1 files changed, 55 insertions, 0 deletions
diff --git a/572/CH12/EX12.11/c12_11.sce b/572/CH12/EX12.11/c12_11.sce
new file mode 100755
index 000000000..12dfd56ac
--- /dev/null
+++ b/572/CH12/EX12.11/c12_11.sce
@@ -0,0 +1,55 @@
+//(12.11)  Moist air at 30C and 50% relative humidity enters a dehumidifier operating at steady state with a volumetric flow rate of 280 m3/min. The moist air passes over a cooling coil and water vapor condenses. Condensate exits the dehumidifier saturated at 10C. Saturated moist air exits in a separate stream at the same temperature. There is no significant loss of energy by heat transfer to the surroundings and pressure remains constant at 1.013 bar. Determine (a) the mass flow rate of the dry air, in kg/min, (b) the rate at which water is condensed, in kg per kg of dry air flowing through the control volume, and (c) the required refrigerating capacity, in tons.
+
+//solution 
+
+//variable initialization
+T1 = 30                                                                         //in degree celcius
+AV1 = 280                                                                       //in m^3/min
+psi1 = .5                                                                       //relative humidity at the inlet
+T2 = 10                                                                         //in degree celcius
+p = 1.013                                                                       //pressure in bar
+
+//part(a)
+//from table A-2
+pg1 = .04246                                                                    //in bar
+pv1 = psi1*pg1                                                                  //in bar
+
+pa1 = p-pv1                                                                     //partial pressure of the dry air in bar
+
+Rbar = 8314                                                                     //universal gas constant
+Ma = 28.97                                                                      //molar mass of air
+madot = AV1/[(Rbar/Ma)*((T1+273)/(pa1*10^5))]                                   //common mass flow rate of the dry air in kg/min
+printf('the mass flow rate of the dry air in kg/min is:  %f',madot)
+
+//part(b)
+omega1 = .622*[pv1/(p-pv1)]
+
+//from table A-2
+pv2 = .01228                                                                    //in bar
+
+omega2 = .622*[pv2/(p-pv2)]
+
+mwdotbymadot = omega1-omega2                                   
+printf('\n\nthe rate at which water is condensed, in kg per kg of dry air flowing through the control volume is:  %f',mwdotbymadot)
+
+//part(c)
+//from table A-2 and A-22
+ha2 = 283.1                                                                     //in kg/kj
+ha1 = 303.2                                                                     //in kg/kj
+hg1 = 2556.3                                                                    //in kg/kj
+hg2 = 2519.8                                                                    //in kg/kj
+hf2 = 42.01                                                                     //in kg/kj
+
+Qcvdot = madot*[(ha2-ha1)-omega1*hg1+omega2*hg2+(omega1-omega2)*hf2]            //in kj/min
+printf('\n\nthe required refrigerating capacity, in tons is:  %f',Qcvdot/211)
+
+
+
+
+
+
+
+
+
+
+