From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3472/CH39/EX39.23/Example39_23.sce | 31 +++++++++++++++++++++++++++++++ 1 file changed, 31 insertions(+) create mode 100644 3472/CH39/EX39.23/Example39_23.sce (limited to '3472/CH39/EX39.23/Example39_23.sce') diff --git a/3472/CH39/EX39.23/Example39_23.sce b/3472/CH39/EX39.23/Example39_23.sce new file mode 100644 index 000000000..5e3f721b6 --- /dev/null +++ b/3472/CH39/EX39.23/Example39_23.sce @@ -0,0 +1,31 @@ +// 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 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS + +// EXAMPLE : 1.23 : +// Page number 706 +clear ; clc ; close ; // Clear the work space and console + +// Given data +hp = 25.0 // Motor rating(hp) +T = 100.0/60 // Heating time constant(hour) +theta = 40.0 // Temperature rise(°C) +t = 0.5 // Time(hour) +n = 0.85 // Motor maximum efficiency + +// Calculations +output = hp*735.5/1000 // Output of motor(kW) +output_max = output*n // Power at maximum efficiency(kW) +theta_f2 = theta/(1-exp(-t/T)) // θ_f2(°C) +loss = 1+(output/output_max)**2 // Losses at 18.4 kW output in terms of W +P = ((theta_f2/theta*loss)-1)**0.5*output_max // Half-hour rating of motor(kW) +P_hp = P*1000/735.5 // Half-hour rating of motor(hp) + +// Results +disp("PART IV - EXAMPLE : 1.23 : SOLUTION :-") +printf("\nHalf-hour rating of motor, P = %.f kW = %.1f hp (metric)\n", P,P_hp) +printf("\nNOTE: ERROR: Calculation mistake from final temperature rise onwards in textbook") -- cgit