blob: 2c55c1fe5eb5fa7dbc6febe420029d2b986a2fe8 (
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
|
//Fluid Systems - By Shiv Kumar
//Chapter 16- Hydraulic Power and Its Transmissions
//Example 16.1
//To Find the Maximum Power Available at the Outlet of Pipe.
clc
clear
//Given Data:-
d=300; //Diameter of the Pipe, mm
l=3000; //Length of the Pipe, m
H=400; //Total Head at Inlet, m
f=0.005;
//Data Required:-
rho=1000; //Density of Water, Kg/m^3
g=9.81; //Acceleration due to gravity, m/s^2
//Computations:-
//Condition for Maximum Power transmission
hf=H/3; //m
V=sqrt(hf*(2*g*d/1000)/(4*f*l)); //m/s
Q=(%pi/4)*(d/1000)^2*V; //Discharge, m^3/s
Pmax=rho*g*Q*(H-hf)/1000; //Maximum Power Available at Outlet of Pipe, kW
//Results:-
printf("The Maximum Power Available at Outlet of Pipe=%.3f kW",Pmax) //The answer vary due to round off error
|
