diff options
Diffstat (limited to '3831/CH13')
| -rw-r--r-- | 3831/CH13/EX13.1/Ex13_1.sce | 15 | ||||
| -rw-r--r-- | 3831/CH13/EX13.10/Ex13_10.sce | 30 | ||||
| -rw-r--r-- | 3831/CH13/EX13.11/Ex13_11.sce | 25 | ||||
| -rw-r--r-- | 3831/CH13/EX13.12/Ex13_12.sce | 18 | ||||
| -rw-r--r-- | 3831/CH13/EX13.13/Ex13_13.sce | 54 | ||||
| -rw-r--r-- | 3831/CH13/EX13.14/Ex13_14.sce | 17 | ||||
| -rw-r--r-- | 3831/CH13/EX13.15/Ex13_15.sce | 36 | ||||
| -rw-r--r-- | 3831/CH13/EX13.16/Ex13_16.sce | 42 | ||||
| -rw-r--r-- | 3831/CH13/EX13.17/Ex13_17.sce | 19 | ||||
| -rw-r--r-- | 3831/CH13/EX13.2/Ex13_2.sce | 23 | ||||
| -rw-r--r-- | 3831/CH13/EX13.3/Ex13_3.sce | 39 | ||||
| -rw-r--r-- | 3831/CH13/EX13.4/Ex13_4.sce | 44 | ||||
| -rw-r--r-- | 3831/CH13/EX13.5/Ex13_5.sce | 44 | ||||
| -rw-r--r-- | 3831/CH13/EX13.6/Ex13_6.sce | 66 | ||||
| -rw-r--r-- | 3831/CH13/EX13.7/Ex13_7.sce | 63 | ||||
| -rw-r--r-- | 3831/CH13/EX13.8/Ex13_8.sce | 68 | ||||
| -rw-r--r-- | 3831/CH13/EX13.9/Ex13_9.sce | 26 |
17 files changed, 629 insertions, 0 deletions
diff --git a/3831/CH13/EX13.1/Ex13_1.sce b/3831/CH13/EX13.1/Ex13_1.sce new file mode 100644 index 000000000..885776754 --- /dev/null +++ b/3831/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,15 @@ +// Example 13_1
+clc;funcprot(0);
+// Given data
+d_in=10;// Diameter of piston in inch
+d_m=0.254;// Diameter of piston in m
+L_in=38.0;// Stroke in inch
+L_m=0.965;// Stroke in m
+mg=291900;// lbf
+h=10.0;// ft
+m=84.0;// lbm
+
+// Calculation
+Duty=mg*h;// ft.lb
+n_T=(Duty/(8.5*10^8))*100;// The thermal efficiency of this engine in %
+printf("\nThe duty=%7.0f ft.lbf \nThe thermal efficiency of this engine=%0.3f percentage",Duty,n_T);
diff --git a/3831/CH13/EX13.10/Ex13_10.sce b/3831/CH13/EX13.10/Ex13_10.sce new file mode 100644 index 000000000..25f126baf --- /dev/null +++ b/3831/CH13/EX13.10/Ex13_10.sce @@ -0,0 +1,30 @@ +// Example 13_10
+clc;funcprot(0);
+// Given data
+PR=2.85;// Pressure ratio
+p_4byp_1=PR;// Pressure ratio
+V_1=0.0110;// m^3
+V_3=3.00*10^-3;// m^3
+m=0.0500;// kg
+R=0.286;// kJ/kg.K
+
+// Calculation
+// (a)
+p_1=0.500;// MPa
+p_2=p_1;// MPa
+T_1=(p_1*1000*V_1)/(m*R);// K
+T_4=T_1;// K
+p_3=p_2*PR;// MPa
+p_4=p_3;// MPa
+V_4=(m*R*T_4)/(p_4*1000);// m^3
+CR=V_1/V_4;// The isentropic compression ratio
+V_2=V_3*CR;// m^3
+// (b)
+p_3=1.43;// MPa
+p_4=p_3;// MPa
+// (d)
+T_2=(p_2*1000*V_2)/(m*R);// K
+T_3=T_2;// K
+// (e)
+n_T_E=(1-(T_2/T_1))*100;// %
+printf("\n(a)The compressor inlet pressure and volume,p_2=%0.3f MPa & V_2=%0.5f m^3 \n(b)The power piston outlet pressure and inlet volume,p_4=%1.2f MPa,V_4=%0.5f m^3 \n(c)The compressor outlet pressure,p_3=%1.2f MPa \n(d)The temperatures at the inlet and outlet of the power and displacer pistons T_1=%3.0f K,T_2=%3.0f K,T_3=%3.0f K,T_4=%3.0f K \n(e)The Ericsson cold ASC thermal efficiency of this engine,n_T=%2.1f percentage",p_2,V_2,p_4,V_4,p_3,T_1,T_2,T_3,T_4,n_T_E);
diff --git a/3831/CH13/EX13.11/Ex13_11.sce b/3831/CH13/EX13.11/Ex13_11.sce new file mode 100644 index 000000000..f589b8050 --- /dev/null +++ b/3831/CH13/EX13.11/Ex13_11.sce @@ -0,0 +1,25 @@ +// Example 13_11
+clc;funcprot(0);
+// Given data
+T_1=800;// R
+T_4=530;// R
+T_3=T_4;// R
+p_4=14.7;// psia
+p_3=p_4;// psia
+p_2s=p_3;// psia
+m=1.00*10^-3;// lbm of air
+R=53.34;// ft.lbf/lbm.R
+k=1.4;// The specific heat ratio
+
+// Calculation
+// (a)
+V_4=(m*R*T_4)/(p_4*144);// ft^3
+V_1=V_4;// ft^3
+p_1=(m*R*T_1)/(V_1*144);// psia
+// (b)
+T_2s=T_1*(p_2s/p_1)^((k-1)/k);// R
+CR=T_2s/T_3;// The isentropic compression ratio
+// (c)
+n_T_L=(1-((k*T_3*(CR-1))/(T_1-T_4)))*100;// The Lenoir cold ASC thermal efficiency in %
+printf("\n(a)The combustion pressure,p_1=%2.1f psia \n(b)The isentropic compression ratio,CR=%1.2f \n(c)The Lenoir cold ASC thermal efficiency,n_T=%1.2f percentage",p_1,CR,n_T_L);
+// The answer vary due to round off error
diff --git a/3831/CH13/EX13.12/Ex13_12.sce b/3831/CH13/EX13.12/Ex13_12.sce new file mode 100644 index 000000000..2582091ca --- /dev/null +++ b/3831/CH13/EX13.12/Ex13_12.sce @@ -0,0 +1,18 @@ +// Example 13_12
+clc;funcprot(0);
+// Given data
+p_4s=0.210;// MPa
+p_1=p_4s;// MPa
+p_3=0.190;// MPa
+p_2s=p_3;// MPa
+k=1.4;// The specific heat ratio
+
+// Calculation
+// (a)
+PR=p_4s/p_3;// The isentropic pressure ratio of a Brayton cycle engine
+// (b)
+CR=(PR)^(1/k);// The isentropic compression ratio of a Brayton cycle engine
+// (c)
+n_T_B=(1-((PR)^((1-k)/k)))*100;// The Brayton cold ASC thermal efficiency
+printf("\n(a)The isentropic pressure ratio,PR=%1.2f \n(b)The isentropic compression ratio,CR=%1.2f \n(c)The Brayton cold ASC thermal efficiency,n_T=%1.2f percentage",PR,CR,n_T_B);
+// The answer vary due to round off error
diff --git a/3831/CH13/EX13.13/Ex13_13.sce b/3831/CH13/EX13.13/Ex13_13.sce new file mode 100644 index 000000000..16ec55ff7 --- /dev/null +++ b/3831/CH13/EX13.13/Ex13_13.sce @@ -0,0 +1,54 @@ +// Example 13_13
+clc;funcprot(0);
+// Given data
+V_inlet=0;// ft/s
+V_exh=1560;// ft/s
+m_exh=270;// lbm/s
+g_c=32.174;// lbm.ft/(lbf.s^2)
+p_1=190;// psia
+T_1=2060;// R
+p_2s=28.0;// psia
+T_2=1350;// R
+p_3=14.7;// psia
+T_3=520;// R
+p_4s=200;// psia
+T_4=1175;// R
+k=1.40;// The specific heat ratio
+
+// Calculation
+// 1.The engine’s static thrust is given directly by Eq. (13.29) as
+T=m_exh*(V_exh-V_inlet)/g_c;// lbf
+// 2a.
+T_4s=T_3*((p_4s/p_3)^((k-1)/k));// °F
+n_s=((T_4s-T_3)/(T_4-T_3))*100;// The compressor’s isentropic efficiency in %
+T_2s=T_1*(p_2s/p_1)^((k-1)/k);// R
+n_s_pm=((T_1-T_2)/(T_1-T_2s))*100;// %
+// 3a.
+n_T_Bc=((T_1-T_2s-(T_4s-T_3))/(T_1-T_4s))*100;// The Brayton cold ASC thermal efficiency in %
+n_T_B=((T_1-T_2-(T_4-T_3))/(T_1-T_4))*100;// The actual thermal efficiency of the engine in %
+// 2b.
+// By using Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics
+p_r4=1.2147*(p_4s/p_3);
+T_4s=1084;// R
+T_4sF=624;// °F
+n_s_c=((T_4s-T_3)/(T_4-T_3))*100;// %
+p_r1=196.16;
+p_r2=p_r1*(p_2s/p_1);
+// By interpolation in Table C.16a,
+T_2s=1261-460;// °F
+T_2s=1261;// R
+n_s_pm2=((T_1-T_2)/(T_1-T_2s))*100;// %
+// 3b.
+// From Table C.16a,
+h_3=124;// Btu/lbm
+h_4s=262;// Btu/lbm
+h_1=521;// Btu/lbm
+h_2s=307;// Btu/lbm
+h_4=284.09;// Btu/lbm
+h_2=329.9;// Btu/lbm
+n_T_Bh=((h_1-h_2s-(h_4s-h_3))/(h_1-h_4s))*100;// %
+n_T_B2=((h_1-h_2-(h_4-h_3))/(h_1-h_4))*100;// %
+// 3c.
+n_T_max=(1-sqrt(T_3/T_1))*100;// The maximum work Brayton cold ASC thermal efficiency in %
+printf("\n(1)The engine’s static thrust is given directly,T=%5.0f lbf \n(2)(a)The compressor and turbine isentropic efficiencies for the Brayton cold air standard cycle,(n_s)_compressor=%2.1f percentage & (n_s)_pm=%2.1f percentage \n (b)The compressor and turbine isentropic efficiencies for the Brayton hot air standard cycle using the gas tables for air,(n_T)_Brayton hot ASC=%2.1f percentage & (n_T)_Brayton actual=%2.1f percentage \n(3)(a)The ASC and actual thermal efficiencies for the Brayton cold air standard cycle,(n_T)_Brayton cold ASC =%2.1f percentage & (n_T)__Brayton actual=%1.1f percentage \n (b)The Brayton hot air standard cycle using the gas tables for air,(n_T)_Brayton hot ASC =%2.1f percentage & (n_T)__Brayton actual=%1.1f percentage \n (c)The maximum work Brayton cold ASC thermal efficiency,(n_T)_max work=%2.1f percentage",T,n_s,n_s_pm,n_s_c,n_s_pm2,n_T_Bc,n_T_B,n_T_Bh,n_T_B2,n_T_max);
+// The answer provided in the text book is wrong
diff --git a/3831/CH13/EX13.14/Ex13_14.sce b/3831/CH13/EX13.14/Ex13_14.sce new file mode 100644 index 000000000..17bcbeffc --- /dev/null +++ b/3831/CH13/EX13.14/Ex13_14.sce @@ -0,0 +1,17 @@ +// Example 13_14
+clc;funcprot(0);
+// Given data
+CR=8.00/1.00;// The isentropic compression ratio
+T_3=70.0;// °F
+p_3=14.7;// psia
+k=1.4;// The specific heat ratio
+
+// Calculation
+// (a)
+T_4s=(T_3+459.67)*(CR^(k-1));// R
+// (b)
+p_4s=p_3*CR^(k);// psia
+// (c)
+n_T_otto=(1-((CR)^(1-k)))*100;// %
+printf("\n(a)The air temperature at the end of the isentropic compression stroke,T_4s=%4.0f R \n(b)The pressure at the end of the isentropic compression stroke before ignition occurs,p_4s=%3.0f psia \n(c)The Otto cold ASC thermal efficiency of this engine,n_T=%2.1f percentage",T_4s,p_4s,n_T_otto);
+// The answer vary due to round off error
diff --git a/3831/CH13/EX13.15/Ex13_15.sce b/3831/CH13/EX13.15/Ex13_15.sce new file mode 100644 index 000000000..fa65adb68 --- /dev/null +++ b/3831/CH13/EX13.15/Ex13_15.sce @@ -0,0 +1,36 @@ +// Example 13_15
+clc;funcprot(0);
+// Given data
+CR=9/1;// The compression ratio
+Qbym=20.0*10^3;// Btu/lbm
+AbyF_mass=16.0/1;
+T_3=60;// °F
+p_3=8.00;// psia
+c_va=0.172;// Btu/(lbm air.R)
+D=260;// The total displacement in in^3
+N=4000;// rpm
+c=2;// The number of crankshaft revolutions per power stroke
+W_Bout=85.0;// hp
+k=1.40;// The specific heat ratio
+R=0.0685;// Btu/(lbm.R)
+
+// Calculation
+// (a)
+n_T_c=(1-(CR^(1-k)))*100;// %
+// (b)
+T_4s=(T_3+459.67)*(CR^(k-1));// R
+T_max=T_4s+(Qbym/(AbyF_mass*c_va));// R
+p_4s=p_3*((T_4s/(T_3+459.67))^(k/(k-1)));// psia
+T_1=T_max;// R
+p_max=p_4s*(T_1/T_4s);// psia
+p_1=p_max;// psia
+// (c)
+W_Iout=((n_T_c/100)*Qbym*D*N*p_1/c)/(AbyF_mass*R*T_1*(CR-1)*12*60);// ft.lbf/s
+W_Iout=W_Iout/550.0;// hp
+// (d)
+n_m=(W_Bout/W_Iout)*100;// The mechanical efficiency of the engine in %
+// (e)
+n_T_act=((n_m/100)*(n_T_c/100))*100;// The actual thermal efficiency of the engine in %
+printf("\n(a)Cold ASC thermal efficiency of the engine,n_T=%2.1f percentage \n(b)Maximum pressure and temperature of the cycle,p_max=%4.0f psia & T_max=%4.0f R \n(c)Indicated power output of the engine,|W_I|_out=%3.0f hp \n(d)Mechanical efficiency of the engine,n_m=%2.1f percentage \n(e)Actual thermal efficiency of the engine,n_T=%2.1f percentage",n_T_c,p_max,T_max,W_Iout,n_m,n_T_act);
+// The answer vary due to round off error
+
diff --git a/3831/CH13/EX13.16/Ex13_16.sce b/3831/CH13/EX13.16/Ex13_16.sce new file mode 100644 index 000000000..cd26329c9 --- /dev/null +++ b/3831/CH13/EX13.16/Ex13_16.sce @@ -0,0 +1,42 @@ +// Example 13_16
+clc;funcprot(0);
+// Given data
+v=3.50;// liter
+p_5=200;// kPa
+T_5=313;// K
+k=1.35;// The specific heat ratio
+HV=43300;// kJ/kg
+AF=15/1;// Air fuel ratio
+CR=8.00/1;// The comprssion ratio
+ER=10.0/1;// An expansion ratio
+R=0.287;// kJ/kg.K
+C_v_air=1;// kJ/kg.K
+
+// Calculation
+V_d=v/4;// L
+V_d=V_d*10^-3;// m^3
+V_c=V_d/(ER-1);// m^3
+V_1=V_c;// m^3
+V_7s=V_1;// m^3
+V_4=V_7s;// m^3
+V_6s=V_d+V_c;// m^3
+V_2s=V_6s;// m^3
+V_3=V_2s;// m^3
+V_5=V_7s*CR;// m^3
+p_6s=p_5*(V_5/V_6s)^k;// kPa
+T_6s=T_5*(V_5/V_6s)^(k-1);// K
+p_7s=p_5*(CR)^k;// kPa
+T_7s=T_5*(CR)^(k-1);// K
+m_air=(p_6s*V_6s)/(R*T_6s);// kg
+m_fuel=m_air/(AF+1);// kg
+Q_comb=m_fuel*HV;// kJ
+T_1=(Q_comb/(m_air*C_v_air))+T_7s;// K
+p_1=(p_7s/10^3)*(T_1/T_7s);// MPa
+p_2s=p_1*10^3*(V_1/V_2s)^k;// MPa
+T_2s=T_1*(V_1/V_2s)^(k-1);// K
+p_3=101;// kPa
+p_exhaust=p_3;// kPa
+T_3=T_2s*(p_3/p_2s);// K
+p_4=p_3;// kPa
+printf("\nThe temperature and pressure at all points of the cycle are given below \nState 5:p_5=%3.0f kPa,T_5=%3.0f K \nState 6:p_6s=%3.0f kPa,T_6s=%3.0f K \nState7s:p_7s=%4.0f kPa,T_7s=%3.0f K \nState 1:p_1=%2.2f MPa,T_1=%4.0f K\nState2s:p_2s=%3.0f kPa,T_2s=%4.0f K \nState 3:p_3=%3.0f kPa,T_3=%3.0f K \nState 4:p_4=%3.0f kPa,T_4=atmospheric temperature",p_5,T_5,p_6s,T_6s,p_7s,T_7s,p_1,T_1,p_2s,T_2s,p_3,T_3,p_3);
+// The answer provided in the textbook is wrong
diff --git a/3831/CH13/EX13.17/Ex13_17.sce b/3831/CH13/EX13.17/Ex13_17.sce new file mode 100644 index 000000000..4fdf74c98 --- /dev/null +++ b/3831/CH13/EX13.17/Ex13_17.sce @@ -0,0 +1,19 @@ +// Example 13_17
+clc;funcprot(0);
+// Given data
+CR=18.0;// The compression ratio
+CO=2.32;// The cut off ratio
+HV=45.5*10^3;// The heating value of a fuel in kJ/kg
+m_fuel=3.35;// The fuel flow rate of rate in kg/s
+W_B=80080;// kW
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+n_T_disel=(1-(((CR)^-0.40)*([((CO)^k)-1]))/(k*(CO-1)))*100;// The Diesel cold ASC thermal efficiency of the engine in %
+// (b)
+Q_fuel=HV*m_fuel;// kW
+n_T_diselact=(W_B/Q_fuel)*100;// The actual thermal efficiency of the engine in %
+// (c)
+n_m=(n_T_diselact/n_T_disel)*100;// The mechanical efficiency of the engine in %
+printf("\n(a)The Diesel cold ASC thermal efficiency of the engine,n_T=%2.1f percentage \n(b)The actual thermal efficiency of the engine,(n_T)_Diesel actual=%2.1f percentage \n(c)The mechanical efficiency of the engine,n_m=%2.1f percentage",n_T_disel,n_T_diselact,n_m);
diff --git a/3831/CH13/EX13.2/Ex13_2.sce b/3831/CH13/EX13.2/Ex13_2.sce new file mode 100644 index 000000000..69c38251f --- /dev/null +++ b/3831/CH13/EX13.2/Ex13_2.sce @@ -0,0 +1,23 @@ +// Example 13_2
+clc;funcprot(0);
+// Given data
+D_piston=2.00;// ft
+W_out=20;// hp
+L=4.00;// ft/stroke
+m_b=4000;// lbf
+d=15.0;// ft
+Duty=35.0*10^6;
+N=18.0;// strokes per minute
+
+// Calculation
+// (a)
+A=(%pi*D_piston^2)/4;// ft^2
+W_out=20*33000;// ft.lbf/min
+p_avg=W_out/(A*L*N);// lbf/ft^2
+p_avg=p_avg/144;// lbf/in^2
+// (b)
+n_T=(Duty/(8.5*10^8))*100;// The actual thermal efficiency of the engine in %
+// (c)
+W_out=20;// hp
+Q_boiler=(W_out*2545)/(n_T/100);// Btu/h
+printf("\n(a)The average pressure of the cycle,p_avg=%2.1f lbf/in^2 \n(b)The actual thermal efficiency of the engine,n_T=%1.2f percentage \n(c)The heat rate produced by the boiler,Q_boiler=%1.2e Btu/h",p_avg,n_T,Q_boiler);
diff --git a/3831/CH13/EX13.3/Ex13_3.sce b/3831/CH13/EX13.3/Ex13_3.sce new file mode 100644 index 000000000..250b3c59e --- /dev/null +++ b/3831/CH13/EX13.3/Ex13_3.sce @@ -0,0 +1,39 @@ +// Example 13_3
+clc;funcprot(0);
+// Given data
+Q_boiler=300;// W
+p_1=20.0;// psia
+p_2s=14.7;// psia
+T_L=671.67;// R
+T_H=687.67;// R
+
+// Solution
+// (a)
+n_T_Carnot=(1-(T_L/T_H))*100;// %
+W_net_Carnot=(n_T_Carnot/100)*Q_boiler;// watts
+// (b)
+// Station 1-Engine inlet
+p_1=20.0;// psia
+x_1=1.00;// The quality of steam at Station 1
+h_1=1156.4;// Btu/lbm
+s_1=1.7322;// Btu/lbm.R
+// Station 2s-Engine exit
+p_2s=14.7;// psia
+s_2s=s_1;// Btu/lbm.R
+s_f2=0.3122;// Btu/lbm.R
+s_fg2=1.4447;// Btu/lbm.R
+x_2s=(s_2s-s_f2)/s_fg2;// The quality of steam at Station 2s
+h_f2=180.1;// Btu/lbm
+h_fg2=970.4;// Btu/lbm
+h_2s=h_f2+(x_2s*h_fg2);// Btu/lbm
+// Station 3-Condenser exit
+p_3=p_2s;// psia
+x_3=0;// The quality of steam at Station 3
+h_3=h_f2;// Btu/lbm
+v_3=0.01672;/// ft^3/lbm
+// Station 4s-Boiler inlet
+p_4s=p_1;// psia
+// s_4s=s_3;
+n_T_max=((h_1-h_2s-(v_3*(p_4s-p_3)))*(144/118.16))/((h_1-h_3-(v_3*(p_4s-p_3)))*(144/118.16));// The isentropic efficiency of the system
+n_T_max=n_T_max*100;// %
+printf("\n(a)The Carnot cycle thermal efficiency,(n_T)_Carnot=%1.2f percentage \n The net power output of the engine,W_net=%1.2f watts \n(b)The isentropic efficiency of the Rankine cycle,n_T_max=%1.2f percentage",n_T_Carnot,W_net_Carnot,n_T_max);
diff --git a/3831/CH13/EX13.4/Ex13_4.sce b/3831/CH13/EX13.4/Ex13_4.sce new file mode 100644 index 000000000..bb8089c27 --- /dev/null +++ b/3831/CH13/EX13.4/Ex13_4.sce @@ -0,0 +1,44 @@ +// Example 13_4
+clc;funcprot(0);
+// Given data
+D=40.0;// inch
+L=10.0;// ft stroke
+W_actual=1400;// hp
+n=36.0;// rpm
+n_s_p=0.650;// The isentropic efficiency of a pump
+n_s_pm=0.550;// The isentropic efficiency of an engine
+d_fw=30.0;// The diameter of the flywheel in ft
+w=56.0;/// tons
+
+// Calculation
+// (a)
+// Station 1-Engine inlet
+p_1=100.0;// psia
+x_1=1.00;// The quality of steam at Station 1
+h_1=1187.8;// Btu/lbm
+s_1=1.6036;// Btu/lbm.R
+// Station 2s-Engine exit
+p_2s=14.7;// psia
+s_2s=s_1;// Btu/lbm.R
+s_f2=0.3122;// Btu/lbm.R
+s_fg2=1.4447;// Btu/lbm.R
+x_2s=(s_2s-s_f2)/s_fg2;// The quality of steam at Station 2s
+h_f2=180.1;// Btu/lbm
+h_fg2=970.4;// Btu/lbm
+h_2s=h_f2+(x_2s*h_fg2);// Btu/lbm
+// Station 3-Condenser exit
+p_3=p_2s;// psia
+x_3=0;// The quality of steam at Station 3
+h_3=h_f2;// Btu/lbm
+v_3=0.01672;// ft^3/lbm
+// Station 4s-Boiler inlet
+p_4s=p_1;// psia
+// s_4s=s_3;
+n_T_max=((h_1-h_2s-(v_3*(p_4s-p_3)))*(144/118.16))/((h_1-h_3-(v_3*(p_4s-p_3)))*(144/118.16));// The maximum isentropic efficiency of the system
+n_T_max=n_T_max*100;// %
+// (b)
+n_T_Rankine=(((h_1-h_2s)*n_s_pm)-((v_3*(p_4s-p_3)/n_s_p)*(144/118.16)))/((h_1-h_3)-((v_3*(p_4s-p_3)/n_s_p)*(144/118.16)));// The isentropic efficiency of the Rankine system
+n_T_Rankine=n_T_Rankine*100;// %
+// (c)
+mdot=(W_actual*2545)/((h_1-h_2s)*n_s_pm);// lbm/h
+printf("\n(a)The maximum isentropic efficiency of the Rankine system,(n_T)_maximum Rankine=%2.1f percentage \n(b)The isentropic efficiency of the Rankine system,(n_T)_Rankine=%1.2f percentage \n(c)The mass flow rate of steam required,mdot=%5.0f lbm/h",n_T_max,n_T_Rankine,mdot);
diff --git a/3831/CH13/EX13.5/Ex13_5.sce b/3831/CH13/EX13.5/Ex13_5.sce new file mode 100644 index 000000000..1db68510d --- /dev/null +++ b/3831/CH13/EX13.5/Ex13_5.sce @@ -0,0 +1,44 @@ +// Example 13_5
+clc;funcprot(0);
+// Given data
+p_1=100;// psia
+T_1=500;// °F
+p_3=1.00;// psia
+
+// Calculation
+// Station 1
+p_1=100.0;// psia
+T_1=500.0;// °F
+h_1=1279.1;// Btu/lbm
+s_1=1.7087;// Btu/lbm.R
+// Station 2s
+
+// Station 3
+p_3=1.00;// psia
+x_3=0.00;// The dryness fraction
+s_3=s_f2;// Btu/lbm.R
+h_3=h_f2;// Btu/lbm
+v_3=0.01614;/// ft^3/lbm
+// Station 4s
+p_4s=p_1;// psia
+s_4s=s_3;// Btu/lbm.R
+h_4s=h_3+(v_3*(p_4s-p_3)*(144/778.16));// Btu/lbm
+
+s_2s=s_1;// Btu/lbm.R
+s_f2=0.1326;// Btu/lbm.R
+s_fg2=1.8455;// Btu/lbm.R
+x_2s=(s_2s-s_f2)/s_fg2;// The dryness fraction
+h_f2=69.7;// Btu/lbm
+h_fg2=1036.0;// Btu/lbm
+h_2s=h_f2+(x_2s*h_fg2);// Btu/lbm
+// (a)
+// The degree of superheat at the outlet of the boiler is determined from Table C.2a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics and Eq. (13.10) as
+T_sat=327.8;// °F
+Dsh=500-T_sat;// Degree of superheat in °F
+// (b)
+T_H=T_1+459.67;// R
+T_L=101.67+459.67;// R
+n_T_carnot=(1-(T_L/T_H))*100;// The Carnot cycle thermal efficiency in %
+// (c)
+n_T_rankine=((h_1-h_2s-((v_3*(p_4s-p_3))*(144/778.16)))/(h_1-h_3-((v_3*(p_4s-p_3))*(144/778.16))))*100;// The isentropic Rankine cycle thermal efficiency in %
+printf("\n(a)The degree of superheat at the boiler outlet=%3.0f°F \n(b)The equivalent Carnot cycle thermal efficiency of the lawn mower,n_T_carnot=%2.1f percentage \n(c)The isentropic Rankine cycle thermal efficiency of the lawn mower,(n_T)_Rankine=%2.1f percentage",Dsh,n_T_carnot,n_T_rankine);
diff --git a/3831/CH13/EX13.6/Ex13_6.sce b/3831/CH13/EX13.6/Ex13_6.sce new file mode 100644 index 000000000..156eb076b --- /dev/null +++ b/3831/CH13/EX13.6/Ex13_6.sce @@ -0,0 +1,66 @@ +// Example 13_6
+clc;funcprot(0);
+// Given data
+p_1=200;// psia
+p_2s=1.00;// psia
+p_4=80.0;// psia
+
+// Calculation
+// (a)
+// Station 1
+p_1=200.0;// psia
+x_1=1.00;// The dryness fraction
+h_1=1199.3;// Btu/lbm
+s_1=1.5466;// Btu/lbm.R
+// Station 2s
+p_2=1.00;// psia
+p_2s=p_2;// psia
+s_2s=s_1;// Btu/(lbm.R)
+s_f2=0.1326;// Btu/(lbm.R)
+s_fg2=1.8455;// Btu/(lbm.R)
+h_f2=69.7;// Btu/lbm
+h_fg2=1036.0;// Btu/lbm
+// Station 3
+p_3=1.00;// psia
+x_3=0.00;// The dryness fraction
+s_3=0.1326;// Btu/(lbm.R)
+h_3=69.7;// Btu/lbm
+v_3=0.01614;/// ft^3/lbm
+// Station 4s
+p_4=200;// psia
+p_4s=p_4;// psia
+s_4s=s_3;// Btu/lbm.R
+h_4s=h_3+(v_3*(p_4s-p_3)*(144/778.16));// Btu/lbm
+x_2s=(s_2s-s_f2)/s_fg2;// The dryness fraction
+h_2s=h_f2+(x_2s*h_fg2);// Btu/lbm
+n_T_Rankine=(((h_1-h_2s)-(h_4s-h_3))/(h_1-h_4s))*100;// The thermal efficiency in %
+// (b)
+// Station 4s
+p_4=200;// psia
+p_4s=p_4;// psia
+s_4s=s_3;// Btu/lbm.R
+h_4s=h_3+(v_3*(p_4s-p_3)*(144/778.16));// Btu/lbm
+// Station 5s
+p_5s=p_4;// psia
+s_5s=s_1;// Btu/(lbm.R)
+s_f5s=0.4535;// Btu/(lbm.R)
+s_fg5s=1.1681;// Btu/(lbm.R)
+x_5s=(s_5s-s_f5s)/s_fg5s;// The dryness fraction
+h_f5s=282.2;// Btu/lbm
+h_fg5s=901.4;// Btu/lbm
+h_5s=h_f5s+(x_5s*h_fg5s);// Btu/lbm
+h_5s=1125.7;// Btu/lbm
+// Station 6
+p_6=80.0;// psia
+x_6=0.00;// The dryness fraction
+s_6=0.4535;// Btu/(lbm.R)
+h_6=282.2;// Btu/lbm
+v_6=0.01757;// ft^3/lbm
+// Station 7s
+p_7=200;// psia
+p_7s=p_7;// psia
+s_7s=s_6;// Btu/(lbm.R)
+h_7s=h_6+(v_6*(p_7-p_6)*(144/778.16));// Btu/lbm
+r=(h_6-h_4s)/(h_5s-h_4s);// The mass fraction of steam
+n_T_reg=(1-(((h_2s-h_3)/(h_1-h_7s))*(1-r)))*100;// %
+printf("\n(a)The isentropic Rankine cycle thermal efficiency of the system without regeneration present,(n_T)_isentropic Rankine=%2.1f percentage.\n(b)The isentropic Rankine cycle thermal efficiency of the system,(n_T)_Rankine cycle with 1 regenerator=%2.1f percentage",n_T_Rankine,n_T_reg);
diff --git a/3831/CH13/EX13.7/Ex13_7.sce b/3831/CH13/EX13.7/Ex13_7.sce new file mode 100644 index 000000000..2822ad70e --- /dev/null +++ b/3831/CH13/EX13.7/Ex13_7.sce @@ -0,0 +1,63 @@ +// Example 13_7
+clc;funcprot(0);
+// Given data
+n_s_pm1=84.0/100;// The isentropic efficiency of the first turbine
+n_s_pm2=80.0/100;// The isentropic efficiency of the second turbine
+n_s_p=61.0/100;// The isentropic efficiency of the boiler feed pump
+n_s_pm=82/100;// The isentropic efficiency of the prime mover
+
+// Calculation
+// (a)
+// Station 1
+p_1=600.0;// psia
+T_1=700.0;// °F
+h_1=1350.6;// Btu/lbm
+s_1=1.5874;// Btu/lbm.R
+// Station 2s
+p_2=100.0;// psia
+p_2s=p_2;// psia
+s_2s=s_1;// Btu/(lbm.R)
+s_f2=0.4745;// Btu/(lbm.R)
+s_fg2=1.1291;// Btu/(lbm.R)
+h_f2=298.6;// Btu/lbm
+h_fg2=889.2;// Btu/lbm
+x_2s=(s_2s-s_f2)/s_fg2;// The dryness fraction
+h_2s=h_f2+(x_2s*h_fg2);// Btu/lbm
+// Station 3
+p_3=100.0;// psia
+T_3=700.0;// °F
+x_3=0.00;// The dryness fraction
+s_3=1.8035;// Btu/(lbm.R)
+h_3=1379.2;// Btu/lbm
+// Station 4s
+p_4=1.00;// psia
+p_4s=p_4;// psia
+s_4s=s_3;// Btu/lbm.R
+s_f4=0.1326;// Btu/(lbm.R)
+s_fg4=1.8455;// Btu/(lbm.R)
+h_f4=69.7;// Btu/lbm
+h_fg4=1036.4;// Btu/lbm
+x_4s=(s_4s-s_f4)/s_fg4;// The dryness fraction
+h_4s=h_f4+(x_4s*h_fg4);// Btu/lbm
+// Station 5
+p_5=1.00;// psia
+x_5=0.00;// The dryness fraction
+s_5=0.1326;// Btu/(lbm.R)
+h_5=69.7;// Btu/lbm
+v_5=0.01614;// ft^3/lbm
+// Station 6s
+p_6=600;// psia
+p_6s=p_6;// psia
+s_6s=s_5;// Btu/(lbm.R)
+h_6s=72.5;// Btu/lbm
+v_6s=0.01614;// ft^3/lbm
+h_7s=h_6s+(v_6s*(p_7-p_6)*(144/778.16));// Btu/lbm
+h_2=h_1-((h_1-h_2s)*n_s_pm1);// Btu/lbm
+h_6=h_5+((v_5*(p_6*p_5)*(144/778.16))/(n_s_p));// Btu/lbm
+n_T_wr=((((h_1-h_2s)*n_s_pm1)+((h_3-h_4s)*n_s_pm2)-((v_5*(p_6-p_5)*(144/778.16))/(n_s_p)))/((h_1-h_6)+(h_3-h_2)))*100;// The Rankine cycle thermal efficiency of the plant with reheat in %
+// (b)
+s_4s=s_1;// Btu/(lbm.R)
+x_4s=(s_4s-s_f4)/s_fg4;// The dryness fraction
+h_4s=h_f4+(x_4s*h_fg4);// Btu/lbm
+n_T_wor=((((h_1-h_4s)*n_s_pm)-((h_6s-h_5)/n_s_pm))/(h_1-h_6))*100;// The Rankine cycle thermal efficiency of the plant without reheat in %
+printf("\n(a)The Rankine cycle thermal efficiency of the plant with reheat,n_T=%2.1f percentage \n(b)The Rankine cycle thermal efficiency of the plant without reheat,n_T=%2.1f percentage",n_T_wr,n_T_wor);
diff --git a/3831/CH13/EX13.8/Ex13_8.sce b/3831/CH13/EX13.8/Ex13_8.sce new file mode 100644 index 000000000..3792d7079 --- /dev/null +++ b/3831/CH13/EX13.8/Ex13_8.sce @@ -0,0 +1,68 @@ +// Example 13_8
+clc;funcprot(0);
+// Given data
+p_1=5000.0;// psia
+T_1=1200.0;// °F
+p_3=1000.0;// psia
+p_5=300.0;// psia
+p_6s=0.400;// psia
+mdot=1.50*10^6;// lbm/h
+W_netout=325;// MW
+
+// Calculation
+// Station 1-Turbine 1 inlet
+p_1=5000.0;// psia
+T_1=1200.0;// °F
+h_1=1530.8;// Btu/lbm
+s_1=1.5068;// Btu/lbm.R
+// Station 2s-Turbine 1 exit
+p_2s=1000;// psia
+s_2s=s_1;// Btu/lbm.R
+h_2s=1316.9;// Btu/lbm
+// (by interpolation in Table C.3a)
+// Station 3-Turbine 2 inlet
+p_3=1000.0;// psia
+T_3=1000.0;// °F
+h_3=1505.9;// Btu/lbm
+s_3=1.6532;// Btu/lbm.R
+// (by interpolation in Table C.3a)
+// Station 4s-Turbine 2 exit
+p_4s=1000;// psia
+s_4s=s_3;// Btu/lbm.R
+h_4s=1343.8;// Btu/lbm
+// Station 5-Turbine 3 inlet
+p_5=300.0;// psia
+T_5=1000.0;// °F
+h_5=1526.4;// Btu/lbm
+s_5=1.7966;// Btu/lbm.R
+// Station 6s-Turbine 3 exit
+p_6s=0.400;// psia
+s_6s=s_5;// Btu/lbm.R
+s_f6s=0.0799;// Btu/lbm.R
+s_fg6s=1.9762;// Btu/lbm.R
+x_6s=(s_6s-s_f6s)/s_fg6s;// The dryness fraction
+h_f6s=40.9;// Btu/lbm
+h_fg6s=1052.4;// Btu/lbm
+h_6s=h_f6s+(x_6s*h_fg6s);// Btu/lbm
+// Station 7-Condenser exit
+p_7=0.400;// psia
+x_7=0.00;// The dryness fraction
+h_7=40.9;// Btu/lbm
+v_7=0.01606;// ft^3/lbm
+// Station 8s-Boiler inlet
+p_8s=p_1;
+// s_8s=s_7;
+h_8s=h_7+((v_7*(p_8s-p_7))*(144/778.16));// Btu/lbm
+// (a)
+n_s_p=1.0;// The isentropic thermal efficiency of this Rankine cycle power plant
+n_s_pm2=n_s_p;// The isentropic thermal efficiency of this Rankine cycle power plant
+n_s_pm1=n_s_pm2;// The isentropic thermal efficiency of this Rankine cycle power plant
+N=(h_1-h_2s)+(h_3-h_4s)+(h_5-h_6s)-(v_7*(p_8s-p_7)*(144/778.16));// The numerator in Btu/lbm
+D=(h_1-h_8s)+(h_3-h_2s)+(h_5-h_4s);// The denominator in Btu/lbm
+n_T=(N/D)*100;// The isentropic thermal efficiency in %
+// (b)
+W_netout=(W_netout*10^3)*3412;// Btu/h
+W_isen=mdot*[(h_1-h_2s)+(h_3-h_4s)+(h_5-h_6s)-(v_7*(p_8s-p_7*(144/778.16)))];// Btu/h
+n_s_tg=(W_netout/W_isen)*100;
+printf("\n(a)The isentropic thermal efficiency of this power plant,(n_T)_s=%2.1f percentage \n(b)The isentropic efficiency of the turbine-generator power unit,(n_s)_turbine generator=%2.1f percentage",n_T,n_s_tg);
+// The answer vary due to round off error
diff --git a/3831/CH13/EX13.9/Ex13_9.sce b/3831/CH13/EX13.9/Ex13_9.sce new file mode 100644 index 000000000..b067f20d8 --- /dev/null +++ b/3831/CH13/EX13.9/Ex13_9.sce @@ -0,0 +1,26 @@ +// Example 13_9
+clc;funcprot(0);
+// Given data
+Pd=0.0110;/// The piston displacement in m^3
+V_4=1.00*10^-3;// m^3
+V_3=V_4;// m^3
+p_1=0.300;// MPa
+p_2=0.100;// MPa
+T_2=30.0;//°C
+R=0.286;// kJ/kg.K
+
+// Calculation
+// (a)
+V_1=Pd-V_3;// m^3
+V_2=V_1;// m^3
+p_3=p_2*(V_2/V_3);// MPa
+// (b)
+V_4=V_3;// m^3
+p_4=p_1*(V_1/V_4);// MPa
+// (c)
+m=((p_2*1000)*V_2)/(R*(T_2+273.15));// kg
+// (d)
+T_1=((p_1*1000)*V_1)/(m*R);// K
+// (e)
+n_T=(1-((T_2+273.15)/T_1))*100;// %
+printf("\n(a)The displacer piston maximum pressure,p_3=%1.2f MPa \n(b)The power piston maximum pressure,p_4=%1.2f MPa\n(c)The mass of air in the engine,m=%0.4f kg \n(d)The heat addition temperature,T_1=%3.0f K \n(e)The Stirling cold ASC thermal efficiency of the engine,n_T=%2.2f percentage",p_3,p_4,m,T_1,n_T);
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