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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3733/CH24 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3733/CH24')
36 files changed, 1393 insertions, 0 deletions
diff --git a/3733/CH24/EX24.1/Ex24_1.sce b/3733/CH24/EX24.1/Ex24_1.sce new file mode 100644 index 000000000..902c6483a --- /dev/null +++ b/3733/CH24/EX24.1/Ex24_1.sce @@ -0,0 +1,30 @@ +// Example 24_1
+clc;funcprot(0);
+//Given data
+P_1=1;// bar
+P_2=5;// bar
+T_1=27+273// K
+T_3=650+273;// K
+C_p=1;// kJ/kg.°C
+//C_p=C_pg=C_pa;
+r=1.4;//The specific heat ratio
+m=5;//kg/s
+//Air-fuel ratio,AF_r=m_air/m_fuel
+AF_r=60/1;
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+
+//Calculation
+//T'2=T_2a;T'4=T_4a;
+T_2a=T_1*(P_2/P_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// Modified equation in K
+T_4a=T_3*(P_1/P_2)^((r-1)/r);// K
+T_4=T_3-(n_t*(T_3-T_4a));// Modified equation in K
+n_th=(((AF_r+1)*(T_3-T_4))-(AF_r*(T_2-T_1)))/((AF_r+1)*(T_3-T_2));
+n_th=n_th*100;// %
+printf('The thermal efficiency of the cycle,n_th=%0.0f percentage\n',n_th);
+W=(C_p*(1+60)*(T_3-T_4))-(C_p*60*(T_2-T_1));//kJ/kg of fuel
+P=(W*m)/1000;// MW
+printf('The power generating capacity of the plant,P=%0.1f MW\n',P);
+// The answer vary due to round off error
+
diff --git a/3733/CH24/EX24.10/Ex24_10.sce b/3733/CH24/EX24.10/Ex24_10.sce new file mode 100644 index 000000000..cf60e019d --- /dev/null +++ b/3733/CH24/EX24.10/Ex24_10.sce @@ -0,0 +1,46 @@ +// Example 24_10
+clc;funcprot(0);
+//Given data
+T_1=20+273;// K
+p_1=1;// bar
+T_6=700+273;// K
+p_r=6;// Pressure ratio
+e=0.7;// The effectiveness of regenerator
+m_air=200;//Air flow through the plant in kg/sec
+n_c=0.82;// Isentropic efficiency of both compressors
+n_t=0.92;// Isentropic efficiency of turbine
+n_com=0.96;// Combustion efficiency
+n_m=0.96;// Mechanical efficiency
+n_g=0.95;// Generation efficiency
+CV=35000;// kJ/kg
+C_p=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+p_2=p_r*p_1;// bar
+p_i=sqrt(p_1*p_2);// bar
+T_2a=T_1*(p_i/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_3=T_1;// K
+T_4a=T_3*(p_2/p_i)^((r-1)/r);// K
+T_4=T_2;// K (as n_c1=n_c2)
+T_7a=T_6*((p_1/p_2)^((r-1)/r));// K
+T_7=T_6-(n_t*(T_6-T_7a));// K
+T_5=(e*(T_7-T_4))+T_4;// K
+function[X]=m_f(y)
+ X(1)=(C_p*(1+y(1))*(T_6-T_5))-(y(1)*CV*n_com);
+endfunction
+y=[0.01]
+z=fsolve(y,m_f);
+m_fuel=z(1);
+m_a=1;// kg/kg of air
+m=(m_a/m_fuel);//Air fuel ratio
+n_th=(((T_6-T_7)-(2*(T_2-T_1)))/(T_6-T_5))*100;// Thermal efficiency
+W=(m_a*(T_6-T_5)*(n_th/100));// Work done per kg of air in kJ
+W_s=W*m_air;// Work done per sec in kJ/sec
+P=W_s/10^3;// Capacity of the plant in MW
+P_a=W_s*n_m*n_t;// Power available at generation terminals in kW
+F=m_air*3600*m_fuel;// Fuel consumption per hour in kg/hr
+Sfc=F/(P_a);// Specific fuel consumption in kg/kW.hr
+printf('\nAir fuel ratio used=%0.0f:1 \nThermal efficiency of the cycle=%0.1f percentage \nFuel consumption per hour=%0.0f kg/hr \nSpecific fuel consumption=%0.3f kg/kW.hr',m,n_th,F,Sfc);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.11/Ex24_11.sce b/3733/CH24/EX24.11/Ex24_11.sce new file mode 100644 index 000000000..f40c1946f --- /dev/null +++ b/3733/CH24/EX24.11/Ex24_11.sce @@ -0,0 +1,40 @@ +// Example 24_11
+clc;funcprot(0);
+//Given data
+P=5;// Power plant capacity in MW
+T_1=27+273;// K
+p_1=1;// bar
+T_4=1000;// K
+p_r=5;// Pressure ratio
+n_c=0.85;// Isentropic efficiency of compressor
+n_t=0.90;// Isentropic efficiency of turbine
+n_com=0.95;// Combustion efficiency
+n_m=0.95;// Mechanical efficiency
+n_g=0.92;// Generation efficiency
+Tl=10;// Transmission losses
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.K
+C_pg=1.1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+m=80;// Air fuel ratio
+Cf_t=5000;// Cost of fuel in Rs./tonne
+Oc=5000;// All other charges in rupees
+
+//Calculation
+n_tt=(1-(Tl/100));// Transmission efficiency
+p_2=p_1*p_r;// bar
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_5a=T_4*(p_1/p_2)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+e_g=P*1000;// The energy generated per second in kJ/sec
+m_a=(e_g)/((((1+(1/m))*C_pg*(T_4-T_5))-(C_pa*(T_2-T_1)))*n_m*n_g*n_tt);// kg/sec
+T_3=T_4-((CV*n_com)/(C_pg*(m+1)));// K
+e=(C_pa*(T_3-T_2))/(C_pg*(1+(1/m))*(T_5-T_2));// Effectiveness of regenerator
+Fc=(m_a*3600*(1/m));// The fuel consumption per hour in kg/hr
+Cf=(Fc/1000)*Cf_t;// Cost of fuel per hour in Rs.
+Tc=Cf+Oc;// Total cost to be charged per hour in Rs.
+E_g=e_g*1;// Energy generated in kW-hr
+Ce=Tc/E_g;// Charges of energy per kW-hr in Rs./kWh
+printf('\nThe mass of air flow through the compressor per second=%0.2f kg/sec \nThe effectiveness of regenerator=%0.3f \nThe charges of energy per kW-hr=Rs.%0.2f/kWh',m_a,e,Ce);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.12/Ex24_12.sce b/3733/CH24/EX24.12/Ex24_12.sce new file mode 100644 index 000000000..aff2f1565 --- /dev/null +++ b/3733/CH24/EX24.12/Ex24_12.sce @@ -0,0 +1,34 @@ +// Example 24_12
+clc;funcprot(0);
+//Given data
+P=10;// Power plant capacity in MW
+T_1=300;// K
+T_4=960;// K
+e=0.7;// The effectiveness of regenerator
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.90;// Isentropic efficiency of turbine
+n_com=0.96;// Combustion efficiency
+n_m=0.95;// Mechanical efficiency
+n_g=0.95;// Generation efficiency
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+Cf_t=4000;// Cost of fuel in Rs./tonne
+Oc=3000;// All other charges in rupees
+Q=90;// Heat developed in combustion chamber in %
+
+//Calculation
+p_r=(n_c*n_t*(T_4/T_1))^(r/(2*(r-1)));// Pressure ratio
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_5a=T_4*(1/p_r)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+m_a=(P*1000)/((C_pa*((T_4-T_5)-(T_2-T_1)))*n_com*n_g);
+T_3=T_2+(e*(T_5-T_2));// K
+m_f=(m_a*C_pa*(T_4-T_3))/(CV*n_com*(Q/100));// kg/sec
+Cf=((m_f*3600)/1000)*Cf_t;// Cost of fuel in Rs./hr
+Tc=Cf+Oc;// Total cost in Rs/hr
+Ce=Tc/(P*1000);// Cost of energy generated in Rs/kWh
+m=m_a/m_f;// Air-fuel ratio
+printf('\nThe cost of energy generated=Rs.%0.2f/kWh',Ce);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.13/Ex24_13.sce b/3733/CH24/EX24.13/Ex24_13.sce new file mode 100644 index 000000000..5238fc3d6 --- /dev/null +++ b/3733/CH24/EX24.13/Ex24_13.sce @@ -0,0 +1,49 @@ +// Example 24_13
+clc;funcprot(0);
+//Given data
+T_1=27+273;// K
+p_1=1;// bar
+p_2=4;// bar
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+e=0.75;// The effectiveness of regenerator
+p_lr=0.1;// Pressure loss in regenerator along air side in bar
+p_lcc=0.05;// Pressure loss in the combustion chamber in bar
+n_com=0.90;// Combustion efficiency
+n_m=0.90;// Mechanical efficiency
+n_g=0.95;// Generation efficiency
+m_a=25;// kg/sec
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.K
+C_pg=1.1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+T_4=700+273;// K
+p_atm=1.03;// bar
+
+// Calculation
+p_i=p_2-(p_lr+p_lcc);// Pressure at the inlet of the turbine in bar
+p_e=p_atm+p_lr;// Pressure at the exit of the turbine in bar
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_5a=T_4*(p_e/p_i)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+// Assume m=(m_a/m_f)
+// m=y(1),T_3=y(2)
+function[X]=airfuelratio(y)
+ X(1)=((y(1)+1)*C_pg*(T_4-y(2)))-(CV*n_com);
+ X(2)=((C_pa*(y(2)-T_2))/(e*C_pg*(T_5-T_2)))-(1+(1/y(1)));
+endfunction
+y=[10 100];
+z=fsolve(y,airfuelratio);
+m=z(1);
+T_3=z(2);// K
+W_c=C_pa*(T_2-T_1);// kJ/kg of air
+W_t=C_pg*(1+(m_a/m))*(T_4-T_5);// kJ/kg of air
+W_a=W_t-W_c;// kJ/kg of air
+W=W_a*n_m*n_g;// Work available per kg of air at the terminals of generator in kJ/kg
+P=(m_a*W)/1000;// Power available at the terminals of generator in kJ/kg
+n_o=((W)/((1/m)*CV))*100;// Over all efficiency
+Fr=m_a*3600*(1/m);// Fuel required per hour in kg/hr
+Sfc=Fr/(P*1000);// Specific fuel consumption in kg/kW.hr
+printf('\nThe over all efficiency of the plant=%0.3f percentage \nSpecific fuel consumption=%0.2f kg/kW.hr',n_o,Sfc);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.14/Ex24_14.sce b/3733/CH24/EX24.14/Ex24_14.sce new file mode 100644 index 000000000..95820516c --- /dev/null +++ b/3733/CH24/EX24.14/Ex24_14.sce @@ -0,0 +1,31 @@ +// Example 24_14
+clc;funcprot(0);
+//Given data
+P=800;// Plant capacity in kW
+T_1=15+273;// K
+p_1=1.01;// bar
+T_4=700+273;// K
+p_r=6;// Pressure ratio
+e=0.75;// The effectiveness of regenerator
+p_lr=0.15;// Pressure drop in regenerator in bar
+p_lcc=0.15;// Pressure drop in the combustion chamber in bar
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+C_p=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+p_2=p_r*p_1;// bar
+p_3=p_2-p_lcc;// Pressure at point 4 in bar
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+p_4=p_1+p_lr;// bar
+T_5a=T_4/(p_3/p_4)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+T_3=T_2+(e*(T_5-T_2));// K
+W_c=C_p*(T_2-T_1);// kJ/kg
+W_t=C_p*(T_4-T_5);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+Q_s=C_p*(T_4-T_3);// kJ/kg
+n_th=(W_n/Q_s)*100;// Thermal Efficiency in percentage
+printf('\nThe thermal efficiency of the plant=%0.1f percentage',n_th);
diff --git a/3733/CH24/EX24.15/Ex24_15.sce b/3733/CH24/EX24.15/Ex24_15.sce new file mode 100644 index 000000000..40d368c7d --- /dev/null +++ b/3733/CH24/EX24.15/Ex24_15.sce @@ -0,0 +1,34 @@ +// Example 24_15
+clc;funcprot(0);
+//Given data
+m_a=10;// kg/sec
+p_r=6;// Pressure ratio
+T_1=300;// K
+p_1=1;// bar
+T_6=1073;// K
+e=0.75;// The effectiveness of regenerator
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+C_pa=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+m=1;// kg
+
+//Calculation
+p_3=p_1*p_r;// bar
+p_2=sqrt(p_1*p_3);// bar
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+//W_c=W_c1+W_c2=2*W_c1 (as intercooling is perfect)
+W_c=2*m*C_pa*(T_2-T_1);// kJ/kg
+// As T_3=T_1 and p_r=(p_2/p_1)=(p_3/p_2)
+T_4=T_2;// K
+T_7a=T_6/(p_3/p_1)^((r-1)/r);// K
+T_7=T_6-(n_t*(T_6-T_7a));// K
+W_t=C_pa*(T_6-T_7);// kJ/kg
+T_5=T_4+(e*(T_7-T_4));// K
+Q_s=m*C_pa*(T_6-T_5);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+P=m_a*W_n;//Power capacity of the plant in kW
+n_th=(W_n/Q_s)*100;// Thermal Efficiency in percentage
+printf('\nPower capacity of the plant=%0.0f kW\nThe thermal efficiency of the plant=%0.1f percentage',P,n_th);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.16/Ex24_16.sce b/3733/CH24/EX24.16/Ex24_16.sce new file mode 100644 index 000000000..7cfa34f85 --- /dev/null +++ b/3733/CH24/EX24.16/Ex24_16.sce @@ -0,0 +1,46 @@ +// Example 24_16
+clc;funcprot(0);
+//Given data
+P=5;// Power plant capacity in MW
+T_1=300;// K
+p_1=1;// bar
+T_5=650+273;// K
+p_r=5;// Pressure ratio
+e=0.7;// The effectiveness of regenerator
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+n_com=0.97;// Combustion efficiency
+n_m=0.98;// Mechanical efficiency of both turbines
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.°C
+C_pg=1.145;// kJ/kg.°C
+r_a=1.4;// Specific heat ratio for air
+r_g=1.35;// Specific heat ratio for gases
+
+//Calculation
+p_2=p_1*p_r;// bar
+p_i=sqrt(p_1*p_2);// The intermediate pressure between two compressors
+T_2a=T_1*(p_i)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_3=T_1;// K
+T_4=T_2;// K
+T_7=T_5;// K
+// Work developed by the compressor turbine = Work required to run the compressor
+T_6=T_5-((2*C_pa*(T_2-T_1))/(C_pg*n_m));// K
+T_6a=T_5-((T_5-T_6)/n_t1);// K
+p_3=p_2/((T_5/T_6a)^(r_g/(r_g-1)));// K
+T_8a=T_7*(p_1/p_3)^((r_g-1)/r_g);// K
+T_8=T_7-(n_t2*(T_7-T_8a));// K
+T_x=T_4+((e*C_pg*(T_8-T_4))/C_pa);// K
+W_net=C_pg*(T_7-T_8)*n_m;//Net Work available per kg of air in kJ/kg of air
+Q_s=C_pg*((T_5-T_x)+(T_7-T_6));// Heat supplied per kg of air kJ/kg of air
+m_f=Q_s/(CV*n_com);// The total mass of fuel in per kg of air flow
+m=1/m_f;// Air fuel ratio
+n_o=(W_net/(m_f*CV))*100;// Over all efficiency
+m_a=(P*1000)/(W_net);// kg/sec
+M_f=m_a*3600*m_f;// Mass of fuel used per hour in kg/hr
+Sfc=M_f/(P*1000*1);// Specific fuel consumption in kg/kW-hr
+printf('\nOver all efficiency of the plant=%0.1f percentage \nMass flow of air through the plant per second=%0.2f kg/sec \nSpecific fuel consumption=%0.3f kg/kW-hr',n_o,m_a,Sfc);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.17/Ex24_17.sce b/3733/CH24/EX24.17/Ex24_17.sce new file mode 100644 index 000000000..488919804 --- /dev/null +++ b/3733/CH24/EX24.17/Ex24_17.sce @@ -0,0 +1,46 @@ +// Example 24_17
+clc;funcprot(0);
+//Given data
+P=1600;// Power plant capacity in kW
+T_1=300;// K
+T_3=1050;// K
+p_1=1;// bar
+T_5=1100;// K
+p_2=5;// bar
+e=0.7;// The effectiveness of regenerator
+n_c=0.8;// Isentropic efficiency of compressor
+n_t1=0.85;// Efficiency of compressor turbine
+n_t2=0.90;// Efficiency of power turbine
+n_com=0.95;// Combustion efficiency
+n_m=0.90;// Mechanical efficiency of both turbines
+n_g=1;// Generation efficiency
+CV=40000;// kJ/kg
+C_pa=1;// kJ/kg.K
+C_pg=1.1;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.35;// Specific heat ratio for gases
+
+//Calculation
+T_2a=T_1*(p_2/p_1)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_4a=T_3*(p_1/p_2)^((r_g-1)/r_g);// K
+T_4=T_3-(n_t1*(T_3-T_4a));// K
+T_6a=T_5*(p_1/p_2)^((r_g-1)/r_g);// K
+T_6=T_5-(n_t2*(T_5-T_6a));// K
+m_a2=P/(C_pg*(T_5-T_6)*n_m*n_g);// kg/sec
+//Power developed by compressor turbine = Power absorbed by compressor
+//m_a1=y(1)
+function[X]=mass(y);
+ X(1)=((y(1)+m_a2)*C_pa*(T_2-T_1))-(y(1)*C_pg*(T_3-T_4)*n_m);
+endfunction
+y=[10];
+z=fsolve(y,mass);
+m_a1=z(1);// kg/sec
+T_y=((m_a1/(m_a1+m_a2))*T_4)+((m_a2/(m_a1+m_a2))*T_6);// The temperature after mixing in °C
+T_x=T_2+((e*C_pg*(T_y-T_2))/C_pa);// K
+m_f=((C_pg*m_a1*(T_3-T_x))+(C_pg*m_a2*(T_5-T_x)))/(CV*n_com);// kg/sec
+n_th=(P/(m_f*CV))*100;// Plant efficiency in percentage
+Sfc=(m_f*3600)/P;// kg/kWh
+Afr=(m_a1+m_a2)/m_f;// Air fuel ratio
+printf('\n(a)Plant efficiency=%0.1f percentage \n(b)Specific fuel consumption=%0.3f kg/kW-hr \n(c)Air fuel ratio=%0.0f:1',n_th,Sfc,Afr);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.18/Ex24_18.sce b/3733/CH24/EX24.18/Ex24_18.sce new file mode 100644 index 000000000..198cca8f8 --- /dev/null +++ b/3733/CH24/EX24.18/Ex24_18.sce @@ -0,0 +1,54 @@ +// Example 24_18
+clc;funcprot(0);
+//Given data
+P=200;// Power plant capacity in MW
+T_6=1000;// K
+T_8=900;// K
+p_1=1;// bar
+T_1=27+273;// K
+p_r=5;// bar
+e=0.7;// The effectiveness of heat exchanger
+n_c=1;// Isentropic efficiency of both compressors
+n_t=0.9;// Efficiency of both turbines
+n_com=0.95;// Combustion efficiency
+n_m=0.92;// Mechanical efficiency of compressor and generator shafts
+CV=40000;// kJ/kg
+C_p=1;// kJ/kg.°C
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+p_2=p_1*p_r;// bar
+p_i=sqrt(p_1*p_2);// bar
+T_7a=T_6*(p_1/p_2)^((r-1)/r);// K
+n_t2=n_t;
+T_7=T_6-(n_t2*(T_6-T_7a));// K
+W_g=C_p*(T_6-T_7)*n_m;//Work done per kg of air in generator-turbine in kJ/kg
+m_2=CV/W_g;// The mass of exhaust gases in kg/sec
+T_2=T_1*(p_i)^((r-1)/r);// K
+W_c=2*C_p*(T_2-T_1);//Work done per kg of air in both compressors in kJ/kg
+T_4=T_2;// K
+// Assume m_1=y(1);T_5=y(2)
+function[X]=massflow(y);
+ X(1)=(m_2*C_p*(y(2)-T_8))-(y(1)*C_p*(T_8-T_4));
+ X(2)=((y(1)*C_p*(T_8-T_4))/(m_2*C_p*(y(2)-T_4)))-(e);
+endfunction
+y=[100 1000];
+z=fsolve(y,massflow);
+T_5=z(2);// K
+m_1=z(1);// kg/sec
+T_9a=T_8/(p_i)^((r-1)/r);// K
+n_t1=n_t;
+T_9=T_8-(n_t1*(T_8-T_9a));// K
+m_c1=(((m_1*(T_8-T_9))/((T_2-T_1)*n_m))-m_1)/2;// Air taken from atmosphere in kg/sec
+m_c2=m_c1+m_1;// kg/sec
+//Assume m_f=y(1)
+function[X]=massoffuel(z);
+ X(1)=((m_c2+z(1))*C_p*(T_5-T_4))/(CV*n_com)-z(1);
+endfunction
+z=[10];
+y=fsolve(z,massoffuel);
+m_f=y(1);// Mass of fuel used per second
+n_o=((P*10^3)/(CV*m_f))*100;// Over all efficiency of the plant in %
+Ctc=(m_1*C_p*(T_8-T_9))/1000;// Compressor-turbine capacity in MW
+printf('\n(a)Air taken from atmosphere per second=%0.0f kg/sec \n(b)Fuel required per second=%0.2f kg/sec \n(c)Over all efficiency of the plant=%0.1f percentage \n(d) Compressor-turbine capacity=%0.0f kW',m_c1,m_f,n_o,Ctc*1000);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.19/Ex24_19.sce b/3733/CH24/EX24.19/Ex24_19.sce new file mode 100644 index 000000000..223ee1487 --- /dev/null +++ b/3733/CH24/EX24.19/Ex24_19.sce @@ -0,0 +1,35 @@ +// Example 24_19
+clc;funcprot(0);
+//Given data
+P=5;// Power plant capacity in MW
+T_1=30+273;// K
+p_1=1;// bar
+T_3=550+273;// K
+p_r=5;// Pressure ratio
+p_3=2.24;// bar
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+C_pa=1;// kJ/kg.°C
+C_pg=1.15;// kJ/kg.°C
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+
+//Calculation
+p_2=p_1*p_r;// bar
+T_5=T_3;// K
+T_2a=T_1*(p_2/p_1)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+W_c=C_pa*(T_2-T_1);// kJ/kg
+T_4a=T_3/(p_2/p_3)^((r_g-1)/r_g);// K
+T_4=T_3-(n_t1*(T_3-T_4a));// K
+T_6a=T_5/(p_3/p_1)^((r_g-1)/r_g);// K
+T_6=T_5-(n_t2*(T_5-T_6a));// K
+W_t=2*C_pg*(T_3-T_4);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+m_a=((P*10^3)/W_n);// kg/sec
+Q_s=(C_pg*(T_3-T_2))+(C_pg*(T_5-T_4));// kJ/kg
+n_o=(W_n/Q_s)*100;// Over all efficiency in %
+printf('\nThe over all efficiency=%0.0f percentage \nThe mass flow rate=%0.1f kg/sec',n_o,m_a);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.2/Ex24_2.sce b/3733/CH24/EX24.2/Ex24_2.sce new file mode 100644 index 000000000..59775490d --- /dev/null +++ b/3733/CH24/EX24.2/Ex24_2.sce @@ -0,0 +1,37 @@ +// Example 24_2
+clc;funcprot(0);
+//Given data
+T_1=300;// K
+P_r=8;// P_r=(p1/p2)
+p_1=1;// bar
+T_4=1080;// K
+m=500;// kg/min
+n_c=0.8;
+n_t=n_c;//Isentropic efficiency of the compressor and turbine
+CV=42000;// kJ/kg
+e=0.6;// The effectiveness of the heat exchanger
+r=1.4;// Specific heat ratio
+C_p=1;// kJ/kg.°C
+//C_p=C_pg=C_pa;
+
+//Calculation
+T_2a=T_1*(P_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// Modified equation in K
+T_5a=T_4*(1/P_r)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+T_3=(e*(T_5-T_2))+T_2;// K
+//m_f=y(1)
+function[X]=Mass(y);
+ X(1)=(y(1)*CV)-(C_p*(1+y(1))*(T_4-T_3));
+endfunction
+y=[0.01]
+z=fsolve(y,Mass);
+m_f=z(1);// kJ/kg of air
+m_a=1;// kg
+q=m_a*(T_3-T_2);//Heat saved in kJ/kg of air
+M=(m*60*q)/CV;// Fuel saved per hour in kg/hr
+W_net=(C_p*(1+m_f)*(T_4-T_5))-(C_p*m_a*(T_2-T_1));// kJ/kg
+P=(m/60)*W_net;// The capacity of the plant in kW
+printf('\nFuel saved per hour=%0.2f kg/hr\nThe capacity of the plant=%0.1f kW',M,P);
+// The answer vary due to round off error
+
diff --git a/3733/CH24/EX24.20/Ex24_20.sce b/3733/CH24/EX24.20/Ex24_20.sce new file mode 100644 index 000000000..b0856b16f --- /dev/null +++ b/3733/CH24/EX24.20/Ex24_20.sce @@ -0,0 +1,58 @@ +// Example 24_20
+clc;funcprot(0);
+//Given data
+P=5;// Power plant capacity in MW
+T_1=15+273;// K
+p_1=1;// bar
+T_4=750+273;// K
+p_r=6;// Pressure ratio
+p_3=2.24;// bar
+e=0.75;// The effectiveness of heat exchanger
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+C_pa=1;// kJ/kg.K
+C_pg=1.15;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+CV=18500;// kJ/kg
+
+//Calculation
+p_2=p_1*p_r;// bar
+p_re=sqrt(p_1*p_2);//Pressure ratio for each turbine in bar
+T_2a=T_1*(p_2/p_1)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_6=T_4;// K
+T_5a=T_4/((p_re)^((r_g-1)/r_g));// K
+T_5=T_4-(n_t1*(T_4-T_5a));// K
+T_7=T_5;// K
+T_3=T_2+(e*(T_7-T_2));// K
+//(i)
+function[X]=massoffuel(y)
+ X(1)=((1+y(1))*C_pg*(T_4-T_3))-(y(1)*CV);
+endfunction
+y=[0.01];
+z=fsolve(y,massoffuel);
+m_f1=z(1);// kg/kg of air
+AF=1/m_f1;// Air fuel ratio
+function[X]=massoffuel1(x)
+ X(1)=(C_pg*((1+m_f1+x(1))*(T_6-T_5)))-(x(1)*CV);
+endfunction
+x=[0.001];
+y=fsolve(x,massoffuel1);
+m_f2=y(1);// kg/kg of air
+W_c=C_pg*(T_2-T_1);// kJ/kg of air
+W_t1=C_pg*(1+(m_f1))*(T_4-T_5);// kJ/kg of air
+W_t2=C_pg*(1+m_f1+m_f2)*(T_6-T_7);// kJ/kg of air
+W_t=W_t1+W_t2;// kJ/kg of air
+W_n=W_t-W_c;// kJ/kg of air
+//(ii)
+Q_s=(m_f1+m_f2)*CV;// kJ/kg of air
+n_th=(W_n/Q_s)*100;// Thermal efficiency of the cycle
+//(iii)
+m_a=((P*10^3)/W_n);// kg/sec
+//(iv)
+F=m_a*(m_f1+m_f2)*3600;// Fuel required per hour in kg/hr
+printf('\n(i)Cycle efficiency=%0.1f percentage \n(ii)Air supplied to the plant=%0.1f kg/sec \n(iii)A:F ratio=%0.1f:1 \n(iv)Fuel consumption of the plant=%0.0f kg/hr',n_th,AF,m_a,F);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.21/Ex24_21.sce b/3733/CH24/EX24.21/Ex24_21.sce new file mode 100644 index 000000000..c3d22f515 --- /dev/null +++ b/3733/CH24/EX24.21/Ex24_21.sce @@ -0,0 +1,39 @@ +// Example 24_21
+clc;funcprot(0);
+//Given data
+T_1=25+273;// K
+p_1=1;// bar
+T_6=1250+273;// K
+p_3=9;// bar
+n_c=0.83;// Isentropic efficiency of both compressors
+n_c1=n_c;
+n_c2=n_c;
+n_t=0.83;// Isentropic efficiency of both turbines
+n_t1=n_t;
+n_t2=n_t;
+C_pa=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+m_a=16.5;// kg/sec
+
+//Calculation
+p_2=sqrt(p_1*p_3);// bar
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c1)+T_1;// K
+T_8=T_6;// K
+T_4=T_2;// K
+T_7a=T_6/(p_3/p_2)^((r-1)/r);// K
+T_7=T_6-(n_t1*(T_6-T_7a));// K
+T_9=T_7;// K
+W_c=2*C_pa*(T_2-T_1);// kJ/kg
+W_t=2*C_pa*(T_6-T_7);// kJ/kg
+W_n=W_t-W_c;// kJ/kg
+T_5=T_7;
+//When the ideal regeneration is given,then
+e=1;// Effectiveness
+Q_s=2*C_pa*(T_6-T_5);// kJ/kg
+//(i)
+n_th=(W_n/Q_s)*100;//The thermal efficiency in %
+//(ii)
+P=W_n*m_a;// Power developed by the plant in kW
+printf('\n(i)The thermal efficiency of the plant=%0.1f percentage \n(ii)Power developed by the plant=%0.2f kW',n_th,P);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.22/Ex24_22.sce b/3733/CH24/EX24.22/Ex24_22.sce new file mode 100644 index 000000000..47916767f --- /dev/null +++ b/3733/CH24/EX24.22/Ex24_22.sce @@ -0,0 +1,37 @@ +// Example 24_22
+clc;funcprot(0);
+//Given data
+T_1=290;// K
+p_1=1.01;// bar
+T_3=650+273;// K
+p_r=8;// Pressure ratio
+n_c=0.8;// Isentropic efficiency of compressor
+n_t1=0.85;// Isentropic efficiency of H.P turbine
+n_t2=0.83;// Isentropic efficiency of L.P turbine
+C_pa=1;// kJ/kg.K
+C_pg=1.15;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+m_a=10;// The air flow through the compressor in kg/sec
+
+//Calculation
+p_2=p_1*p_r;// bar
+T_2a=T_1*(p_r)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+W_c=1*C_pa*(T_2-T_1);//Work input to the compressor in kJ/kg
+W_t1=W_c;// kJ/kg
+T_4=T_3-(W_t1/C_pg);// K
+T_4a=T_3-((T_3-T_4)/n_t1);// K
+p_3=p_2/((T_3/T_4a)^(r_g/(r_g-1)));// bar
+p_re=p_3/p_1;// The pressure ratio of expansion in the power turbine
+T_5a=T_4/(p_3/p_1)^((r_g-1)/r_g);// K
+dT_45=n_t2*(T_4-T_5a);// (dT_45=T_4-T_5) K
+W_t2=C_pg*(dT_45);//Work developed by power turbine in kJ/kg
+W_net=W_t2;// The net work done per kg of air in kJ/kg
+W_t=W_t1+W_t2;// Total work done per in kJ/kg
+W_r=W_t2/W_t;// Work ratio
+Q_s=C_pa*(T_3-T_2);// kJ/kg
+n_th=(W_t2/Q_s)*100;// Thermal efficiency in %
+P=W_t2*m_a;// Power capacity of the plant in kW
+printf('\nThe power developed by the unit=%0.0f kW \nThe thermal efficiency=%0.0f percentage \nWork ratio=%0.1f',P,n_th,W_r);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.23/Ex24_23.sce b/3733/CH24/EX24.23/Ex24_23.sce new file mode 100644 index 000000000..05337e968 --- /dev/null +++ b/3733/CH24/EX24.23/Ex24_23.sce @@ -0,0 +1,35 @@ +// Example 24_23
+clc;funcprot(0);
+//Given data
+p_1=1;// bar
+p_2=5;// bar
+p_3=2.5;// bar
+T_1=300;// K
+T_3=900;// K
+T_5=T_3;// K
+m_a=10;// kg/sec
+CV=33500;// kJ/kg
+C_p=1;// kJ/kg.°C
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+T_2=T_1*(p_2/p_1)^((r-1)/r);// K
+T_4=T_3/(p_2/p_3)^((r-1)/r);// K
+T_6=T_5/(p_2/p_3)^((r-1)/r);// K
+function[X]=massoffuel(y)
+ X(1)=((1+y(1))*C_p*(T_3-T_2))-(y(1)*CV);
+endfunction
+y=[0.01];
+z=fsolve(y,massoffuel);
+m_f1=z(1);// kg/kg of air
+function[X]=massoffuel1(x)
+ X(1)=(C_p*((1+m_f1+x(1))*(T_5-T_4)))-(x(1)*CV);
+endfunction
+x=[0.001];
+y=fsolve(x,massoffuel1);
+m_f2=y(1);// kg/kg of air
+W_n=((m_a*(1+m_f1)*C_p*(T_3-T_4)))+((m_a*(1+m_f1+m_f2)*C_p*(T_5-T_6)))-(m_a*C_p*(T_2-T_1));// kW
+n_g=100;//The generator efficiency is considered 100%
+n_th=(W_n/(m_a*(m_f1+m_f2)*CV))*100;// The efficiency of the plant in %
+printf('\nThe thermal efficiency of the plant=%0.1f percentage \nPower generating capacity=%0.0f kW',n_th,W_n);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.24/Ex24_24.sce b/3733/CH24/EX24.24/Ex24_24.sce new file mode 100644 index 000000000..80988f573 --- /dev/null +++ b/3733/CH24/EX24.24/Ex24_24.sce @@ -0,0 +1,30 @@ +// Example 24_24
+clc;funcprot(0);
+//Given data
+T_1=300;// K
+p_1=1;// bar
+T_4=870+273;// K
+p_r=6;// Pressure ratio
+e=0.65;// The effectiveness of heat exchanger
+n_c=0.8;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+n_g=0.95// Generator efficiency
+m_a=5;// kg/sec
+C_p=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+//(a)
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_5a=T_4/(p_r)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+W_n=m_a*C_p*((T_4-T_5)-(T_2-T_1))*n_g;// kW
+//(b)
+T_3=T_2+(e*(T_5-T_2));// K
+n_th=((C_p*((T_4-T_5)-(T_2-T_1)))/(C_p*(T_4-T_3)))*100;// Thermal efficiency of the plant in %
+T_6=T_5-(T_3-T_2);// K
+//(c)
+Q=(m_a*60)*C_p*(T_6-T_1);// KJ/min
+printf('\n(a)Power output of the plant=%0.2f kW \n(b)Thermal efficiency of the plant=%0.1f percentage \n(c)Heat carried by the exhaust gases=%0.0f kJ/min',W_n,n_th,Q);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.25/Ex24_25.sce b/3733/CH24/EX24.25/Ex24_25.sce new file mode 100644 index 000000000..5582f703d --- /dev/null +++ b/3733/CH24/EX24.25/Ex24_25.sce @@ -0,0 +1,33 @@ +// Example 24_25
+clc;funcprot(0);
+//Given data
+p_r=4.5;// Pressure ratio
+m_a=82;// kg/min
+m_f=1.4;// kg/min
+W_o=200;// kW
+W_c=230// kW
+p_1=1;// bar
+T_1=15+273;// K
+T_3=765+273;// K
+r_c=1.4;// The index of compression
+r_e=1.34;// The index of expansion
+C_pa=1;// kJ/kg.K
+C_pg=1.13;// kJ/kg.K
+n_m=0.98;// Mechanical efficiency of the compressor
+
+//Calculation
+W_t=(W_o+W_c)/n_m;// kW
+m_a=(m_a)/60;// kg/sec
+m_f=(m_f)/60;// kg/sec
+AF=m_a/m_f;// Air fuel ratio
+//(a)
+T_2a=T_1*(p_r)^((r_c-1)/r_c);// K
+n_c=(m_a*C_pa*((T_2a-T_1)/W_c))*100;// Isentropic efficiency of compressor in %
+//(b)
+T_4a=T_3/(p_r)^((r_e-1)/r_e);// K
+n_t=(W_t/((m_a+m_f)*C_pg*(T_3-T_4a)))*100;// Isentropic efficiency of turbine in %
+//(c)
+T_2=T_1+((T_2a-T_1)/(n_c/100));// K
+n_o=(W_o/((m_a+m_f)*C_pg*(T_3-T_2)))*100;// The over all efficiency of the plant in %
+printf('\n(a)Isentropic efficiency of compressor=%0.0f percentage \n(b)Isentropic efficiency of turbine=%0.1f percentage \n(c) The over all efficiency of the plant=%0.1f percentage',n_c,n_t,n_o);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.26/Ex24_26.sce b/3733/CH24/EX24.26/Ex24_26.sce new file mode 100644 index 000000000..07ffe9cd4 --- /dev/null +++ b/3733/CH24/EX24.26/Ex24_26.sce @@ -0,0 +1,25 @@ +// Example 24_26
+clc;funcprot(0);
+//Given data
+T_1=303;// K
+p_1=0.9;// bar
+p_2=4.5;// bar
+T_3=1000+273;// K
+p_3=1.1;// bar
+e=0.8;// Effectiveness of heat exchanger
+n_c=0.85;// Isentropic efficiency of compressor
+n_t=0.80;// Isentropic efficiency of turbine
+m_a=5;// kg/sec
+C_p=1.005;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_4a=T_3/(p_2/p_3)^((r-1)/r);// K
+T_4=T_3-(n_t*(T_3-T_4a));// K
+T_5=T_2+(e*(T_4-T_2));// K
+n_th=(((T_3-T_4)-(T_2-T_1))/(T_3-T_5))*100;// The thermal efficiency of the system in %
+P=m_a*C_p*((T_3-T_4)-(T_2-T_1));// The power developed by the system in kW
+printf('\nThe thermal efficiency of the system=%0.0f percentage \nThe power developing capacity of the system=%0.1f kW',n_th,P);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.27/Ex24_27.sce b/3733/CH24/EX24.27/Ex24_27.sce new file mode 100644 index 000000000..314c88714 --- /dev/null +++ b/3733/CH24/EX24.27/Ex24_27.sce @@ -0,0 +1,34 @@ +// Example 24_27
+clc;funcprot(0);
+//Given data
+T_1=21+273;// K
+T_4=925+273;// K
+n_c=0.86;// Isentropic efficiency of compressor
+n_t1=0.85;// Isentropic efficiency of H.P turbine
+n_t2=0.87;// Isentropic efficiency of L.P turbine
+e=0.75;// Effectiveness of heat exchanger
+n_com=0.98;// Combustion efficiency
+n_m=0.99;// Mechanical efficiency of compressor and H.P turbine assembly
+P=2040;// kW
+C_pa=1.005;// kJ/kg.K
+r=1.4;// Specific heat ratio for air
+m=1;// kg
+
+//Calculation
+// p_r=p_1/p_2;
+p_r1=7;// Pressure ratio
+T_2a=T_1*(p_r1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+W_c=1*C_pa*(T_2-T_1);//Work input to the compressor in kJ/kg
+W_t1=W_c/n_m;// kJ/kg
+T_5=T_4-(W_t1/(m*C_pa))// K
+T_5a=T_4-((T_4-T_5)/n_t1);// K
+p_r2=(T_4/T_5a)^(r/(r-1));// Pressure ratio(p_2/p_3)
+p_r3=(1/p_r1)*(p_r2);// Pressure ratio(p_3/p_1)
+T_6a=T_5*(p_r3)^((r-1)/r);// K
+T_6=T_5-((T_5-T_6a)*n_t2);// K
+T_3=T_2+(e*(T_6-T_2));// K
+m_a=(P/(C_pa*(T_5-T_6)));// kg/sec
+n_th=(P)/(m_a*C_pa*(T_4-T_3)*n_com)*100;// The thermal efficiency of the plant in %
+printf('\nThe air flow rate=%0.2f kg/sec \nThe thermal efficiency of the plant=%0.1f percentage',m_a,n_th);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.28/Ex24_28.sce b/3733/CH24/EX24.28/Ex24_28.sce new file mode 100644 index 000000000..593272ded --- /dev/null +++ b/3733/CH24/EX24.28/Ex24_28.sce @@ -0,0 +1,42 @@ +// Example 24_28
+clc;funcprot(0);
+//Given data
+T_1=15+273;// K
+p_1=1;// bar
+T_3=680+273;// K
+p_2=5;// bar
+n_c=0.76;// Isentropic efficiency of compressor
+n_t=0.86;// Isentropic efficiency of both turbines
+m_a=23;// kg/sec
+C_pa=1.005;// kJ/kg.K
+C_pg=1.128;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.34;// Specific heat ratio for gases
+CV=42000;// kJ/kg
+
+//Calculation
+//First considering C-TB_1
+T_2a=T_1*(p_2/p_1)^((r_a-1)/r_a);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+// Assume m_r1=m_a1/m_f1
+m_r1=(CV/(C_pg*(T_3-T_2)))-1;
+T_4a=T_3/(p_2/p_1)^((r_g-1)/r_g);// K
+T_4=T_3-((T_3-T_4a)*n_t);// K
+m_f1=(m_a*C_pa*(T_2-T_1))/((m_r1+1)*C_pg*(T_3-T_4));// kg/sec
+m_a1=m_r1*m_f1;// kg/sec
+m_a2=m_a-m_a1;// kg/sec
+// Now considering G-TB_2
+//m_f2=y(1)
+function[X]=massoffuel(y)
+ X(1)=((m_a2+y(1))*C_pg*(T_3-T_2))-(y(1)*CV);
+endfunction
+y=[0.01];
+z=fsolve(y,massoffuel);
+m_f2=z(1);// kg/kg of air
+m_r2=m_a2/m_f2;
+W_2=(m_a2+m_f2)*C_pg*(T_3-T_4);//Work developed by TB_2 kW
+W_1=m_a1*C_pa*(T_2-T_1);// The capacity of TB_1 to run the compressor in kW
+m_f=(m_f1+m_f2)*60;// kg/min
+n_th=(W_2/((m_f/60)*CV))*100;// The thermal efficiency of the plant in %
+printf('\n\The power output of the plant=%0.0f kW \nThe thermal efficiency of the plant=%0.1f percentage',W_2,n_th);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.29/Ex24_29.sce b/3733/CH24/EX24.29/Ex24_29.sce new file mode 100644 index 000000000..bbb4c542d --- /dev/null +++ b/3733/CH24/EX24.29/Ex24_29.sce @@ -0,0 +1,55 @@ +// Example 24_29
+clc;funcprot(0);
+//Given data
+T_1=15+273;// K
+p_1=1;// bar
+T_5=1000;// K
+dp_in=0.07;// bar
+dp_re=0.1;// bar
+R_c1=2;// Compression ratio
+n_c=0.80;// Efficiency of compressor
+n_c1=n_c;
+n_c2=n_c;
+dp_com=0.15;// bar
+dp_rh=0.1;// bar
+n_t1=0.87;// Efficiency of turbine 1
+n_t2=0.7;// Efficiency of turbine 2
+e=0.75;// Effectiveness of heat exchanger
+n_com=0.98;// Combustion efficiency
+n_m=0.99;// Mechanical efficiency of compressor-turbine
+m_a=20;// kg/sec
+C_pa=1;// kJ/kg.K
+C_pg=1.1;// kJ/kg.K
+r_a=1.4;// Specific heat ratio for air
+r_g=1.33;// Specific heat ratio for gases
+CV=43.5;// MJ/kg
+
+//Calculation
+p_2=p_1*R_c1;// bar
+p_3=p_2-dp_in;// bar
+p_4=2*p_3;// bar
+p_5=p_4-dp_in-dp_re;// bar
+p_8=1+dp_rh;// bar
+T_2=T_1+((T_1/n_c1)*(((R_c1)^((r_a-1)/r_a))-1));// K
+T_3=T_1;// K
+T_4=T_3+((T_3/n_c2)*(((R_c1)^((r_a-1)/r_a))-1));// K
+// as T_4-T_3=T_2-T_1
+W_1=2*m_a*C_pa*(T_2-T_1);// Power required to run the compressor in kW
+W_t1=W_1/n_m;// Power developed by compressor turbine in kW
+W_t1=W_t1/m_a;// The work developed by the turbine per kg of air in kJ/kg
+dT_56=W_t1/C_pg;//(dT_56=T_5-T_6) K
+R_t1=1/(1-((dT_56/(T_5*n_t1))))^(r_a/(r_a-1));
+p_6=p_5/R_t1;// bar
+p_7=p_6-dp_rh;// bar
+R_t2=p_7/p_8;// bar
+T_7=T_5;// K
+dT_78=T_7*n_t2*(1-((1/R_t2)^((r_a-1)/r_a)));// K
+T_8=T_7-dT_78;// K
+W=m_a*C_pa*(T_7-T_8);// Net output of the plant in kW
+T_9=T_4+(e*(T_8-T_4));// K
+Q_s=C_pa*((T_5-T_9)+(dT_56));// The total heat supplied in the plant per kg of air in kJ/kg
+m_f=((m_a*Q_s)/(CV*10^3*n_com))*3600;// The mass of fuel supplied in kg/hr
+Sfc=m_f/W;// Specific fuel consumption in kg/kWh
+n_th=(W/(Q_s*m_a))*100;// Thermal efficiency in %
+printf('\nThe specific fuel consumption of the plant=%0.2f kg/kWh \nPlant capacity=%0.0f kW \nOver all efficiency of the plant=%0.1f',Sfc,W,n_th);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.3/Ex24_3.sce b/3733/CH24/EX24.3/Ex24_3.sce new file mode 100644 index 000000000..0ddb909bc --- /dev/null +++ b/3733/CH24/EX24.3/Ex24_3.sce @@ -0,0 +1,29 @@ +// Example 24_3
+clc;funcprot(0);
+//Given data
+T_1=288;// K
+P_r=6;// P_r=p1/p2
+T_3=1000;// K
+m=2;// tonnes/hr
+n_c=0.85;
+n_t=0.90;//Isentropic efficiencies of the compressor and turbine
+CV=46500;// kJ/kg
+e=0.6;// The effectiveness of the heat exchanger
+r=1.4;// Specific heat ratio
+C_p=1;// kJ/kg.°C
+//C_p=C_pg=C_pa
+
+//Calculation
+T_2a=T_1*(P_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_4a=T_3/(P_r)^((r-1)/r);// K
+T_4=T_3-(n_t*(T_3-T_4a));// K
+W_c=C_p*(T_2-T_1);// kJ/kg
+W_t=C_p*(T_3-T_4);// kJ/kg
+Q_a=C_p*(T_3-T_2);// kJ/kg
+n_th=((W_t-W_c)/Q_a)*100;//Cycle efficiency
+W_s=W_t-W_c;// kJ/kg
+P=((m*1000)/3600)*CV*n_th/100*n_t*n_c;// kW
+P=P/1000;//MW
+printf('\n Cycle efficiency=%0.1f percentage \nThe specific work output=%0.0f kJ/kg',n_th,W_s);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.30/Ex24_30.sce b/3733/CH24/EX24.30/Ex24_30.sce new file mode 100644 index 000000000..57bfe52cd --- /dev/null +++ b/3733/CH24/EX24.30/Ex24_30.sce @@ -0,0 +1,51 @@ +// Example 24_30
+clc;funcprot(0);
+//Given data
+T_1=15+273;// K
+p_1=1;// bar
+R_c=5;//Compression ratio
+T_3=800+273;// K
+T_9=265+273;// K
+W=625;// kW
+e=0.75;// Effectiveness of heat exchanger
+n_c=0.86;// Isentropic efficiency of compressor
+n_t=0.86;// Isentropic efficiency of both turbine
+n_t1=n_t;
+n_t2=n_t;
+m_a=5.85;// kg/sec
+C_p=1;// kJ/kg.K
+C_pa=C_p;
+C_pg=C_p;
+r=1.4;// Specific heat ratio
+
+//Calculation
+R_t1=R_c;
+R_t2=R_c;
+dT_21=(T_1/n_c)*(((R_c)^((r-1)/r))-1);// K
+T_2=T_1+dT_21;// K
+W_c=m_a*C_pa*(T_2-T_1);// The work done in the compressor in kW
+dT_34=T_3*n_t1*(1-((1/R_t1)^((r-1)/r)));//(T_3-T_4) K
+m_a1=W_c/(dT_34);// kg/sec
+P_ta=(m_a1/m_a)*100;// Percentage of total air supplied to turbine 1 in %
+m_a2=m_a-m_a1;// kg/sec
+// Assume T_7=y(1); T_8=y(2);
+function[X]=Temperature8(y)
+ X(1)=(m_a*C_pg*(y(1)-T_2))-(m_a*C_pg*(y(2)-T_9));
+ X(2)=((y(1)-T_2)/(y(2)-T_2))-e;
+endfunction
+y=[100 100];
+z=fsolve(y,Temperature8);
+T_8=z(2);// K
+T_7=z(1);// K
+// Assume T_5=x(1); T_6=x(2);
+function[Y]=Temperature5(x)
+ Y(1)=(x(1)*n_t2*(1-((1/R_t2)^((r-1)/r))))-(x(1)-x(2));
+ Y(2)=(m_a2*C_pa*(x(1)-x(2)))-W;
+endfunction
+x=[100 100];
+q=fsolve(x,Temperature5);
+T_5=q(1);// K
+T_6=q(2);// K
+n_th=(W/(((m_a1*C_pa*(T_3-T_7))+(m_a2*C_pa*(T_5-T_6)))))*100;//The over all efficiency of the plant in %
+printf('\nPercentage of total air passed to the compressor turbine=%0.1f percentage \nThe combined temperature of of the exhaust gases entering into the heat exchanger,T_8=%0.0f K \nThe temperature of gases entering into the power turbine,T_5=%0.0f K \nThe over all efficiency of the plant=%0.1f percentage',P_ta,T_8,T_5,n_th);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.31/Ex24_31.sce b/3733/CH24/EX24.31/Ex24_31.sce new file mode 100644 index 000000000..797df8e47 --- /dev/null +++ b/3733/CH24/EX24.31/Ex24_31.sce @@ -0,0 +1,47 @@ +// Example 24_31
+clc;funcprot(0);
+//Given data
+T_1=288;// K
+p_1=1;// bar
+R_c=2.5;// Pressure ratio of each compressor stage
+R_c1=R_c;
+R_c2=R_c;
+T_3=300// K
+T_5=1000;// K
+W_2=100;// kW/kg of air
+p_l1=0.2;// Pressure loss in air side of H.P and main combustion chamber in bar
+p_l2=0.1;// Pressure loss in reheat combustion chamber in bar
+p_l3=0.05;// Pressure loss in intercooler in bar
+n_c=0.85;// Isentropic efficiency of compressor
+n_c1=n_c;
+n_c2=n_c;
+n_t1=0.88;// Isentropic efficiency of turbine 1
+n_t2=0.85;// Isentropic efficiency of turbine 2
+m_a=5.85;// kg/sec
+C_p=1;// kJ/kg.K
+n_o=0.30;// The over all efficiency of the plant
+r=1.4;// Specific heat ratio
+
+//Calculation
+T_2=T_1+(T_1/n_c1)*(((R_c1)^((r-1)/r))-1);// K
+p_2=R_c*p_1;// bar
+p_3=p_2-p_l3;// bar
+T_4=T_3+(T_3/n_c2)*(((R_c1)^((r-1)/r))-1);// K
+p_4=p_3*p_2;// ba
+T_1=T_3;
+W_1=C_p*((T_2-T_1)+(T_4-T_3));//The work required to compress one kg of air in kJ/kg
+n_m=1;// Mechanical efficiency (Assumed)
+T_6=T_5-(W_1/C_p);// K
+R_t1=1/(1-(((T_5-T_6)/(T_5*n_t1))))^(r/(r-1));// Pressure ratio in turbine 1
+p_5=p_4-p_l1;// bar
+p_6=p_5/R_t1;// bar
+p_7=p_6-p_l2;// bar
+T_7=T_5;// K
+T_8=T_7-(W_2/C_p);// K
+R_t2=1/(1-(((T_7-T_8)/(T_7*n_t2))))^(r/(r-1));// Pressure ratio in turbine 2
+p_8=p_7/R_t2;// bar
+p_m=p_8-p_1;// Maximum pressure loss in H.E towards gas side in bar
+T_9=T_5-(((T_7-T_8)/(n_o))-(T_7-T_6));// K
+e=(T_9-T_4)/(T_8-T_4);// The effectiveness of heat exchanger
+printf('\nThe effectiveness of heat exchanger=%0.3f \nMaximum pressure loss in H.E towards gas side=%0.2f bar',e,p_m);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.32/Ex24_32.sce b/3733/CH24/EX24.32/Ex24_32.sce new file mode 100644 index 000000000..93d2464d8 --- /dev/null +++ b/3733/CH24/EX24.32/Ex24_32.sce @@ -0,0 +1,37 @@ +// Example 24_32
+clc;funcprot(0);
+//Given data
+T_1=15+273;// K
+p_1=1;// bar
+p_r=6;// Pressure ratio
+T_4=750+273;// K
+e=0.75;// Effectiveness of heat exchanger
+n_c=0.80;// Isentropic efficiency of compressor
+n_t=0.85;// Isentropic efficiency of turbine
+C_pa=1;// kJ/kg.K
+C_pg=1;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+p_2=p_1*p_r;// bar
+p_3=sqrt(p_1*p_2);// bar
+p_r1=p_2/p_3;// Pressure ratio
+p_r2=p_r1;
+T_5a=T_4/(p_r1)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+T_6=T_4;// K
+T_7=T_5;// K
+T_3=T_2+(e*(T_7-T_2));// K
+W_c=C_pa*(T_2-T_1);// The work of compression in kJ/kg
+W_t=2*C_pg*(T_4-T_5);// The work developed by both turbines in kJ/kg
+W_n=W_t-W_c;// Net work in kJ/kg
+Q_1=C_pg*(T_4-T_3);// kJ/kg
+Q_2=C_pa*(T_6-T_5);// kJ/kg
+Q_s=Q_1+Q_2;// The total heat supplied in kJ/kg
+W_r=W_n/W_t;// Work ratio
+n_p=(W_n/Q_s)*100;// The plant efficiency in %
+printf('\nEfficiency of the plant=%0.1f percentage \nWork ratio=%0.4f',n_p,W_r);
+// The answer vary due to round off error
+
diff --git a/3733/CH24/EX24.33/Ex24_33.sce b/3733/CH24/EX24.33/Ex24_33.sce new file mode 100644 index 000000000..3292987a1 --- /dev/null +++ b/3733/CH24/EX24.33/Ex24_33.sce @@ -0,0 +1,41 @@ +// Example 24_33
+clc;funcprot(0);
+//Given data
+p_1=1;// bar
+p_2=9;// bar
+T_1=25+273;// K
+T_6=1250+273;// K
+e=0.83;// The effectiveness of regenerator
+n_c=0.83;// Isentropic efficiency of both compressors
+n_t=0.83;// Isentropic efficiency of both turbines
+n_com=0.95;// Combustion efficiency
+CV=42;// MJ/kg
+C_p=1;// kJ/kg.K
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+T_8=T_6;// K
+n_c1=n_c;
+n_c2=n_c;
+n_t1=n_t;
+n_t2=n_t;
+p_i=sqrt(p_1*p_2);// bar
+T_2a=T_1*(p_i/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_3=T_1;// K
+T_4=T_2;// K
+T_7a=T_6/(p_2/p_i)^((r-1)/r);// K
+T_7=T_6-(n_t1*(T_6-T_7a));// K
+T_8=T_6;// K
+T_9=T_7;// K
+T_5=T_2+(e*(T_9-T_4));// K
+W_c=2*C_p*(T_2-T_1);// The work developed by both compressors in kJ/kg
+W_t=2*C_p*(T_6-T_7);// The work developed by both turbines in kJ/kg
+W_n=W_t-W_c;// Net work in kJ/kg
+W_r=W_n/W_t;// Work ratio
+Q_1=C_p*(T_6-T_5);// kJ/kg
+Q_2=C_p*(T_8-T_7);// kJ/kg
+Q_s=Q_1+Q_2;// The total heat supplied in kJ/kg
+n=(W_n/Q_s)*100;// The plant efficiency in %
+printf('\nThermal efficiency of the plant=%0.0f percentage \nWork ratio=%0.2f',n,W_r);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.34/Ex24_34.sce b/3733/CH24/EX24.34/Ex24_34.sce new file mode 100644 index 000000000..0bcc40438 --- /dev/null +++ b/3733/CH24/EX24.34/Ex24_34.sce @@ -0,0 +1,35 @@ +// Example 24_34
+clc;funcprot(0);
+//Given data
+p_1=1;// bar
+p_2=8;// bar
+T_1=300;// K
+T_3=1000;// K
+CV=40;// MJ/kg
+W_2=500;// kW
+C_pa=1;// kJ/kg.°C
+C_pg=1;// kJ/kg.°C
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+p_r=(p_2/p_1);// Pressure ratio
+T_2=T_1*(p_r)^((r-1)/r);// K
+T_4=T_3/(p_r)^((r-1)/r);// K
+// Assume m_a=y(1);m_f=y(2);// m_g1=y(3);m_g2=y(4)
+function[X]=mass(y)
+ X(1)=(y(1)+y(2))-(y(3)+y(4));
+ X(2)=(y(4)*C_pg*(T_3-T_4))-(W_2);
+ X(3)=(y(1)*C_pa*(T_2-T_1))-(y(3)*C_pg*(T_3-T_4));
+ X(4)=(y(2)*CV*10^3)-((y(1)+y(2))*C_pg*(T_3-T_2));
+endfunction
+y=[1 0.1 1 1];
+z=fsolve(y,mass);
+m_a=z(1)*60;// kg/min
+m_f=z(2)*3600;// kg/hr
+m_g1=z(3);// kg/sec
+m_g2=z(4);// kg/sec
+Sfc=(m_f/W_2);// kg/kWh
+AF=(m_a/60)/(m_f/3600);// Air fuel ratio
+n_th=(W_2/((m_f/3600)*CV*10^3))*100;// Thermal efficiency in %
+printf('\nThe mass of air consumed by the plant=%0.1f kg/min \nA:F ratio used=%0.0f \nSpecific fuel consumption=%0.2f kg/kWh \nThermal efficiency of the plant=%0.1f percentage',m_a,AF,Sfc,n_th);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.35/Ex24_35.sce b/3733/CH24/EX24.35/Ex24_35.sce new file mode 100644 index 000000000..4245d72c7 --- /dev/null +++ b/3733/CH24/EX24.35/Ex24_35.sce @@ -0,0 +1,38 @@ +// Example 24_35
+clc;funcprot(0);
+//Given data
+P=2000;// kW
+p_r=8;// Pressure ratio
+T_1=300;// K
+T_3=1000;// K
+T_3a=900;// K
+CV=42*10^3;// kJ/kg
+n_com=0.95;// Combustion efficiency
+C_pa=1;// kJ/kg.K
+C_pg=1;// kJ/kg.K
+r=1.4;// Specific heat ratio for air and gases
+
+// Calculation
+T_2=T_1*(p_r)^((r-1)/r);// K
+T_4=T_3/(p_r)^((r-1)/r);// K
+T_4a=T_3a/(p_r)^((r-1)/r);// K
+// Assume m_a=y(1);m_a1=y(2); m_a2=y(3);m_f1=y(4);m_f2=y(5);
+function[X]=mass(y)
+ X(1)=(y(1)*C_pa*(T_2-T_1))-((y(2)+y(4))*C_pa*(T_3-T_4));
+ X(2)=y(1)-(y(2)+y(3));
+ X(3)=P-((y(3)+y(5))*C_pg*(T_3a-T_4a));
+ X(4)=(y(4)*CV*n_com)-((y(2)+y(4))*C_pg*(T_3-T_2));
+ X(5)=(y(5)*CV*n_com)-((y(3)+y(5))*C_pg*(T_3-T_2));
+endfunction
+y=[1 1 1 0.01 0.01];
+z=fsolve(y,mass);
+m_a=z(1)*60;// kg/min
+m_a1=z(2)*3600;// kg/hr
+m_a2=z(3)*3600;// kg/hr
+m_f1=z(4)*3600;// kg/hr
+m_f2=z(5)*3600;// kg/hr
+m_f=m_f1+m_f2;// kg/hr
+Sfc=m_f/P;// kg/kW-hr
+n_th=(3600/(Sfc*CV))*100;// Thermal efficiency of the plant in %
+printf('\nFuel consumed by the plant=%0.1f kg/hr \nSpecific fuel consumption=%0.3f kg/kW-hr \nThermal efficiency of the plant=%0.1f percentage \nMass of air compressed=%0.0f kg/min',m_f,Sfc,n_th,m_a);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.36/Ex24_36.sce b/3733/CH24/EX24.36/Ex24_36.sce new file mode 100644 index 000000000..884254949 --- /dev/null +++ b/3733/CH24/EX24.36/Ex24_36.sce @@ -0,0 +1,49 @@ +// Example 24_36
+clc;funcprot(0);
+//Given data
+P=25;// MW
+p_r=8;// Pressure ratio
+T_1=300;// K
+p_1=1;// bar
+T_8=700;// K
+AF_1=80;// Air fuel ratio
+AF_2=70;// Air fuel ratio
+e=0.7;// Effectiveness of heat exchanger
+CV=40*10^3;// kJ/kg
+C_pa=1;// kJ/kg.K
+C_pg=1;// kJ/kg.K
+r=1.4;// Specific heat ratio for air and gases
+
+//Calculation
+p_2=p_1*p_r;// bar
+T_2=T_1*(p_r)^((r-1)/r);// K
+function[Y]=temperature(x)
+ Y(1)=(e*(x(1)-T_2))-(x(1)-T_8);
+endfunction
+x=[100];
+T=fsolve(x,temperature);
+T_7=T(1);// K
+T_2a=(T_7-T_8)+T_2;// K
+//Assume m_f1=y(1);T_3=y(2);m_f2=y(3);T_5=y(4);T_6=y(5);T_4=y(6)
+function[X]=massoffuel(y)
+ X(1)=(((80*y(1))+y(1))*C_pg*(y(2)-T_2a))-(y(1)*CV);
+ X(2)=(((70*y(3))+y(3))*C_pg*(y(4)-T_2a))-(y(3)*CV);
+ X(3)=(((70*y(3))+y(3))*C_pg*(y(4)-y(5)))-(P*10^3);
+ X(4)=y(5)-((y(4))/((p_r)^((r-1)/r)));
+ X(5)=(((80*y(1))+y(1))*((y(2))-y(6)))-(((80*y(1))+(70*y(3)))*(T_2-T_1));
+ X(6)=y(6)-((y(2))/((p_r)^((r-1)/r)));
+endfunction
+y=[0.1 1000 0.1 1000 100 100 ];
+z=fsolve(y,massoffuel);
+m_f1=z(1);// kg/sec
+m_f2=z(3);// kg/sec
+T_3=z(2);// K
+T_4=z(6);// K
+T_5=z(4);// K
+T_6=z(5);// K
+m_f=(m_f1+m_f2)*3600;// Total mass of fuel consumed per hour in kg/hr
+m_a=((m_f1*AF_1)+(m_f2*AF_2))*60;// Mass of air compressed per minute in kg/hr
+Sfc=(m_f)/(P*10^3);// Specific fuel consumption in kg/kW-hr
+n_th=((P*10^3)/((m_f1+m_f2)*CV))*100;// Thermal efficiency in %
+printf('\n(a)Total mass of fuel consumed per hour=%0.0f kg/hr \n(b)Mass of air compressed per minute=%0.0f kg/hr \n(c)Specific fuel consumption=%0.3f kg/kW-hr \n(d)Thermal efficiency=%0.0f percentage',m_f,m_a,Sfc,n_th);
+// The answers provided in the textbook is wrong
diff --git a/3733/CH24/EX24.4/Ex24_4.sce b/3733/CH24/EX24.4/Ex24_4.sce new file mode 100644 index 000000000..bc4aec58c --- /dev/null +++ b/3733/CH24/EX24.4/Ex24_4.sce @@ -0,0 +1,29 @@ +// Example 24_4
+clc;funcprot(0);
+//Given data
+p_1=101;//kN/m^2
+p_2=606;//kN/m^2
+e=0.65;//Effectiveness of regenerative heat exchanger
+T_1=15+273;// K
+n_c=0.85;// The compressor efficiency
+n_t=0.80;// The turbine efficiency
+m=4;// Air flow rate in kg/s
+T_3=870+273;// K
+// P_r=(P_1/P_2)=(P_3/P_4)
+P_r=6;// Pressure ratio
+C_p=1.005;// kJ/kg K
+r=1.4;// Specific heat ratio
+
+//Calculation
+T_2a=T_1*(P_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_4a=T_3/(P_r)^((r-1)/r);// K
+T_4=T_3-(n_t*(T_3-T_4a));// K
+P=m*C_p*((T_3-T_4)-(T_2-T_1));// kW
+T_5=(e*(T_4-T_2))+T_2;// K
+// T_4-T_6=T_5-T_2, neglecting,the weight of the fuel
+T_6=T_4+T_2-T_5;// K
+n_th1=(((T_3-T_4)-(T_2-T_1))/(T_3-T_5))*100;//%
+n_th2=(((T_3-T_4)-(T_2-T_1))/(T_3-T_2))*100;// %
+printf('\nEfficiency of the plant with regeneration=%0.1f percentage \nEfficiency without heat exchanger=%0.1f percentage',n_th1,n_th2);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.5/Ex24_5.sce b/3733/CH24/EX24.5/Ex24_5.sce new file mode 100644 index 000000000..c836bf522 --- /dev/null +++ b/3733/CH24/EX24.5/Ex24_5.sce @@ -0,0 +1,32 @@ +// Example 24_5
+clc;funcprot(0);
+//Given data
+T_1=19+273;// K
+p_1=100;//kN/m^2
+p_2=800;// kN/m^2
+n_c=0.85;// The isentropic efficiency of compressor
+n_t=0.88;// The isentropic efficiency of turbine
+n_pt=0.86;// The isentropic efficiency of power turbine
+m=7;//Air flow rate in kg/s
+T_3=980+273;// K
+C_p=1.006;// kJ/kg.K
+r=1.4;// Specific heat ratio
+
+//Calculation
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+//(1)For the first turbine
+// Compressor work= Turbine work
+T_4=T_3-(T_2-T_1);// Turbine exit temperature in K
+T_4a=T_3-((T_3-T_4)/(n_t));// K
+p_3=p_2;// bar
+p_4a=(p_3)/((T_3/T_4a)^(r/(r-1)));// kN/m^2
+p_4=p_4a;//kN/m^2
+//(2)For the power turbine
+p_5=p_1;// bar
+T_5a=T_4*(p_5/p_4)^((r-1)/r);// K
+T_5=T_4-(n_pt*(T_4-T_5a));// K
+P=(m*C_p*(T_4-T_5));// kW
+n_th=(C_p*(T_4-T_5))/(C_p*(T_3-T_2));// Thermal efficiency
+printf('\n1.The condition of air at the exit of the first turbine:T_4=%0.0f K & p_4=%0.0f kN/m^2 \n2.The power output of the turbine=%0.0f kW\nThe thermal efficiency of the plant=%0.3f or %0.1f percentage',T_4,p_4,P,n_th,n_th*100 );
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.6/Ex24_6.sce b/3733/CH24/EX24.6/Ex24_6.sce new file mode 100644 index 000000000..8fb65e868 --- /dev/null +++ b/3733/CH24/EX24.6/Ex24_6.sce @@ -0,0 +1,35 @@ +// Example 24_6
+clc;funcprot(0);
+//Given data
+T_1=288;// K
+p_1=1.03;// bar
+p_2=6;// bar
+p_3=p_2-0.1;// bar
+n_c=80/100;// The isentropic efficiency of compressor
+n_t=n_c;// The isentropic efficiency of turbine
+n_com=90/100;// Combustion efficiency
+W=1.1*1000;// kW
+m=7;//Air flow rate in kg/s
+T_3=750+273;// K
+//C_p=C_pa=C_pg
+C_p=1.0;// kJ/kg.K
+r=1.4;// Specific heat ratio
+CV=20000;// kJ/kg
+
+//Calculation
+//Applying isentropic law to the process 1-2
+T_2a=T_1*(p_2/p_1)^((r-1)/r);// K
+T_2=T_1+((T_2a-T_1)/n_c);// K
+// m=m_a/m_f
+m=((CV*n_com)/(T_3-T_2))-1;
+//Applying isentropic law to the process 3-4'
+T_4a=T_3/((p_3/p_1)^((r-1)/r));// K
+T_4=T_3-(n_t*(T_3-T_4a));//K
+m_a=W/(((1+(1/m))*C_p*(T_3-T_4))-(C_p*(T_2-T_1)));// kg/sec
+m_f=m_a/37;// kg/sec
+m_g=m_a+m_f;// kg/sec
+W_t=m_g*C_p*(T_3-T_4);// kW
+W_r=W/W_t;// Work ratio
+n_th=W/(m_g*C_p*(T_3-T_2));// Thermal efficiency of the plant
+printf('\n(i)Flow of air and flow of gases per second,m_a=%0.1f kg/sec & m_g=%0.2f kg/sec \n(ii)Work ratio=%0.4f \n(iii) Thermal efficiency of the plant=%0.3f (or)%0.1f percentage',m_a,m_g,W_r,n_th,n_th*100);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.7/Ex24_7.sce b/3733/CH24/EX24.7/Ex24_7.sce new file mode 100644 index 000000000..03ba73962 --- /dev/null +++ b/3733/CH24/EX24.7/Ex24_7.sce @@ -0,0 +1,38 @@ +// Example 24_7
+clc;funcprot(0);
+//Given data
+p_r=6.5// Pressure ratio
+T_1=300;// K
+p_1=1;// bar
+T_5=850;// K
+P=10;//The power plant capacity in MW
+CV=45000;// kJ/kg
+r=1.4;// Spcific heat ratio for air and gases
+C_p=1;// kJ/kg-k for air and gases
+C_pg=C_p;
+
+//Calculation
+p_2=p_1*p_r;// bar
+p_i=sqrt(p_1*p_2);//The required intermediate pressure in bar
+T_2=T_1*(p_i/p_1)^((r-1)/r);// K
+T_7=T_1*(p_2/p_1)^((r-1)/r);// K
+T_3=T_1;// K
+T_4=T_1*(p_2/p_i)^((r-1)/r);// K
+W_wi=2*C_p*(T_2-T_1);//The workdone per kg of air with perfect inter cooling in kJ/kg
+W_woi=C_p*(T_7-T_1);//The workdone per kg of air without inter cooling in kJ/kg
+W_s=W_woi-W_wi;// Work saved per kg of air compressed due to intercooling in kJ/kg
+// Assume m=m_a/m_f
+m=(CV/(C_pg*(T_5-T_4)))-1;
+T_6=T_5*(p_1/p_2)^((r-1)/r);// K
+W_e=C_pg*(T_5-T_6);// Work done per kg of exhaust gases in turbine in kJ/kg
+//When 1 kg of fuel used,m_f=1
+m_a=m*1;// The mass of air supplied in kg
+W_net=((1+m_a)*W_e)-(m_a*W_wi);// Net work available in kJ/kg of fuel
+m_f=(P*10^3)/W_net;// kg /sec
+m_f=m_f*3600;// kg/hr
+W_si=W_s*m_f*m_a;// kJ/hr
+W_si=W_si/3600;// kJ/hr
+P_woi=P-(W_si/1000);// MW
+n_th=((((m+1)*(T_5-T_6))-(2*m*(T_2-T_1)))/((m+1)*(T_5-T_4)))*100;// Thermal efficiency of the plant
+printf('\n Thermal efficiency of the plant=%0.1f percentage \n Fuel consumption per hour=%0.1f kg/hr \n Work saved per hour due to inter cooling=%0.0f kW',n_th,m_f,W_si);
+// The answer provided in the textbook is wrong
diff --git a/3733/CH24/EX24.8/Ex24_8.sce b/3733/CH24/EX24.8/Ex24_8.sce new file mode 100644 index 000000000..800238ee5 --- /dev/null +++ b/3733/CH24/EX24.8/Ex24_8.sce @@ -0,0 +1,29 @@ +// Example 24_8
+clc;funcprot(0);
+//Given data
+p_1=1;// bar
+p_2=5;// bar
+p_i=2.5;// bar
+T_1=27+273;// K
+T_3=900;// K
+T_5=T_3;// K
+r=1.4;// Specific heat ratio
+CV=40000;// kJ/kg
+r=1.4;// Spcific heat ratio for air and gases
+C_p=1;// kJ/kg-k for air and gases
+m_a=10;// kg/sec
+C_pg=C_p;
+C_pa=C_p
+
+//Calculation
+T_2=T_1*(p_2/p_1)^((r-1)/r);// K
+T_4=T_3*(p_i/p_2)^((r-1)/r);// K
+T_6=T_5*(p_1/p_i)^((r-1)/r);// K
+m_f1=1/((CV/(T_3-T_2))-1);// kg/kg of air
+m_f2=1/((CV/(T_5-T_4))-(1+m_f1));// kg/kg of air
+W_net=(C_pg*(1+m_f1)*(T_3-T_4))+(C_pg*(1+m_f1+m_f2)*(T_5-T_6))-(C_pa*(T_2-T_1));//Net work done per kg of air flow in kJ/kg of air
+Q_net=(m_f1+m_f2)*CV;// Net heat supplied per kg of air passing through the system in kJ.
+n_th=(W_net/Q_net)*100;// Thermal efficiency in %
+P=m_a*W_net;// Capacity of the plant in kW
+printf('\nThermal efficiency=%0.1f percentage \nPlant capacity=%0.1f MW',n_th,P/10^3);
+// The answer vary due to round off error
diff --git a/3733/CH24/EX24.9/Ex24_9.sce b/3733/CH24/EX24.9/Ex24_9.sce new file mode 100644 index 000000000..c5456cc3f --- /dev/null +++ b/3733/CH24/EX24.9/Ex24_9.sce @@ -0,0 +1,33 @@ +// Example 24_9
+clc;funcprot(0);
+//Given data
+W=2;// Work done in MW
+p_1=1;// bar
+p_r=5;// Pressure ratio in bar
+p_i=2.5;// bar
+T_1=27+273;// K
+r=1.4;// Specific heat ratio
+CV=40000;// kJ/kg
+n_c=85/100;// The isentropic efficiency of the compressor
+n_t=85/100;// The isentropic efficiency of the turbine
+Q_a=80;// Heat absorbed in kJ/kg of air
+m_f=0.01;// kg per kg of air
+m_a=1;// kg
+r=1.4;// Spcific heat ratio for air and gases
+C_p=1;// kJ/kg-k for air and gases
+C_pg=C_p;
+C_pa=C_p
+
+//Calculation
+T_2a=T_1*(p_r)^((r-1)/r);// K
+T_2=((T_2a-T_1)/n_c)+T_1;// K
+T_3=T_2+(Q_a/(C_pa*m_a));// K
+T_4=((m_f*CV)/((1+m_f)*C_p))+T_3;// K
+T_5a=T_4*(1/p_r)^((r-1)/r);// K
+T_5=T_4-(n_t*(T_4-T_5a));// K
+n_th=(((T_4-T_5)-(T_2-T_1))/(T_4-T_3))*100;// Thermal efficiency in %
+Q=(W*10^3)/(n_th/100);//Heat supplied in kJ/sec
+F=(Q/CV)*3600;// Fuel required per hour in kg/hr
+n_cp=(1-(1/(p_r)^((r-1)/r)))*100;//Efficiency of normal constant pressure cycle
+printf('\nThe thermal efficiency of the plant=%0.1f percentage \nEfficiency of normal constant pressure cycle=%0.0f percentage \nFuel consumption per hour=%0.0f kg/hr',n_th,n_cp,F);
+// The answer provided in the textbook is wrong
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