Question

Data Point Number or Index Location Crank Angle [degrees] Volume [m3] Pressure [kPa]

1 -360 6.12E-05 107.3112488

2 -359.8 6.12E-05 107.3112488

3 -359.6 6.12E-05 107.6164246

4 -359.4 6.12E-05 106.3957214

5 -359.2 6.12E-05 107.6164246

6 -359 6.12E-05 106.7008972

7 -358.8 6.13E-05 106.0905457

8 -358.6 6.13E-05 106.7008972

9 -358.4 6.13E-05 106.0905457

10 -358.2 6.14E-05 106.7008972

11 -358 6.14E-05 106.3957214

12 -357.8 6.15E-05 106.7008972

13 -357.6 6.15E-05 106.3957214

14 -357.4 6.16E-05 105.7853699

15 -357.2 6.16E-05 105.7853699

16 -357 6.17E-05 105.4801941

17 -356.8 6.18E-05 106.3957214

18 -356.6 6.18E-05 105.1750183

19 -356.4 6.19E-05 104.8698425

20 -356.2 6.20E-05 105.1750183

21 -356 6.21E-05 105.4801941

22 -355.8 6.22E-05 104.259491

23 -355.6 6.23E-05 104.8698425

24 -355.4 6.24E-05 104.8698425

25 -355.2 6.25E-05 104.259491

26 -355 6.26E-05 104.5646667

27 -354.8 6.27E-05 103.3439636

28 -354.6 6.28E-05 103.9543152

29 -354.4 6.30E-05 103.3439636

30 -354.2 6.31E-05 103.6491394

31 -354 6.32E-05 103.3439636

32 -353.8 6.34E-05 103.3439636

33 -353.6 6.35E-05 96.63009644

34 -353.4 6.37E-05 103.0387878

35 -353.2 6.38E-05 102.7336121

36 -353 6.40E-05 111.8888855

37 -352.8 6.41E-05 103.0387878

38 -352.6 6.43E-05 101.5129089

39 -352.4 6.45E-05 102.4284363

40 -352.2 6.46E-05 102.7336121

41 -352 6.48E-05 103.6491394

42 -351.8 6.50E-05 98.15597534

43 -351.6 6.52E-05 103.0387878

44 -351.4 6.54E-05 102.1232605

45 -351.2 6.56E-05 98.15597534

46 -351 6.58E-05 101.8180847

47 -350.8 6.60E-05 98.7663269

48 -350.6 6.62E-05 99.98703003

49 -350.4 6.64E-05 99.07150269

50 -350.2 6.66E-05 99.07150269

51 -350 6.69E-05 95.71456909

52 -349.8 6.71E-05 102.7336121

53 -349.6 6.73E-05 91.74728394

54 -349.4 6.76E-05 94.79904175

55 -349.2 6.78E-05 95.10421753

56 -349 6.80E-05 102.4284363

57 -348.8 6.83E-05 96.01974487

58 -348.6 6.86E-05 93.57833862

59 -348.4 6.88E-05 92.3576355

60 -348.2 6.91E-05 96.63009644

61 -348 6.93E-05 92.66281128

62 -347.8 6.96E-05 95.10421753

63 -347.6 6.99E-05 93.57833862

64 -347.4 7.02E-05 93.8835144

65 -347.2 7.05E-05 92.3576355

66 -347 7.08E-05 92.66281128

67 -346.8 7.11E-05 91.13693237

68 -346.6 7.14E-05 92.96798706

69 -346.4 7.17E-05 92.05245972

70 -346.2 7.20E-05 88.69552612

71 -346 7.23E-05 91.44210815

72 -345.8 7.26E-05 90.52658081

73 -345.6 7.29E-05 89.30587769

74 -345.4 7.32E-05 89.91622925

75 -345.2 7.36E-05 88.69552612

76 -345 7.39E-05 89.91622925

77 -344.8 7.42E-05 87.77999878

78 -344.6 7.46E-05 88.39035034

79 -344.4 7.49E-05 87.16964722

80 -344.2 7.53E-05 87.474823

81 -344 7.56E-05 87.16964722

82 -343.8 7.60E-05 87.16964722

83 -343.6 7.64E-05 86.55929565

84 -343.4 7.67E-05 85.64376831

85 -343.2 7.71E-05 85.64376831

86 -343 7.75E-05 85.64376831

87 -342.8 7.79E-05 83.81271362

88 -342.6 7.83E-05 84.1178894

89 -342.4 7.86E-05 84.72824097

90 -342.2 7.90E-05 82.5920105

91 -342 7.94E-05 83.81271362

92 -341.8 7.98E-05 82.5920105

93 -341.6 8.02E-05 81.67648315

Task 1 (46 points +4 for comments): Write computer code to perform Trapezoidal Rule Integration of the data in the attached excel file to determine the thermodynamic work in Joules over the whole engine cycle a) b) c) Calculate the work in Joules over the whole engine cycle. (Note: remember from Thermodynamics that the Work is the integral of pressure over volume) In addition to the work over the whole cycle, calculate the work during only the compression and expansion process combined, which is the period defined by the index points 901 to 2700 Now calculate the work to pump the air in and out of the cylinder during the intake and exhaust strokes by subtracting the compression and expansion work from the total work. We call this the pumping work. Assuming that 1000 Joules of energy were added to the thermodynamic cycle when the fuel was burned during combustion, what fraction of the total energy input went to the total thermodynamic work (i.e., what is the efficiency of the cycle)? What fraction of the input energy went to pumping the air in and out of the cylinder? Do you think that this estimation of the work would benefit from the higher order/accuracy integration techniques that we've discussed in class? Why or why not? d) e) Data Point Number or Index Location Crank Angle [degrees] Volume m3] Pressure [kPa] 6.12E-05 107.3112488 6.12E-05 107.3112488 6.12E-05 6.12E-05 106.3957214 6.12E-05 6.12E-05 106.7008972 6.13E- 106.0905457 6.13E-05 106.7008972 6,13E-0S 106.0905457 6.14E-05 106.7008972 6,14E-0 106.3957214 6.151,05 106.7008972 6.1SE-0S 106.3957214 6.16E-05 105.7853699 6.16E-0S 05.7853699 6.17E-05 105,4801941 6.18E-0S 106.3957214 6.18E-05 105.1750183 6.19E-0S 104.8698425 6.20E-05 105.1750183 6.21E-0S 105.4801941 6.22E-0S 6.23Е-05 104.8698425 6.24E-05 104.8698425 6.25E-05 6.26E-05 104.5646667 6.27E-05 103.3439636 6.28E-05 103.9543152 6.30E-05 6.31E-05 03.6491394 6.32E-05 03.3439636 6.34Е-05 103.3439636 6.35E-05 96.63009644 6.37E-05 103.0387878 6.38E-05 6.40E-05 1.8888855 6.41E-05 6.43E-05 101.5129089 6,4SE-05 102.4284363 6.46E-05 102.7336121 648E-0 103.6491394 6.50E-05 98.15597534 6.52E-0 103.0387878 6.54E-05 102.123260S 6.56E-0S 98.15597534 .58E-05101.8180847 6.60E-05 6.62E-05 99.98703003 664E-0599.07150269 6,66E-05 99,07150269 -359.8 359.6 359.4 -359.2 107.6164246 107.6164246 358.8 358.6 357.8 357.6 357.4 357.2 -357 356.8 356.6 356.4 356.2 355.8 355.6 355.4 355.2 104.259491 104.259491 354.8 354.6 354.4 354.2 103.3439636 353.8 353.6 353.4 353.2 102.7336121 103.0387878 352.6 352.4 351.8 351.6 3514 351.2 350.8 350.6 350.4 350.2 98.7663269

Answer 1

Data is much longer than provided data. Hence not able to produce results.

Producing result for this much data only will be absurd.

But you can use this code on full data which you must be having.

% Reading data from excel file & dividing into catergories &num:- numerical data, txt :- text data, raw :- raw data % Make sure to save engine-cycle, data.x13x ln MATLAB folder num,txt,raw]xlsread ("engine_cycle_data.xlsx") % We are working on numerical data num P-10-3 * num (:, 4); % Extracting Pressure (4th column) (in Pa) % work over whole cycle in J % Extracting Volume (3rd column) % Integration using trapezoidal rule % work for P-comp-exp V-comp-exp N-comp-exp comp & expansion (index 901 to 2700) = P(901:2700); = V(901:2700); -tr apz ( P-comp-exp, V-comp-exp ) ; % Pressure values for comp & expa only % Pumping Work pump-w-cycle -w-comp-exp; % Total cycle work-work for comp & exp % fraction of input energy went into work efficiency W cycle/1000 % fraction of input enery went into pumping fracpump/1000:

Text :-

% Reading data from excel file & dividing into catergories
% num :- numerical data,  txt :- text data,  raw :- raw data
% Make sure to save engine_cycle_data.xlsx in MATLAB folder
[num,txt,raw] = xlsread("engine_cycle_data.xlsx");

% We are working on numerical data num
V = num(:,3);           % Extracting Volume (3rd column)
P = 10^3 * num(:,4);    % Extracting Pressure(4th column) (in Pa)

% Work over whole cycle in J
W_cycle = trapz(P,V);       % Integration using trapezoidal rule

% Work for comp & expansion (index 901 to 2700)
P_comp_exp = P(901:2700);       % Pressure values for comp & expa only
V_comp_exp = V(901:2700);
W_comp_exp = trapz(P_comp_exp,V_comp_exp);

% Pumping Work
W_pump = W_cycle - W_comp_exp;  % Total cycle work - work for comp & exp

% fraction of input energy went into work
efficiency = W_cycle/1000;

% fraction of input enery went into pumping
frac = W_pump / 1000;