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Diffstat (limited to '572/CH12/EX12.7/c12_7.sce')
| -rwxr-xr-x | 572/CH12/EX12.7/c12_7.sce | 69 |
1 files changed, 69 insertions, 0 deletions
diff --git a/572/CH12/EX12.7/c12_7.sce b/572/CH12/EX12.7/c12_7.sce new file mode 100755 index 000000000..a3beb9d4b --- /dev/null +++ b/572/CH12/EX12.7/c12_7.sce @@ -0,0 +1,69 @@ +//(12.7) A 1 kg sample of moist air initially at 21C, 1 bar, and 70% relative humidity is cooled to 5C while keeping the pressure constant. Determine (a) the initial humidity ratio, (b) the dew point temperature, in C, and (c) the amount of water vapor that condenses, in kg.
+
+//solution
+
+//variable initialization
+m =1 //mass of sample in kg
+T1 = 21 //initial temperature in degree celcius
+psi1 = .7 //initial relative humidity
+T2 = 5 //final temperature in degree celcius
+
+//part(a)
+//from table A-2
+pg = .02487 //in bar
+
+pv1 = psi1*pg //partial pressure of water vapor in bar
+
+omega1 = .622*(.2542)/(14.7-.2542)
+printf('the initial humidity ratio is: %f',omega1)
+
+//part(b)
+//The dew point temperature is the saturation temperature corresponding to the partial pressure, pv1. Interpolation in Table A-2 gives
+T = 15.3 //the dew point temperature in degree celcius
+printf('\n\nthe dew point temperature in degree celcius is: %f',T)
+
+//part(c)
+mv1 = 1/[(1/omega1)+1] //initial amount of water vapor in the sample in kg
+ma = m-mv1 //mass of dry air present in kg
+
+//the partial pressure of the water vapor remaining in the system at the final state is the saturation pressure corresponding to 5C:
+pg = .00872 //in bar
+omega2 = .622*(pg)/(1.01325-pg) //humidity ratio after cooling
+
+mv2 = omega2*ma //The mass of the water vapor present at the final state
+mw = mv1-mv2
+printf('\n\n the amount of water vapor that condenses, in kg. is: %f',mw)
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