Question

Question 2 Table Q2 show the working pressure inside an engine which moves the piston at different positions inside the engine cylinder. Determine the total work done during the piston movement, 15 inches inside the cylinder. The piston head diameter was measure to be 8 inches. Given that the total work done relation is ; Work = A Loºp (x)dx , where A is the area of the piston. Item 130 1 2. 3 4 5 6 7 Table Q2 Piston head Piston Pressure position from (p), psi top center of the engine cylinder (y), cm 0 110 1 2 140 2.5 170 4 185 5 210 7 215 8 185 11 135 100 13 14 70 15 40 00 12 9 10 11 12 13 90

Answer 1

hii

Taking all the positions in inches (in question it is given in inches )

taking the pressure units as psi = pound-force per square inches

we can calculate the work done as below by representing the given data in a graph

us 220 210 200 185 180 (60 140 1135 (30 ho 119 8 90 80 20 60 40 H A1 Alo Al AL As I Ag A2 At A5, A6 AB AIR 13 2-5 14 8 0 1 15 2 D area under the grophe cause calculated as the areas of trapesiums. Each internal contains a trapezium. Let us see o Sum of both sides dis farmee between them. between oland! f- area = (10+130) (0 Ag - (130+1400 Ag = (185+210(1) (105+133X3) A8 - 2 2 2 (10 +17040.52 AD (210+215)(2) (135+ (OX0 A3 - Aq 2 2 (1 70+185)(105) Au = Az (245+1800) Alo z (11049070 2 Au = (90+70) ) 2 A2 = (70+40)(1) So we get Sumo od all areas as in sum of areas = 2858015 pound force (Puch) Now area of pirtono Td.? where d=diameter = 8inches 4 & area = .167 cius? to weger & Wondone a 16 T1 x 2258.75 Work = (1353701585 1.1353 810 Scanned with CamScanner

The similar work can be done easily using Matlab as below

d = 8;
Area = (pi*d^2)/4
x = [0 1 2 2.5 4 5 7 8 11 12 13 14 15];
p = [110 130 140 170 185 210 215 185 135 100 90 70 40];
plot(x,p)
title('Pressure vs head position')
xlabel('inches of head position')
ylabel('pounds-force per square inch')
Integral = trapz(x,p);
workdone = Integral*Area

The graph of pressure vs distance

Pressure vs head position 220 200 180 160 140 pounds-force per square inch 20 100 80 50 40 5 10 15 inches of head position

Area 11 50.2655 workdone = 1.1303e+05

hence the work done will be 1.1303 * 10^5 inch- pounds