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Diffstat (limited to '572/CH12/EX12.14/c12_14.sce')
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diff --git a/572/CH12/EX12.14/c12_14.sce b/572/CH12/EX12.14/c12_14.sce new file mode 100755 index 000000000..b4e154919 --- /dev/null +++ b/572/CH12/EX12.14/c12_14.sce @@ -0,0 +1,47 @@ +//(12.14) A stream consisting of 142 m3/min of moist air at a temperature of 5C and a humidity ratio of 0.002 kg(vapor)kg(dry air) is mixed adiabatically with a second stream consisting of 425 m3/min of moist air at 24C and 50% relative humidity. The pressure is constant throughout at 1 bar. Using the psychrometric chart, determine (a) the humidity ratio and (b) the temperature of the exiting mixed stream, in C.
+
+//solution
+
+//variable initialization
+AV1 = 142 //in m^3/min
+T1 = 5 //in degree celcius
+omega1 = .002
+AV2 = 425 //in m^3/min
+T2 = 24 //in degree celcius
+psi2 = .5
+p = 1 //in bar
+
+
+//part(a)
+//from the psychrometric chart, Fig. A-9.
+va1 = .79 //in m^3/kg
+va2 = .855 //in m^3/kg
+omega2 = .0094
+
+ma1dot = AV1/va1 //in kg/min
+ma2dot = AV2 /va2 //in kg/min
+
+omega3 = (omega1*ma1dot+omega2*ma2dot)/(ma1dot + ma2dot)
+printf('the humidity ratio is: %f',omega3)
+
+//part(b)
+//Reduction of the energy rate balance gives
+//(ha + omega*hv)3 = [ma1dot*(ha + omega*hv)1 + ma2dot*(ha + omega*hv)2]/(ma1dot+ma2dot)
+//with (ha + omega*hv)1 = 10kj/kg and (ha + omega*hv)2 = 47.8kj/kg from figure A-9
+LHS = (ma1dot*10+ma2dot*47.8)/(ma1dot + ma2dot)
+
+//This value for the enthalpy of the moist air at the exit, together with the previously determined value for omega3, fixes the state of the exiting moist air. From inspection of Fig. A-9,
+T3 = 19 //in degree celcius
+printf('\n\nthe temperature of the exiting mixed stream in degree celcius is: %f',T3)
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