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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 /530/CH1/EX1.3/example_1_3.sce | |
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initial commit / add all books
Diffstat (limited to '530/CH1/EX1.3/example_1_3.sce')
| -rwxr-xr-x | 530/CH1/EX1.3/example_1_3.sce | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/530/CH1/EX1.3/example_1_3.sce b/530/CH1/EX1.3/example_1_3.sce new file mode 100755 index 000000000..d8d73c9ef --- /dev/null +++ b/530/CH1/EX1.3/example_1_3.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 1
+// Introduction
+
+
+//Example 1.3
+// Page 16
+printf("Example 1.3, Page 16\n\n");
+
+//Solution:
+// Given
+v_i=10; // [m/s]
+q=1000; // [W]
+d_i=0.04; // [m]
+d_o=0.06; // [m]
+
+// From appendix table A.2
+rho1=0.946; // [kg/m^3] at 100 degree C
+C_p=1009; // [J/kg K]
+
+mdot=rho1*(%pi/4)*(d_i^2)*v_i; // [kg/s]
+
+
+// In this case (dW/dt)_shaft=0 and (z_o - z_i)=0
+// From eqn 1.4.15 , q=mdot*(h_o-h_i)
+// Let dh = (h_o-h_i)
+dh=q/mdot; // [J/kg]
+// Let T_o be the outlet temperature
+T_o=dh/C_p+100;
+
+rho2=0.773; // [kg/m^3] at T_o = 183.4 degree C
+// From eqn 1.4.6
+v_o=mdot/(rho2*(%pi/4)*(d_o)^2); // [m/s]
+
+dKE_kg=(v_o^2-v_i^2)/2; // [J/kg]
+
+
+printf("Exit Temperature is %f degree C \n",T_o);
+printf("Exit velocity is %f m/s \n",v_o);
+printf("Change in Kinetic Energy per kg = %f J/kg",dKE_kg);
\ No newline at end of file |