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Solution to Exercise 6.3

12 October, 2015 - 14:06
    clc
t=linspace(0,35,8)                  % Data entry for time [min]
m=[50 48.25 46 42.5 37.5 30.5 19 9] % Data entry for mass flow [kg/min]
% Calculate time intervals
dt=[t(2)-t(1),t(3)-t(2),t(4)-t(3),...
t(5)-t(4),t(6)-t(5),t(7)-t(6),t(8)-t(7)]
% Calculate mass out
dm=[0.5*(m(2)+m(1)),0.5*(m(3)+m(2)),0.5*(m(4)+m(3)),0.5*(m(5)+...
m(4)),0.5*(m(6)+m(5)),0.5*(m(7)+m(6)),0.5*(m(8)+m(7))]
% Calculate differential areas
da=dt.*dm;
% Tabulate time and mass flow
[t',m']
% Tabulate time intervals, mass out and differential areas
[dt',dm',da']
% Calculate the amount of oil drained [kg] in 35 minutes
Oil_Drained=sum(da)

The output is:

ans =
 
      0     50.0000
 5.0000     48.2500
10.0000     46.0000
15.0000     42.5000
20.0000     37.5000
25.0000     30.5000
30.0000     19.0000
35.0000      9.0000
 
ans =
 
5.0000      49.1250     245.6250
5.0000      47.1250     235.6250
5.0000      44.2500     221.2500
5.0000      40.0000     200.0000
5.0000      34.0000     170.0000
5.0000      24.7500     123.7500
5.0000      14.0000      70.0000
 
Oil_Drained =
 
    1.2663e+003