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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /608/CH21/EX21.25 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '608/CH21/EX21.25')
| -rwxr-xr-x | 608/CH21/EX21.25/21_25.sce | 16 |
1 files changed, 16 insertions, 0 deletions
diff --git a/608/CH21/EX21.25/21_25.sce b/608/CH21/EX21.25/21_25.sce new file mode 100755 index 000000000..60e1dee74 --- /dev/null +++ b/608/CH21/EX21.25/21_25.sce @@ -0,0 +1,16 @@ +//Problem 21.25: A 250 V series motor draws a current of 40 A. The armature resistance is 0.15 ohm and the field resistance is 0.05ohm . Determine the maximum efficiency of the motor.
+
+//initializing the variables:
+I = 40; // in Amperes
+Rf = 0.05; // in ohms
+Ra = 0.15; // in ohm
+V = 250; // in Volts
+
+//calculation:
+//However for a series motor, If = 0 and the Ia*Ia*Ra loss needs to be I*I*(Ra + Rf)
+//For maximum efficiency I*I*(Ra + Rf) = C
+//Efficiency =((V*I -2*Ia*Ia*Ra)/(V*I))*100%
+eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n efficiency is %.1f",eff)
\ No newline at end of file |