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
|
//Fluid Systems - By - Shiv Kumar
//Chapter 11- Centrifugal Pumps
//Example 11.20
//To Determine (i)Head generated by the Pump (ii)Shaft Power required to run the Pump.
clc
clear
//Given Data:-
n=3; //Number of Stages
Do=400; //Diameter of the Impeller at Outlet, mm
bo=20; //Width of Impeller at outlet, mm
beta_o=45; //degrees
Area_per=10; //Percentage of Total Area which is reduced.
eta_o=80/100; //Overall Efficiency
eta_man=90/100; //Manometric Efficiency
N=1000; //Speed, rpm
Q=0.05; //Discharge, m^3/s
//Data Used: -
rho=1000; //Density of water, kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations:-
Do=Do/1000; //m
bo=bo/1000; //m
A=%pi*Do*bo*(1-Area_per/100); //Actual Area of Flow, m^2
uo=%pi*Do*N/60; //Tangential Velocity of Impeller at Outlet, m/s
Vfo=Q/A; //Velocity of Flow, m/s
Vfi= Vfo;
Vwo=uo-Vfo/tand(beta_o); //m/s (Value given in book is wrong due to incorrect value of beta_o is used)
// (i)Head generated by the Pump , H_Tm
Hm=eta_man*Vwo*uo/g; //m
H_Tm=n*Hm; //m
//(ii) Shaft Power required to run the Pump , P
P=rho*Q*g*H_Tm/(eta_o*1000); //kW
//Results:-
printf(" (i)Head generated by the Pump , H_Tm=%.2f m \n",H_Tm) //The answer provided in the textbook is wrong
printf(" (ii) Shaft Power required to run the Pump , P =%.2f kW \n",P) //The answer provided in the textbook is wrong
|
