blob: cdca63f25f921f59f4c25c41bc800320ba541dae (
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
|
// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART IV : UTILIZATION AND TRACTION
// CHAPTER 4: ILLUMINATION
// EXAMPLE : 4.9 :
// Page number 758-759
clear ; clc ; close ; // Clear the work space and console
// Given data
b = 27.45 // Breadth of hall(m)
l = 45.75 // Length of hall(m)
I_avg = 108.0 // Average illumination(lumens/sq.m)
h = 0.75 // Height(m)
cou = 0.35 // Co-efficient of utilization
pf = 0.9 // Pereciation factor
P_fl = 80.0 // Fluorescent lamp power(W)
n_100 = 13.4 // Luminous efficiency for 100W filament lamp(lumens/watt)
n_200 = 14.4 // Luminous efficiency for 200W filament lamp(lumens/watt)
n_80 = 30.0 // Luminous efficiency for 80W fluorescent lamp(lumens/watt)
// Calculations
area = b*l // Area to be illuminated(Sq.m)
I_total = area*I_avg // Total illumination on working plane(lumens)
gross_lumen = I_total/(cou*pf) // Gross lumens required
P_required = gross_lumen/n_200 // Power required for illumination(W)
P_required_kW = P_required/1000 // Power required for illumination(kW)
no_lamp = P_required/200 // Number of lamps
P_required_new = gross_lumen/n_80 // Power required when fluorescent lamp used(W)
P_required_new_kW = P_required_new/1000 // Power required when fluorescent lamp used(kW)
P_saving = P_required_kW-P_required_new_kW // Saving in power(kW)
// Results
disp("PART IV - EXAMPLE : 4.9 : SOLUTION :-")
printf("\nSuitable scheme: Whole area divided into %.f rectangles & 200-watt fitting is suspended at centre of each rectangle", no_lamp)
printf("\nSaving in power consumption = %.1f kW", P_saving)
|
