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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /572/CH10/EX10.4/c10_4.sce | |
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Diffstat (limited to '572/CH10/EX10.4/c10_4.sce')
| -rwxr-xr-x | 572/CH10/EX10.4/c10_4.sce | 54 |
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diff --git a/572/CH10/EX10.4/c10_4.sce b/572/CH10/EX10.4/c10_4.sce new file mode 100755 index 000000000..a3e164f21 --- /dev/null +++ b/572/CH10/EX10.4/c10_4.sce @@ -0,0 +1,54 @@ +//(10.4) Air enters the compressor of an ideal Brayton refrigeration cycle at 1 bar, 270K, with a volumetric flow rate of 1.4 m3/s. If the compressor pressure ratio is 3 and the turbine inlet temperature is 300K, determine (a) the net power input, in kW, (b) the refrigeration capacity, in kW, (c) the coefficient of performance
+
+//solution
+
+//variable initialization
+p1 = 1 //in bar
+T1 = 270 //in kelvin
+AV = 1.4 //in m^3/s
+r = 3 //compressor pressure ratio
+T3 = 300 //turbine inlet temperature in kelvin
+
+//analysis
+//From Table A-22,
+h1 = 270.11 //in kj/kg
+pr1 = .9590
+pr2 = r*pr1
+//interpolating in Table A-22,
+h2s = 370.1 //in kj/kg
+//From Table A-22,
+h3 = 300.19 //in kj/kg
+pr3 = 1.3860
+pr4 = pr3/r
+//Interpolating in Table A-22, we obtain
+h4s = 219 //in kj/kg
+
+//part(a)
+R = 8.314 //universal gas constant, in SI units
+M = 28.97 //molar mass of air in grams
+mdot = (AV*p1)/((R/M)*T1)*10^2 //mass flow rate in kg/s
+
+Wcycledot = mdot*((h2s-h1)-(h3-h4s))
+printf('the net power input in kw is: %f',Wcycledot)
+
+//part(b)
+Qindot = mdot*(h1-h4s) //refrigeration capacity in kw
+printf('\nthe refregeration capacity in kw is: %f',Qindot)
+
+//part(c)
+beta = Qindot/Wcycledot //coefficient of performance
+printf('\nthe coefficient of performance is: %f',beta)
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