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
46
47
48
49
50
51
52
53
54
55
56
57
58
59
|
//Fluid Systems - By - Shiv Kumar
//Chapter 11- Centrifugal Pumps
//Example 11.6
//To Calculate the Blade angle at Outlet, Power Required and Overall Efficiency of Pump.
clc
clear
//Given Data:-
Do=80; //Outer Diameter of the Impeller, cm
Q=1; //Discharge, m^3/s
H=80; //Head, m
N=1000; //Speed, rpm
bo=8; //Width at Outlet, cm
Delta_Q_per=3; //Percentage of Leakage Loss(of the Discharge)
Delta_P=10; //Mechanical Loss, kW
eta_H=80/100; //Hydraulic Efficiency
//Data Used:-
rho=1000; //Density of water, kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations:-
Do=Do/100; //m
bo=bo/100; //m
uo=%pi*Do*N/60; //m/s
Vfo=Q/(%pi*Do*bo); //m/s
Vwo=g*H/(uo*eta_H); //m/s
Vrwo=uo-Vwo; //m/s
//(a)
beta_o=atand(Vfo/Vrwo); //Blade Angle at Outlet, degrees
//Result1
printf(" Blade Angle at Outlet, beta_o=%.2f Degrees \n",beta_o) //The answer vary due to round off error
//(b)Power Required
Pi=rho*(1+Delta_Q_per/100)*Q*Vwo*uo; //Power delivered by the Impeller, W
P=Pi/1000+Delta_P; //Power required, kW
//Result2
printf(" Power Required, P =%.3f kW \n",P) //The answer vary due to round off error
//(c)Overall Efficiency, eta_o
eta_V=1/(1+Delta_Q_per/100); //Volumetric Efficiency
eta_m=(P-Delta_P)/P; //Mechanical Efficiency
eta_o=eta_H*eta_V*eta_m*100; //In Percentage
//Result3
printf(" Overall Efficiency, eta_o =%.2f Percent \n",eta_o) //The answer vary due to round off error
//Also, Overall Efficiency
eta_o=rho*Q*g*H/(P*1000)*100; //In Percentage
printf("Also, Overall Efficiency, eta_o=%.2f Percent\n",eta_o)
|
