diff options
Diffstat (limited to '608/CH38')
-rwxr-xr-x | 608/CH38/EX38.06/38_06.sce | 37 |
1 files changed, 19 insertions, 18 deletions
diff --git a/608/CH38/EX38.06/38_06.sce b/608/CH38/EX38.06/38_06.sce index e0cdac3ae..4c850ce53 100755 --- a/608/CH38/EX38.06/38_06.sce +++ b/608/CH38/EX38.06/38_06.sce @@ -1,19 +1,20 @@ -//Problem 38.06: The core of a transformer operating at 50 Hz has an eddy current loss of 100 W/m3 and the core laminations have a thickness of 0.50 mm. The core is redesigned so as to operate with the same eddy current loss but at a different voltage and at a frequency of 250 Hz. Assuming that at the new voltage the maximum flux density is one-third of its original value and the resistivity of the core remains unaltered, determine the necessary new thickness of the laminations.
-
-//initializing the variables:
-Pe = 100; // in W/m3
-f1 = 50; // in Hz
-t1 = 0.0005; // in m
-x = 1/3;
-f2 = 250; // in Hz
-
-//calculation:
-//Pe = ke*(Bm1*f1*t1)^2
-//Hence, at 50 Hz frequency
-ke = Pe/(Bm1*f1*t1)^2
-//At 250 Hz frequency
-Bm2 = x*Bm1
-t2 = ((Pe/ke)^0.5)/(Bm2*f2)
-
-printf("\n\n Result \n\n")
+//Problem 38.06: The core of a transformer operating at 50 Hz has an eddy current loss of 100 W/m3 and the core laminations have a thickness of 0.50 mm. The core is redesigned so as to operate with the same eddy current loss but at a different voltage and at a frequency of 250 Hz. Assuming that at the new voltage the maximum flux density is one-third of its original value and the resistivity of the core remains unaltered, determine the necessary new thickness of the laminations. + +//initializing the variables: +Pe = 100; // in W/m3 +f1 = 50; // in Hz +t1 = 0.0005; // in m +x = 1/3; +f2 = 250; // in Hz +//Assigning a sample value to Bm1 +Bm1 = 1; +//calculation: +//Pe = ke*(Bm1*f1*t1)^2 +//Hence, at 50 Hz frequency +ke = Pe/(Bm1*f1*t1)^2 +//At 250 Hz frequency +//Bm2 = x*Bm1 +t2 = ((Pe/ke)^0.5)/(Bm2*f2) + +printf("\n\n Result \n\n") printf("\nlamination thickness is %.2Em",t2)
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