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| author | prashantsinalkar | 2017-10-10 12:38:01 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:38:01 +0530 |
| commit | f35ea80659b6a49d1bb2ce1d7d002583f3f40947 (patch) | |
| tree | eb72842d800ac1233e9d890e020eac5fd41b0b1b /608/CH22 | |
| parent | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (diff) | |
| download | Scilab-TBC-Uploads-f35ea80659b6a49d1bb2ce1d7d002583f3f40947.tar.gz Scilab-TBC-Uploads-f35ea80659b6a49d1bb2ce1d7d002583f3f40947.tar.bz2 Scilab-TBC-Uploads-f35ea80659b6a49d1bb2ce1d7d002583f3f40947.zip | |
updated the code
Diffstat (limited to '608/CH22')
| -rwxr-xr-x | 608/CH22/EX22.09/22_09.sce | 50 |
1 files changed, 25 insertions, 25 deletions
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)
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