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Diffstat (limited to '3681/CH4/EX4.26')
| -rw-r--r-- | 3681/CH4/EX4.26/Ans4_26.PNG | bin | 0 -> 5395 bytes | |||
| -rw-r--r-- | 3681/CH4/EX4.26/Ex4_26.sce | 19 |
2 files changed, 19 insertions, 0 deletions
diff --git a/3681/CH4/EX4.26/Ans4_26.PNG b/3681/CH4/EX4.26/Ans4_26.PNG Binary files differnew file mode 100644 index 000000000..26074580a --- /dev/null +++ b/3681/CH4/EX4.26/Ans4_26.PNG diff --git a/3681/CH4/EX4.26/Ex4_26.sce b/3681/CH4/EX4.26/Ex4_26.sce new file mode 100644 index 000000000..08bd149d3 --- /dev/null +++ b/3681/CH4/EX4.26/Ex4_26.sce @@ -0,0 +1,19 @@ +// Calculating the temperature rise of tank
+clc;
+disp('Example 4.26, Page No. = 4.52')
+// Given Data
+MVA = 15;// MVA rating of transformer
+Q_iron = 80;// Iron losses (in kW)
+Q_copper = 120;// Copper losses (in kW)
+T_water = 15;// Temperature rise of water (in degree celsius)
+Vw = 3;// Amount of water (in litre per second)
+Dimensions = 3.5*3.0*1.4;// Tank dimensions (in meter)
+l = 10;// Specific loss dissipation from tank walls (in Watt per degree celsius per meter square)
+// Calculation of the temperature rise of tank
+Q_total = Q_iron+Q_copper;// Total losses (in kW)
+Q = Vw*T_water/0.24;// Heat taken away by water (in kW)
+Q_walls = Q_total-Q;// Loss dissipated by walls (in kW)
+S = 2*3.5*(3+1.14);// Area of tank walls by neglecting top and bottom surfaces
+T = Q_walls*10^(3)/(S*l);// Temperature rise of tank (in degree celsius)
+disp(T,'Temperature rise of tank (degree celsius)=');
+//in book answer is 40.6 (degree celsius). The provided in the textbook is wrong
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