blob: a5a7cb17365517b5e023dc1d550ca32c169a6f5d (
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
|
// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART I : GENERATION
// CHAPTER 3: HYDRO-ELECTRIC STATIONS
// EXAMPLE : 3.4 :
// Page number 41-42
clear ; clc ; close ; // Clear the work space and console
// Given data
load_factor = 0.15 // Load factor
P = 10.0*10**3 // Rated installed capacity(kW)
H = 50.0 // Head of plant(m)
n = 0.8 // Efficiency of plant
//Calculation
units_day = P*load_factor // Total units generated daily on basis of load factor(kWhr)
units_week = units_day*24.0*7 // Total units generated for one week(kWhr)
Q = units_week/(9.81*H*n*24*7) // Minimum flow of water(cubic mt/sec)
//Result
disp("PART I - EXAMPLE : 3.4 : SOLUTION :-")
printf("\nMinimum flow of river water to operate the plant, Q = %.3f cubic mt/sec", Q)
|