// Given:- // Part(a) v = 0.4646 // 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 = 0.0211 // in m^3/kmol Rbar = 8314.0 // universal gas constant in N.m/kmol.K T = 513.0 // 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 // Result printf( ' The value of delpbydelT in kj/(m^3*K) is: %.2f',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 = 240.0 // in degree celcius // At p =1, 1.5, 3, 5, 7, 10 bar respectively y = [7.994, 7.805, 7.477, 7.230, 7.064, 6.882] x = [2.359, 1.570, 0.781, 0.4646, 0.3292, 0.2275] plot(x,y) xlabel("Specific volume") ylabel("Specific entropy") // The pressure at the desired state is 5 bar.The corresponding slope is delsbydelv = 1 // in kj/m^3.K printf( ' From the data of the table,delsbydelv = %.2f',delsbydelv);