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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2705/CH15 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '2705/CH15')
| -rwxr-xr-x | 2705/CH15/EX15.1/Ex15_1.sce | 17 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.10/Ex15_10.sce | 32 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.11/Ex15_11.sce | 37 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.12/Ex15_12.sce | 48 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.13/Ex15_13.sce | 79 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.14/Ex15_14.sce | 77 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.15/Ex15_15.sce | 41 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.16/Ex15_16.sce | 48 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.17/Ex15_17.sce | 36 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.18/Ex15_18.sce | 22 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.2/Ex15_2.sce | 29 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.3/Ex15_3.sce | 69 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.4/Ex15_4.sce | 72 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.5/Ex15_5.sce | 28 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.6/Ex15_6.sce | 23 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.7/Ex15_7.sce | 68 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.8/Ex15_8.sce | 65 | ||||
| -rwxr-xr-x | 2705/CH15/EX15.9/Ex15_9.sce | 72 |
18 files changed, 863 insertions, 0 deletions
diff --git a/2705/CH15/EX15.1/Ex15_1.sce b/2705/CH15/EX15.1/Ex15_1.sce new file mode 100755 index 000000000..153a3eea6 --- /dev/null +++ b/2705/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+disp('Example 15.1');
+
+// aim : To determine
+// the thermal efficiency of the cycle
+
+// given values
+T1 = 273+400;// temperature limit, [K]
+T3 = 273+70;// temperature limit, [K]
+
+// solution
+// using equation [15] of section 15.3
+n_the = (T1-T3)/T1*100;// thermal efficiency
+mprintf('\n The thermal efficiency of the cycle is = %f percent\n',n_the);
+
+// End
diff --git a/2705/CH15/EX15.10/Ex15_10.sce b/2705/CH15/EX15.10/Ex15_10.sce new file mode 100755 index 000000000..5748bff66 --- /dev/null +++ b/2705/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+disp('Example 10');
+
+// aim : To determine
+// (a) the maximum temperature attained during the cycle
+// (b) the thermal efficiency of the cycle
+
+// given value
+rva =7.5;// volume ratio of adiabatic expansion
+rvc =15;// volume ratio of compression
+P1 = 98;// initial pressure, [kn/m^2]
+T1 = 273+44;// initial temperature, [K]
+P4 = 258;// pressure at the end of the adiabatic expansion, [kN/m^2]
+Gama = 1.4;// heat capacity ratio
+
+// solution
+// by seeing diagram
+// for process 4-1, P4/T4=P1/T1
+T4 = T1*(P4/P1);// [K]
+// for process 3-4
+T3 = T4*(rva)^(Gama-1);
+mprintf('\n (a) The maximum temperature during the cycle is = %f C\n',T3-273);
+
+// (b)
+
+// for process 1-2,
+T2 = T1*(rvc)^(Gama-1);// [K]
+n_the = 1-(T4-T1)/((Gama)*(T3-T2));// thermal efficiency
+mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the*100);
+
+// End
diff --git a/2705/CH15/EX15.11/Ex15_11.sce b/2705/CH15/EX15.11/Ex15_11.sce new file mode 100755 index 000000000..9045f1cd3 --- /dev/null +++ b/2705/CH15/EX15.11/Ex15_11.sce @@ -0,0 +1,37 @@ +clear;
+clc;
+disp('Example 15.11');
+
+// aim : To determine
+// (a) the thermal efficiency of the cycle
+// (b) the indicared power of the cycle
+
+// given values
+// taking basis one second
+rv = 11;// volume ratio
+P1 = 96;// initial pressure , [kN/m^2]
+T1 = 273+18;// initial temperature, [K]
+Gama = 1.4;// heat capacity ratio
+
+// solution
+// taking reference Fig. 15.24
+// (a)
+Beta = 2;// ratio of V3 and V2
+TE = 1-(Beta^(Gama)-1)/((rv^(Gama-1))*Gama*(Beta-1));// thermal efficiency
+mprintf('\n (a) the thermal efficiency of the cycle is = %f percent\n ',TE*100);
+
+// (b)
+// let V1-V2=.05, so
+V2 = .05*.1;// [m^3]
+// from this
+V1 = rv*V2;// [m^3]
+V3 = Beta*V2;// [m^3]
+V4 = V1;// [m^3]
+P2 = P1*(V1/V2)^(Gama);// [kN/m^2]
+P3 = P2;// [kn/m^2]
+P4=P3*(V3/V4)^(Gama);// [kN/m^2]
+// indicated power
+W = P2*(V3-V2)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// indicated power, [kW]
+mprintf('\n (c) The indicated power of the cycle is = %f kW\n',W);
+
+// End
diff --git a/2705/CH15/EX15.12/Ex15_12.sce b/2705/CH15/EX15.12/Ex15_12.sce new file mode 100755 index 000000000..43bddcdf8 --- /dev/null +++ b/2705/CH15/EX15.12/Ex15_12.sce @@ -0,0 +1,48 @@ +clear;
+clc;
+disp('Example 15.12');
+
+// aim : To determine
+// (a) the pressure and temperature at the end of compression
+// (b) the pressure and temperature at the end of the constant volume process
+// (c) the temperature at the end of constant pressure process
+
+// given values
+P1 = 103;// initial pressure, [kN/m^2]
+T1 = 273+22;// initial temperature, [K]
+rv = 16;// volume ratio of the compression
+Q = 244;//heat added, [kJ/kg]
+Gama = 1.4;// heat capacity ratio
+cv = .717;// heat capacity, [kJ/kg k]
+
+// solution
+// taking reference as Fig.15.26
+// (a)
+// for compression
+// rv = V1/V2
+P2 = P1*(rv)^Gama;// pressure at end of compression, [kN/m^2]
+T2 = T1*(rv)^(Gama-1);// temperature at end of compression, [K]
+mprintf('\n (a) The pressure at the end of compression is = %f MN/m^2\n',P2*10^-3);
+mprintf('\n The temperature at the end of compression is = %f C\n',T2-273);
+
+// (b)
+// for constant volume process,
+// Q = cv*(T3-T2), so
+T3 = T2+Q/cv;// temperature at the end of constant volume, [K]
+
+// so for constant volume, P/T=constant, hence
+P3 = P2*(T3/T2);// pressure at the end of constant volume process, [kN/m^2]
+mprintf('\n (b) The pressure at the end of constant volume process is = %f MN/m^2\n ',P3*10^-3);
+mprintf('\n The temperature at the end of constant volume process is = %f C\n',T3-273);
+
+// (c)
+S = rv-1;// stroke
+// assuming
+V3 = 1;// [volume]
+//so
+V4 = V3+S*.03;// [volume]
+// also for constant process V/T=constant, hence
+T4 = T3*(V4/V3);// temperature at the end of constant presure process, [k]
+mprintf('\n (c) The temperature at the end of constant pressure process is = %f C\n',T4-273);
+
+// End
diff --git a/2705/CH15/EX15.13/Ex15_13.sce b/2705/CH15/EX15.13/Ex15_13.sce new file mode 100755 index 000000000..50608fe00 --- /dev/null +++ b/2705/CH15/EX15.13/Ex15_13.sce @@ -0,0 +1,79 @@ +clear;
+clc;
+disp('Example 15.13');
+
+// aim : To determine
+// (a) the pressure, volume and temperature at cycle process change points
+// (b) the net work done
+// (c) the thermal efficiency
+// (d) the heat received
+// (e) the work ratio
+// (f) the mean effective pressure
+// (g) the carnot efficiency
+
+
+// given values
+rv = 15;// volume ratio
+P1 = 97*10^-3;// initial pressure , [MN/m^2]
+V1 = .084;// initial volume, [m^3]
+T1 = 273+28;// initial temperature, [K]
+T4 = 273+1320;// maximum temperature, [K]
+P3 = 6.2;// maximum pressure, [MN/m^2]
+cp = 1.005;// specific heat capacity at constant pressure, [kJ/kg K]
+cv = .717;// specific heat capacity at constant volume, [kJ/kg K]
+
+// solution
+// taking reference Fig. 15.27
+// (a)
+R = cp-cv;// gas constant, [kJ/kg K]
+Gama = cp/cv;// heat capacity ratio
+// for process 1-2
+V2 = V1/rv;// volume at stage2, [m^3]
+// using PV^(Gama)=constant for process 1-2
+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2]
+T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K]
+
+// for process 2-3
+// since volumee is constant in process 2-3 , so using P/T=constant, so
+T3 = T2*(P3/P2);// volume at stage 3, [K]
+V3 = V2;// volume at stage 3, [MN/m^2]
+
+// for process 3-4
+P4 = P3;// pressure at stage 4, [m^3]
+// since in stage 3-4 P is constant, so V/T=constant,
+V4 = V3*(T4/T3);// temperature at stage 4,[K]
+
+// for process 4-5
+V5 = V1;// volume at stage 5, [m^3]
+P5 = P4*(V4/V5)^(Gama);// pressure at stage5,. [MN/m^2]
+T5 = T4*(V4/V5)^(Gama-1);// temperature at stage 5, [K]
+
+mprintf('\n (a) P1 = %f kN/m^2, V1 = %f m^3, t1 = %f C,\n P2 = %f MN/m^2, V2 = %f m^3, t2 = %f C,\n P3 = %f MN/m^2, V3 = %f m^3, t3 = %f C,\n P4 = %f MN/m^2, V4 = %f m^3, t4 = %f C,\n P5 = %fkN/m^2, V5 = %fm^3, t5 = %fC\n',P1*10^3,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273,P5*10^3,V5,T5-273);
+
+
+// (b)
+W = (P3*(V4-V3)+((P4*V4-P5*V5)-(P2*V2-P1*V1))/(Gama-1))*10^3;// work done, [kJ]
+mprintf('\n (b) The net work done is = %f kJ\n',W);
+
+// (c)
+TE = 1-(T5-T1)/((T3-T2)+Gama*(T4-T3));// thermal efficiency
+mprintf('\n (c) The thermal efficiency is = %f percent\n',TE*100);
+
+// (d)
+Q = W/TE;// heat received, [kJ]
+mprintf('\n (d) The heat received is = %f kJ\n',Q);
+
+// (e)
+PW = P3*(V4-V3)+(P4*V4-P5*V5)/(Gama-1)
+WR = W*10^-3/PW;// work ratio
+mprintf('\n (f) The work ratio is = %f\n',WR);
+
+// (e)
+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]
+mprintf('\n (e) The mean effective pressure is = %f kN/m^2\n',Pm);
+
+// (f)
+CE = (T4-T1)/T4;// carnot efficiency
+mprintf('\n (f) The carnot efficiency is = %f percent\n',CE*100);
+
+// End
diff --git a/2705/CH15/EX15.14/Ex15_14.sce b/2705/CH15/EX15.14/Ex15_14.sce new file mode 100755 index 000000000..ebf067b5b --- /dev/null +++ b/2705/CH15/EX15.14/Ex15_14.sce @@ -0,0 +1,77 @@ +clear;
+clc;
+disp('Example 15.14');
+
+// aim : To determine
+// (a) the thermal efficiency
+// (b) the heat received
+// (c) the heat rejected
+// (d) the net work
+// (e) the work ratio
+// (f) the mean effective pressure
+// (g) the carnot efficiency
+
+
+// given values
+P1 = 101;// initial pressure , [kN/m^2]
+V1 = 14*10^-3;// initial volume, [m^3]
+T1 = 273+15;// initial temperature, [K]
+P3 = 1850;// maximum pressure, [kN/m^2]
+V2 = 2.8*10^-3;// compressed volume, [m^3]
+Gama = 1.4;// heat capacity
+R = .29;// gas constant, [kJ/kg k]
+
+// solution
+// taking reference Fig. 15.29
+// (a)
+// for process 1-2
+// using PV^(Gama)=constant for process 1-2
+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2]
+T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K]
+
+// for process 2-3
+// since volumee is constant in process 2-3 , so using P/T=constant, so
+T3 = T2*(P3/P2);// volume at stage 3, [K]
+
+// for process 3-4
+P4 = P1;
+T4 = T3*(P4/P3)^((Gama-1)/Gama);// temperature
+
+TE = 1-Gama*(T4-T1)/(T3-T2);// thermal efficiency
+mprintf('\n (a) The thermal efficiency is = %f percent\n',TE*100);
+
+// (b)
+cv = R/(Gama-1);// heat capacity at copnstant volume, [kJ/kg k]
+m = P1*V1/(R*T1);// mass of gas, [kg]
+Q1 = m*cv*(T3-T2);// heat received, [kJ/cycle]
+mprintf('\n (b) The heat received is = %f kJ/cycle\n',Q1);
+
+// (c)
+cp = Gama*cv;// heat capacity at constant at constant pressure, [kJ/kg K]
+Q2 = m*cp*(T4-T1);// heat rejected, [kJ/cycle]
+mprintf('\n (c) The heat rejected is = %f kJ/cycle\n',Q2);
+
+// (d)
+W = Q1-Q2;// net work , [kJ/cycle]
+mprintf('\n (d) The net work is = %f kJ/cycle\n',W);
+
+// (e)
+// pressure is constant for process 1-4, so V/T=constant
+V4 = V1*(T4/T1);// volume, [m^3]
+V3 = V2;// for process 2-3
+P4 = P1;// for process 1-4
+PW = (P3*V3-P1*V1)/(Gama-1);// positive work done, [kJ/cycle]
+WR = W/PW;// work ratio
+mprintf('\n (e) The work ratio is = %f\n',WR);
+
+// (f)
+Pm = W/(V4-V2);// mean effective pressure, [kN/m^2]
+mprintf('\n (f) The mean effefctive pressure is = %f kN/m^2\n',Pm);
+
+// (g)
+CE = (T3-T1)/T3;// carnot efficiency
+mprintf('\n (g) The carnot efficiency is = %f percent\n',CE*100);
+
+// there is minor variation in answer reported in the book
+
+// End
diff --git a/2705/CH15/EX15.15/Ex15_15.sce b/2705/CH15/EX15.15/Ex15_15.sce new file mode 100755 index 000000000..d6b24e902 --- /dev/null +++ b/2705/CH15/EX15.15/Ex15_15.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+disp('Example 15.15');
+
+// aim : To determine
+// (a) the net work done
+// (b) the ideal thermal efficiency
+// (c) the thermal efficiency if the process of generation is not included
+
+// given values
+P1 = 110;// initial pressure, [kN/m^2)
+T1 = 273+30;// initial temperature, [K]
+V1 = .05;// initial volume, [m^3]
+V2 = .005;// volume, [m^3]
+T3 = 273+700;// temperature, [m^3]
+R = .289;// gas constant, [kJ/kg K]
+cv = .718;// heat capacity, [kJ/kg K]
+
+// solution
+// (a)
+m = P1*V1/(R*T1);// mass , [kg]
+W = m*R*(T3-T1)*log(V1/V2);// work done, [kJ]
+mprintf('\n (a) The net work done is = %f kJ\n',W);
+
+// (b)
+n_the = (T3-T1)/T3;// ideal thermal efficiency
+mprintf('\n (b) The ideal thermal efficiency is = %f percent\n',n_the*100);
+
+// (c)
+V4 = V1;
+V3 = V2;
+T4 = T3;
+T2 = T1;
+
+Q_rej = m*cv*(T4-T1)+m*R*T1*log(V1/V2);// heat rejected
+Q_rec = m*cv*(T3-T2)+m*R*T3*log(V4/V3);// heat received
+
+n_th = (1-Q_rej/Q_rec);// thermal efficiency
+mprintf('\n (c) the thermal efficiency if the process of regeneration is not included is = %f percent\n',n_th*100);
+
+// End
diff --git a/2705/CH15/EX15.16/Ex15_16.sce b/2705/CH15/EX15.16/Ex15_16.sce new file mode 100755 index 000000000..5417fa898 --- /dev/null +++ b/2705/CH15/EX15.16/Ex15_16.sce @@ -0,0 +1,48 @@ +clear;
+clc;
+disp('Example 15.16');
+
+// aim : To determine
+// (a) the maximum temperature
+// (b) the net work done
+// (c) the ideal thermal efficiency
+// (d) the thermal efficiency if the process of regeneration is not included
+
+// given values
+P1 = 100;// initial pressure, [kN/m^2)
+T1 = 273+20;// initial temperature, [K]
+V1 = .08;// initial volume, [m^3]
+rv = 5;// volume ratio
+R = .287;// gas constant, [kJ/kg K]
+cp = 1.006;// heat capacity, [kJ/kg K]
+V3_by_V2 = 2;
+
+// solution
+// (a)
+// using Fig.15.33
+// process 1-2 is isothermal
+T2 = T1;
+// since process 2-3 isisobaric, so V/T=constant
+T3 = T2*(V3_by_V2);// maximumtemperature, [K]
+mprintf('\n (a) The maximum temperature is = %f C\n',T3-273);
+
+// (b)
+m = P1*V1/(R*T1);// mass , [kg]
+W = m*R*(T3-T1)*log(rv);// work done, [kJ]
+mprintf('\n (b) The net work done is = %f kJ\n',W);
+
+// (c)
+TE = (T3-T1)/T3;// ideal thermal efficiency
+mprintf('\n (c) The ideal thermal efficiency is = %f percent\n',TE*100);
+
+// (d)
+T4 = T3;
+T2 = T1;
+
+Q_rej = m*cp*(T4-T1)+m*R*T1*log(rv);// heat rejected
+Q_rec = m*cp*(T3-T2)+m*R*T3*log(rv);// heat received
+
+n_th = (1-Q_rej/Q_rec);// thermal efficiency
+mprintf('\n (d) the thermal efficiency if the process of regeneration is not included is = %f percent\n',n_th*100);
+
+// End
diff --git a/2705/CH15/EX15.17/Ex15_17.sce b/2705/CH15/EX15.17/Ex15_17.sce new file mode 100755 index 000000000..5182839fe --- /dev/null +++ b/2705/CH15/EX15.17/Ex15_17.sce @@ -0,0 +1,36 @@ +clear;
+clc;
+disp('Example 15.17');
+
+// aim : To determine
+// (a) the net work done
+// (b) thethermal efficiency
+
+// given values
+m = 1;// mass of air, [kg]
+T1 = 273+230;// initial temperature, [K]
+P1 = 3450;// initial pressure, [kN/m^2]
+P2 = 2000;// pressure, [kN/m^2]
+P3 = 140;// pressure, [kN/m^2]
+P4 = P3;
+Gama = 1.4; // heat capacity ratio
+cp = 1.006;// heat capacity, [kJ/kg k]
+
+// solution
+T2 =T1;// isothermal process 1-2
+// process 2-3 and 1-4 are adiabatic so
+T3 = T2*(P3/P2)^((Gama-1)/Gama);// temperature, [K]
+T4 = T1*(P4/P1)^((Gama-1)/Gama);// [K]
+R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K]
+Q1 = m*R*T1*log(P1/P2);// heat received, [kJ]
+Q2 = m*cp*(T3-T4);// heat rejected
+
+//hence
+W = Q1-Q2;// work done
+mprintf('\n (a) The net work done is = %f kJ\n',W);
+
+// (b)
+TE = 1-Q2/Q1;// thermal efficiency
+mprintf('\n (b) The thermal efficiency is = %f percent\n',TE*100);
+
+// End
diff --git a/2705/CH15/EX15.18/Ex15_18.sce b/2705/CH15/EX15.18/Ex15_18.sce new file mode 100755 index 000000000..2932ac9d0 --- /dev/null +++ b/2705/CH15/EX15.18/Ex15_18.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+disp('Example 15.18');
+
+// aim : To determine
+// thermal eficiency
+// carnot efficiency
+
+// given values
+rv = 5;// volume ratio
+Gama = 1.4;// heat capacity ratio
+
+// solution
+// under given condition
+
+TE = 1-(1/Gama*(2-1/rv^(Gama-1)))/(1+2*((Gama-1)/Gama)*log(rv/2));// thermal efficiency
+mprintf('\n The thermal efficiency is = %f percent\n',TE*100);
+
+CE = 1-1/(2*rv^(Gama-1));// carnot efficiency
+mprintf('\n The carnot efficiency is = %f \n',CE*100);
+
+// End
diff --git a/2705/CH15/EX15.2/Ex15_2.sce b/2705/CH15/EX15.2/Ex15_2.sce new file mode 100755 index 000000000..2d6440345 --- /dev/null +++ b/2705/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+disp('Example 15.2');
+
+// aim : To determine
+// (a) the volume ratios of the isothermal and adiabatic processes
+// (b) the thermal efficiency of the cycle
+
+// given values
+T1 = 273+260;// temperature, [K]
+T3 = 273+21;// temperature, [K]
+er = 15;// expansion ratio
+Gama = 1.4;// heat capacity ratio
+
+// solution
+// (a)
+T2 = T1;
+T4 = T3;
+// for adiabatic process
+rva = (T1/T4)^(1/(Gama-1));// volume ratio of adiabatic
+rvi = er/rva;// volume ratio of isothermal
+mprintf('\n (a) The volume ratio of the adiabatic process is = %f\n',rva);
+mprintf('\n The volume ratio of the isothermal process is = %f\n',rvi);
+
+// (b)
+n_the = (T1-T4)/T1*100;// thermal efficiency
+mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the);
+
+// End
diff --git a/2705/CH15/EX15.3/Ex15_3.sce b/2705/CH15/EX15.3/Ex15_3.sce new file mode 100755 index 000000000..8cab979fa --- /dev/null +++ b/2705/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,69 @@ +clear;
+clc;
+disp('Example 15.3');
+
+// aim : To determine
+// (a) the pressure, volume and temperature at each corner of the cycle
+// (b) the thermal efficiency of the cycle
+// (c) the work done per cycle
+// (d) the work ratio
+
+// given values
+m = 1;// mass of air, [kg]
+P1 = 1730;// initial pressure of carnot engine, [kN/m^2]
+T1 = 273+300;// initial temperature, [K]
+R = .29;// [kJ/kg K]
+Gama = 1.4;// heat capacity ratio
+
+// solution
+// taking reference Fig. 15.15
+// (a)
+// for the isothermal process 1-2
+// using ideal gas law
+V1 = m*R*T1/P1;// initial volume, [m^3]
+T2 = T1;
+V2 = 3*V1;// given condition
+// for isothermal process, P1*V1=P2*V2, so
+P2 = P1*(V1/V2);// [MN/m^2]
+// for the adiabatic process 2-3
+V3 = 6*V1;// given condition
+T3 = T2*(V2/V3)^(Gama-1);
+// also for adiabatic process, P2*V2^Gama=P3*V3^Gama, so
+P3 = P2*(V2/V3)^Gama;
+// for the isothermal process 3-4
+T4 = T3;
+// for both adiabatic processes, the temperataure ratio is same,
+// T1/T4 = T2/T3=(V4/V1)^(Gama-1)=(V3/V2)^(Gama-1), so
+V4 = 2*V1;
+// for isothermal process, 3-4, P3*V3=P4*V4, so
+P4 = P3*(V3/V4);
+disp('(a) At line 1');
+mprintf('\n V1 = %f m^3, t1 = %f C, P1 = %f kN/m^2\n',V1,T1-273,P1);
+
+disp('At line 2');
+mprintf('\n V2 = %f m^3, t2 = %f C, P2 = %f kN/m^2\n',V2,T2-273,P2);
+
+disp('At line 3');
+mprintf('\n V3 = %f m^3, t3 = %f C, P3 = %f kN/m^2\n',V3,T3-273,P3);
+
+
+disp('At line 4');
+mprintf('\n V4 = %f m^3, t4 = %f C, P4 = %f kN/m^2\n',V4,T4-273,P4);
+
+
+// (b)
+n_the = (T1-T3)/T1;// thermal efficiency
+mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the*100);
+
+// (c)
+W = m*R*T1*log(V2/V1)*n_the;// work done, [J]
+mprintf('\n (c) The work done per cycle is = %f kJ\n',W);
+
+// (d)
+wr = (T1-T3)*log(V2/V1)/(T1*log(V2/V1)+(T1-T3)/(Gama-1));// work ratio
+mprintf('\n (d) The work ratio is = %f\n',wr);
+
+// there is calculation mistake in the book so answer is not matching
+
+// End
+
diff --git a/2705/CH15/EX15.4/Ex15_4.sce b/2705/CH15/EX15.4/Ex15_4.sce new file mode 100755 index 000000000..e13c274f1 --- /dev/null +++ b/2705/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,72 @@ +clear;
+clc;
+disp('Example 15.4');
+
+// aim : To determine
+// (a) the pressure, volume and temperature at cycle state points
+// (b) the heat received
+// (c) the work done
+// (d) the thermal efficiency
+// (e) the carnot efficiency
+// (f) the work ration
+// (g) the mean effective pressure
+
+// given values
+ro = 8;// overall volume ratio;
+rv = 6;// volume ratio of adiabatic compression
+P1 = 100;// initial pressure , [kN/m^2]
+V1 = .084;// initial volume, [m^3]
+T1 = 273+28;// initial temperature, [K]
+Gama = 1.4;// heat capacity ratio
+cp = 1.006;// specific heat capacity, [kJ/kg K]
+
+// solution
+// taking reference Fig. 15.18
+// (a)
+V2 = V1/rv;// volume at stage2, [m^3]
+V4 = ro*V2;// volume at stage 4;[m^3]
+// using PV^(Gama)=constant for process 1-2
+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2]
+T2 = T1*(V1/V2)^(Gama-1);// [K]
+
+P3 = P2;// pressure at stage 3, [kN/m^2]
+V3 = V4/rv;// volume at stage 3, [m^3]
+// since pressure is constant in process 2-3 , so using V/T=constant, so
+T3 = T2*(V3/V2);// temperature at stage 3, [K]
+
+// for process 1-4
+T4 = T1*(V4/V1);// temperature at stage4, [K
+P4 = P1;// pressure at stage4, [kN/m^2]
+
+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)
+R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K]
+m = P1*V1/(R*T1);// mass of gas, [kg]
+Q = m*cp*(T3-T2);// heat received, [kJ]
+mprintf('\n (b) The heat received is = %f kJ\n',Q);
+
+// (c)
+W = P2*(V3-V2)-P1*(V4-V1)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// work done, [kJ]
+mprintf('\n (c) The work done is = %f kJ\n',W);
+
+// (d)
+TE = 1-T1/T2;// thermal efficiency
+mprintf('\n (d) The thermal efficiency is = %f percent\n',TE*100);
+
+// (e)
+CE = (T3-T1)/T3;// carnot efficiency
+mprintf('\n (e) The carnot efficiency is = %f percent\n',CE*100);
+
+// (f)
+PW = P2*(V3-V2)+(P3*V3-P4*V4)/(Gama-1);// positive work done, [kj]
+WR = W/PW;// work ratio
+mprintf('\n (f) The work ratio is = %f\n',WR);
+
+// (g)
+Pm = W/(V4-V2);// mean effective pressure, [kN/m^2]
+mprintf('\n (g) The mean effective pressure is = %f kN/m^2\n',Pm);
+
+// there is minor variation in answer reported in the book
+
+// End
diff --git a/2705/CH15/EX15.5/Ex15_5.sce b/2705/CH15/EX15.5/Ex15_5.sce new file mode 100755 index 000000000..7de39d424 --- /dev/null +++ b/2705/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+disp('Example 15.5');
+
+// aim : To determine
+// (a) the actual thermal efficiency of the turbine
+// (b) the specific fuel consumption of the turbine in kg/kWh
+
+// given values
+P2_by_P1 = 8;
+n_tur = .6;// ideal turbine thermal efficiency
+c = 43*10^3;// calorific value of fuel, [kJ/kg]
+Gama = 1.4;// heat capacity ratio
+
+// solution
+// (a)
+rv = P2_by_P1;
+n_tur_ide = 1-1/(P2_by_P1)^((Gama-1)/Gama);// ideal thermal efficiency
+ate = n_tur_ide*n_tur;// actual thermal efficiency
+mprintf('\n (a) The actual thermal efficiency of the turbine is = %f percent\n',ate*100);
+
+// (b)
+ewf = c*ate;// energy to work fuel, [kJ/kg]
+kWh = 3600;// energy equivalent ,[kJ]
+sfc = kWh/ewf;// specific fuel consumption, [kg/kWh]
+mprintf('\n (b) The specific fuel consumption of the turbine is = %f kg/kWh',sfc);
+
+// End
diff --git a/2705/CH15/EX15.6/Ex15_6.sce b/2705/CH15/EX15.6/Ex15_6.sce new file mode 100755 index 000000000..d0bad01c2 --- /dev/null +++ b/2705/CH15/EX15.6/Ex15_6.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+disp('Example 15.6');
+
+// aim : To determine
+// the relative efficiency of the engine
+
+// given values
+d = 80;// bore, [mm]
+l = 85;// stroke, [mm]
+V1 = .06*10^6;// clearence volume, [mm^3]
+ate = .22;// actual thermal efficiency of the engine
+Gama = 1.4;// heat capacity ratio
+
+// solution
+sv = %pi*d^2/4*l;// stroke volume, [mm^3]
+V2 = sv+V1;// [mm^3]
+rv = V2/V1;
+ite = 1-(1/rv)^(Gama-1);// ideal thermal efficiency
+re = ate/ite;// relative thermal efficiency
+mprintf('\n The relative efficiency of the engine is = %f percent\n',re*100);
+
+// End
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
diff --git a/2705/CH15/EX15.8/Ex15_8.sce b/2705/CH15/EX15.8/Ex15_8.sce new file mode 100755 index 000000000..a37208dcc --- /dev/null +++ b/2705/CH15/EX15.8/Ex15_8.sce @@ -0,0 +1,65 @@ +clear;
+clc;
+disp('Example 15.8');
+
+// aim : To determine
+// (a) the pressure, volume and temperature at cycle state points
+// (b) the thermal efficiency
+// (c) the theoretical output
+// (d) the mean effective pressure
+// (e) the carnot efficiency
+
+// given values
+rv = 9;// volume ratio
+P1 = 101;// initial pressure , [kN/m^2]
+V1 = .003;// initial volume, [m^3]
+T1 = 273+18;// initial temperature, [K]
+P3 = 4500;// maximum pressure, [kN/m^2]
+N = 3000;
+cp = 1.006;// specific heat capacity at constant pressure, [kJ/kg K]
+cv = .716;// specific heat capacity at constant volume, [kJ/kg K]
+
+// solution
+// taking reference Fig. 15.20
+// (a)
+// for process 1-2
+Gama = cp/cv;// heat capacity ratio
+R = cp-cv;// gas constant, [kJ/kg K]
+V2 = V1/rv;// volume at stage2, [m^3]
+// using PV^(Gama)=constant for process 1-2
+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2]
+T2 = T1*(V1/V2)^(Gama-1);// [K]
+
+// for process 2-3
+V3 = V2;// volume at stage 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 process 3-4
+V4 = V1;// volume at stage 4
+// using PV^(Gama)=constant for process 3-4
+P4 = P3*(V3/V4)^(Gama);// pressure at stage2,. [kN/m^2]
+T4 = T3*(V3/V4)^(Gama-1);// temperature at stage 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)
+TE = 1-(T4-T1)/(T3-T2);// thermal efficiency
+mprintf('\n (b) The thermal efficiency is = %f percent\n',TE*100);
+
+// (c)
+m = P1*V1/(R*T1);// mass os gas, [kg]
+W = m*cv*((T3-T2)-(T4-T1));// work done, [kJ]
+Wt = W*N/60;// workdone per minute, [kW]
+mprintf('\n (c) The theoretical output is = %f kW\n',Wt);
+
+// (d)
+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]
+mprintf('\n (g) The mean effefctive pressure is = %f kN/m^2\n',Pm);
+
+// (e)
+CE = (T3-T1)/T3;// carnot efficiency
+mprintf('\n (e) The carnot efficiency is = %f percent\n',CE*100);
+
+
+// End
diff --git a/2705/CH15/EX15.9/Ex15_9.sce b/2705/CH15/EX15.9/Ex15_9.sce new file mode 100755 index 000000000..7482cbe06 --- /dev/null +++ b/2705/CH15/EX15.9/Ex15_9.sce @@ -0,0 +1,72 @@ +clear;
+clc;
+disp('Example 15.9');
+
+// aim : To determine
+// (a) the pressure and temperature at cycle process change points
+// (b) the work done
+// (c) the thermal efficiency
+// (d) the work ratio
+// (e) the mean effective pressure
+// (f) the carnot efficiency
+
+
+// given values
+rv = 16;// volume ratio of compression
+P1 = 90;// initial pressure , [kN/m^2]
+T1 = 273+40;// initial temperature, [K]
+T3 = 273+1400;// maximum temperature, [K]
+cp = 1.004;// specific heat capacity at constant pressure, [kJ/kg K]
+Gama = 1.4;// heat capacoty ratio
+
+// solution
+cv = cp/Gama;// specific heat capacity at constant volume, [kJ/kg K]
+R = cp-cv;// gas constant, [kJ/kg K]
+// for one kg of gas
+V1 = R*T1/P1;// initial volume, [m^3]
+// taking reference Fig. 15.22
+// (a)
+// for process 1-2
+// using PV^(Gama)=constant for process 1-2
+// also rv = V1/V2
+P2 = P1*(rv)^(Gama);// pressure at stage2,. [kN/m^2]
+T2 = T1*(rv)^(Gama-1);// temperature at stage 2, [K]
+
+// for process 2-3
+P3 = P2;// pressure at stage 3, [kN/m^2]
+V2 = V1/rv;//[m^3]
+// since pressure is constant in process 2-3 , so using V/T=constant, so
+V3 = V2*(T3/T2);// volume at stage 3, [m^3]
+
+// for process 1-4
+V4 = V1;// [m^3]
+P4 = P3*(V3/V4)^(Gama)
+// since in stage 1-4 volume is constant, so P/T=constant,
+T4 = T1*(P4/P1);// temperature at stage 4,[K]
+
+mprintf('\n (a) P1 = %f kN/m^2, t1 = %f C,\n P2 = %f kN/m^2, t2 = %f C,\n P3 = %f kN/m^2, t3 = %f C,\n P4 = %f kN/m^2, t4 = %f C\n',P1,T1-273,P2,T2-273,P3,T3-273,P4,T4-273);
+
+// (b)
+W = cp*(T3-T2)-cv*(T4-T1);// work done, [kJ]
+mprintf('\n (b) The work done is = %f kJ\n',W);
+
+// (c)
+TE = 1-(T4-T1)/((T3-T2)*Gama);// thermal efficiency
+mprintf('\n (c) The thermal efficiency is = %f percent\n',TE*100);
+
+// (d)
+PW = cp*(T3-T2)+R*(T3-T4)/(Gama-1);// positive work done
+WR = W/PW;// work ratio
+mprintf('\n (d) The work ratio is = %f\n',WR);
+
+// (e)
+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]
+mprintf('\n (e) The mean effefctive pressure is = %f kN/m^2\n',Pm);
+
+// (f)
+CE = (T3-T1)/T3;// carnot efficiency
+mprintf('\n (f) The carnot efficiency is = %f percent\n',CE*100);
+
+// value of t2 printed in the book is incorrect
+
+// End
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