From f35ea80659b6a49d1bb2ce1d7d002583f3f40947 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:38:01 +0530
Subject: updated the code

---
 608/CH22/EX22.09/22_09.sce | 50 +++++++++++++++++++++++-----------------------
 1 file changed, 25 insertions(+), 25 deletions(-)

(limited to '608/CH22')

diff --git a/608/CH22/EX22.09/22_09.sce b/608/CH22/EX22.09/22_09.sce
index 715fdcc5f..f2aab4c26 100755
--- a/608/CH22/EX22.09/22_09.sce
+++ b/608/CH22/EX22.09/22_09.sce
@@ -1,26 +1,26 @@
-//Problem 22.09: The speed of the induction motor of Problem 22.08 is reduced to 35% of its synchronous speed by using external rotor resistance. If the torque and stator losses are unchanged, determine (a) the rotor copper loss, and (b) the efficiency of the motor.
-
-//initializing the variables:
-Psi = 32000; // in Watts
-Psl = 1200; // in Watts
-Pfl = 750; // in Watts
-x = 0.35;
-
-//calculation:
-nr = x*ns
-//The slip, s
-s = ((ns - nr)/ns)
-//Input power to rotor = stator input power - stator losses
-Pi =  Psi - Psl
-//slip = rotor copper loss/rotor input
-Pl = s*Pi
-//Total mechanical power developed by the rotor = rotor input power - rotor losses
-Pr = Pi - Pl
-//Output power of motor = power developed by the rotor - friction and windage losses
-Po = Pr - Pfl
-//Efficiency of induction motor = (output power/input power)*100
-eff = (Po/Psi)*100 // in percent
-
-printf("\n\n Result \n\n")
-printf("\n(a) rotor copper loss is %.0f Watt",Pl)
+//Problem 22.09: The speed of the induction motor of Problem 22.08 is reduced to 35% of its synchronous speed by using external rotor resistance. If the torque and stator losses are unchanged, determine (a) the rotor copper loss, and (b) the efficiency of the motor.
+
+//initializing the variables:
+Psi = 32000; // in Watts
+Psl = 1200; // in Watts
+Pfl = 750; // in Watts
+x = 0.35;
+
+//calculation:
+//nr = x*ns
+//The slip, s
+s = ((1 - 0.35)/1)
+//Input power to rotor = stator input power - stator losses
+Pi =  Psi - Psl
+//slip = rotor copper loss/rotor input
+Pl = s*Pi
+//Total mechanical power developed by the rotor = rotor input power - rotor losses
+Pr = Pi - Pl
+//Output power of motor = power developed by the rotor - friction and windage losses
+Po = Pr - Pfl
+//Efficiency of induction motor = (output power/input power)*100
+eff = (Po/Psi)*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n(a) rotor copper loss is %.0f Watt",Pl)
 printf("\n(b) efficiency of induction motor is %.2f percent",eff)
\ No newline at end of file
-- 
cgit