blob: 1c382287ea5708146a173ca633946dd87518ac07 (
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
|
//Fluid Systems - By Shiv Kumar
//Chapter 16- Hydraulic Power and Its Transmissions
//Example 16.3
//To Determine the Minimum Number of Pipes.
clc
clear
//Given Data:-
l=7500; //Length of each Pipe, m
d=125; //Diameter of each Pipe, mm
Pr=6000; //Pressure at Discharge End, kPa
eta=85/100; //Efficiency
P=156; //Power Delivered, kW
f=0.006;
//Data Required:-
rho=1000; //Density of Water, Kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations:-
H_minus_hf=Pr*10^3/(rho*g); //H-hf, m
H=H_minus_hf/eta; //m
hf=H-H_minus_hf; //m
Q=P*1000/(rho*g*(H-hf)); //m^3/s
q=sqrt((hf*2*g*%pi^2*(d/1000)^5)/(64*f*l)); //Discharge in each Pipe, m^3/s
n=Q/q; //Number of Pipes
//Results:-
printf("The Minimum Number of Pipes Required=%.f\n",n)
|
