blob: 9edcdea6e86befcde22e3dc871a694883fa52ac8 (
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
|
clc;
clear;
//Example 5.24
ho=200; //[W/sq m.K]
hi=1500; //[W/sq m.K]
Cpw=4.2; //Sp heat of Water in [kJ/(kg.K)]
Cpo=2.1; //Sp heat of Oil in [kJ/(kg.K)]
E=0.8; //Effectiveness
k=46; //[W/m.K]
m_dot=0.167; //[kg/s]
mCp_oil=2*m_dot*Cpo*1000 //For oil [W/K]
//mCp_oil is wrongly calculated as 710.4
mCp_water=m_dot*Cpw*1000 //For water [W/K]
//mCp_oil is wrongly calculated as 710.4
//NOTE:The above two values are wrongly calculated in book as 710.4
//so we take here:
mCp_small=710.4 //[W/K]
//Since both mCp_water and mCp_oil are equal ,therefore:
C=1;
deff('[x]=f(ntu)','x=E-(ntu/(1+ntu))');
ntu=fsolve(1,f)
id=20; //Internal diameter in [mm]
od=25; //External diameter in [mm]
hio=hi*id/od //[W/sq m.K]
Dw=(od-id)/log(od/id) //[mm]
Dw=Dw/1000 //[m]
x=(od-id)/2 //[mm]
x=x/1000 //[m]
Do=0.025 //External dia in [m]
L=2.5; //Length of tube in [m]
Uo=1/(1/ho+1/hio+(x/k)*(Dw/Do)) //[W/sq m.K]
A=ntu*mCp_small/Uo //Heat transfer area in [sq m]
n=A/(%pi*Do*L) //No of tubes
printf("\nNo. of tubes required = %d",round(n+1));
|
