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Diffstat (limited to '572/CH3/EX3.6/c3_6.sce')
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1 files changed, 39 insertions, 0 deletions
diff --git a/572/CH3/EX3.6/c3_6.sce b/572/CH3/EX3.6/c3_6.sce new file mode 100755 index 000000000..074496fa1 --- /dev/null +++ b/572/CH3/EX3.6/c3_6.sce @@ -0,0 +1,39 @@ +// (3.6) A closed, rigid tank filled with water vapor, initially at 20 MPa, 520C, is cooled until its temperature reaches 400C. Using the compressibility chart, determine. (a) the specific volume of the water vapor in m3/kg at the initial state.(b) the pressure in MPa at the final state.Compare the results of parts (a) and (b) with the values obtained from the superheated vapor table, Table A-4.
+
+//solution
+
+//variable initialization
+p1 = 20 //initial pressure in MPa
+T1 = 520 // initial temperature in degree celcius
+T2 = 400 // final temperature in degree celcius
+
+//part(a)
+//from table A-1
+Tc = 647.3 //critical temperature in kelvin
+pc = 22.09 //critical pressure in MPa
+
+Tr = (T1+273)/Tc //reduced temperature
+Pr = p1/pc //reduced pressure
+Z1 = .83 //compressibility factor
+R = 8.314 //universal gas constant in SI unit
+n = 1000/18.02 //number of moles in a kg of water
+v1 = (Z1*n*R*(T1+273))/(p1*10^6)
+printf('the specific volume in state1 in m3/Kg is:\n\t v1 = %f',v1)
+printf('\n and the corresponding value obtained from table A-4 is .01551 m^3/Kg')
+
+//part(b)
+vr = v1*(pc*10^6)/(n*R*Tc)
+Tr2 = (T2+273)/Tc
+//at above vr and Tr2
+PR = .69
+P2 = pc*PR
+printf('\n\n the pressure in MPa in the final state is: \n\t P2 = %f',P2)
+printf('\n and the corresponding value from the table is 15.16Mpa')
+
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