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Diffstat (limited to '572/CH7/EX7.5/c7_5.sce')
| -rwxr-xr-x | 572/CH7/EX7.5/c7_5.sce | 38 |
1 files changed, 38 insertions, 0 deletions
diff --git a/572/CH7/EX7.5/c7_5.sce b/572/CH7/EX7.5/c7_5.sce new file mode 100755 index 000000000..20e742b33 --- /dev/null +++ b/572/CH7/EX7.5/c7_5.sce @@ -0,0 +1,38 @@ +//Superheated water vapor enters a valve at 3.0 MPa, 320C and exits at a pressure of 0.5 MPa. The expansion is a throttling process. Determine the specific flow exergy at the inlet and exit and the exergy destruction per unit of mass flowing, each in kJ/kg. Let T0 = 25C, p0= 1 atm.
+
+//solution
+
+//variable initialization
+p1 = 3 //entry pressure in Mpa
+p2 = .5 //exit pressure in Mpa
+T1 = 320 //entry temperature in degree celcius
+T0 = 25 //in degree celcius
+p0 = 1 //in atm
+
+
+//from table A-4
+h1 = 3043.4 //in kj/kg
+s1 = 6.6245 //in kj/kg.k
+
+h2 = h1 //from reduction of the steady-state mass and energy rate balances
+
+s2 = 7.4223 //Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k
+
+//from table A-2
+h0 = 104.89 //in kj/kg
+s0 = 0.3674 //in kj/kg.k
+
+ef1 = h1-h0-(T0+273)*(s1-s0) //flow exergy at the inlet
+ef2 = h2-h0-(T0+273)*(s2-s0) //flow exergy at the exit
+
+//from the steady-state form of the exergy rate balance
+Ed = ef1-ef2 //the exergy destruction per unit of mass flowing is
+
+printf(' the specific flow exergy at the inlet in kj/kg is :\n\t ef1 =%f',ef1)
+printf('\nthe specific flow exergy at the exit in kj/kg is:\n\t ef2 = %f', ef2)
+printf('\nthe exergy destruction per unit of mass flowing in kj/kg is:\n\t = %f',Ed)
+
+
+
+
+
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