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author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
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committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /629/CH10/EX10.8/example10_8.sce | |
parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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diff --git a/629/CH10/EX10.8/example10_8.sce b/629/CH10/EX10.8/example10_8.sce new file mode 100644 index 000000000..7a055baa3 --- /dev/null +++ b/629/CH10/EX10.8/example10_8.sce @@ -0,0 +1,27 @@ +clear
+clc
+//Example 10.8 PRESSURE DROP IN AN HVAC DUCT
+g=9.81; //[m/s^2]
+rho=1.2; //[kg/m^3]
+L=50; //[m]
+b=0.6; //[m]
+h=0.3; //[m]
+A=b*h //area of cross section [m^2]
+Q=2.5; //[m^3/s]
+V=Q/A //[m/s]
+v=15.1*10^-6; //[m^2/s]
+//Hydraulic perimeter
+Dh=4*A/(2*(b+h)) //[m]
+//Reynolds number
+Re=V*Dh/v
+//Thus, flow is turbulent
+ks=0.000046;
+//Relative roughness
+Rr=ks/Dh
+//From Moody diagram for Re and Rr,
+f=0.015;
+//Darcy-Weisbach equation
+hf=f*(L/Dh)*(V^2/(2*g)) //[m]
+//1 inch H20=249.7 Pa
+delp=rho*g*hf/249.7 //pressure drop in inch H2O
+printf("\nThe pressure drop in inches of water per 50 m of duct = %.3f inch H2O.\n",delp)
\ No newline at end of file |