1 files changed, 91 insertions, 0 deletions
diff --git a/572/CH11/EX11.3/c11_3.sce b/572/CH11/EX11.3/c11_3.sce
new file mode 100755
index 000000000..9efadfd0e
--- /dev/null
+++ b/572/CH11/EX11.3/c11_3.sce
@@ -0,0 +1,91 @@
+//(11.3)  Evaluate the partial derivative (dels/delv)T for water vapor at a state fixed by a temperature of 240C and a specific volume of 0.4646 m3/kg. (a) Use the Redlich–Kwong equation of state and an appropriate Maxwell relation. (b) Check the value obtained using steam table data.
+
+//solution
+
+//part(a)
+v = .4646                                                                       //specific volume in in m^3/kg
+M = 18.02                                                                       //molar mass of water in kg/kmol
+//At the specified state, the temperature is 513 K and the specific volume on a molar basis is
+vbar = v*M                                                                      //in m^3/kmol
+//From Table A-24
+a = 142.59                                                                      //(m^3/kmol)^2 * K^.5
+b = .0211                                                                       //in m^3/kmol
+
+Rbar = 8314                                                                     //universal gas constant in N.m/kmol.K
+T = 513                                                                         //in kelvin
+delpbydelT = (Rbar/(vbar-b) + a/[2*vbar*(vbar+b)*T^1.5]*10^5)/10^3          //in kj/(m^3*K)
+
+//by The Maxwell relation
+delsbydelv = delpbydelT
+printf('the value of delpbydelT in kj/(m^3*K) is:  %f ',delpbydelT)
+
+//part(b)
+//A value for (dels/delv)T can be estimated using a graphical approach with steam table data, as follows: At 240C, Table A-4 provides the values for specific entropy s and specific volume v tabulated below
+T = 240                                                                         //in degree celcius
+//at p =1 bar
+s(1,1) = 7.9949                                                                 //in kj/kg.k
+v(1,1) = 2.359                                                                  //in m^3/kg
+//at p= 1.5 bar
+s(2,1) = 7.8052                                                                 //in kj/kg.k
+v(2,1) = 1.570                                                                  //in m^3/kg
+//at p = 3 bar
+s(3,1) = 7.4774                                                                 //in kj/kg.k
+v(3,1) = .781                                                                   //in m^3/kg
+//at p = 5 bar
+s(4,1) = 7.2307                                                                 //in kj/kg.k
+v(4,1) = .4646                                                                  //in m^3/kg
+//at p =7 bar
+s(5,1) = 7.0641                                                                 //in kj/kg.k
+v(5,1) = .3292                                                                  //in m^3/kg
+//at p = 10 bar
+s(6,1) = 6.8817                                                                 //in kj/kg.k
+v(6,1) = .2275                                                                  //in m^3/kg
+plot(v,s)
+xtitle("","Specific volume, m3/kg","Specific entropy, kJ/kg·K")
+//The pressure at the desired state is 5 bar.The corresponding slope is
+delsbydelv = 1                                                                  //in kj/m^3.K
+printf('\n\nfrom the data of the table,delsbydelv = %f',delsbydelv)             
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+