summaryrefslogtreecommitdiff
path: root/572/CH3/EX3.6/c3_6.sce
blob: 074496fa113a160dae4258d1cfee7b4e3a931250 (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
// (3.6) A closed, rigid tank filled with water vapor, initially at 20 MPa, 520C, is cooled until its temperature reaches 400C. Using the compressibility chart, determine. (a) the specific volume of the water vapor in m3/kg at the initial state.(b) the pressure in MPa at the final state.Compare the results of parts (a) and (b) with the values obtained from the superheated vapor table, Table A-4.

//solution

//variable initialization
p1 = 20                        //initial pressure in MPa
T1 = 520                       // initial temperature in degree celcius
T2 = 400                       // final temperature in degree celcius

//part(a)
//from table A-1
Tc = 647.3                     //critical temperature in kelvin
pc = 22.09                     //critical pressure in MPa

Tr = (T1+273)/Tc               //reduced temperature
Pr = p1/pc                     //reduced pressure 
Z1 = .83                       //compressibility factor
R = 8.314                      //universal gas constant in SI unit 
n = 1000/18.02                 //number of moles in a kg of water
v1 = (Z1*n*R*(T1+273))/(p1*10^6)
printf('the specific volume in state1 in m3/Kg is:\n\t v1 = %f',v1)
printf('\n and the corresponding value obtained from table A-4 is .01551 m^3/Kg')

//part(b)
vr = v1*(pc*10^6)/(n*R*Tc)
Tr2 = (T2+273)/Tc
//at above vr and Tr2
PR = .69
P2 = pc*PR
printf('\n\n the pressure in MPa in the final state is: \n\t P2 = %f',P2)
printf('\n and the corresponding value from the table is 15.16Mpa')