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Diffstat (limited to '572/CH4/EX4.11/c4_11.sce')
-rwxr-xr-x | 572/CH4/EX4.11/c4_11.sce | 38 |
1 files changed, 38 insertions, 0 deletions
diff --git a/572/CH4/EX4.11/c4_11.sce b/572/CH4/EX4.11/c4_11.sce new file mode 100755 index 000000000..c08cf5837 --- /dev/null +++ b/572/CH4/EX4.11/c4_11.sce @@ -0,0 +1,38 @@ +//(4.11) A tank having a volume of 0.85 m3 initially contains water as a two-phase liquid—vapor mixture at 260C and a quality of 0.7. Saturated water vapor at 260C is slowly withdrawn through a pressure-regulating valve at the top of the tank as energy is transferred by heat to maintain the pressure constant in the tank. This continues until the tank is filled with saturated vapor at 260C. Determine the amount of heat transfer, in kJ. Neglect all kinetic and potential energy effects.
+
+
+//solution
+
+//variable initialization
+V = .85 //volume of tank in m^3
+T1 = 260 //initial temperature of the tank in degree celcius
+X1 = .7 //initial quality
+
+//from table A-2
+uf1 = 1128.4 //in kg/kg
+ug1 = 2599 //in kg/kg
+
+vf1 = 1.2755e-3 //in m^3/kg
+vg1 = .04221 //in m^3/kg
+
+u1 = uf1 + X1*(ug1-uf1) //in kj/kg
+v1 = vf1 + X1*(vg1-vf1) //in m^3/kg
+
+m1 = V/v1 //initial mass in kg
+
+//for final state, from table A-2,
+u2 = 2599 // units in KJ/kg
+v2 = 42.21e-3 //units in m^3/Kg
+he = 2796.6 //units in KJ/kg
+m2 = V/v2 //final mass in kg
+U2 = m2*u2 //final internal energy in KJ
+U1 = m1*u1 //initial internal energy in KJ
+Qcv = (U2-U1) - he*(m2-m1)
+printf('the amount of heat transfer in KJ is : \n\t Qcv = %f',Qcv)
+
+
+
+
+
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+
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