cylinder % P5 is the Pressure in the receiver tank %---- Steady-State 1 at P5 = 164 psi % % Done by hand (from Data.xls). Power_1 = 5.25; Voltage_1 = 490; Current_1 = 11.5; % Done by computer (from KAB_COMP_LAB.xls). T1_1 = 76.*ones(1,13); T4_1 = 78.*ones(1,13); T7_1 = 25.*ones(1,13); T9_1 = [27, 27, 27, 28, 27, 27, 27, 27, 27, 27, 27, 27, 27]; P1_1 = 51.*ones(1,13); P2_1 = 49.*ones(1,13); P3_1 = 146.*ones(1,13); P5_1 = 164.*ones(1,13); Ch15_1 = [2.83, 2.94, 3.37, 2.78, 3, 3.36, 2.76, 3.04, 3.34, 2.73, ... 3.09, 3.3, 2.71]; %---- Steady-State 2 at P5 = 208 psi % Done by hand (from Data.xls). Power_2 = 6; Voltage_2 = 490; Current_2 = 12; % Done by computer (from KAB_COMP_LAB.xls). T1_2 = 76.*ones(1,12); T4_2 = [81, 80, 81, 81, 81, 81, 81, 81, 81, 81, 81, 81]; T7_2 = 25.*ones(1,12); T9_2 = 30.*ones(1,12); P1_2 = 53.*ones(1,12); P2_2 = 52.*ones(1,12); P3_2 = [190 188 190 188 188 190 190 188 193 190 188 188]; P5_2 = [208 208 208 208 208 208 208 208 208 210 208 208]; Ch15_2 = [1.88 2.02 2.22 1.87 2.04 2.19 1.87 2.1 2.17 1.86 2.1 2.16]; %----- Steady-State 3 at P5 = 247 psi % Done by hand (from Data.xls). Power_3 = 6.5; Voltage_3 = 490; Current_3 = 12.5; % Done by computer (from KAB_COMP_LAB.xls). T1_3 = [77 77 77 77 77 78 77 77 77 77 77 77 77 77]; T4_3 = [84 83 83 83 84 83 83 83 84 84 84 84 84 84]; T7_3 = 25.*ones(1,14); T9_3 = [31 31 30 30 30 30 30 30 30 30 30 30 30 30]; P1_3 = [56 56 56 56 56 56 56 56 56 56 57 56 56 56 ]; P2_3 = 55.*ones(1,14); P3_3 = [227 227 227 227 227 227 227 227 227 227 227 227 229 227]; P5_3 = [247 247 244 247 247 247 247 247 247 247 247 247 247 249]; Ch15_3 = [1.44 1.85 1.6 1.44 1.85 1.6 1.43 1.85 1.62 1.43 1.84 1.62 1.43 1.83]; %----- Steady-State 4 at P5 = 286 psi % Done by hand (from Data.xls). Power_4 = 6.75; Voltage_4 = 490; Current_4 = 13; % Done by computer (from KAB_COMP_LAB.xls). T1_4 = [78 78 78 78 78 78 78 78 78 77 78]; T4_4 = 86.*ones(1,11); T7_4 = 25.*ones(1,11); T9_4 = [32 32 32 32 32 32 32 31 31 31 31]; P1_4 = 59.*ones(1,11); P2_4 = 58.*ones(1,11); P3_4 = [266 266 264 266 266 266 266 266 266 266 266]; P5_4 = [283 283 286 283 283 286 286 286 286 286 286]; Ch15_4 = [1.23 1.55 1.2 1.21 1.54 1.22 1.19 1.54 1.26 1.17 1.53]; %----- Steady-State 5 at P5 = 325 psi % Done by hand (from Data.xls). Power_5 = 7.25; Voltage_5 = 490; Current_5 = 14; % Done by computer (from KAB_COMP_LAB.xls). T1_5 = 78.*ones(1,12); T4_5 = 89.*ones(1,12); T7_5 = 25.*ones(1,12); T9_5 = [32 32 32 32 32 32 32 32 31 32 32 32]; P1_5 = 62.*ones(1,12); P2_5 = 61.*ones(1,12); P3_5 = [305 305 305 305 305 305 305 308 305 305 305 305]; P5_5 = 325.*ones(1,12); Ch15_5 = [1.17 0.96 1.23 1.22 0.98 1.17 1.24 1.01 1.12 1.26 1.04 1.07];
Calculations
% 1. Get average values for the variables T1, T4, T7, T9, P1, P2, P3, P5, % and Ch15 for each of the five states measured. %Means for steady-state 1, P5_1 = 164 psi = 1130.8 kPa T1_1 = mean(T1_1); T4_1 = mean(T4_1); T7_1 = mean(T7_1); T9_1 = mean(T9_1); P1_1 = mean(P1_1); P2_1 = mean(P2_1); P3_1 = mean(P3_1); P5_1 = mean(P5_1); Ch15_1 = mean(Ch15_1); %Means for steady-state 2, P5_2 = 208 psi = 1435.3 kPa T1_2 = mean(T1_2); T4_2 = mean(T4_2); T7_2 = mean(T7_2); T9_2 = mean(T9_2); P1_2 = mean(P1_2); P2_2 = mean(P2_2); P3_2 = mean(P3_2); P5_2 = mean(P5_2); Ch15_2 = mean(Ch15_2); %Means for steady-state 3, P5_3 = 247 psi = 1702.6 kPa T1_3 = mean(T1_3); T4_3 = mean(T4_3); T7_3 = mean(T7_3); T9_3 = mean(T9_3); P1_3 = mean(P1_3); P2_3 = mean(P2_3); P3_3 = mean(P3_3); P5_3 = mean(P5_3); Ch15_3 = mean(Ch15_3); %Means for steady-state 4, P5_4 = 286 psi = 1964.4 kPa T1_4 = mean(T1_4); T4_4 = mean(T4_4); T7_4 = mean(T7_4); T9_4 = mean(T9_4); P1_4 = mean(P1_4); P2_4 = mean(P2_4); P3_4 = mean(P3_4); P5_4 = mean(P5_4); Ch15_4 = mean(Ch15_4); %Means for steady-state 5, P5_5 = 325 psi = 2240.9 kPa T1_5 = mean(T1_5); T4_5 = mean(T4_5); T7_5 = mean(T7_5); T9_5 = mean(T9_5); P1_5 = mean(P1_5); P2_5 = mean(P2_5); P3_5 = mean(P3_5); P5_5 = mean(P5_5); Ch15_5 = mean(Ch15_5); T1 = [T1_1; T1_2; T1_3; T1_4; T1_5]; T4 = [T4_1; T4_2; T4_3; T4_4; T4_5]; T7 = [T7_1; T7_2; T7_3; T7_4; T7_5]; T9 = [T9_1; T9_2; T9_3; T9_4; T9_5]; P1 = [P1_1; P1_2; P1_3; P1_4; P1_5]; P2 = [P2_1; P2_2; P2_3; P2_4; P2_5]; P3 = [P3_1; P3_2; P3_3; P3_4; P3_5]; P5 = [P5_1; P5_2; P5_3; P5_4; P5_5]; Ch15 = [Ch15_1; Ch15_2; Ch15_3; Ch15_4; Ch15_5]; Power_sup = [Power_1; Power_2; Power_3; Power_4; Power_5]; % 2. Get Pi, Ti, Pd, and Td for both the low pressure cylinder and % the high pressure cylinder. % Low pressure cylinder: Ti_low = T_room; Td_low = T1; % Pi_low = P_atm Pi_low = 14.7; % given that P_atm = 29.92 inHg = 14.7 psi Pd_low = P1; % High pressure cylinder: Ti_high = T7; Td_high = T4; Pi_high = P2; Pd_high = P3; % 3. Convert all values to metric. L = L/39.37; % Meters P5 = 6.895.*P5; % kPa T9 = 6.895.*T9; % kPa % Low pressure cylinder: Ti_low = T_room + 273.15; % Kelvin Td_low = T1 + 273.15; % Kelvin Pi_low = 101; % given that P_atm = 29.92 inHg = 101kPa Pd_low = 6.895.*P1; % kPa d_low = d_low/39.37; % Meters % High pressure cylinder: Ti_high = T7 + 273.15; % Kelvin Td_high = T4 + 273.15; % Kelvin Pi_high = 6.895.*P2; % kPa Pd_high = 6.895.*P3; % kPa d_high = d_high/39.37; % Meters % 4. Find n, Vi, and Vd. % Rearrange Td/Ti = (Pd/Pi)^((n-1)/n) to get % n = -1/(ln(Td/Ti)/ln(Pd/Pi) - 1)... (eqn 1). % Volume displaced by Piston/stroke (D): D = (L*pi*d^2)/4 ... (eqn 2). % Assume: Vc = 0.05*Vi ... (eqn 3). % Given: D = Vi - Vc ... (eqn 4). % Combine (eqn 3) and (eqn 4): D = .95*Vi, Vi = D/.95 ... (eqn 5) % Definition of polytropic process: Vi/Vd = (Pd/Pi)^(1/n) ... (eqn 6) fprintf('%%%%%%4. First, get Vi through (eqn 2) and (eqn 5).%%%%%% \n') D_low = (L*pi*d_low^2)/4 % units: m*m^2 = m^3 D_high = (L*pi*d_high^2)/4 % units: m*m^2 = m^3 Vi_low = D_low/.95 % units: m^3 Vi_high = D_high/.95 % units: m^3 fprintf('%%%%%%4. Second, use (eqn 6) and (eqn 1) to find Vd.%%%%%% \n') n_low = -1./(log(Td_low./Ti_low)./log(Pd_low./Pi_low) - 1) n_high = -1./(log(Td_high./Ti_high)./log(Pd_high./Pi_high) - 1) Vd_low = Vi_low./(Pd_low./Pi_low).^(1./n_low) % units: kPa Vd_high = Vi_high./(Pd_high./Pi_high).^(1./n_high) % units: kPa % 5. Obtain constants (C's) fprintf('%%%%%%5. Use P*V^n = C to solve for the constant C.%%%%%% \n') C_low = Pi_low.*Vi_low.^n_low % Check both i and d values to make C_low = Pd_low.*Vd_low.^n_low % sure they lead to the same C values. C_high = Pi_high.*Vi_high.^n_high % Check both i and d values to make C_high = Pd_high.*Vd_high.^n_high % sure they lead to the same C values.
%%%4. First, get Vi through (eqn 2) and (eqn 5).%%%
D_low =
0.0032
D_high =
6.2844e-004
Vi_low =
0.0033
Vi_high =
6.6152e-004
%%%4. Second, use (eqn 6) and (eqn 1) to find Vd.%%%
n_low =
1.1520
1.1468
1.1428
1.1391
1.1339
n_high =
1.1763
1.1535
1.1448
1.1393
1.1374
Vd_low =
0.0011
0.0011
0.0010
0.0010
0.0009
Vd_high =
1.0e-003 *
0.2615
0.2159
0.1916
0.1739
0.1606
%%%5. Use P*V^n = C to solve for the constant C.%%%
C_low =
0.1408
0.1450
0.1484
0.1516
0.1561
C_low =
0.1408
0.1450
0.1484
0.1516
0.1561
C_high =
0.0615
0.0771
0.0869
0.0954
0.1018
C_high =
0.0615
0.0771
0.0869
0.0954
0.1018
Graph of 3 Polytropic processes from i to d for the low pressure pump
% Three processes have bean chosen to establish the general trend for how % P5 (receiver tank pressure) values effect the polytropic processes. The % three P5 values chosen are the smallest, middle, and larges values. V_domain_1 = [Vd_low(1,1):0.01*Vd_low(1,1):Vi_low(1,1)]; P_range_1 = C_low(1,1)./(V_domain_1.^n_low(1,1)); V_domain_3 = [Vd_low(3,1):0.01*Vd_low(3,1):Vi_low(1,1)]; P_range_3 = C_low(3,1)./(V_domain_3.^n_low(3,1)); V_domain_5 = [Vd_low(5,1):0.01*Vd_low(5,1):Vi_low(1,1)]; P_range_5 = C_low(5,1)./(V_domain_5.^n_low(5,1)); % Using "x"'s and "."'s because regulare lines will not appear in the % graph (they will be hidden underneith the other lines). plot(V_domain_1, P_range_1, '.', ... V_domain_3, P_range_3, 'x', V_domain_5, P_range_5,'r'), ... grid, axis([0.0005 .004 50 500]), ... title('Graph of LP Pump PV^n = C for three P5 values'), ... xlabel('Volume (m^3)'), ylabel('Pressure (N/m^2)'), ... legend('P5 = 1130.8 kPa', 'P5 = 1702.6 kPa', 'P5 = 2240.9 kPa')
Graph of Polytropic processes from i to d for the high pressure pump
% Three processes have bean chosen to establish the general trend for how % P5 (receiver tank pressure) values effect the polytropic processes. The % three P5 values chosen are the smallest, middle, and larges values. V_domain_1 = [Vd_high(1,1):0.01*Vd_high(1,1):Vi_high(1,1)]; P_range_1 = C_high(1,1)./(V_domain_1.^n_high(1,1)); V_domain_3 = [Vd_high(3,1):0.01*Vd_high(3,1):Vi_high(1,1)]; P_range_3 = C_high(3,1)./(V_domain_3.^n_high(3,1)); V_domain_5 = [Vd_high(5,1):0.01*Vd_high(5,1):Vi_high(1,1)]; P_range_5 = C_high(5,1)./(V_domain_5.^n_high(5,1)); plot(V_domain_1, P_range_1, V_domain_3, P_range_3, V_domain_5, P_range_5), ... grid, axis([0.0001 .00075 250 2200]), ... title('Graph of HP Pump PV^n = C for three P5 values'), ... xlabel('Volume (m^3)'), ylabel('Pressure (N/m^2)'), ... legend('P5 = 1130.8 kPa', 'P5 = 1702.6 kPa', 'P5 = 2240.9 kPa')
Find the work done in a single, double-acting cylinder
% polytropic work: Wid = (PdVd - PiVi)/(1-n) - Pi(Vd - Vi) ... (eqn 7) % Pc = Pd ... (eqn 8) % Work done per stroke: Wdc = Pc(Vc - Vd) - Pi(Vc - Vd) ... (eqn 9) % Total work for complete cycle: W = Wid + Wdc ... (eqn 10) % W(low-pressure): % use (eqn 7) Wid_low = ((Pd_low.*Vd_low - Pi_low.*Vi_low)./(1-n_low) - ... Pi_low.*(Vd_low - Vi_low)).*1000 % J per isentropic compression % use (eqn 3) Vc_low = 0.05.*Vi_low; % use (eqn 8) Pc_low = Pd_low; % use (eqn 9) Wdc_low = (Pc_low.*(Vc_low - Vd_low) - ... Pi_low.*(Vc_low - Vd_low)).*1000 % J/Stroke % use (eqn 10) W_low = Wdc_low + Wid_low % J % W(high-pressure) % use (eqn 7) Wid_high = ((Pd_high.*Vd_high - Pi_high.*Vi_high)./(1-n_high) - ... Pi_high.*(Vd_high - Vi_high)).*1000 % J per isentropic compression % use (eqn 3) Vc_high = 0.05.*Vi_high; % use (eqn 8) Pc_high = Pd_high; % use (eqn 9) Wdc_high = (Pc_high.*(Vc_high - Vd_high) - ... Pi_high.*(Vc_high - Vd_high)).*1000 % J/Stroke % use (eqn 10) W_high = Wdc_high + Wid_high % J
Wid_low = -172.8936 -182.6275 -197.0478 -210.4010 -223.7432 Wdc_low = -240.1193 -242.1145 -245.4795 -247.9820 -249.4018 W_low = -413.0129 -424.7420 -442.5273 -458.3831 -473.1450 Wid_high = -90.2243 -130.0337 -162.2201 -193.5202 -224.1315 Wdc_high = -152.7613 -172.9690 -188.2101 -201.7448 -214.7157 W_high = -242.9855 -303.0027 -350.4302 -395.2650 -438.8471
Find the total work and power done in a 2-stage double-acting compressor
% Total work done by both compressers: W_tot = W_low + W_high ... (eqn 11) % Power consumed: P = W_tot * 2strokes/rev * N / (60min/sec) / (1000J/kJ) ... (eqn 12) % use (eqn 11) W_tot = W_low + W_high % J % use (eqn 12) P_tot = W_tot.*2.*N./60./1000 % kW
W_tot = -655.9984 -727.7447 -792.9575 -853.6481 -911.9922 P_tot = -3.2800 -3.6387 -3.9648 -4.2682 -4.5600
Compare the supplied power consumed to the calculated one.
% Fit the data to a line P_tot_line = polyfit(P5, abs(P_tot), 1) P_sup_line = polyfit(P5, abs(Power_sup), 1) fprintf('P_calc: Power = %g*P5 + %g \n', P_tot_line) fprintf('P_sup: Power = %g*P5 + %g \n', P_sup_line) eval = [P5(1,1):10:P5(5,1)]; P_calc_line = polyval(P_tot_line, eval); P_sup_line = polyval(P_sup_line, eval); % Graph Power Consumed as a function of Receiver Tank Pressure (P5) plot(P5, abs(P_tot), 'o', eval, P_calc_line, P5, Power_sup, 'x', eval, P_sup_line), ... xlabel('P5, Receiver Tank Pressure (kPa)'), ... ylabel('Power Consumed (kW)'), ... text(1600, 6.0, 'Power sup = 0.00173413*P5 + 3.411 '), text(1600, 3.6, 'Power calc = 0.00116031*P5 + 1.97585 '), ... title('Power Supplied and Calculated vs. Receiver Pressure'), ... legend('Power calc.', 'Line fit', 'Power supplied', 'Line fit', 'Location', 'NorthWest')
P_tot_line =
0.0012 1.9759
P_sup_line =
0.0017 3.4110
P_calc: Power = 0.00116031*P5 + 1.97585
P_sup: Power = 0.00173413*P5 + 3.411
