blob: 83f0cc59cc8c508ca184007ab799af6096d1d7c7 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
|
clear;
clc;
disp('Example 5.26');
// aim : To determine
// the volume of the pressure vessel and the volume of the gas before transfer
// Given values
P1 = 1400;// initial pressure,[kN/m^2]
T1 = 273+85;// initial temperature,[K]
P2 = 700;// final pressure,[kN/m^2]
T2 = 273+60;// final temperature,[K]
m = 2.7;// mass of the gas passes,[kg]
cp = .88;// [kJ/kg]
cv = .67;// [kJ/kg]
// solution
// steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W [1],
// given, there is no kinetic energy change and neglecting potential energy term
W = 0;// no external work done
// so final equation is,u1+P1*v1+Q=u2 [2]
// also u2-u1=cv*(T2-T1)
// hence Q=cv*(T2-T1)-P1*v1 [3]
// and for unit mass P1*v1=R*T1=(cp-cv)*T1 [4]
// so finally
Q = cv*(T2-T1)-(cp-cv)*T1;// [kJ/kg]
// so total heat transferred is
Q = m*Q;// [kJ]
// using eqn [4]
v1 = (cp-cv)*T1/P1;// [m^3/kg]
// Total volume is
V1 = m*v1;// [m^3]
// using ideal gas equation P1*V1/T1=P2*V2/T2
V2 = P1*T2*V1/(P2*T1);// final volume,[m^3]
mprintf('\n The volume of gas before transfer is = %f m^3\n',V1);
mprintf('\n The volume of pressure vessel is = %f m^3\n',V2);
// End
|
