1 files changed, 68 insertions, 0 deletions
diff --git a/2705/CH15/EX15.7/Ex15_7.sce b/2705/CH15/EX15.7/Ex15_7.sce
new file mode 100755
index 000000000..0f950939f
--- /dev/null
+++ b/2705/CH15/EX15.7/Ex15_7.sce
@@ -0,0 +1,68 @@
+clear;
+clc;
+disp('Example 15.7');
+
+// aim : To determine
+// (a) the pressure, volume and temperature at each cycle process change points
+// (b) the heat transferred to air
+// (c) the heat rejected by the air
+// (d) the ideal thermal efficiency
+// (e) the work done 
+// (f) the mean effective pressure
+
+// given values
+m = 1;// mass of air, [kg]
+rv = 6;// volume ratio of adiabatic compression
+P1 = 103;// initial pressure , [kN/m^2]
+T1 = 273+100;// initial temperature, [K]
+P3 = 3450;// maximum pressure, [kN/m^2]
+Gama = 1.4;// heat capacity ratio
+R = .287;// gas constant, [kJ/kg K]
+
+// solution
+// taking reference  Fig. 15.20
+// (a)
+// for point 1
+V1 = m*R*T1/P1;// initial volume, [m^3]
+
+// for point 2
+V2 = V1/rv;// volume at point 2, [m^3] 
+// using PV^(Gama)=constant for process 1-2
+P2 = P1*(V1/V2)^(Gama);// pressure at point 2,. [kN/m^2]
+T2 = T1*(V1/V2)^(Gama-1);// temperature at point 2,[K]
+
+// for point 3
+V3 = V2;// volume at point 3, [m^3]
+// since volume is constant in process 2-3 , so using P/T=constant, so
+T3 = T2*(P3/P2);// temperature at stage 3, [K]
+
+// for point 4
+V4 = V1;//  volume at point 4, [m^3]
+P4 = P3*(V3/V4)^Gama;// pressure at point 4, [kN/m^2] 
+// again  since volume is constant in process 4-1 , so using P/T=constant, so
+T4 = T1*(P4/P1);// temperature at point 4, [K]
+
+mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4  =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);
+
+//  (b)
+cv = R/(Gama-1);//  specific heat capacity, [kJ/kg K]
+Q23 = m*cv*(T3-T2);// heat transferred, [kJ]
+mprintf('\n (b) The heat transferred to the air is  =  %f  kJ\n',Q23);
+
+// (c) 
+Q34 = m*cv*(T4-T1);// heat rejected by air, [kJ]
+mprintf('\n (c) The heat rejected by the air is  =  %f  kJ\n',Q34);
+
+// (d)
+TE = 1-Q34/Q23;// ideal thermal efficiency
+mprintf('\n (d) The ideal thermal efficiency is  =  %f percent\n',TE*100);
+
+// (e)
+W = Q23-Q34;// work done ,[kJ]
+mprintf('\n (e) The work done  is  =  %f  kJ\n',W);
+
+// (f)
+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]
+mprintf('\n (f)  The mean effefctive pressure is  =  %f kN/m^2\n',Pm);
+
+//  End