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Diffstat (limited to '3681/CH4/EX4.12')
| -rw-r--r-- | 3681/CH4/EX4.12/Ans4_12.PNG | bin | 0 -> 6694 bytes | |||
| -rw-r--r-- | 3681/CH4/EX4.12/Ex4_12.sce | 20 |
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diff --git a/3681/CH4/EX4.12/Ans4_12.PNG b/3681/CH4/EX4.12/Ans4_12.PNG Binary files differnew file mode 100644 index 000000000..27758af4c --- /dev/null +++ b/3681/CH4/EX4.12/Ans4_12.PNG diff --git a/3681/CH4/EX4.12/Ex4_12.sce b/3681/CH4/EX4.12/Ex4_12.sce new file mode 100644 index 000000000..6ff5f1e2e --- /dev/null +++ b/3681/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,20 @@ +// Estimating the final steady temperature rise of coil and its time constant
+clc;
+disp('Example 4.12, Page No. = 4.21')
+// Given Data
+S = 0.15;// Heat dissipating surface (in meter square)
+l = 1;// Length of mean turn in meter
+Sf = 0.56;// Space Factor
+A = 100*50;// Area of cross-section (in mm square)
+Q = 150;// Dissipating loss (in Watts)
+emissivity = 34;// Emissivity (in Watt per degree celsius per meter square)
+h = 390;// Specific heat of copper (in J per kg per degree celsius)
+// Calculation of the final steady temperature rise of coil and its time constant
+V = l*A*Sf*10^(-6);// Volume of copper (in meter cube)
+G = V*8900;// Since copper weighes 8900 kg per meter cube. Weight of copper(in kg)
+Tm = Q/(S*emissivity);// Final steady temperature rise (in degree celsius)
+Th = G*h/(S*emissivity);// Heating time constant (in seconds)
+disp(Tm,'Final steady temperature rise (degree celsius))=');
+disp(Th,'Heating time constant (seconds)=');
+//in book final steady temperature rise (in degree celsius) is equal to 29.4 and heating time constant (in seconds) is equal to 1906. The answers vary due to round off error
+
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