diff options
Diffstat (limited to '3843/CH9')
| -rw-r--r-- | 3843/CH9/EX9.1/Ex9_1.sce | 17 | ||||
| -rw-r--r-- | 3843/CH9/EX9.10/Ex9_10.sce | 23 | ||||
| -rw-r--r-- | 3843/CH9/EX9.11/Ex9_11.sce | 14 | ||||
| -rw-r--r-- | 3843/CH9/EX9.12/Ex9_12.sce | 30 | ||||
| -rw-r--r-- | 3843/CH9/EX9.13/Ex9_13.sce | 23 | ||||
| -rw-r--r-- | 3843/CH9/EX9.14/Ex9_14.sce | 39 | ||||
| -rw-r--r-- | 3843/CH9/EX9.15/Ex9_15.sce | 18 | ||||
| -rw-r--r-- | 3843/CH9/EX9.16/Ex9_16.sce | 20 | ||||
| -rw-r--r-- | 3843/CH9/EX9.2/Ex9_2.sce | 22 | ||||
| -rw-r--r-- | 3843/CH9/EX9.3/Ex9_3.sce | 21 | ||||
| -rw-r--r-- | 3843/CH9/EX9.4/Ex9_4.sce | 26 | ||||
| -rw-r--r-- | 3843/CH9/EX9.5/Ex9_5.sce | 35 | ||||
| -rw-r--r-- | 3843/CH9/EX9.6/Ex9_6.sce | 23 | ||||
| -rw-r--r-- | 3843/CH9/EX9.7/Ex9_7.sce | 16 | ||||
| -rw-r--r-- | 3843/CH9/EX9.8/Ex9_8.sce | 25 | ||||
| -rw-r--r-- | 3843/CH9/EX9.9/Ex9_9.sce | 15 |
16 files changed, 367 insertions, 0 deletions
diff --git a/3843/CH9/EX9.1/Ex9_1.sce b/3843/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..df9309215 --- /dev/null +++ b/3843/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,17 @@ +// Example 9_1
+clc;funcprot(0);
+// Given data
+m=20;// The mass flow rate of air in kg/min
+P_2=1600;// kPa
+T_1=20+273;// K
+P_1=100;// kPa
+n=0.90;// The efficiency of the compressor
+c_p=1.006;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+// Assume T_2'=T_2a
+T_2a=T_1*(P_2/P_1)^((k-1)/k);// K
+T_2=T_1+((1/n)*(T_2a-T_1));// K
+W_comp=(m/60)*c_p*(T_2-T_1);// The required power in kW
+printf("\nThe power required to drive the adiabatic compressor,W_comp=%3.1f kW",W_comp);
diff --git a/3843/CH9/EX9.10/Ex9_10.sce b/3843/CH9/EX9.10/Ex9_10.sce new file mode 100644 index 000000000..ed93ddb3a --- /dev/null +++ b/3843/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,23 @@ +// Example 9_10
+clc;funcprot(0);
+// Given data
+P_1=100;// kPa
+T_1=25+273;// K
+T_2a=472.0;// K
+r_p=5;// The pressure ratio
+T_3=850+273;// The maximum temperature in K
+T_4a=709.1;// K
+k=1.4;// The specific heat ratio
+c_p=1.00;// kJ/kg.K
+n_comp=0.80;// The isentropic efficiency of the compressor
+n_turb=0.80;// The isentropic efficiency of the turbine
+
+// Calculation
+w_comp=(c_p/n_comp)*(T_2a-T_1);// kJ/kg
+w_turb=n_turb*c_p*(T_3-T_4a);// kJ/kg
+w_r=w_comp/w_turb;// The back work ratio
+T_2=(w_comp/c_p)+T_1;// K
+w_net=w_turb-w_comp;// kJ/kg
+q_in=c_p*(T_3-T_2);// kJ/kg
+n=w_net/q_in;// The thermal efficiency of the cycle
+printf("\nThe back work ratio=%0.3f or %2.1f percentage. \nThe thermal efficiency of the cycle,n=%0.3f or %2.1f percentage.",w_r,w_r*100,n,n*100);
diff --git a/3843/CH9/EX9.11/Ex9_11.sce b/3843/CH9/EX9.11/Ex9_11.sce new file mode 100644 index 000000000..3ebd9700e --- /dev/null +++ b/3843/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,14 @@ +// Example 9_11
+clc;funcprot(0);
+// Given data
+// From example 9.9
+P_1=100;// kPa
+T_1=25+273;// K
+r_p=5;// The pressure ratio
+T_4=850+273;// The maximum temperature in K
+k=1.4;// The specific heat ratio
+
+// Calculation
+n=1-((T_1/T_4)*(r_p)^((k-1)/k));// The thermal efficiency
+w_r=0.420;// The back work ratio
+printf("\nThe thermal efficiency,n=%0.3f or %2.1f percentage \nThe back work ratio,w_comp/w_turb=%0.3f",n,n*100,w_r);
diff --git a/3843/CH9/EX9.12/Ex9_12.sce b/3843/CH9/EX9.12/Ex9_12.sce new file mode 100644 index 000000000..aaba81941 --- /dev/null +++ b/3843/CH9/EX9.12/Ex9_12.sce @@ -0,0 +1,30 @@ +// Example 9_12
+clc;funcprot(0);
+// Given data
+// From example 9.9
+P_1=100;// kPa
+P_4=500;// kPa
+T_1=25+273;// K
+T_6=850+273;// The maximum temperature in K
+c_p=1.00// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+P_2=sqrt(P_1*P_4);// The intermediate pressure in kPa
+T_2=T_1*(P_2/P_1)^((k-1)/k);// K
+T_8=T_6;// K
+P_7=P_2;// kPa
+P_6=P_4;// kPa
+T_7=T_6*(P_7/P_6)^((k-1)/k);// K
+T_9=T_7;// K
+T_5=T_7;// K
+T_4=T_2;// K
+T_3=T_1;// K
+w_turb=(c_p*(T_6-T_7))+(c_p*(T_8-T_9));// kJ/kg
+w_comp=(c_p*(T_2-T_1))+(c_p*(T_4-T_3));// kJ/kg
+w_out=w_turb-w_comp;// kJ/kg
+q_C=c_p*(T_6-T_5);// kJ/kg
+q_R=c_p*(T_8-T_7);// kJ/kg
+q_in=q_C+q_R;// kJ/kg
+n=w_out/q_in;// The thermal efficiency of the cycle
+printf("\nThe thermal efficiency of the cycle,n=%0.3f or %2.1f percentage",n,n*100);
diff --git a/3843/CH9/EX9.13/Ex9_13.sce b/3843/CH9/EX9.13/Ex9_13.sce new file mode 100644 index 000000000..3f7f813f9 --- /dev/null +++ b/3843/CH9/EX9.13/Ex9_13.sce @@ -0,0 +1,23 @@ +// Example 9_13
+clc;funcprot(0);
+// Given data
+m=100;// lbm/sec
+P_1=5;// psia
+P_2=50;// psia
+T_1=-50+460;// °R
+T_3=2000+460;// °R
+V_1=600;// ft/sec
+c_p=0.24// Btu/lbm-°R
+k=1.4;// The specific heat ratio
+
+// Calculation
+T_2=T_1*(P_2/P_1)^((k-1)/k);// °R
+T_4=T_3-(T_2-T_1);// °R
+P_3=P_2;// psia
+P_5=P_1;// psia
+P_4=P_3*(T_4/T_3)^(k/(k-1));// psia
+T_5=T_4*(P_5/P_4)^((k-1)/k);// °R
+V_5=[2*c_p*778*32.2*(T_4-T_5)]^(1/2);// ft/sec
+T=(m/32.2)*(V_5-V_1);// lbf
+hp=(T*V_1)/550;// hp
+printf("\nThe thrust developed by the engine,T=%4.0f lbf \nThe horse power developed by the engine,hp=%4.0f hp",T,hp);
diff --git a/3843/CH9/EX9.14/Ex9_14.sce b/3843/CH9/EX9.14/Ex9_14.sce new file mode 100644 index 000000000..6ce8e5d68 --- /dev/null +++ b/3843/CH9/EX9.14/Ex9_14.sce @@ -0,0 +1,39 @@ +// Example 9_14
+clc;funcprot(0);
+// Given data
+P_1=10;// kPa
+P_3=4;// MPa
+P_5=100;// kPa
+W_ST=100;// The power output from the turbine in MW
+T_5=25+273;// K
+r_p=5;// The pressure ratio
+T_7=850+273;// K
+T_9=350;// K
+c_p=1.00// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+h_1=192;// kJ/kg
+h_2=h_1;// kJ/kg
+// At 400°C and 4 MPa
+h_3=3214;// kJ/kg
+s_3=6.7698;// kJ/kg.K
+s_4=s_3;// kJ/kg.K
+s_f4=0.6491;// kJ/kg.K
+s_fg4=7.5019;// kJ/kg.K
+x=(s_4-s_f4)/s_fg4;// The quality of steam
+h_f4=192;// kJ/kg
+h_fg4=2393;// kJ/kg
+h_4=h_f4+(x*h_fg4);// kJ/kg
+h_3=3214;// kJ/kg
+m_s=(W_ST*10^3)/(h_3-h_4);// kg/s
+T_6=T_5*(r_p)^((k-1)/k);// K
+T_8=T_7*(1/r_p)^((k-1)/k);// K
+h_2=192;// kJ/kg
+m_a=(m_s*(h_3-h_2))/(c_p*(T_8-T_9));// kg/s
+W_turb=m_a*c_p*(T_7-T_8);// kJ/kg
+W_comp=m_a*c_p*(T_6-T_5);// kJ/kg
+W_GT=(W_turb-W_comp)/10^3;// The net gas turbine output in MW
+Q_in=(m_a*c_p*(T_7-T_6))/10^3;// MW
+n=(W_ST+W_GT)/Q_in;// The combined cycle efficiency
+printf("\nThe efficiency of the combined Brayton-Rankine cycle,n=%0.3f or %2.1f percentage.",n,n*100);
diff --git a/3843/CH9/EX9.15/Ex9_15.sce b/3843/CH9/EX9.15/Ex9_15.sce new file mode 100644 index 000000000..9fde5451a --- /dev/null +++ b/3843/CH9/EX9.15/Ex9_15.sce @@ -0,0 +1,18 @@ +// Example 9_15
+clc;funcprot(0);
+// Given data
+T_2=-10+273;// K
+T_4=30+273;// K
+r=10;// The compression ratio
+c_p=1.00// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+T_3=T_2*(r)^((k-1)/k);// K
+T_1=T_4*(1/r)^((k-1)/k);// K
+T_1C=T_1-273;// The minimum cycle temperature in °C
+q_in=c_p*(T_2-T_1);// kJ/kg
+w_comp=c_p*(T_3-T_2);// kJ/kg
+w_turb=c_p*(T_4-T_1);// kJ/kg
+COP=q_in/(w_comp-w_turb);// The coefficient of performance
+printf("\nThe minimum cycle temperature,T_1=%3.0f°C \nThe coefficient of performance,COP=%1.2f",T_1C,COP);
diff --git a/3843/CH9/EX9.16/Ex9_16.sce b/3843/CH9/EX9.16/Ex9_16.sce new file mode 100644 index 000000000..5cf138d0e --- /dev/null +++ b/3843/CH9/EX9.16/Ex9_16.sce @@ -0,0 +1,20 @@ +// Example 9_16
+clc;funcprot(0);
+// Given data
+T_3=-10+273;// K
+T_2=-40+273;// K
+r=10;// The compression ratio
+c_p=1.00// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+T_4=T_3*(r)^((k-1)/k);// K
+T_5=T_3;// K
+T_6=T_2;// K
+T_1=T_6*(1/r)^((k-1)/k);// K
+T_1C=T_1-273;// The minimum cycle temperature in °C
+q_in=c_p*(T_2-T_1);// kJ/kg
+w_comp=c_p*(T_4-T_3);// kJ/kg
+w_turb=c_p*(T_6-T_1);// kJ/kg
+COP=q_in/(w_comp-w_turb);// The coefficient of performance
+printf("\nThe minimum cycle temperature,T_1=%3.0f°C \nThe coefficient of performance,COP=%0.3f",T_1C,COP);
diff --git a/3843/CH9/EX9.2/Ex9_2.sce b/3843/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..7dd0ad003 --- /dev/null +++ b/3843/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,22 @@ +// Example 9_2
+clc;funcprot(0);
+// Given data
+m=20;// The mass flow rate of air in kg/min
+P_4=1600;// kPa
+T_1=20+273;// K
+P_1=100;// kPa
+n=0.90;// The efficiency of the compressor
+c_p=1.00;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+P_2=sqrt(P_1*P_4);// kPa
+T_3=T_1;// K
+T_2a=T_1*(P_2/P_1)^((k-1)/k);// K
+// Assume T_2'=T_2a
+// P_4/P_3=P_2/P_1
+T_4a=T_3*(P_2/P_1)^((k-1)/k);// K
+T_2=T_1+((1/n)*(T_2a-T_1));// K
+T_4=T_2;// K
+W_comp=((m/60)*c_p*(T_2-T_1))+((m/60)*c_p*(T_4-T_3));// The required power in kW
+printf("\nThe power required to drive the two-stage adiabatic compressor,W_comp=%3.0f kW",W_comp);
diff --git a/3843/CH9/EX9.3/Ex9_3.sce b/3843/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..caa754aae --- /dev/null +++ b/3843/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,21 @@ +// Example 9_3
+clc;funcprot(0);
+// Given data
+r=12;// The compression ratio
+P_1=200;// kPa
+P_3=10000;// kPa
+k=1.4;// The specific heat ratio
+
+// Calculation
+// (a)
+c=(1/(12-1))*100;// The percent clearance in %
+// (b)
+// r=V_1/V_2
+P_2=P_1*(r)^k;// kPa
+// V_3/V_4=V_2/V_1
+P_4=P_3*(1/r)^k;// kPa
+// W_cycle=20070*V_2;.............(1)
+// W_cycle=MEP*(12V_2-V_2);.......(2)
+// Solving equations (1)&(2) we get,
+MEP=20070/11;// kPa
+printf("\n(a)The percent clearance,c=%1.2f percentage \n(b)MEP=%4.0f kPa",c,MEP);
diff --git a/3843/CH9/EX9.4/Ex9_4.sce b/3843/CH9/EX9.4/Ex9_4.sce new file mode 100644 index 000000000..2ad6d2e05 --- /dev/null +++ b/3843/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,26 @@ +// Example 9_4
+clc;funcprot(0);
+// Given data
+r=10;// The compression ratio
+T_1=200+273;// K
+P_1=200;// kPa
+w_net=1000;// kJ/kg
+c_v=0.717;// kJ/kg.K
+R=0.287;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+T_2=T_1*(r)^(k-1);// K
+function[X]=temperature(y)
+ X(1)=w_net-((c_v*(T_1-T_2))+(c_v*(y(1)-y(2))))
+ X(2)=y(1)-(y(2)*(r)^(k-1));
+endfunction
+y=[1000 1000];
+z=fsolve(y,temperature);
+T_3=z(1);// K
+T_4=z(2);// K
+n_carnot=(1-(T_1/T_3));// %
+v_1=(R*T_1)/P_1;// m^3/kg
+// v_2=v_1/r;
+MEP=w_net/(0.9*v_1);// kPa
+printf("\nThe maximum possible efficiency,n_carnot=%0.3f or %2.1f percentage.\nMEP=%4.0f kPa",n_carnot,n_carnot*100,MEP);
diff --git a/3843/CH9/EX9.5/Ex9_5.sce b/3843/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..05c679025 --- /dev/null +++ b/3843/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,35 @@ +// Example 9_5
+clc;funcprot(0);
+// Given data
+r=18;// The compression ratio
+T_1=200+273;// K
+P_1=200;// kPa
+w_net=1000;// kJ/kg
+c_p=1.00;// kJ/kg.K
+c_v=0.717;// kJ/kg.K
+R=0.287;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+v_1=(R*T_1)/P_1;// m^3/kg
+v_2=v_1/r;// m^3/kg
+T_2=T_1*(r)^(k-1);// K
+P_2=P_1*(r)^(k);// kPa
+function[X]=temperature(y)
+ X(1)=w_net-((c_p*(y(1)-T_2))+(c_v*(T_1-y(2))));
+ v_4=v_1;// m^3/kg
+ X(2)=y(2)-(y(1)*(y(3)/v_4)^(k-1));
+ X(3)=(y(1)/y(3))-(T_2/v_2);
+endfunction
+y=[1000 1000 0.01];
+z=fsolve(y,temperature);
+T_3=z(1);// K
+T_4=z(2);// K
+v_3=z(3);// m^3/kg
+r_c=v_3/v_2;// The cut off ratio
+n=(1-((1/(r^(k-1)))*(((r_c^k)-1)/(k*(r_c-1)))));// The thermal efficiency
+MEP=w_net/(v_1-v_2);// kPa
+r_otto=v_1/v_3;// The compression ratio for otto cycle
+n_otto=(1-(1/(r^(k-1))));// The thermal efficiency for otto cycle
+printf("\nThe thermal efficiency,n=%0.3f or %2.1f percentage.\nMEP=%3.0f kPa. \nThe thermal efficiency of an otto cycle operating with the same maximum pressure,n_otto=%0.3f or %2.1f percentage.",n,n*100,MEP,n_otto,n_otto*100);
+// The answer provided in the text book is wrong
diff --git a/3843/CH9/EX9.6/Ex9_6.sce b/3843/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..100fe7f7b --- /dev/null +++ b/3843/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,23 @@ +// Example 9_6
+clc;funcprot(0);
+// Given data
+r=16;// The compression ratio
+T_1=200+273;// K
+P_1=200;// kPa
+r_c=2;// The cut off ratio
+r_p=1.3;// The pressure ratio
+c_p=1.00;// kJ/kg.K
+c_v=0.717;// kJ/kg.K
+R=0.287;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+n=1-((1/(r^(k-1)))*(((r_p*r_c^k)-1)/((k*r_p*(r_c-1))+(r_p-1))));// The thermal efficiency
+T_2=T_1*(r)^(k-1);// K
+T_3=T_2*r_p;// K
+T_4=T_3*r_c;// K
+q_in=(c_v*(T_3-T_2))+(c_p*(T_4-T_3));// kJ/kg
+w_out=n*q_in;// kJ/kg
+v_1=(R*T_1)/P_1;// m^3/kg
+MEP=w_out/(v_1*(1-(1/r)));// kPa
+printf("\nThe thermal efficiency,n=%0.3f or %2.1f percentage. \nThe heat input,q_in=%4.0f kJ/kg. \nThe work output,w_out=%4.0f kJ/kg. \nThe MEP=%4.0f kPa",n,n*100,q_in,w_out,MEP);
diff --git a/3843/CH9/EX9.7/Ex9_7.sce b/3843/CH9/EX9.7/Ex9_7.sce new file mode 100644 index 000000000..b4ee9c256 --- /dev/null +++ b/3843/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,16 @@ +// Example 9_7
+clc;funcprot(0);
+// Given data
+r=10;// The compression ratio
+P_1=30;// psia
+T_1=200+460;// °R
+T_3=1000+460;// °R
+R=53;// Btu/lbm°R
+
+// Calculation
+w_34=R*T_3*log(r);// ft-lbf/lbm
+w_12=R*T_1*log(1/r);// ft-lbf/lbm
+w_out=w_34+w_12;// The work output in ft-lbf/lbm
+n=1-(T_1/T_3);// The thermal efficiency
+q_in=(w_out/778)/n;// The heat input in Btu/lbm
+printf("\nThe work output,w_out=%5.0f ft-lbf/lbm \nThe heat input,q_in=%3.0f Btu/lbm",w_out,q_in);
diff --git a/3843/CH9/EX9.8/Ex9_8.sce b/3843/CH9/EX9.8/Ex9_8.sce new file mode 100644 index 000000000..180e13eac --- /dev/null +++ b/3843/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,25 @@ +// Example 9_8
+clc;funcprot(0);
+// Given data
+r=10;// The compression ratio
+P_1=200;// kPa
+T_1=100+273;// K
+T_3=600+273;// K
+R=0.287;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+v_1=(R*T_1)/P_1;// m^3/kg
+T_4=T_3;// K
+v_4=(T_4/T_1)*v_1;// m^3/kg
+v_2=v_4/r;// m^3/kg
+T_2=T_1;// K
+P_2=(R*T_2)/v_2;// kPa
+P_3=P_2;// kPa
+v_3=(R*T_3)/P_3;// m^3/kg
+w_out=(R*T_1*log(v_2/v_1))+(P_2*(v_3-v_2))+(R*T_3*log(v_4/v_3))+(P_1*(v_1-v_4));// The work output in kJ/kg
+T_L=T_1;// K
+T_H=T_3;// K
+n=1-(T_L/T_H);// The thermal efficiency
+q_in=w_out/n;// The heat input in kJ/kg
+printf("\nThe work output,w_out=%3.0f kJ/kg \nThe heat input,q_in=%3.0f kJ/kg",w_out,q_in);
diff --git a/3843/CH9/EX9.9/Ex9_9.sce b/3843/CH9/EX9.9/Ex9_9.sce new file mode 100644 index 000000000..b98e4e3d6 --- /dev/null +++ b/3843/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,15 @@ +// Example 9_9
+clc;funcprot(0);
+// Given data
+P_1=100;// kPa
+T_1=25+273;// K
+r_p=5;// The pressure ratio
+T_3=850+273;// The maximum temperature in K
+k=1.4;// The specific heat ratio
+
+// Calculation
+T_2=T_1*(r_p)^((k-1)/k);// K
+T_4=T_3*(1/r_p)^((k-1)/k);// K
+w_r=(T_2-T_1)/(T_3-T_4);// The back work ratio
+n=1-(r_p)^((1-k)/k);// The thermal efficiency
+printf("\nThe back work ratio,w_comp/w_turb=%0.3f or %2.0f percentage. \nThe thermal efficiency,n=%0.3f(%2.1f percentage)",w_r,w_r*100,n,n*100);
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