blob: 9f863200e282a6e99ded92d8f8a428cb8eeede8d (
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
33
34
35
36
37
38
39
40
|
// 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 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION
// EXAMPLE : 7.3 :
// Page number 73
clear ; clc ; close ; // Clear the work space and console
// Given data
cap_baseload = 400.0*10**3 // Installed capacity of base load plant(kW)
cap_standby = 50.0*10**3 // Installed capacity of standby unit(kW)
output_baseload = 101.0*10**6 // Annual baseload station output(kWh)
output_standby = 87.35*10**6 // Annual standby station output(kWh)
peakload_standby = 120.0*10**3 // Peak load on standby station(kW)
hours_use = 3000.0 // Hours of standby station use/year(hrs)
// Calculations
// Case(i)
LF_1 = output_standby*100/(peakload_standby*hours_use) // Annual load factor(%)
hours_year = 365.0*24 // Total hours in a year
CF_1 = output_standby*100/(cap_standby*hours_year) // Annual capacity factor(%)
// Case(ii)
peakload_baseload = peakload_standby // Peak load on baseload station(kW)
LF_2 = output_baseload*100/(peakload_baseload*hours_use) // Annual load factor on baseload station(%)
hours_year = 365.0*24 // Total hours in a year
CF_2 = output_baseload*100/(cap_baseload*hours_year) // Annual capacity factor on baseload station(%)
// Results
disp("PART I - EXAMPLE : 7.3 : SOLUTION :-")
printf("\nCase(i) : Standby Station")
printf("\n Annual load factor = %.2f percent", LF_1)
printf("\n Annual capacity factor = %.2f percent\n", CF_1)
printf("\nCase(ii): Base load Station")
printf("\n Annual load factor = %.2f percent", LF_2)
printf("\n Annual capacity factor = %.2f percent\n", CF_2)
printf("\nNOTE: Incomplete solution in the textbook") ;
|
