blob: bc665751f601588d1c42093added26d1b9c62605 (
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
|
clear;
clc;
printf("\t\t\tProblem Number 3.16\n\n\n");
// Chapter 3 : The First Law Of Thermodynamics
// Problem 3.16 (page no. 116)
// Solution
//In problem 3.15,
p1=100; //Unit:psia //Initial pressure
t1=950; //Unit:Fahrenheit //Temperature at pressure p1
p2=76; //Unit:psia //Final pressure
t2=580; //Unit:Fahrenheit //Temperature at pressure p2
v1=4; //Unit:ft^3/LBm //Specific volume at inlet conditions
v2=3.86; //Unit:ft^3/LBm //Specific volume at outlet conditions
Cv=0.32; //Unit:Btu/(LBm*R) //Specific heat for constant volume process
T1=t1+460; //Unit:R //Temperature at pressure p1
T2=t2+460; //Unit:R //Temperature at pressure p2
J=778; //J=Conversion factor
gc=32.174; //Unit: (LBm*ft)/(LBf*s^2) //gc is constant of proportionality
g=gc; //Unit:ft/s^2 //g=The local gravity
//Z1=Inlet position //Unit:m
//V1=Inlet velocity //Unit:m/s
//Z2=Outlet position //Unit:m
//V2=Outlet velocity Unit:m/s
//u1=internal energy //energy in
//u2=internal energy //energy out
//Energy equation is given by
//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + u1 + ((p1*v1)/J) + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; //Unit:Btu/LBm
//In 3.15, the elevation of the pipe at section 1 makes Z1 = 0
// Also no work crosses the boundaries of the system, the energy equation is reduced to
//u1 + ((p1*v1)/J) + q = u2 + ((p2*v2)/J) + ((Z2/J)*(g/gc))
//In problrm 3.16,
Z2=100; //Given //Unit:ft //Outlet position
//u2-u1=Cv*(T2-T1) //For a constant volume process //u2-u1=Chnage in internal energy
//So,
q=Cv*(T2-T1) + (p2*v2*144)/J - (p1*v1*144)/J + ((Z2/J)*(g/gc)) ; //q=heat transfer //1 ft^2=144 in^2 //Unit:Btu/LBm
printf("%f Btu/LBm heat is transferred from the gas \n",q);
//For this problem , neglecting the elevation term leads to an insignificant error
|
