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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3685/CH18/EX18.4/Ex18_4.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3685/CH18/EX18.4/Ex18_4.sce')
| -rw-r--r-- | 3685/CH18/EX18.4/Ex18_4.sce | 19 |
1 files changed, 19 insertions, 0 deletions
diff --git a/3685/CH18/EX18.4/Ex18_4.sce b/3685/CH18/EX18.4/Ex18_4.sce new file mode 100644 index 000000000..376c63658 --- /dev/null +++ b/3685/CH18/EX18.4/Ex18_4.sce @@ -0,0 +1,19 @@ +clc
+h = 17.4 // Convective heat transfer coefficient in W/m^2K
+K = 52.2 // Thermal conductivity in W/mK
+t = 120 // Heat reservoir wall temperature in degree celcius
+t0 = 35 // Ambient temperature in degree celcius
+L = 0.4 // Lenght of rod in m
+b = .050 // width of rod in mm
+H = .050 // Heigth of rod in mm
+
+printf("\n Example 18.4\n")
+l= L/2
+A = b*H
+m = sqrt(4*h*b/(K*b*H))
+t1 = (t-t0)/cosh(m*l) + t0 // Midway temperature of rod
+Q1 = 2*5.12*K*A*(t-t0)*tanh(m*l) // Heat loss rate
+printf("\n Midway temperature of rod is %f degree Celcius",t1)
+printf("\n Heat loss rate is %fW",Q1)
+//The answers vary due to round off error
+
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