From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3733/CH24/EX24.1/Ex24_1.sce | 30 ++++++++++++++++++++++ 3733/CH24/EX24.10/Ex24_10.sce | 46 ++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.11/Ex24_11.sce | 40 +++++++++++++++++++++++++++++ 3733/CH24/EX24.12/Ex24_12.sce | 34 +++++++++++++++++++++++++ 3733/CH24/EX24.13/Ex24_13.sce | 49 ++++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.14/Ex24_14.sce | 31 +++++++++++++++++++++++ 3733/CH24/EX24.15/Ex24_15.sce | 34 +++++++++++++++++++++++++ 3733/CH24/EX24.16/Ex24_16.sce | 46 ++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.17/Ex24_17.sce | 46 ++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.18/Ex24_18.sce | 54 ++++++++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.19/Ex24_19.sce | 35 ++++++++++++++++++++++++++ 3733/CH24/EX24.2/Ex24_2.sce | 37 +++++++++++++++++++++++++++ 3733/CH24/EX24.20/Ex24_20.sce | 58 +++++++++++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.21/Ex24_21.sce | 39 +++++++++++++++++++++++++++++ 3733/CH24/EX24.22/Ex24_22.sce | 37 +++++++++++++++++++++++++++ 3733/CH24/EX24.23/Ex24_23.sce | 35 ++++++++++++++++++++++++++ 3733/CH24/EX24.24/Ex24_24.sce | 30 ++++++++++++++++++++++ 3733/CH24/EX24.25/Ex24_25.sce | 33 ++++++++++++++++++++++++ 3733/CH24/EX24.26/Ex24_26.sce | 25 +++++++++++++++++++ 3733/CH24/EX24.27/Ex24_27.sce | 34 +++++++++++++++++++++++++ 3733/CH24/EX24.28/Ex24_28.sce | 42 +++++++++++++++++++++++++++++++ 3733/CH24/EX24.29/Ex24_29.sce | 55 ++++++++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.3/Ex24_3.sce | 29 ++++++++++++++++++++++ 3733/CH24/EX24.30/Ex24_30.sce | 51 +++++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.31/Ex24_31.sce | 47 +++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.32/Ex24_32.sce | 37 +++++++++++++++++++++++++++ 3733/CH24/EX24.33/Ex24_33.sce | 41 ++++++++++++++++++++++++++++++ 3733/CH24/EX24.34/Ex24_34.sce | 35 ++++++++++++++++++++++++++ 3733/CH24/EX24.35/Ex24_35.sce | 38 ++++++++++++++++++++++++++++ 3733/CH24/EX24.36/Ex24_36.sce | 49 ++++++++++++++++++++++++++++++++++++ 3733/CH24/EX24.4/Ex24_4.sce | 29 ++++++++++++++++++++++ 3733/CH24/EX24.5/Ex24_5.sce | 32 ++++++++++++++++++++++++ 3733/CH24/EX24.6/Ex24_6.sce | 35 ++++++++++++++++++++++++++ 3733/CH24/EX24.7/Ex24_7.sce | 38 ++++++++++++++++++++++++++++ 3733/CH24/EX24.8/Ex24_8.sce | 29 ++++++++++++++++++++++ 3733/CH24/EX24.9/Ex24_9.sce | 33 ++++++++++++++++++++++++ 36 files changed, 1393 insertions(+) create mode 100644 3733/CH24/EX24.1/Ex24_1.sce create mode 100644 3733/CH24/EX24.10/Ex24_10.sce create mode 100644 3733/CH24/EX24.11/Ex24_11.sce create mode 100644 3733/CH24/EX24.12/Ex24_12.sce create mode 100644 3733/CH24/EX24.13/Ex24_13.sce create mode 100644 3733/CH24/EX24.14/Ex24_14.sce create mode 100644 3733/CH24/EX24.15/Ex24_15.sce create mode 100644 3733/CH24/EX24.16/Ex24_16.sce create mode 100644 3733/CH24/EX24.17/Ex24_17.sce create mode 100644 3733/CH24/EX24.18/Ex24_18.sce create mode 100644 3733/CH24/EX24.19/Ex24_19.sce create mode 100644 3733/CH24/EX24.2/Ex24_2.sce create mode 100644 3733/CH24/EX24.20/Ex24_20.sce create mode 100644 3733/CH24/EX24.21/Ex24_21.sce create mode 100644 3733/CH24/EX24.22/Ex24_22.sce create mode 100644 3733/CH24/EX24.23/Ex24_23.sce create mode 100644 3733/CH24/EX24.24/Ex24_24.sce create mode 100644 3733/CH24/EX24.25/Ex24_25.sce create mode 100644 3733/CH24/EX24.26/Ex24_26.sce create mode 100644 3733/CH24/EX24.27/Ex24_27.sce create mode 100644 3733/CH24/EX24.28/Ex24_28.sce create mode 100644 3733/CH24/EX24.29/Ex24_29.sce create mode 100644 3733/CH24/EX24.3/Ex24_3.sce create mode 100644 3733/CH24/EX24.30/Ex24_30.sce create mode 100644 3733/CH24/EX24.31/Ex24_31.sce create mode 100644 3733/CH24/EX24.32/Ex24_32.sce create mode 100644 3733/CH24/EX24.33/Ex24_33.sce create mode 100644 3733/CH24/EX24.34/Ex24_34.sce create mode 100644 3733/CH24/EX24.35/Ex24_35.sce create mode 100644 3733/CH24/EX24.36/Ex24_36.sce create mode 100644 3733/CH24/EX24.4/Ex24_4.sce create mode 100644 3733/CH24/EX24.5/Ex24_5.sce create mode 100644 3733/CH24/EX24.6/Ex24_6.sce create mode 100644 3733/CH24/EX24.7/Ex24_7.sce create mode 100644 3733/CH24/EX24.8/Ex24_8.sce create mode 100644 3733/CH24/EX24.9/Ex24_9.sce (limited to '3733/CH24') 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 -- cgit