diff options
Diffstat (limited to '3831')
207 files changed, 4619 insertions, 0 deletions
diff --git a/3831/CH1/EX1.1/Ex1_1.sce b/3831/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..a64922c00 --- /dev/null +++ b/3831/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,13 @@ +// Example 1_1
+clc;funcprot(0);
+// Given data
+V_final=120;// mph
+V_initial=85;// mph
+t=5;// seconds
+
+// Calculation
+a=(V_final-V_initial)/t;// miles/(hour/seconds)
+V_initial=V_initial*(5280/3600);// feet/second
+V_final=V_final*(5280/3600);// feet/second
+a=(V_final-V_initial)/t;// The acceleration in ft/s^2
+printf("\nThe acceleration of the car,a=%2.1f ft/s^2",a);
diff --git a/3831/CH1/EX1.2/Ex1_2.sce b/3831/CH1/EX1.2/Ex1_2.sce new file mode 100644 index 000000000..9295a872d --- /dev/null +++ b/3831/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,27 @@ +// Example 1_2
+clc;funcprot(0);
+// Given data
+T=55;// °F
+
+// Calculation
+// (a)
+T_w=0;// The freezing point of water in °N
+T_wF=32;// The freezing point of water in °F
+T_bh=12;// The temperature of the body in °N
+T_bhF=98.6;// The temperature of the body in °F
+x=T_bh*(1-((T_bhF-T)/(T_bhF-T_wF)));// The temperature in °N
+printf("\n(a)The temperature on the Newton scale,x=%1.2f °N",x);
+// (b)
+T_w=0;// The temperature in °Re
+T_wF=80;// The temperature in °Re
+T_bh=32;// The temperature in °F
+T_bhF=212;// The temperature in °F
+r=T_wF*(1-((T_bhF-T)/(T_bhF-T_bh)));// The temperature in °Re
+printf("\n(b)The temperature on the Reaumur scale,r=%2.1f °Re",r);
+// (c)
+T_w=273.15;// The temperature in K
+T_wF=373.15;// The temperature in K
+T_bh=32;// The temperature in °F
+T_bhF=212;// The temperature in °F
+z=T_wF-((T_wF-T_w)*(((T_bhF-T)/(T_bhF-T_bh))));// The temperature in K
+printf("\n(c)The Kelvin temperature,z=%3.1f K",z);
diff --git a/3831/CH1/EX1.3/Ex1_3.sce b/3831/CH1/EX1.3/Ex1_3.sce new file mode 100644 index 000000000..2f36ecbcd --- /dev/null +++ b/3831/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,14 @@ +// Example 1_3
+clc;funcprot(0);
+// Given data
+g_c=32.174;// ft/s^2
+F=1;// lbf
+m=1;// lbm
+
+// Solution
+onechunk=1;// (lbf.s^2)/lbm
+// In the Engineering English units system, 1 lbf accelerates 1 lbm at a rate of
+a=(F*g_c)/m;// ft/s^2
+onechunk=32.174;// ft/s^2
+onechunk=32.174/3.281;// m
+printf("\n 1 chunk=%1.3f m",onechunk);
diff --git a/3831/CH1/EX1.4/Ex1_4.sce b/3831/CH1/EX1.4/Ex1_4.sce new file mode 100644 index 000000000..aaa40e7b1 --- /dev/null +++ b/3831/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,15 @@ +// Example 1_4
+clc;funcprot(0);
+// Given data
+W=25000;// Weight in lbf
+V=5000;// mph
+g=32.174;// ft/s^2
+g_orbit=2.50;// ft/s^2
+
+// Calculation
+// (a)
+g_c=32.174;// ft/s^2
+// (b)
+m=(W*g_c)/g;// The mass in lbm
+W_orbit=(m*g_orbit)/g_c;// lbf
+printf("\n(a)The value of g_c in this orbit,g_c=%2.3f ft/s^2 \n(b)The weight in Earth orbit,W_orbit=%4.0f lbf",g_c,W_orbit);
diff --git a/3831/CH1/EX1.5/Ex1_5.sce b/3831/CH1/EX1.5/Ex1_5.sce new file mode 100644 index 000000000..817f351d9 --- /dev/null +++ b/3831/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,14 @@ +// Example 1_5
+clc;funcprot(0);
+// Given data
+D=0.07;// The diameter in m
+R=D/2;// The radius in m
+h=0.15;// The height in m
+L=(3/4)*h;// m
+rho=1000;// The density in kg/m^3
+M=18;// kg/kg mole
+
+// Calculation
+m=(%pi*R^2*L)*rho;// The mass of water in the glass in kg
+n=m/M;// The number of moles in kg moles
+printf("\nThe number of kilogram moles of water in the glass,n=%0.3f kgmole",n);
diff --git a/3831/CH1/EX1.6/Ex1_6.sce b/3831/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..af887a45f --- /dev/null +++ b/3831/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,8 @@ +// Example 1_6
+clc;funcprot(0);
+// Given data
+D=2.5;// The inside diameter of a circular water pipe in inches
+
+// Calculation
+A=(%pi*D^2)/4;// in^2
+printf("\nThe cross-sectional area of the pipe,A=%1.4f in^2",A);
diff --git a/3831/CH1/EX1.7/Ex1_7.sce b/3831/CH1/EX1.7/Ex1_7.sce new file mode 100644 index 000000000..757057a86 --- /dev/null +++ b/3831/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,21 @@ +// Example 1_7
+clc;funcprot(0);
+// Given data
+W=2000;// lbf
+F=W;// lbf
+Z=8.00;// ft
+g_c=1;// Dimensionless
+g=9.81;// m/s^2
+
+// Calculation
+m=((F/0.2248)*g_c)/(g);// kg
+Z=Z/3.281;// m
+PE=(m*g*Z)/g_c;// kJ
+// In the Engineering English units system, we have
+Z=8.00;// m
+g_c=32.174;// lbm.ft/lbf.s^2
+g=32.174;// ft/s^2
+m=(F*g_c)/g;// lbm
+PE=(m*g*Z)/g_c;// ft.lbf
+PE=PE/778.17;// Btu
+printf("\nThe potential energy of an automobile,PE=%2.1f Btu",PE);
diff --git a/3831/CH1/EX1.8/Ex1_8.sce b/3831/CH1/EX1.8/Ex1_8.sce new file mode 100644 index 000000000..68ed87868 --- /dev/null +++ b/3831/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,17 @@ +// Example 1_8
+clc;funcprot(0);
+// Given data
+V_ft=3000;// ft/s
+m=10.0;// grams
+g_c=1;// Dimensionless
+
+// Calculation
+m=m/1000;// kg
+V=V_ft/3.281;// m/s
+KE=(m*V_ft^2)/(2*g_c);// kJ
+m=(m*2.205);// lbm
+// In the Engineering English units system, we have
+g_c=32.174;// lbm.ft/(lbf.s^2)
+KE=(m*V_ft^2)/(2*g_c);// ft.lbf
+KE=KE/778.17;// Btu
+printf("\nThe translational kinetic energy of a bullet,KE=%1.2f Btu",KE)
diff --git a/3831/CH1/EX1.9/Ex1_9.sce b/3831/CH1/EX1.9/Ex1_9.sce new file mode 100644 index 000000000..803db7236 --- /dev/null +++ b/3831/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,13 @@ +// Example 1_9
+clc;funcprot(0);
+// Given data
+m=10.0;// lbm
+omega=1800;// rpm
+d=4.00;// inches
+R=d/2;// inches
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+I=(m*(R/12)^2)/2;// lbm.ft^2
+KE_rot=(I*((2*%pi*omega)/60)^2)/(2*g_c);// lbf
+printf("\nThe rotational kinetic energy in the armature of an electric motor,KE_rot=%2.1f ft.lbf",KE_rot);
diff --git a/3831/CH10/EX10.1/Ex10_1.sce b/3831/CH10/EX10.1/Ex10_1.sce new file mode 100644 index 000000000..12ee45519 --- /dev/null +++ b/3831/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,25 @@ +// Example 10_1
+clc;funcprot(0);
+// Given data
+p_0=0.101;// MPa
+T=10;// °C
+T_0=20+273;// K
+L=0.150;// m
+D=0.0700;// m
+R=D/2;// m
+rho=1000;// kg/m^3
+Z=0.762;// m
+g=9.81;// m/s^2
+g_c=1;// The gravitational constant
+
+// Calculation
+m=%pi*R^2*((3/4)*L)*rho;// kg
+// From Table C.1b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics
+u=42.0;// kJ/kg
+u_0=83.9;// kJ/kg
+v=0.001000;// m^3/kg
+v_0=0.001002;// m^3/kg
+s=0.1510;// kJ/kg.K
+s_0=0.2965;// kJ/kg.K
+A=m*[(u-u_0)+((p_0*10^3)*(v-v_0))-(T_0*(s-s_0))+0+((g*Z)/g_c)];// kJ
+printf("\nThe total availability of the water in the glass relative to the floor,A=%1.2f kJ",A);
diff --git a/3831/CH10/EX10.12/Ex10_12.sce b/3831/CH10/EX10.12/Ex10_12.sce new file mode 100644 index 000000000..936e59efd --- /dev/null +++ b/3831/CH10/EX10.12/Ex10_12.sce @@ -0,0 +1,21 @@ +// Example 10_12
+clc;funcprot(0);
+// Given data
+Q_H=1.00*10^6;// kJ/s
+T_0=5.00;// ºC
+T_H=700;// ºC
+p_0=0.101;// MPa
+Q_L=7.00*10^5;// kJ/s
+T_L=40.0;// ºC
+W_net=3.00*10^5;// kJ/s
+
+// Calculation
+// (a)
+n=(W_net/Q_H)*100;// %
+// (b)
+A_bin=(1-((T_0+273.15)/(T_H+273.15)))*Q_H;// kJ/s
+// (c)
+A_cin=(1-((T_0+273.15)/(T_L+273.15)))*Q_L;// kJ/s
+// (d)
+E_HE=(W_net/(A_bin-A_cin))*100;// %
+printf("\n(a)The first law thermal efficiency of the power plant,n=%2.1f percentage \n(b)The rate at which available energy enters the boiler,A_boiler input=%1.2e kJ/s \n(c)The rate at which available energy enters the condenser,A_boiler output=%0.2e kJ/s \n(d)The second law efficiency of the power plant,E_He=%2.1f percentage",n,A_bin,A_cin,E_HE);
diff --git a/3831/CH10/EX10.13/Ex10_13.sce b/3831/CH10/EX10.13/Ex10_13.sce new file mode 100644 index 000000000..92b72d252 --- /dev/null +++ b/3831/CH10/EX10.13/Ex10_13.sce @@ -0,0 +1,18 @@ +// Example 10_13
+clc;funcprot(0);
+// Given data
+Q_H=30.0*10^3;// Btu/h
+W_in=1.50;// hp
+T_0=30.0+459.67;// K
+T_H=70.0+459.67;// K
+
+// Calculation
+// (a)
+COP_act_hp=Q_H/(W_in*2545);// The actual COP of heat pump
+n_T=COP_act_hp;// The first law thermal efficiency of the heat pump
+// (b)
+E_HP=((1-(T_0/T_H))*COP_act_hp)*100;// The second law availability efficiency of the heat pump
+T_L=T_0;// °F
+COP_Carnot_hp=T_H/(T_H-T_L);// The COP of Carnot heat pump
+E_HP=(COP_act_hp/COP_Carnot_hp)*100;// The second law availability efficiency of the heat pump
+printf("\n(a)The first law thermal efficiency of the heat pump,n_T=%1.2f \n(b)The second law availability efficiency of the heat pump,E_HP=%2.1f percentage",n_T,E_HP);
diff --git a/3831/CH10/EX10.14/Ex10_14.sce b/3831/CH10/EX10.14/Ex10_14.sce new file mode 100644 index 000000000..d3a431797 --- /dev/null +++ b/3831/CH10/EX10.14/Ex10_14.sce @@ -0,0 +1,12 @@ +// Example 10_14
+clc;funcprot(0);
+// Given data
+T_L=20+273.15;// K
+T_0=T_L;// K
+T_H=35.0+273.15;// K
+COP_act=8.92;// Actual Coefficient of Performance
+
+// Calculation
+COP_Carnot=T_L/(T_H-T_L);// The coefficient of performance of a Carnot refrigerator or air conditioner
+epsilon_RAC=(COP_act/COP_Carnot)*100;// The second law efficiency in %
+printf("\nThe second law availability efficiency of this air conditioner,epsilon_R/AC=%2.1f percentage",epsilon_RAC);
diff --git a/3831/CH10/EX10.15/Ex10_15.sce b/3831/CH10/EX10.15/Ex10_15.sce new file mode 100644 index 000000000..bdf39775a --- /dev/null +++ b/3831/CH10/EX10.15/Ex10_15.sce @@ -0,0 +1,31 @@ +// Example 10_15
+clc;funcprot(0);
+// Given data
+m_air=0.800;// kg/s
+T_air_in=20.0;// °C
+T_0=20.0;// °C
+p_ein=1.10;// atm
+p_eout=1.0;// atm
+p_ain=1.50;// atm
+p_aout=1.40;// atm
+p_0=1.00;// atm
+T_ein=500;// °C
+T_eout=400;// °C
+c_p_exh=0.990;// kJ/(kg.K)
+c_p_air=1.004;// kJ/(kg.K)
+m_exh=m_air;// kg/s
+m_cold=m_air;// kg/s
+R_exh=0.272;// kJ/(kg.K)
+R_air=0.286;// kJ/(kg.K)
+
+// Calculation
+// (a)
+c_p_cold=c_p_air;// kJ/(kg.K)
+T_air_out=T_air_in+(((m_exh*c_p_exh)/(m_cold*c_p_cold))*(T_ein-T_eout));// °C
+// (b)
+a_f_inexh=(c_p_exh*(T_ein-T_0))-((T_0+273.15)*[(c_p_exh*log((T_ein+273.15)/(T_0+273.15)))-(R_exh*log(p_ein/p_0))])+0+0;// kJ/kg
+a_f_outexh=(c_p_exh*(T_eout-T_0))-((T_0+273.15)*[(c_p_exh*log((T_eout+273.15)/(T_0+273.15)))-(R_exh*log(p_eout/p_0))])+0+0;// kJ/kg
+a_f_inair=(c_p_air*(T_air_in-T_0))-((T_0+273.15)*[(c_p_air*log((T_air_in+273.15)/(T_0+273.15)))-(R_air*log(p_ain/p_0))])+0+0;// kJ/kg
+a_f_outair=(c_p_air*(T_air_out-T_0))-((T_0+273.15)*[(c_p_air*log((T_air_out+273.15)/(T_0+273.15)))-(R_air*log(p_aout/p_0))])+0+0;// kJ/kg
+E_nmHX=((m_air*(a_f_outair-a_f_inair))/(m_exh*(a_f_inexh-a_f_outexh)))*100;// The second law availability efficiency in %
+printf("\n(a)The exit temperature of the inlet air,(T_out)_air=%3.0f°C \n(b)The second law availability efficiency of the preheater,E_nonmixingHX=%2.1f percentage",T_air_out,E_nmHX);
diff --git a/3831/CH10/EX10.16/Ex10_16.sce b/3831/CH10/EX10.16/Ex10_16.sce new file mode 100644 index 000000000..e85d71d2a --- /dev/null +++ b/3831/CH10/EX10.16/Ex10_16.sce @@ -0,0 +1,23 @@ +// Example 10_16
+clc;funcprot(0);
+// Given data
+m_H=0.180;// lbm/s
+T_H=130;// °F
+m_C=0.270;// lbm/s
+T_C=60.0;// °F
+T_0=55.0;// °F
+p_0=14.7;// psia
+C_w=1.00;// Btu/(lbm.R)
+
+// Calculation
+// (a)
+T_M=((m_H*(T_H+459.67))+(m_C*(T_C+459.67)))/(m_H+m_C);// R
+T_M=T_M-459.67;// °F
+// (b)
+a_fH=(C_w*(T_H-T_0))-(C_w*(T_0+459.67)*log((T_H+459.67)/(T_0+459.67)));// Btu/lbm
+a_fC=(C_w*(T_C-T_0))-(C_w*(T_0+459.67)*log((T_C+459.67)/(T_0+459.67)));// Btu/lbm
+a_fm=(C_w*(T_M-T_0))-(C_w*(T_0+459.67)*log((T_M+459.67)/(T_0+459.67)));// Btu/lbm
+m_m=m_H+m_C;// lbm/s
+gamma=m_H/m_m;// The second law availability efficiency
+epsilon_mixingHX=(((1-gamma)*(a_fm-a_fC))/(gamma*(a_fH-a_fm)))*100;// %
+printf("\n(a)The temperature of the mixed water in the sink,T_M=%2.0f°F \n(b)The second law availability efficiency of the sink as a mixing-type heat exchanger,epsilon_mixingHX=%2.1f percentage",T_M,epsilon_mixingHX)
diff --git a/3831/CH10/EX10.2/Ex10_2.sce b/3831/CH10/EX10.2/Ex10_2.sce new file mode 100644 index 000000000..317534203 --- /dev/null +++ b/3831/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,20 @@ +// Example 10_2
+clc;funcprot(0);
+// Given data
+p_0=0.101;// MPa
+T_0=20.0+273;// K
+p=1.500; // MPa
+T=20+273;// K
+C_v=0.781;// kJ/kg.K
+C_p=1.004;// kJ/kg.K
+R=0.286;// kJ/kg.K
+g=9.81;// m/s^2
+g_c=1;// The gravitational constant
+
+// Calculation
+// Assume deltau=u-u_0;deltav=v-v_0;deltas=s-s_0;
+deltau=C_v*(T-T_0);// kJ/kg
+deltav=R*((T/(p*10^3))-(T_0/(p_0*10^3)));// kJ/kg
+deltas=(C_p*log(T/T_0))-(R*log((p*10^3)/(p_0*10^3)));// kJ/kg
+a=deltau+(p_0*10^3*deltav)-(T_0*deltas)+0+0;// kJ/kg
+printf("\nThe specific available energy,a=%3.0f kJ/kg",a);
diff --git a/3831/CH10/EX10.3/Ex10_3.sce b/3831/CH10/EX10.3/Ex10_3.sce new file mode 100644 index 000000000..4a2cbdcc8 --- /dev/null +++ b/3831/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,41 @@ +// Example 10_3
+clc;funcprot(0);
+// Given data
+T_1=50.0;// °F
+V_1=500;// mph
+Z=30.0*10^3;// ft
+T_0=70.0;// °F
+p_0=136.12;// psia
+m=5.00;// lbm
+g=32.174;// ft/s^2
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+// State 1 (flying)
+x_1=0.00;// The quality of steam
+T_1=50.0;// °F
+v_1=0.0128;// ft^3/lbm
+u_1=24.04;// Btu/lbm
+s_1=0.0519;// Btu/lbm.R
+V_1=500;// mph
+Z_1=30000;// ft
+// State 2 (landed)
+p_2=100;// psia
+T_2=400;// °F
+v_2=1.046;// ft^3/lbm
+u_2=154.77;// Btu/lbm
+s_2=0.31464;// Btu/lbm.R
+V_2=0;// mph
+Z_2=0;// ft
+// Ground state
+x_0=0.00;// The quality of steam
+T_0=70.0;// °F
+v_0=0.01325;// ft^3/lbm
+u_0=29.78;// Btu/lbm
+s_0=0.06296;// Btu/lbm.R
+p_0=136.12;// psia
+A_1=(m*[(u_1-u_0)+((p_0)*(144/778.16)*(v_1-v_0))])-(m*(T_0+459.67)*(s_1-s_0))+(m*[(([V_1*5280*(1/3600)]^2)/(2*g_c*778.16))+((g*Z_1)/(g_c*778.16))]);// Btu
+A_2=(m*[(u_2-u_0)+((p_0)*(144/778.16)*(v_2-v_0))])-(m*(T_0+459.67)*(s_2-s_0))+(m*[(([V_2*5280*(1/3600)]^2)/(2*g_c*778.16))+((g*Z_2)/(g_c*778.16))]);// Btu
+// (b)
+dA=A_2-A_1;// Btu
+printf("\n(a)The total availability of the refrigerant before and after the aircraft lands,A_1=%3.0f Btu & A_2=%2.1f Btu. \n(b)The change in total availability during the landing,A_2-A_1=%3.0f Btu",A_1,A_2,dA);
diff --git a/3831/CH10/EX10.4/Ex10_4.sce b/3831/CH10/EX10.4/Ex10_4.sce new file mode 100644 index 000000000..912c9395d --- /dev/null +++ b/3831/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,38 @@ +// Example 10_4
+clc;funcprot(0);
+// Given data
+m=1.00;// kg
+T_0=20.0;// °C
+p_0=0.101;// MPa
+T_s=130.0+273;// K
+x_1=0.00;// The quality of steam at state 1
+T_1=120.0;// °C
+x_2=0.500;// The quality of steam at state 1
+
+// Calculation
+// State 1
+x_1=0;// The quality of steam at state 1
+T_1=120.0+273;// K
+v_f=0.001060;// m^3/kg
+v_1=v_f;// m^3/kg
+u_f=503.5;// kJ/kg
+u_1=u_f;// kJ/kg
+s_f=1.5280;// kJ/kg.K
+s_1=s_f;// kJ/kg.K
+// State 2
+x_2=0.500;// The quality of steam at state 2
+p_sat=198.5;// kN/m^2
+p_1=p_sat;// kN/m^2
+p_2=p_1;// kN/m^2
+v_2=0.44648;// m^3/kg
+u_2=1516.4;// kJ/kg
+s_2=4.3292;// kJ/kg.K
+// Ground state
+T_0=20.0+273;// K
+p_0=0.101;// MPa
+a_2minusa_1=(u_2-u_1)+(p_0*10^3*(v_2-v_1))-(T_0*(s_2-s_1));// kJ/kg
+W_12=m*p_2*(v_2-v_1);// kJ
+Q_12=(m*(u_2-u_1))+W_12;// kJ
+I_12=((1-(T_0/T_s))*Q_12)-W_12+(p_0*10^3*(v_2-v_1));// kJ
+printf("\nThe irreversibility of the process,I_12=%2.1f kJ",I_12)
+// The answer provided in the textbook is wrong
diff --git a/3831/CH10/EX10.5/Ex10_5.sce b/3831/CH10/EX10.5/Ex10_5.sce new file mode 100644 index 000000000..0177ce161 --- /dev/null +++ b/3831/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,14 @@ +// Example 10_5
+clc;funcprot(0);
+// Given data
+Q=-100;// W
+W=-100;// W
+T_b=24.0+273;// K
+p_0=0.101;// MPa
+T_0=15.0+273;// K
+
+// Calculation
+dVbydt=0;
+dAbydt=0;
+I=((1-(T_0/T_b))*Q)-W+(p_0*dVbydt)-dAbydt;// W
+printf("\nThe irreversibility rate within the room,I=%2.1f W",I);
diff --git a/3831/CH10/EX10.6/Ex10_6.sce b/3831/CH10/EX10.6/Ex10_6.sce new file mode 100644 index 000000000..fe8b72de6 --- /dev/null +++ b/3831/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,18 @@ +// Example 10_6
+clc;funcprot(0);
+// Given data
+T_w=50.0+459.67;// R
+V_w=3.00;// ft
+Z_w=4.00;// ft
+T_0=70.0+459.67;// R
+p_0=14.7;// psia
+c_w=1.00;// Btu/lbm
+g=32.174;// ft/s^2
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+v_w=0.01602;// ft^3/lbm
+p_sat=0.1780;// lbf/in^2
+p_w=p_0-p_sat;// lbf/in^2
+a_f=(c_w*(T_w-T_0))+(v_w*(p_w-p_0)*(144/778.16))-(c_w*T_0*log(T_w/T_0))+((V_w^2)/(2*g_c*778.16))+((g*Z_w)/(g_c*778.16));// Btu/lbm
+printf("\nThe specific flow availability at the exit of a garden hose,a_f=%0.3f Btu/lbm",a_f);
diff --git a/3831/CH10/EX10.7/Ex10_7.sce b/3831/CH10/EX10.7/Ex10_7.sce new file mode 100644 index 000000000..fc7c2032c --- /dev/null +++ b/3831/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,31 @@ +// Example 10_7
+clc;funcprot(0);
+// Given data
+p_1=5000;// psia
+T_1=1000;// °F
+V_1=50.0;// ft/s
+V_2=300.0;// ft/s
+x_0=0.00;// The quality of steam
+T_0=70.0;// °F
+g_c=32.174;// lbm.ft/lbf.s^2
+g=32.174;// ft/s^2
+
+// Calculation
+// Station 1
+p_1=5000;// psia
+T_1=1000;// °F
+h_1=1363.4;// Btu/lbm
+s_1=1.3990;// Btu/lbm.R
+// Station 2
+p_2=14.696;// psia
+h_2=h_1+((V_1^2-V_2^2)/(2*g_c*778.16));// Btu/lbm
+h_0=38.1;// Btu/lbm
+s_0=0.0746;// Btu/lbm.R
+Z_1=0;// ft
+Z_2=Z_1;// ft
+T_2=655;// °F
+s_2=1.9981;// Btu/lbm.R
+a_f1=(h_1-h_0)-((T_0+459.67)*(s_1-s_0))+((V_1^2)/(2*g_c*778.16))+((g*Z_1)/g_c);// Btu/lbm
+a_f2=(h_2-h_0)-((T_0+459.67)*(s_2-s_0))+((V_2^2)/(2*g_c*778.16))+((g*Z_2)/g_c);// Btu/lbm
+Ibym=a_f1-a_f2;// Btu/lbm
+printf("\nThe specific flow availabilities at the inlet and outlet of the crack,a_f1=%3.0f Btu/lbm & a_f2=%3.0f Btu/lbm \nThe irreversibility per unit mass of steam exiting the crack,I/m=%3.0f Btu/lbm",a_f1,a_f2,Ibym);
\ No newline at end of file diff --git a/3831/CH10/EX10.8/Ex10_8.sce b/3831/CH10/EX10.8/Ex10_8.sce new file mode 100644 index 000000000..76195566b --- /dev/null +++ b/3831/CH10/EX10.8/Ex10_8.sce @@ -0,0 +1,43 @@ +// Example 10_8
+clc;funcprot(0);
+// Given data
+m=2.80;// lbm/s
+// Station 1
+p_1=100;// psia
+T_1=500;// °F
+h_1=1279.1;// Btu/lbm
+s_1=1.7087;// Btu/lbm.R
+// Station 2
+p_2=10.0;// psia
+p_2s=p_2;// psia
+s_2f=0.2836;// Btu/lbm.R
+s_2fg=1.5043;// Btu/lbm.R
+s_2=s_1;// Btu/lbm.R
+h_2f=161.4;// Btu/lbm
+h_2fg=982.1;// Btu/lbm
+h_2s=1091.6;// Btu/lbm
+// Ground state
+x_0=0;// The quality of steam
+T_0=70.0;// °F
+s_0=0.0746;// Btu/lbm.R
+h_0=38.1;// Btu/lbm
+g_c=32.174;// lbm.ft/lbf.s^2
+g=32.174;// ft/s^2
+
+// Calculation
+// (a)
+V_1=0;// ft/s
+Z_2=0;// ft
+Z_1=Z_2;// ft
+V_2s=[2*g_c*(h_1-h_2s)*778.16]^(1/2);// ft/s
+V_2=(95/100)*V_2s;// ft/s
+h_2=h_1-((V_2^2)/(2*g_c*778.16));// Btu/lbm
+x_2=(h_2-h_2f)/h_2fg;// The quality of steam
+s_2=s_2f+(x_2*s_2fg);// Btu/lbm.R
+a_f1=(h_1-h_0)-((T_0+459.67)*(s_1-s_0))+(V_1^2/(2*g_c*778.16))+((g*Z_1)/g_c);// Btu/lbm
+// (b)
+a_f2=(h_2-h_0)-((T_0+459.67)*(s_2-s_0))+(V_2^2/(2*g_c*778.16))+((g*Z_2)/g_c);// Btu/lbm
+// (c)
+I=m*(a_f1-a_f2);// Btu/s
+printf("\n(a)The inlet specific flow availability,a_f1=%3.0f Btu/lbm \n(b)The exit specific flow availability,a_f2=%3.0f Btu/lbm \n(c)The irreversibility rate inside the nozzle,I=%2.1f Btu/s",a_f1,a_f2,I);
+// The answer vary due to round off error
diff --git a/3831/CH10/EX10.9/Ex10_9.sce b/3831/CH10/EX10.9/Ex10_9.sce new file mode 100644 index 000000000..3358fd04d --- /dev/null +++ b/3831/CH10/EX10.9/Ex10_9.sce @@ -0,0 +1,32 @@ +// Example 10_9
+clc;funcprot(0);
+// Given data
+m=18.0;// kg/s
+T_b=350.0;// °C
+W=20*10^3;// kW
+// Station 1
+T_1=500.0;// °C
+p_1=3.00;// MPa
+h_1=3456.5;// kJ/kg
+s_1=7.2346;// kJ/kg.K
+// Station 2
+p_2=0.0100;// MPa
+x_2=0.960;// The quality of steam
+h_2f=191.8;// kJ/kg
+h_2fg=2392.8;// kJ/kg
+h_2=h_2f+(x_2*h_2fg);// kJ/kg
+s_2f=0.6491;// kJ/kg.K
+s_2fg=7.5019;// kJ/kg.K
+s_2=s_2f+(x_2*s_2fg);// kJ/kg.K
+// Ground state
+x_0=0.00;// The quality of steam
+T_0=20.0;// °C
+h_0=83.9;// kJ/kg
+s_0=0.2965;// kJ/kg.K
+
+// Calculation
+a_f1=(h_1-h_0)-((T_0+273.15)*(s_1-s_0));// kJ/kg
+a_f2=(h_2-h_0)-((T_0+273.15)*(s_2-s_0));// kJ/kg
+Q=(W+(m*(a_f2-a_f1)))/(1-((T_0+273.15)/(T_b+273.15)));// kW
+printf("\nThe rate of heat loss from the surface of the turbine,Q=%4.0f kW",Q);
+// The answer vary due to round off error
diff --git a/3831/CH11/EX11.10/Ex11_10.sce b/3831/CH11/EX11.10/Ex11_10.sce new file mode 100644 index 000000000..d0515bdbf --- /dev/null +++ b/3831/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,12 @@ +// Example 11_10
+clc;funcprot(0);
+// Given data
+T=20.0;// °C
+beta=0.207*10^-6;// K^-1
+k=45.9*10^-11;// m^2/N
+
+// Solution
+v_f=0.001002;// m^3/kg
+v=v_f;// m^3/kg
+c_pminusc_v=(((T+273.15)*beta^2*v)/k)*10^-3;// kJ/(kg.K)
+printf("\nThe difference between c_p and c_v for saturated liquid water,c_p-c_v=%1.2e kJ/(kg.K)",c_pminusc_v);
diff --git a/3831/CH11/EX11.12/Ex11_12.sce b/3831/CH11/EX11.12/Ex11_12.sce new file mode 100644 index 000000000..a3a969613 --- /dev/null +++ b/3831/CH11/EX11.12/Ex11_12.sce @@ -0,0 +1,21 @@ +// Example 11_12
+clc;funcprot(0);
+// Given data
+T=60.0;// °F
+p_1=14.7;// psia
+r_c=19.2:1;// Compression ratio
+
+// Solution
+// From Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics,we find that, at 60.0°F = 520.R,
+u_1=88.62;// Btu/lbm
+p_r1=1.2147;
+v_r1=158.58;
+v_2byv_1=1/19.2;
+v_r2=v_r1*v_2byv_1;
+// Scanning down the v_r column in Table C.16a, we find that vr = 8.26 at about
+T_2=1600-459.67;// °F
+u_2=286.06;// Btu/lbm
+p_r2=71.73;
+p_2=p_1*(p_r2/p_r1);// psia
+W_12bym=u_1-u_2;// Btu/lbm
+printf("\nThe final temperature and pressure of the air at the end of the compression stroke,T_2=%4.0f°F and p_2=%3.1f psia. \nThe work required per lbm of air present,1W2/m=%3.2f Btu/lbm",T_2,p_2,W_12bym);
diff --git a/3831/CH11/EX11.13/Ex11_13.sce b/3831/CH11/EX11.13/Ex11_13.sce new file mode 100644 index 000000000..31120dd5e --- /dev/null +++ b/3831/CH11/EX11.13/Ex11_13.sce @@ -0,0 +1,25 @@ +// Example 11_13
+clc;funcprot(0);
+// Given data
+m=8.20;// lbm
+V=1.00;// ft^3
+T=-78.0;// °F
+
+// Solution
+// From Table C.12a, we find that
+T_c=240;// R
+p_c=507;// psia
+v_c=1.49/28.011;// ft^3/lbm
+// Also, from Table C.13a, we find that
+R=0.0709;// Btu/lbm.R
+T_R=(T+460)/T_c;
+v=V/m;// ft^3/lbm
+// Assume 'a' instead of '
+v_ca=(R*(T_c*778.16))/(p_c*144);// ft^3/lbm
+v_Ra=v/v_ca;
+// Using T_R = T/Tc = 1.60 and v′R = v/v′c = 0:67, we find from Figure 11.6 that
+p_R=2.10;
+Z=0.850;
+p=p_c*p_R;// psia
+printf("\nThe pressure exerted by 8.20 lbm of the carbon monoxide,p=%4.0f psia",p);
+// The answer is different due to round off error.
diff --git a/3831/CH11/EX11.14/Ex11_14.sce b/3831/CH11/EX11.14/Ex11_14.sce new file mode 100644 index 000000000..f4286bc10 --- /dev/null +++ b/3831/CH11/EX11.14/Ex11_14.sce @@ -0,0 +1,24 @@ +// Example 11_14
+clc;funcprot(0);
+// Given data
+V=0.100;// m^3
+p=20.0;// MPa
+m=15.6;// kg
+T=1000;// °C
+
+// Solution
+// From Table C.12b, we find the critical state properties of methane to be
+T_c=191.1;// K
+p_c=4.64;// MPa
+v=V/m;// m^3/kg
+v_1=v;// m^3/kg
+v_2=v_1;// m^3/kg
+// Table C.13b, gives the gas constant for methane as
+R=0.518;// kJ/kg.K
+// Assume 'a' instead of '
+v_Ra=(v*p_c*10^3)/(R*T_c);
+T_R=(T+273.15)/T_c;
+p_R=32.0;
+p_2byp_c=p_R;
+p_2_worstcase=p_R*p_c;// MPa
+printf("\nThe maximum pressure in the CNG tank at this worst case temperature,(p_2)_worstcase=%3.0f MPa",p_2_worstcase);
diff --git a/3831/CH11/EX11.15/Ex11_15.sce b/3831/CH11/EX11.15/Ex11_15.sce new file mode 100644 index 000000000..088cdb68e --- /dev/null +++ b/3831/CH11/EX11.15/Ex11_15.sce @@ -0,0 +1,23 @@ +// Example 11_15
+clc;funcprot(0);
+// Given data
+p_1=20.0;// MPa
+T_1=150;// °C
+p_2=0.101;// MPa
+
+// Solution
+// From Table C.12b, we find the critical temperature and pressure for CO2 are
+T_c=304.2;// K
+p_c=7.39;// MPa
+M_CO2=44.01;// kg/kg mole
+c_p=0.845;// kJ/kg.K
+p_R1=p_1/p_c;
+T_R1=(T_1+273.15)/T_c;
+// Assume s_1=[(h*-h)/T_c]_1
+s_1=14.0;// kJ/kgmole·K
+p_R2=p_2/p_c;
+// Assume s_2=[(h*-h)/T_c]_2
+// h_2-h_1=0
+T_2=(T_1+273.15)-((s_1/c_p)*(T_c/M_CO2));// K
+T_2=T_2-273.15;// °C
+printf("\nThe exit temperature of the throttle,T_2=%2.1f°C",T_2);
diff --git a/3831/CH11/EX11.16/Ex11_16.sce b/3831/CH11/EX11.16/Ex11_16.sce new file mode 100644 index 000000000..6e3b97503 --- /dev/null +++ b/3831/CH11/EX11.16/Ex11_16.sce @@ -0,0 +1,54 @@ +// Example 11_16
+clc;funcprot(0);
+// Given data
+p_1=150;// psia
+p_2=15.0*10^3;// psia
+T_1=80.0;// °F
+T_2=T_1;// °F
+
+// Calculation
+// (a)
+// The properties of ethylene at its critical state and its molecular mass are found in Table C.12a as
+T_c=508.3;// R
+p_c=742;// psia
+M=28.05;// lbm/lbmoles
+p_R1=p_1/p_c;
+T_R1=(T_1+459.67)/T_c;
+p_R2=p_2/p_c;
+T_R2=T_R1;
+// Using p_R1 and T_R1, Figure 11.9 gives the enthalpy correction for state 1 as
+// Assume s_1=[(h*-hbar)/T_c]_1
+s_1=1.50;// kJ/kgmole.K
+s_1=s_1*(1/4.1865);// Btu/(lbmole.R)
+// Using p_R2 and T_R2, Figure 11.9 gives the enthalpy correction for state 2 as
+// Assume s_2=[(h*-hbar)/T_c]_2
+s_2=31.5;// kJ/kgmole.K
+s_2=s_2*(1/4.1865);// Btu/(lbmole.R)
+// h*2-h*1=0;
+// dh=h_2-h_1;
+dh=0-([s_2-s_1]*(T_c/M));// Btu/lbm
+// (b)
+p_R1=0.202;
+T_R1=1.06;
+Z_1=0.940;
+p_R2=20.2;
+T_R2=T_R2;
+Z_2=2.15;
+R=55.1;// ft.lbf/(lbm.R)
+v_1=(Z_1*R*(T_1+459.67))/(p_1*144);// ft^3/lbm
+v_2=(Z_2*R*(T_2+459.67))/(p_2*144);// ft^3/lbm
+du=dh-(((p_2*144)*v_2*(1/778.16))-((p_1*144)*v_1*(1/778.16)));// Btu/lbm
+// (c)
+// s*2-s*1=dS;
+dS=(c_p*log(T_2/T_1))-((R/778.16)*log(p_2/p_1));// Btu/lbm.R
+// Using p_R1 and T_R1, Figure 11.11 gives the entropy correction for state 1 as
+// Assume (s*bar-sbar)_1=S_1
+S_1=1.50;// kJ/kgmole.K
+S_1=S_1*(1/4.1865);// Btu/(lbmole.R)
+// Using p_R2 and T_R2, Figure 11.11 gives the entropy correction for state 2 as
+S_2=2.22;// kJ/kgmole.K
+S_2=S_2*(1/4.1865);// Btu/(lbmole.R)
+// d_s=S_1-S_2
+ds=dS-([S_2-S_1]*(1/M));// Btu/(lbm.R)
+printf("\n(a)The change in specific enthalpy,h_2-h_1=%3.0f Btu/lbm \n(b)The change in specific internal energy,u_2-u_1=%3.0f Btu/lbm \n(c)The change in specific entropy of the ethylene,s_2-s_1=%0.3f Btu/lbm.R",dh,du,ds);
+// The answer vary due to round off error
diff --git a/3831/CH11/EX11.17/Ex11_17.sce b/3831/CH11/EX11.17/Ex11_17.sce new file mode 100644 index 000000000..2b34f4c8d --- /dev/null +++ b/3831/CH11/EX11.17/Ex11_17.sce @@ -0,0 +1,15 @@ +// Example 11_17
+clc;funcprot(0);
+// Given data
+m_A=1.00;// lbm
+p_A=1.00;// psia
+T_A=200;// °F
+m_B=5.00;// lbm
+p_B=5.00;// psia
+T_B=400;// °F
+
+// Calculation
+T_2=((m_A*(T_A+459.67))+(m_B*(T_B+459.67)))/(m_A+m_B);// R
+T_2=T_2-459.67;// °F
+p_2=((m_A+m_B)*(T_2+459.67))/(((m_A*(T_A+459.67))/p_A)+((m_B*(T_B+459.67))/p_B));// psia
+printf("\nThe final temperature,T_2=%3.0f°F \nThe final pressure,p_2=%1.2f psia",T_2,p_2);
diff --git a/3831/CH11/EX11.2/Ex11_2.sce b/3831/CH11/EX11.2/Ex11_2.sce new file mode 100644 index 000000000..5456fc8ea --- /dev/null +++ b/3831/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,14 @@ +// Example 11_2
+clc;funcprot(0);
+// Given data
+p=200;// psia
+T=400;// °F
+
+// Solution
+// From Table C.3a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at this state,
+u=1123.5;// Btu/lbm
+h=1210.8;// Btu/lbm
+s=1.5602;// Btu/lbm.R
+f=u-((T+459.67)*s);// Btu/lbm
+g=h-((T+459.67)*s);// Btu/lbm
+printf("\nThe value of the specific Helmholtz function for superheated water vapor,f=%3.0f Btu/lbm \nThe value of the specific Gibbs function for superheated water vapor,g=%3.0f Btu/lbm",f,g);
diff --git a/3831/CH11/EX11.3/Ex11_3.sce b/3831/CH11/EX11.3/Ex11_3.sce new file mode 100644 index 000000000..a1fb34060 --- /dev/null +++ b/3831/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,11 @@ +// Example 11_3
+clc;funcprot(0);
+// Given data
+p=1.00;// MPa
+
+// Solution
+// From Table C.2b at p = 1.00 MPa, we find that,
+h_fg=2015.3;// kJ/kg
+T_sat=179.90;// °C
+s_fg=h_fg/(T_sat+273.15);// kJ/kg .K
+printf("\nThe phase change entropy for water,s_fg=%1.4f kJ/kg.K",s_fg);
diff --git a/3831/CH11/EX11.6/Ex11_6.sce b/3831/CH11/EX11.6/Ex11_6.sce new file mode 100644 index 000000000..e3108f37c --- /dev/null +++ b/3831/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,11 @@ +// Example 11_6
+clc;funcprot(0);
+// Given data
+L_0=0.0700;// m
+L=0.200;// m
+T=20.0;// °C
+K=0.150;// N/K
+
+// Solution
+Q_12=(-K*(T+273.15)*L_0*((L/L_0)-1)^3)/3;// N.m
+printf("\n(c)The required heat transfer,Q_12=%1.2f N.m",Q_12);
diff --git a/3831/CH11/EX11.7/Ex11_7.sce b/3831/CH11/EX11.7/Ex11_7.sce new file mode 100644 index 000000000..3e172a8d2 --- /dev/null +++ b/3831/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,16 @@ +// Example 11_7
+clc;funcprot(0);
+// Given data
+// ln p_sat=14.05-(6289.78/T_sat)-(913998.2/T_sat^2);
+// T_sat=°F + 461.2
+T=212;// °F
+R=0.1102;// Btu/(lbm.R)
+
+// Solution
+T_sat=T+461.2;// R
+// From Table C.13a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics
+h_fg=[(6289.78)+((1827997.8)/(T_sat))]*R;// Btu/lbm
+// Table C.1a gives
+h_fg_212F=970.4;// Btu/lbm
+p_error=((h_fg-h_fg_212F)/h_fg_212F)*100;// %
+printf("\nThus, the value obtained from Rankine’s equation is in error by only %1.2f percentage.",p_error);
diff --git a/3831/CH12/EX12.1/Ex12_1.sce b/3831/CH12/EX12.1/Ex12_1.sce new file mode 100644 index 000000000..7416d9f61 --- /dev/null +++ b/3831/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,21 @@ +// Example 12_1
+clc;funcprot(0);
+// Given data
+w_propane=0.500;// The mass fraction
+w_air=0.500;// The mass fraction
+R=8.3143;// kJ/kgmole.K
+
+// Calculation
+// (a)
+// The molecular masses of the components are found in Table C.13 in Thermodynamic Tables to accompany Modern Engineering Thermodynamics as
+M_propane=44.09;// kg/kgmole
+M_air=28.97;// kg/kgmole
+M_m=1/((w_propane/M_propane)+(w_air/M_air));// kg/kgmole
+// (b)
+X_propane=w_propane*(M_m/M_propane);// The molar value for propane
+X_air=w_air*(M_m/M_air);// The molar value for air
+w_m=w_propane+w_air;// The mass of the mixture
+X_m=X_propane+X_air;// The molar value of the mixture
+// (c)
+R_m=R/M_m;// The equivalent gas constant in kJ/kg.K
+printf("\n(a)The equivalent molecular mass of the mixture,M_m=%2.1f kg/kgmole \n(b)The mixture composition on a molar basis,X_propane=%0.3f & X_air=%0.3f \n(c)The equivalent gas constant of the mixture,R_m=%0.3f kJ/kg.K",M_m,X_propane,X_air,R_m);
diff --git a/3831/CH12/EX12.10/Ex12_10.sce b/3831/CH12/EX12.10/Ex12_10.sce new file mode 100644 index 000000000..9066a0a45 --- /dev/null +++ b/3831/CH12/EX12.10/Ex12_10.sce @@ -0,0 +1,18 @@ +// Example 12_10
+clc;funcprot(0);
+// Given data
+T_DB=50.0+273.15;// K
+T_ref=0+273.15;// K
+phi=40.0/100;// The relative humidity
+p_m=0.101;// MPa
+c_p=1.004;// kJ/(kg.K)
+
+// Calculation
+h_a=c_p*(T_DB-T_ref);// kJ/(kg dry air)
+// From Table C.1b,
+p_sat=0.01235;// MPa
+w=0.622*((phi*p_sat)/(p_m-(phi*p_sat)));// kg watervapor/kg dry air
+p_w=phi*p_sat;// MPa
+h_w=2593.6;// kJ/kg water vapor
+h=h_a+(w*h_w);// kJ/kg dry air
+printf("\nThe value of h#=%3.0f kJ/kg dry air",h);
diff --git a/3831/CH12/EX12.11/Ex12_11.sce b/3831/CH12/EX12.11/Ex12_11.sce new file mode 100644 index 000000000..7f05c6ca2 --- /dev/null +++ b/3831/CH12/EX12.11/Ex12_11.sce @@ -0,0 +1,13 @@ +// Example 12_11
+clc;funcprot(0);
+// Given data
+T_DB1=35.0;// °C
+phi_1=80/100;// The relative humidity
+T_DB3=20.0;// °C
+phi_2=40.0/100;// The relative humidity
+
+// Calculation
+h_1=110;// kJ/kg dry air
+h_3=35;// kJ/kg dry air
+Qbym_a=h_3-h_1;// kJ/kg dry air
+printf("\n The heat transfer rate per unit mass flow rate of dry air required to carry out this process,Q/m_a=%2.0f kJ/kg dry air",Qbym_a);
diff --git a/3831/CH12/EX12.12/Ex12_12.sce b/3831/CH12/EX12.12/Ex12_12.sce new file mode 100644 index 000000000..a278ad620 --- /dev/null +++ b/3831/CH12/EX12.12/Ex12_12.sce @@ -0,0 +1,28 @@ +// Example 12_12
+clc;funcprot(0);
+// Given data
+// State 1
+T_DB1=25.0;// °C
+phi_1=80.0/100;// The relative humidity
+h_1=67;// kg/(kg da)
+w_1=0.016;// kg H2O/(kg da)
+// State 2
+T_DB2=6.0;// °C
+phi_2=100/100;// The relative humidity
+h_2=21;// kg/(kg da)
+h_f2=25.2;// kg/(kg da)
+w_2=0.0056;// kg H2O/(kg da)
+// State 3
+T_DB3=20.0;// °C
+phi_3=40/100;// The relative humidity
+h_3=35;// kg/(kg da)
+w_3=w_2;// kg H2O/(kg da)
+
+// Calculation
+// (a)
+dw=w_1-w_2;// kg H2O/(kg dry air)
+// (b)
+Q_cbym_da=(h_2-h_1)+((w_1-w_2)*h_f2);// kJ/(kg dry air)
+// (c)
+Q_rbym_da=h_3-h_2;// kJ/(kg dry air)
+printf("\n(a)The amount of water removed per unit mass of dry air,w_1-w_2=%0.3f kg H2O/(kg dry air) \n(b)The amount of cooling required per unit mass of dry air,Q_cooling/m_dry air=%2.1f kJ/(kg dry air) \n(c)The reheating heat transfer rate,Q_reheating/m_dry air=%2.0f kJ/(kg dry air)",dw,Q_cbym_da,Q_rbym_da);
diff --git a/3831/CH12/EX12.13/Ex12_13.sce b/3831/CH12/EX12.13/Ex12_13.sce new file mode 100644 index 000000000..ee219a047 --- /dev/null +++ b/3831/CH12/EX12.13/Ex12_13.sce @@ -0,0 +1,37 @@ +// Example 12_13
+clc;funcprot(0);
+// Given data
+V_a1=2000;// ft^3/min
+T_DB1=50.0+459.67;// R
+phi_1=80.0/100;// The relative humidity
+V_a2=1000;// ft^3/min
+T_DB2=100.0+459.67;// R
+phi_2=40.0/100;// The relative humidity
+R_a=53.34;// ft.lbf/(lbm.R)
+p_m=14.7// lbf/in^2
+
+// Calculation
+p_sat1=0.178;// psia
+p_w1=phi_1*p_sat1;// psia
+p_a1=p_m-p_w1;// lbf/in^2
+v_a1=(R_a*T_DB1)/(p_a1*144);// ft^3/(lbm dry air)
+p_sat2=0.9503;// psia
+p_w2=phi_2*p_sat2;// psia
+p_a2=p_m-p_w2;// lbf/in^2
+v_a2=(R_a*T_DB2)/(p_a2*144);// ft^3/(lbm dry air)
+m_a1=V_a1/v_a1;// lbmdry air/min
+m_a2=V_a2/v_a2;// lbmdry air/min
+m_a3=m_a1+m_a2;// lbmdry air/min
+// Then, from the psychrometric chart (Chart D.5), we find
+w_1=44/7000;// lbm water vapor/(lbm dry air)
+w_2=115/7000;// lbm water vapor/(lbm dry air)
+h_1=19;// Btu/(lbm dry air)
+h_2=42;// Btu/(lbm dry air)
+w_3=((m_a1/m_a3)*w_1)+((m_a2/m_a3)*w_2);// grains of water vapor/(lbm dry air)
+h_3=((m_a1/m_a3)*h_1)+((m_a2/m_a3)*h_2);// Btu/(lbm dry air)
+// From the point where the lines ω = 65.8 grains/(lbm dry air) = constant and h = 26 Btu/(lbm dry air) = constant intersect on the psychrometric chart, we can read from this chart that
+T_DB=63;// °F
+T_WB=59;// °F
+phi=75;// %
+T_DP=56;// °F
+printf("\nThe dry bulb temperature of the outlet mixture,T_DB=%2.0f°F \nThe relative humidity of the outlet mixture,phi=%2.0f percentage",T_DB,phi);
diff --git a/3831/CH12/EX12.14/Ex12_14.sce b/3831/CH12/EX12.14/Ex12_14.sce new file mode 100644 index 000000000..c70e6c58c --- /dev/null +++ b/3831/CH12/EX12.14/Ex12_14.sce @@ -0,0 +1,38 @@ +// Example 12_14
+clc;funcprot(0);
+// Given data
+m_methane=3.00;// lbm
+m_propane=4.00;// lbm
+V_m=1.00;// ft^3
+T_m=240.0+459.67;// R
+R=1545.35;// ft.lbf/(lb mole.R)
+
+// Calculation
+m_m=m_methane+m_propane;// lbm
+w_methane=m_methane/m_m;// The mass fraction
+w_propane=m_propane/m_m;// The mass fraction
+// The molecular masses of the components are found in Table C.12a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics as
+M_methane=16.043;// lbm/lbmole
+M_propane=44.097;// lbm/lbmole
+M_m=1/((w_methane/M_methane)+(w_propane/M_propane));// lbm/lb mole
+// From Tables C.12a and C.13a, we find that
+p_c_methane=673;// psia
+p_c_propane=617;// psia
+T_c_methane=343.9;// R
+T_c_propane=665.9;// R
+R_methane=96.3;// ft.lbf/(lbm.R)
+R_propane=35.0;// ft.lbf/(lbm.R)
+v_m=V_m/m_m;// ft^3/lbm
+T_R_methane=T_m/T_c_methane;// The reduced temperature for methane
+v_R_methane=(v_m/w_methane)*((p_c_methane*144)/(R_methane*T_c_methane));// The reduced pseudospecific volume for methane
+T_R=2.03;// The reduced temperature for methane
+v_R=0.975;// The reduced pseudospecific volume for methane
+Z_D_methane=0.975;// The Dalton compressibility factor for methane
+T_R=1.05;// The reduced temperature for propane
+v_R=0.95;// The reduced pseudospecific volume for propane
+Z_D_propane=0.720;// The Dalton compressibility factor for propane
+Z_Dm=(((w_methane*M_m)/M_methane)*Z_D_methane)+(((w_propane*M_m)/M_propane)*Z_D_propane);// The Dalton compressibility factor for the mixture
+R_m=R/M_m;// ft.lbf/lbm.R
+p_m=(Z_Dm*m_m*R_m*T_m)/V_m;// lbf/ft^2
+p_m=p_m/144;// psia
+printf("\nThe total pressure in the tank,p_m=%4.0f psia",p_m);
diff --git a/3831/CH12/EX12.15/Ex12_15.sce b/3831/CH12/EX12.15/Ex12_15.sce new file mode 100644 index 000000000..ac0ea4925 --- /dev/null +++ b/3831/CH12/EX12.15/Ex12_15.sce @@ -0,0 +1,45 @@ +// Example 12_15
+clc;funcprot(0);
+// Given data
+T_m=500;// K
+p_m=20.0;// MPa
+R=8.3143;// kJ/(kg mole.K)
+
+// Calculation
+// Assume a-ammonia,cl-chlorine,no-nitrous oxide
+m_a=1.00;// kg
+m_cl=1.00;// kg
+m_no=1.00;// kg
+m_m=m_a+m_cl+m_no;// kg
+// The mass fractions are
+w_a=m_a/m_m;
+w_cl=m_cl/m_m;
+w_no=m_no/m_m;
+M_a=17.030;// kg/kgmole
+M_cl=70.906;// kg/kgmole
+M_no=44.013;// kg/kgmole
+M_m=1/((w_a/M_a)+(w_cl/M_cl)+(w_no/M_no));// The molecular mass of the mixture in kg/kgmole
+p_c_a=11.280;// MPa
+T_c_a=405.5;// K
+p_c_cl=7.710;// MPa
+T_c_cl=417.0;// K
+p_c_no=7.270;// MPa
+T_c_no=309.7;// K
+R_a=R/M_a;// kJ/kg.K
+R_cl=R/M_cl;// kJ/kg.K
+R_no=R/M_no;// kJ/kg.K
+// The reduced temperatures and pressures are
+T_R_a=T_m/T_c_a;
+p_R_a=p_m/p_c_a;
+T_R_cl=T_m/T_c_cl;
+p_R_cl=p_m/p_c_cl;
+T_R_no=T_m/T_c_no;
+p_R_no=p_m/p_c_no;
+// Using these values on Figure 7.6 gives the following Amagat compressibility factors:
+Z_A_a=0.64;
+Z_A_cl=0.55;
+Z_A_no=0.86;
+Z_Am=(((w_a*M_m)/M_a)*Z_A_a)+(((w_cl*M_m)/M_cl)*Z_A_cl)+(((w_no*M_m)/M_no)*Z_A_no);// The Amagat compressibility factor for the mixture
+R_m=R/M_m;// kJ/kg.K
+V_m=(Z_Am*m_m*R_m*T_m)/(p_m*1000);// m^3
+printf("\nThe total volume occupied by the mixture,V_m=%0.4f m^3",V_m);
diff --git a/3831/CH12/EX12.16/Ex12_16.sce b/3831/CH12/EX12.16/Ex12_16.sce new file mode 100644 index 000000000..f00ad9afc --- /dev/null +++ b/3831/CH12/EX12.16/Ex12_16.sce @@ -0,0 +1,22 @@ +// Example 12_16
+clc;funcprot(0);
+// Given data
+X_N_2=0.7809;// The mole fraction for nitrogen
+X_O_2=0.2095;// The mole fraction for oxygen
+X_Ar=0.00930;// The mole fraction for Argon
+X_CO_2=0.0003;// The mole fraction for Carbondioxide
+
+// Calculation
+// Using Eqs. (12.39) and (12.40), the composition data given in Example 12.2 and the critical point data given in Table C.12b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics give
+p_c_N_2=3.39;// MPa
+p_c_O_2=5.08;// MPa
+p_c_Ar=4.86;// MPa
+p_c_CO_2=7.39;// MPa
+p_c_air=(X_N_2*p_c_N_2)+(X_O_2*p_c_O_2)+(X_CO_2*p_c_CO_2);// MPa
+T_c_N_2=126.2;// K
+T_c_O_2=154.8;// K
+T_c_Ar=151;// K
+T_c_CO_2=304.2;// K
+T_c_air=(X_N_2*T_c_N_2)+(X_O_2*T_c_O_2)+(X_CO_2*T_c_CO_2);// K
+printf("\nThe critical pressure for air,(p_c)_air=%1.2f MPa \nThe critical temperature for air,(T_c)_air=%3.0f K",p_c_air,T_c_air);
+// The answer vary due to round off error
diff --git a/3831/CH12/EX12.17/Ex12_17.sce b/3831/CH12/EX12.17/Ex12_17.sce new file mode 100644 index 000000000..02fde25eb --- /dev/null +++ b/3831/CH12/EX12.17/Ex12_17.sce @@ -0,0 +1,53 @@ +// Example 12_17
+clc;funcprot(0);
+// Given data
+T_m=-100;// °F
+p_m=1500;// psia
+R=1545.35;// ft.lbf/(lbmole.R)
+v_ma=1.315;// ft^3/lb mole
+
+// Calculation
+// (a)For ideal gas mixture behavior,
+v_m=(R*(T_m+459.67))/(p_m*144);// ft^3/lb mole
+Error_a=((v_m-v_ma)/v_ma)*100;// % high
+printf("Results: \n(a)v_m=%1.2f ft^3/mole \n Percentage error=%2.1f percentage high",v_m,Error_a);
+// (b)The Dalton compressibility factor
+// From Table C.12a, we find
+p_c_N_2=492;// psia
+T_c_N_2=227.1;// R
+p_c_CH_4=673;// psia
+T_c_CH_4=343.9;// R
+x_N_2=0.300;// The mole fraction for Nitrogen
+x_CH_4=0.700;// The mole fraction for methane
+vbar_m=1.51;// ft^3/lb mole
+v_R_N_2=(vbar_m*p_c_N_2*144)/(x_N_2*R*T_c_N_2);// The reduced pseudospecific volume for Nitrogen
+v_R_CH_4=(vbar_m*p_c_CH_4*144)/(x_CH_4*R*T_c_CH_4);// The reduced pseudospecific volume for methane
+T_R_N_2=(T_m+459.67)/T_c_N_2;// The reduced temperature for Nitrogen
+T_R_CH_4=(T_m+459.67)/T_c_CH_4;// The reduced temperature for methane
+// From Figure 7.6 in Chapter 7, we find that, for these values
+Z_D_N_2=0.91;// The Dalton compressibility factor for Nitrogen
+Z_D_CH_4=0.39;// The Dalton compressibility factor for methane
+Z_D_m=(x_N_2*Z_D_N_2)+(x_CH_4*Z_D_CH_4);// The Dalton compressibility factor for the mixture
+vbar_m=(Z_D_m*(R*(T_m+459.67)))/(p_m*144);// ft^3/lbmole
+Error_b=((vbar_m-v_ma)/v_ma)*100;// % high
+printf("\n(b)vbar_m=%1.2f ft^3/mole \n Percentage error=%2.1f percentage high",vbar_m,Error_b);
+// (c) The Amagat compressibility factor
+p_R_N_2=p_m/p_c_N_2;// The reduced pressure for Nitrogen
+T_R_N_2=(T_m+459.67)/T_c_N_2;// The reduced temperature for Nitrogen
+p_R_CH_4=p_m/p_c_CH_4;// The reduced pressure for methane
+T_R_CH_4=(T_m+459.67)/T_c_CH_4;// The reduced temperature for nitrogen
+Z_A_N_2=0.84;// The Amagat compressibility factor
+Z_A_CH_4=0.35;// The Amagat compressibility factor
+Z_Am=(x_N_2*Z_A_N_2)+(x_CH_4*Z_A_CH_4);// The Amagat compressibility factor
+vbar_m=(Z_Am*R*(T_m+459.67))/(p_m*144);// ft^3/lbmole
+Error_c=((vbar_m-v_ma)/v_ma)*100;// % high
+printf("\n(c)vbar_m=%1.2f ft^3/mole \n Percentage error=%2.1f percentage high",vbar_m,Error_c);
+// (d) Using Kay’s law, Eqs. (12.39) and (12.40), we get
+p_cm=(x_N_2*p_c_N_2)+(x_CH_4*p_c_CH_4);// psia
+T_cm=(x_N_2*T_c_N_2)+(x_CH_4*T_c_CH_4);// R
+p_Rm=p_m/p_cm;// The reduced pressure
+T_Rm=(T_m+459.67)/T_cm;;// The reduced temperature
+Z_Km=0.51;// The compressibility factor
+vbar_m=(Z_Km*R*(T_m+459.67))/(p_m*144);// ft^3/lbmole
+printf("\n(d)vbar_m=%1.2f ft^3/mole",vbar_m);
+// The answer vary due to round off error
diff --git a/3831/CH12/EX12.2/Ex12_2.sce b/3831/CH12/EX12.2/Ex12_2.sce new file mode 100644 index 000000000..086a34f4f --- /dev/null +++ b/3831/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,25 @@ +// Example 12_2
+clc;funcprot(0);
+// Given data
+X_N_2=78.09;// %
+X_O_2=20.95;// %
+X_Ar=0.930;// %
+X_CO_2=0.0300;// %
+R=8.3143;// kJ/(kg.mole.K)
+
+// Calculation
+// (a)
+M_N_2=28.02;// kg/kgmole
+M_O_2=32.00;// kg/kgmole
+M_Ar=39.94;// kg/kgmole
+M_CO_2=44.01;// kg/kgmole
+M_air=((X_N_2/100)*M_N_2)+((X_O_2/100)*M_O_2)+((X_Ar/100)*M_Ar)+((X_CO_2/100)*M_CO_2);// kg/kgmole
+// (b)
+R_air=R/M_air;// kJ/kg.K
+// (c)
+// Equation (12.13) can be used to determine the corresponding mass or weight fraction composition as
+w_N_2=(X_N_2/100)*(M_N_2/M_air)*100;// % by mass
+w_O_2=(X_O_2/100)*(M_O_2/M_air)*100;// % by mass
+w_Ar=(X_Ar/100)*(M_Ar/M_air)*100;// % by mass
+w_CO_2=(X_CO_2/100)*(M_CO_2/M_air)*100;// % by mass
+printf("\n(a)The equivalent molecular mass,M_air=%2.2f kg/kgmole \n(b)The gas constant for the mixture,R_air=%0.3f kJ \n(c)The composition of air on a mass (or weight) basis,w_N2=%2.2f percentage by mass \n w_O2=%2.2f percentage by mass \n w_Ar=%1.2f percentage by mass \n w_CO2=%0.4f percentage by mass",M_air,R_air,w_N_2,w_O_2,w_Ar,w_CO_2);
diff --git a/3831/CH12/EX12.3/Ex12_3.sce b/3831/CH12/EX12.3/Ex12_3.sce new file mode 100644 index 000000000..b5f51b875 --- /dev/null +++ b/3831/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,27 @@ +// Example 12_3
+clc;funcprot(0);
+// Given data
+X_CO_2=9.51;// %
+X_H_2O=19.01;// %
+X_N_2=71.48;// %
+M_N_2=28.02;// kg/kgmole
+M_CO_2=44.01;// kg/kgmole
+M_H_2O=18.02;// kg/kgmole
+p_m=14.7;// psia
+
+// Calculation
+// (a)
+// For ideal gas behavior, Eq. (12.23) tells us that the mole fractions, volume fractions, and the pressure fractions are all the same, or
+Shi_CO_2=X_CO_2;// The volume fraction in %
+Shi_H_2O=X_H_2O;// The volume fraction in %
+Shi_N_2=X_N_2;// The volume fraction in %
+pi_CO_2=X_CO_2;// The pressure fraction in %
+pi_H_2O=X_H_2O;// The pressure fraction in %
+pi_N_2=X_N_2;// The pressure fraction in %
+M_m=(X_CO_2*M_CO_2)+(X_N_2*M_N_2)+(X_H_2O*M_H_2O);// The equivalent molecular mass of this ideal gas mixture in kg/kgmole
+w_CO_2=Shi_CO_2*(M_CO_2/M_m);// The mass fraction in %
+w_H_2O=Shi_H_2O*(M_H_2O/M_m);// The mass fraction in %
+w_N_2=Shi_N_2*(M_N_2/M_m);// The mass fraction in %
+// (b)
+p_H_2O=p_m*X_H_2O/100;// The partial pressure of the water vapor in the exhaust gas mixture in psia
+printf("\n(a)The volume fraction compostion of the mixture,Shi_CO_2=%1.2f percentage \n Shi_H_2O=%2.2f percentage \n Shi_N_2=%2.2f percentage \n(b)The partial pressure of the water vapor in the exhaust gas mixture,p_H2O=%1.2f psia",Shi_CO_2,Shi_H_2O,Shi_N_2,p_H_2O);
diff --git a/3831/CH12/EX12.4/Ex12_4.sce b/3831/CH12/EX12.4/Ex12_4.sce new file mode 100644 index 000000000..31488a79e --- /dev/null +++ b/3831/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,33 @@ +// Example 12_4
+clc;funcprot(0);
+// Given data
+X_O_2=0.2095;// The mole fraction for oxygen
+p_m=0.1013;// MN/m^2
+d=100;// m
+M_O_2=32.00;// The molecular mass of oxygen
+M_He=4.003;// The molecular mass of helium
+R=8.3143;// kJ/(kgmole.K)
+
+// Calculation
+// (a)
+p_O_2=X_O_2*p_m;// MN/m^2
+p_m=1.08;// MN/m^2
+X_O_2=p_O_2/p_m;// The mole fraction for oxygen
+Shi_O_2=X_O_2;// The volume fraction for oxygen
+pi_O_2=X_O_2;// The pressure fraction for oxygen
+X_He=1-X_O_2;// The mole fraction for helium
+Shi_He=X_He;// The volume fraction for oxygen
+M_m=(X_O_2*M_O_2)+(X_He*M_He);// kg/kgmole
+w_O_2=X_O_2*(M_O_2/M_m);// The mass fraction for oxygen
+w_He=1-w_O_2;// The mass fraction for helium
+printf("\n(a)The mole and volume fraction of oxygen,X_O2=Shi_O2=pi_O2=%0.4f \n The helium mole and volume fractions,X_He=Shi_He=%0.3f \n The mixture equivalent molecular mass,M_m=%1.2f kg/kgmole",X_O_2,X_He,M_m);
+// (b)
+R_m=R/M_m;// kJ/(kg.K)
+c_vO_2=0.657;// kJ/(kg.K)
+c_vHe=3.123;// kJ/(kg.K)
+c_pO_2=0.917;// kJ/(kg.K)
+c_pHe=5.200;// kJ/(kg.K)
+c_vm=(w_O_2*c_vO_2)+(w_He*c_vHe);// kJ/(kg.K)
+c_pm=(w_O_2*c_pO_2)+(w_He*c_pHe);// kJ/(kg.K)
+k_m=c_pm/c_vm;// The specific heat ratio of the mixture
+printf("\n(b)The mixture equivalent gas constant,R_m=%1.2f kJ/(kg.K) \n The mixture specific heats,c_vm=%1.2f kJ/(kg.K) & c_pm=%1.2f kJ/(kg.K) \n The specific heat ratio of the mixture,k_m=%1.2f",R_m,c_vm,c_pm,k_m);
diff --git a/3831/CH12/EX12.5/Ex12_5.sce b/3831/CH12/EX12.5/Ex12_5.sce new file mode 100644 index 000000000..818e2e88a --- /dev/null +++ b/3831/CH12/EX12.5/Ex12_5.sce @@ -0,0 +1,15 @@ +// Example 12_5
+clc;funcprot(0);
+// Given data
+T_m1=20.0+273.15;// K
+p_m1=0.101;// MN/m^2
+p_m2=1.08;// MN/m^2
+k_m=1.66;// The specific heat ratio of the mixture
+c_pm=4.61;// kJ/kg.K
+
+// Calculation
+T_m2=T_m1*((p_m2/p_m1)^((k_m-1)/k_m));// K
+T_m2C=T_m2-273.15;// °C
+Wbym_m=c_pm*(T_m1-T_m2);// kJ/kg
+printf("\nThe power per unit mass flow rate required to isentropically compress the helium-oxygen mixture,Wdot/mdot_m=%4.0f kJ/kg",Wbym_m);
+// The answer provided in the textbook is wrong
diff --git a/3831/CH12/EX12.6/Ex12_6.sce b/3831/CH12/EX12.6/Ex12_6.sce new file mode 100644 index 000000000..e2f349ef9 --- /dev/null +++ b/3831/CH12/EX12.6/Ex12_6.sce @@ -0,0 +1,21 @@ +// Example 12_6
+clc;funcprot(0);
+// Given data
+T=25.0;// °C
+p_m=.101*10^3;// MPa
+phi=56.8/100;// The relative humidity
+
+// Calculation
+// (a)
+// From Table C.1b, we find that
+p_sat=0.003169;// MPa
+p_w=phi*p_sat*10^3;// kPa
+// (b)
+// From Dalton’s law for partial pressure, we can find the partial pressure of the dry air in the mixture as
+p_a=p_m-p_w;// kPa
+w=0.622*(p_w/p_a);// kg H2O per kg of dry air
+// (c)
+// Using Eq. (12.27) and Table C.2b, we find the dew point temperature to be
+T_sat=15.8;// °C
+T_DP=T_sat;// °C
+printf("\n(a)The partial pressure of the water vapor in the atmosphere,p_w=%1.2f kPa \n(b)The humidity ratio of the atmosphere,w=%0.4f kg H2O per kg of dry air \n(c)The dew point temperature of the atmosphere,T_DP=%2.1f°C",p_w,w,T_DP);
diff --git a/3831/CH12/EX12.8/Ex12_8.sce b/3831/CH12/EX12.8/Ex12_8.sce new file mode 100644 index 000000000..fde011a2f --- /dev/null +++ b/3831/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,19 @@ +// Example 12_8
+clc;funcprot(0);
+// Given data
+T_WB=60.0;// °F
+T_DB=70.0;// °F
+
+// Calculation
+// From Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find
+h_g1=1092.0;// Btu/lbm
+h_fg2=1059.6;// Btu/lbm
+h_f2=28.1;// Btu/lbm
+p_sat=0.2563;// psia
+p_w3=p_sat;// psia
+p_m=14.7;// psia
+w_3=0.622*((p_w3)/(p_m-p_w3));// lbm water per lbm of dry air
+c_pa=0.240;// Btu/lbm.R
+w_1=((c_pa*(T_WB-T_DB))+(w_3*h_fg2))/(h_g1-h_f2);// lbm water per lbm of dry air
+w_1=w_1*7000;// grains of water per lbm of dry air
+printf("\nThe humidity ratio (ω) in the room,w_1=%2.1f grains of water per lbm of dry air",w_1);
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);
diff --git a/3831/CH14/EX14.1/Ex14_1.sce b/3831/CH14/EX14.1/Ex14_1.sce new file mode 100644 index 000000000..ad4f35593 --- /dev/null +++ b/3831/CH14/EX14.1/Ex14_1.sce @@ -0,0 +1,14 @@ +// Example 14_1
+clc;funcprot(0);
+// Given data
+T_L=20.0+273.15;// K
+T_H=200.0+273.15;// K
+
+// Solution
+// (a)
+n_T_carnot=(1-(T_L/T_H))*100;// The thermal efficiency of a Carnot engine in %
+// (b)
+COP_Carnot_HP=T_H/(T_H-T_L);// The coefficient of performance of a Carnot heat pump
+// (c)
+COP_Carnot_RAC=T_L/(T_H-T_L);// The coefficient of performance of a Carnot refrigerator
+printf("\n(a)The thermal efficiency of a Carnot engine,(n_T)_Carnot=%2.0f percentage \n(b)The coefficient of performance of a Carnot heat pump,COP_Carnot HP=%1.2f \n(c)The coefficient of performance of a Carnot refrigerator or air conditioner,COP_Carnot R/AC=%1.2f",n_T_carnot,COP_Carnot_HP,COP_Carnot_RAC);
diff --git a/3831/CH14/EX14.10/Ex14_10.sce b/3831/CH14/EX14.10/Ex14_10.sce new file mode 100644 index 000000000..dfe7a8af1 --- /dev/null +++ b/3831/CH14/EX14.10/Ex14_10.sce @@ -0,0 +1,22 @@ +// Example 14_10
+clc;funcprot(0);
+// Given data
+PR=2.00;// The pressure ratio
+T_1=70+460;// R
+T_3=80.0+459.67;// R
+n_s_e=65/100;// The isentropic efficiency of the expander
+n_s_c=65/100;// The isentropic efficiency of the compressor
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+COP_rBa=(((PR)^((k-1)/k))-1)^-1;// The COP for a reversed Brayton R/AC ASC
+T_4s=T_3/((PR)^((k-1)/k));// R
+T_4s=T_4s-459.67;// °F
+// (b)
+T_4=T_3-((T_3-(T_4s+459.67))*(n_s_e));// R
+T_4=T_4-459.67;// °F
+T_2s=(T_1*T_3)/(T_4s+459.67);// R
+COP_rB=(T_1-(T_4+459.67))/(((T_2s-T_1)/(n_s_c))-((T_3-(T_4s+459.67))*(n_s_e)));// The COP for a reversed Brayton cycle R/AC
+printf("\n(a)The COP for a reversed Brayton R/AC ASC,COP=%1.2f \n(b)The COP for a reversed Brayton cycle R/AC,COP=%0.3f",COP_rBa,COP_rB);
+// The answer provided in the textbook is wrong
diff --git a/3831/CH14/EX14.11/Ex14_11.sce b/3831/CH14/EX14.11/Ex14_11.sce new file mode 100644 index 000000000..721ce3f6e --- /dev/null +++ b/3831/CH14/EX14.11/Ex14_11.sce @@ -0,0 +1,27 @@ +// Example 14_11
+clc;funcprot(0);
+// Given data
+m=4.00;// lbm/s
+T_1=530;// R
+p_1=1;// psia
+p_2s=3.00// psia
+p_3=3.00;// psia
+p_4s=1;// psia
+T_3=600;// R
+c_p=0.240;// Btu/(lbm.R)
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+T_4s=T_3*((p_4s/p_3)^((k-1)/k));// R
+W_expander=m*c_p*(T_3-T_4s);// Btu/s
+// (b)
+T_2s=T_1*((p_2s/p_1)^((k-1)/k));// R
+W_compressor=m*c_p*(T_1-T_2s);// Btu/s
+// (c)
+PR=p_3/p_4s;// Pressure ratio
+COP=(((PR)^((k-1)/k))-1)^-1;// The coefficient of performance of the unit
+// (d)
+W_net=(abs(W_compressor)-W_expander);// Btu/s
+Q_L=(COP*(W_net)*60*(1/200));// The refrigeration capacity of the unit in tons
+printf("\n(a)The expander power,W_expander=%3.0f Btu/s \n(b)The compressor power,W_compressor=%3.0f Btu/s \n(c)The coefficient of performance of the unit,COP=%1.2f \n(d)The refrigeration capacity of the unit in tons,Q_L=%2.1f tons of refrigeration",W_expander,W_compressor,COP,Q_L);
diff --git a/3831/CH14/EX14.12/Ex14_12.sce b/3831/CH14/EX14.12/Ex14_12.sce new file mode 100644 index 000000000..b6ed9d236 --- /dev/null +++ b/3831/CH14/EX14.12/Ex14_12.sce @@ -0,0 +1,15 @@ +// Example 14_12
+clc;funcprot(0);
+// Given data
+T_H=22.0+273.15;// K
+T_L=65;// K
+Q_cooling=0.100;// J/s
+W_input=3.00;// W
+V=12;// V
+
+// Calculation
+// (a)
+COP_rS=T_L/(T_H-T_L);// The Stirling ASC coefficient of performance of a refrigeration unit
+// (b)
+COP_rSact=Q_cooling/W_input;// The actual coefficient of performance of the unit
+printf("\n(a)The Stirling ASC coefficient of performance of a refrigeration unit,COP_reversed Stirling ASC R/AC=%0.3f \n(b)The actual coefficient of performance of the unit,COP_reversed Stirling actual R/AC=%0.4f",COP_rS,COP_rSact);
diff --git a/3831/CH14/EX14.13/Ex14_13.sce b/3831/CH14/EX14.13/Ex14_13.sce new file mode 100644 index 000000000..20152442c --- /dev/null +++ b/3831/CH14/EX14.13/Ex14_13.sce @@ -0,0 +1,15 @@ +// Example 14_13
+clc;funcprot(0);
+// Given data
+mu_j=0.0300;// °F/psi
+p_1=300;// psia
+p_2=14.7;// psia
+T_1=70.0;// °F
+n_s_c=90.0/100;// The isentropic efficiency of the air compressor
+k=1.40;// The specific heat ratio
+
+// Calculation
+dT=mu_j*(p_2-p_1);// °F
+T_2=T_1+dT;// °F
+COP_RAC=(mu_j*(p_2-p_1))/(((T_1+459.67)*[((p_2/p_1)^((k-1)/k))-1])/n_s_c);// The coefficient of performance
+printf("\nThe outlet temperature,T_2=%2.1f°F \nCOP of a Joule-Thomson expansion throttling device,COP=%0.4f",T_2,COP_RAC);
diff --git a/3831/CH14/EX14.14/Ex14_14.sce b/3831/CH14/EX14.14/Ex14_14.sce new file mode 100644 index 000000000..d4a6a6371 --- /dev/null +++ b/3831/CH14/EX14.14/Ex14_14.sce @@ -0,0 +1,60 @@ +// Example 14_14
+clc;funcprot(0);
+// Given data
+m_ref=0.500;// kg/s
+T_0=25.0;// °C
+n_c=70.0;// The isentropic efficiency of compressor
+// Using Figure 14.36 as the illustration for this example, the properties at the four stations can be found in Tables C.7e, C.7f, and C.8d as
+// Station 1
+// Compressor inlet
+x_1=1.00;// The quality of steam
+T_1=-20.0;// °C
+h_1=235.31;// kJ/kg
+s_1=0.9332;// kJ/kg.K
+p_1=132.99;// kPa
+// Station 2
+// Compressor outlet
+p_2s=800;// kPa
+s_2=s_1;// kJ/kg.K
+h_2s=271.10;// kJ/kg
+T_2s=39.8;// °C
+// Station 3
+// Condenser outlet
+x_3=0.00;// The quality of steam
+p_3=725;// kPa
+h_3=87.46;// kJ/kg
+s_3=0.3257;// kJ/kg.K
+T_3=27.9;// °C
+// Station 4h
+// Expansion valve outlet
+h_4h=h_3;// kJ/kg
+p_4h=160;// kPa
+h_4h=87.46;// kJ/kg
+x_4h=0.280;// The quality of steam
+s_4h=0.3449;// kJ/kg.K
+T_4h=-15.6;// °C
+T_e=-15.6;// °C
+
+// Calculation
+// (a)
+h_2=((h_2s-h_1)/(n_c/100))+h_1;// kJ/kg
+p_2=p_2s;// kPa
+//Interpolation in Table C.7f in Thermodynamic Tables to accompany Modern Engineering Thermodynamics (or through the use of an appropriate computer program) gives the following additional properties at this state:
+s_2=0.9814;// kJ/kg.K
+T_2=54.97;// °C
+Q_condenser=m_ref*(h_3-h_2);// kJ/s
+Q_evaporator=m_ref*(h_1-h_4h);// kJ/s
+Q_compressor=m_ref*(h_2-h_1);// kJ/s
+I_ac=m_ref*(T_0+273.15)*(s_2-s_1);// kW
+I_con=(T_0+273.15)*((m_ref*(s_3-s_2))-(Q_condenser/(T_0+273.15)));// kW
+I_ev=m_ref*(T_0+273.15)*(s_4h-s_3);// kW
+I_e=(T_0+273.15)*((m_ref*(s_1-s_4h))-(Q_evaporator/(T_e+273.15)));// kW
+I_total=I_ac+I_con+I_ev+I_e;// kW
+W_compressor=Q_compressor;// kW
+// (b)
+COP=Q_evaporator/W_compressor;// The system coefficient of performance
+T_L=T_e;// °C
+COP_act=2.85;// The second law efficiency for a refrigeration system
+E_RAC=(abs(1-((T_0+273.15)/(T_e+273.15)))*COP_act)*100;// %
+printf("\n(a)The irreversibility rate of each component in the system are given below: \n I_adiabatic compressor=%1.2f kW \n I_condenser=%1.2f kW \n I_expansion valve=%1.2f kW \n I_evaporator=%1.2f kW \n The total irreversibility rate of the system,I_total=%2.0f kW \n(b)The system coefficient of performance,COP=%1.2f \n The second law efficiency for a refrigeration system,E_RAC=%2.1f percentage",I_ac,I_con,I_ev,I_e,I_total,COP,E_RAC);
+// The answer provided in the text book is wrong(The value of h_2 changed little bit)
diff --git a/3831/CH14/EX14.2/Ex14_2.sce b/3831/CH14/EX14.2/Ex14_2.sce new file mode 100644 index 000000000..8222d8ec1 --- /dev/null +++ b/3831/CH14/EX14.2/Ex14_2.sce @@ -0,0 +1,12 @@ +// Example 14_2
+clc;funcprot(0);
+// Given data
+V=2.50*10^16;// m^3
+T=0.00;// °C
+t=24.0;// s
+rho=917;// kg/m^3
+
+// Solution
+m_ice=V*rho*2.2046;// lbm
+Q=m_ice/(2*10^3);// tons of refrigeration
+printf("\nThe tons of refrigeration produced,Q=%1.2e tons of refrigeration",Q);
diff --git a/3831/CH14/EX14.3/Ex14_3.sce b/3831/CH14/EX14.3/Ex14_3.sce new file mode 100644 index 000000000..fb994c0b5 --- /dev/null +++ b/3831/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,60 @@ +// Example 14_3
+clc;funcprot(0);
+// Given data
+T_H=20.0+273.15;// K
+T_L=-15.0+273.15;// K
+
+// Solution
+// (a)
+COP_Cr=T_L/(T_H-T_L);// COP of a reversed Carnot cycle
+// From Table C.9b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, the thermodynamic data at the monitoring stations shown in the schematic are
+// Station 1
+T_1=-15.0;// °C
+s_1=0.89973;// kJ/(kg.K)
+s_2s=s_1;// kJ/(kg.K)
+x_1=0.9395;// The quality of steam
+h_1=231.0;// kJ/kg
+s_f1=0.11075;// kJ/(kg.K)
+s_fg1=0.83977;// kJ/(kg.K)
+h_f1=27.33;// kJ/kg
+h_fg1=216.79;// kJ/kg
+// Station 2
+T_2s=20.0;// °C
+x_2s=1.00;// The quality of steam
+h_2s=256.5;// kJ/kg
+s_2s=0.89973;// kJ/(kg.K)
+p_2s=909.9;// kPa
+// Station 3
+T_3=20.0;// °C
+x_3=0.00;// The quality of steam
+h_3=68.67;// kJ/kg
+s_3=0.25899;// kJ/(kg.K)
+p_3=p_2s;// kPa
+// Station 4
+T_4s=T_1;// °C
+s_4s=s_3;// kJ/(kg.K)
+x_4s=0.1765;// The quality of steam
+h_4s=65.6;// kJ/kg
+s_f4=0.11075;// kJ/(kg.K)
+s_fg4=0.83977;// kJ/(kg.K)
+h_f4=h_f1;// kJ/kg
+h_fg4=h_fg1;// kJ/kg
+x_1=(s_2s-s_f1)/s_fg1;
+h_1=h_f1+(x_1*h_fg1);// kJ/kg
+// where we have calculated
+x_4s=(s_3-s_f4)/s_fg4;// The quality of steam
+h_4s=h_f4+(x_4s*h_fg4);// kJ/kg
+Q_L=h_1-h_4s;// kJ/kg
+W_c=h_2s-h_1;// kJ/kg
+W_t=h_3-h_4s;// kJ/kg
+COP_et=Q_L/(W_c-W_t);// COP for isentropic vapour compressor cycle with expansion turbine
+// (c)
+// Station 4h
+T_4h=T_1;// °C
+h_4h=h_3;// kJ/kg
+x_4h=(h_4h-h_f4)/h_fg4;// The quality of steam
+s_4h=s_f4+(x_4h*s_fg4);// kJ/(kg.K)
+Q_L=h_1-h_4h;// kJ/kg
+W_c=h_2s-h_1;// kJ/kg
+COP_tv=Q_L/W_c;// COP for isentropic vapor-compression cycle with throttling valve
+printf("\n(a)COP_carnot refrigerator=%1.2f \n(b)COP_isentropic vapour compressor cycle with expansion turbine=%1.2f \n(c)COP_isentropic vapor-compression cycle with throttling valve=%1.2f",COP_Cr,COP_et,COP_tv);
diff --git a/3831/CH14/EX14.4/Ex14_4.sce b/3831/CH14/EX14.4/Ex14_4.sce new file mode 100644 index 000000000..c3731ece2 --- /dev/null +++ b/3831/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,31 @@ +// Example 14_4
+clc;funcprot(0);
+// Given data
+// Station 1
+x_1=1.00;// The dryness fraction
+T_1=-15.0;// °C
+h_1=244.13;// kJ/kg
+s_1=0.95052;// kJ/kg.K
+// Station 2
+p_2s=909.9;// kPa
+s_2s=0.95052;// kJ/kg.K
+s_2s=s_1;// kJ/kg.K
+h_2s=271.92;// kJ/kg
+T_2s=39.3;// °C
+// Station 3
+T_3=20.0;// °C
+x_3=0.00;// The dryness fraction
+h_3=68.67;// kJ/kg
+s_3=0.25899;// kJ/kg.K
+// Station 4h
+T_4h=T_1;// °C
+h_4h=h_3;// kJ/kg
+x_4h=0.1910;// The quality of steam
+s_4h=0.27088;// kJ/kg.K
+n_c=75.0/100;// The isentropic efficiency of compressor
+
+// Calculation
+Q=h_1-h_4h;// kJ/kg
+W_c=(h_2s-h_1)/n_c;// kJ/kg
+COP_vc=Q/W_c;// The coefficient of performance of vapor compression cycle
+printf("\nCOP_vapor compression cycle R/AC=%1.2f",COP_vc);
diff --git a/3831/CH14/EX14.6/Ex14_6.sce b/3831/CH14/EX14.6/Ex14_6.sce new file mode 100644 index 000000000..94ecc7294 --- /dev/null +++ b/3831/CH14/EX14.6/Ex14_6.sce @@ -0,0 +1,60 @@ +// Example 14_6
+clc;funcprot(0);
+// Given data
+// Loop A
+// Station 1A
+// Compressor A in
+x_1A=1.00;// The dryness fraction
+T_1A=-20.0;// °C
+h_1A=242.05;// kJ/kg
+s_1A=0.95927;// kJ/kg.K
+p_1A=244.8;// kPa
+// Station 2sA
+// Compressor A out
+p_2A=1500;// kPa
+s_2A=s_1A;// kJ/kg.K
+h_2sA=289.08;// kJ/kg
+T_2sA=71.07;// °C
+// Station 3A
+// Condenser A out
+x_3A=0.00;// The dryness fraction
+T_3A=25.0;// °C
+h_3A=74.91;// kJ/kg
+// Station 4hA
+// Expansion A out
+h_4hA=h_3A;// kJ/kg
+// Loop B
+// Station 1B
+// Compressor B in
+x_1B=1.00;// The dryness fraction
+T_1B=-50.0;// °C
+h_1B=228.51;// kJ/kg
+s_1B=1.02512;// kJ/kg.K
+p_1B=63.139;// kPa
+// Station 2sB
+// Compressor B out
+p_2B=300;// kPa
+s_2B=s_1B;// kJ/kg.K
+h_2sB=264.05;// kJ/kg
+T_2sB=15.0;// °C
+// Station 3B
+// Condenser B out
+x_3B=0.00;// The dryness fraction
+T_3B=-20.0;// °C
+h_3B=21.73;// kJ/kg
+// Station 4hB
+// Expansion B out
+h_4hB=h_3B;// kJ/kg
+Q_L=40.0;// tons of refrigeration
+n_s=80/100;// The isentropic efficiencies of both compressors
+
+// (a)
+m_B=(Q_L*210*1/60)/(h_1B-h_4hB);// kg/s
+h_2B=((h_2sB-h_1B)/n_s)+h_1B;// kJ/kg
+m_A=m_B*((h_2B-h_3B)/(h_1A-h_4hA));// kg/s
+// (b)
+COP_dc=(m_B*(h_1B-h_4hB))/(((m_A*(h_2sA-h_1A))/n_s)+((m_B*(h_2sB-h_1B))/n_s));// The coeeficient of performance
+// (c)
+PR_cA=p_2A/p_1A;// The compressor pressure ratio
+PR_cB=p_2B/p_1B;// The compressor pressure ratio
+printf("\n(a)The mass flow rate of refrigerant in loops A and B,m_A=%1.2f kg/s & m_B=%0.3f kg/s \n(b)The system’s coefficient of performance,COP_dual cascade=%1.2f \n(c)The pressure ratios across each of the compressors,PR_compressorA=%1.2f & PR_compressorA=%1.2f",m_A,m_B,COP_dc,PR_cA,PR_cB);
diff --git a/3831/CH14/EX14.7/Ex14_7.sce b/3831/CH14/EX14.7/Ex14_7.sce new file mode 100644 index 000000000..e076d2bf8 --- /dev/null +++ b/3831/CH14/EX14.7/Ex14_7.sce @@ -0,0 +1,65 @@ +// Example 14_7
+clc;funcprot(0);
+// Given data
+// Loop A
+// Station 1A
+// Compressor A inlet
+p_1A=500;// kPa
+h_1A=265.60;// kJ/kg
+s_1A=0.9486;// kJ/kg.K
+// Station 2sA
+// Compressor A outlet
+p_2sA=1600;// kPa
+s_2A=s_1A;// kJ/kg.K
+h_2A=256.60;// kJ/kg
+// Station 3A
+// Condenser A outlet
+x_3A=0.00;// The quality of steam
+p_3A=1600;// kPa
+h_3A=134.02;// kJ/kg
+// Station 4hA
+// Expansion valve A outlet
+h_4hA=h_3A;// kJ/kg
+// Loop B
+// Station 1B
+// Compressor B inlet
+x_1B=1.00;// The quality of steam
+p_1B=100;// kPa
+h_1B=231.35;// kJ/kg
+s_1B=0.9395;// kJ/kg.K
+
+// Station 2sB
+// Compressor B outlet
+p_2B=500;// kPa
+s_2sB=s_1B;// kJ/kg.K
+h_2sB=264.25;// kJ/kg
+T_2sB=15.0;// °C
+// Station 3B
+// Condenser B outlet
+x_3B=0.00;// The quality of steam
+p_3B=500;// kPa
+h_3B=71.33;// kJ/kg
+// Station 4hB
+// Expansion B outlet
+h_4hB=h_3B;// kJ/kg
+Q_L=10.0;// tons
+n_s_c_A=80/100;// The isentropic efficiency of compressor A
+n_s_c_B=80/100;// The isentropic efficiency of compressor B
+
+// Calculation
+// (a)
+h_2B=((h_2sB-h_1B)/n_s_c_B)+h_1B;// kJ/kg
+h_f_500kPa=71.33;// kJ/kg
+h_f_1600kPa=134.02;// kJ/kg
+h_fg_500kPa=184.74;// kJ/kg
+x_flash=((h_f_1600kPa-h_f_500kPa)/h_fg_500kPa)*100;// The quality of the vapor exiting the flash chamber
+h_gflash=252.07;// kJ/kg
+h_1A=((x_flash/100)*h_gflash)+((1-(x_flash/100))*h_2B);// kJ/kg
+m_B=(Q_L*210*1/60)/(h_1B-h_4hB);// kg/s
+m_A=m_B/(1-(x_flash/100));// kg/s
+// (b)
+h_2sA=292.33;// kJ/kg
+COP_ds=[(1-(x_flash/100))*(h_1B-h_4hB)]/[((h_2sA-h_1A)/n_s_c_A)+((1-(x_flash/100))*((h_2B-h_1B)/n_s_c_B))];// The system’s coefficient of performance.
+// (c)
+W_comp=m_A*[((h_2sA-h_1A)/n_s_c_A)+((1-(x_flash/100))*((h_2B-h_1B)/n_s_c_B))];// The total compressor power in kW
+printf("\n(a)The mass flow rate of the two refrigerants,m_A=%0.3f kg/s & m_B=%0.3f kg/s \n(b)The system’s coefficient of performance,COP_dual stage=%1.2f \n(c)The total power required by the compressors,W_compressors=%2.1f kW",m_A,m_B,COP_ds,W_comp);
diff --git a/3831/CH14/EX14.8/Ex14_8.sce b/3831/CH14/EX14.8/Ex14_8.sce new file mode 100644 index 000000000..7324a8d5a --- /dev/null +++ b/3831/CH14/EX14.8/Ex14_8.sce @@ -0,0 +1,10 @@ +// Example 14_8
+clc;funcprot(0);
+// Given data
+T_g=100;// °C
+T_e=5.00;// °C
+T_a=20.0;// °C
+
+// Calculation
+COP_Car=((T_e+273.15)/(T_g+273.15))*((T_g-T_a)/(T_a-T_e));// The Carnot absorption refrigeration coefficient of performance
+printf("\nThe Carnot absorption refrigeration coefficient of performance,(COP)_Carnot absorption refrigerator=%1.2f",COP_Car);
diff --git a/3831/CH14/EX14.9/Ex14_9.sce b/3831/CH14/EX14.9/Ex14_9.sce new file mode 100644 index 000000000..cd16f05e0 --- /dev/null +++ b/3831/CH14/EX14.9/Ex14_9.sce @@ -0,0 +1,41 @@ +// Example 14_9
+clc;funcprot(0);
+// Given data
+Q_R=422;// kJ/h
+Q_F=422;// kJ/h
+// Station 1- Compressor inlet
+x_1=1.00;// The quality of steam
+T_1=-18.0;// °C
+h_1=236.53;// kJ/kg
+s_1=0.9315;// kJ/kg.K
+// Station 2- Compressor outlet
+s_2=s_1;// kJ/kg.K
+p_sat=0.770;// MPa
+p_3=p_sat;// MPa
+p_2s=p_3;// MPa
+h_2s=271.0;// kJ/kg
+// Station 3- Condenser outlet
+x_3=0.00;// The quality of steam
+T_3=30.0;// °C
+p_3=0.770;// MPa
+h_3=91.49;// kJ/kg
+// Station 4h-Refrigerator evaporator inlet
+h_4h=h_3;// kJ/kg
+T_4h=4.00;// °C
+h_f=55.35;// kJ/kg
+h_fg=194.19;// kJ/kg
+// Station 5-Refrigerator evaporator outlet
+T_5=T_4h;// °C
+// Station 6h-Freezer evaporator outlet
+T_6h=-18.0;// °C
+n_s_c=0.80;// The isentropic efficency of the compressor
+
+// Calculation
+// (a)
+COP=(h_1-h_4h)/((h_2s-h_1)/n_s_c);// The coefficient of performance
+// (b)
+m_ref=((Q_R+Q_F)*(1/60))/(h_1-h_4h);// kg/min
+// (c)
+h_5=h_4h+((Q_R*1/60)/m_ref);// kJ/kg
+x_5=((h_5-h_f)/h_fg)*100;// The quality at the exit of the refrigeration evaporator
+printf("\n(a)The coefficient of performance,COP=%1.2f \n(b)The mass flow rate of refrigerant required,m_ref=%0.4f kg/min \n(c)The quality at the outlet of the refrigeration evaporator,x_5=%2.1f percentage",COP,m_ref,x_5);
diff --git a/3831/CH15/EX15.10/Ex15_10.sce b/3831/CH15/EX15.10/Ex15_10.sce new file mode 100644 index 000000000..afc29fb37 --- /dev/null +++ b/3831/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,35 @@ +// Example 15_10
+clc;funcprot(0);
+// Given data
+T=25.0;// °C
+p=0.100;// MPa
+
+// Calculation
+// (a)
+h_f_C8H18=-249.952;// MJ/kgmole
+h_fuel=h_f_C8H18;// MJ/kgmole
+h_f_CO2=-393.522;// MJ/kgmole
+h_f_H2O_g=-241.827;// MJ/kgmole
+h_f_N_2=0;// MJ/kgmole
+h_fi=(8*h_f_CO2)+(9*h_f_H2O_g)+(47*h_f_N_2);// MJ/kgmole of C8H18
+c_p_CO2=0.05818;// MJ/kgmole.K
+c_p_H2O=0.04250;// MJ/kgmole.K
+c_p_N2=0.03118;// MJ/kgmole.K
+c_pi_avg=(8*c_p_CO2)+(9*c_p_H2O)+(47*c_p_N2);// MJ/kgmole of C8H18.K
+T_Aos=((h_f_C8H18-(h_fi))/c_pi_avg)*1.8;// °F
+printf("\n(a)The open system (constant pressure) adiabatic flame temperature burning with 100.percent theoretical air,T_A|open system=%4.0f°F",T_Aos);
+// (b)
+c_p_O2=0.03299;// MJ/(kgmole).K
+c_pi_avg=(8*c_p_CO2)+(9*c_p_H2O)+(12.5*c_p_O2)+(94*c_p_N2);// MJ/kgmole of C8H18.K
+T_Aos=(((h_f_C8H18-(h_fi))/c_pi_avg)+T)*1.8;// °F
+printf("\n(b)The open system (constant pressure) adiabatic flame temperature burning with 200.percent theoretical air,T_A|open system=%4.0f°F",T_Aos)
+// (c)
+R=0.0083143;// // MJ/kgmole.K
+N=h_fuel-h_fi-((R*(T+273.15))*[1+(12.4*4.76)-(8+9+47)]);// The numerator in MJ/kgmole of C8H18.K
+c_v_CO2=0.04987;// MJ/kgmole.K
+c_v_H2O=0.03419;// MJ/kgmole.K
+c_v_N2=0.02287;// MJ/kgmole.K
+c_vi_avg=(8*c_v_CO2)+(9*c_v_H2O)+(47*c_v_N2);// MJ/kgmole.K
+T_Acs=T+(N/c_vi_avg);// The denominator in MJ/kgmole of C8H18.K
+T_Acs=(T_Acs*1.8)+32;// °F
+printf("\n(c)The closed system (constant volume) adiabatic flame temperature burning with 100.percent theoretical air,T_A|closed system=%4.0f°F",T_Acs);
diff --git a/3831/CH15/EX15.11/Ex15_11.sce b/3831/CH15/EX15.11/Ex15_11.sce new file mode 100644 index 000000000..6c634b591 --- /dev/null +++ b/3831/CH15/EX15.11/Ex15_11.sce @@ -0,0 +1,34 @@ +// Example 15_11
+clc;funcprot(0);
+// Given data
+T=25+273.15;// K
+m_f=0.0100;// kg
+M_octane=114;// kg/kg mole
+R=1545.35;// ft.lbf/(lbmole.R)
+V_p=50.0*10^-3;// ft^3
+R_u=0.0083143;// MJ/kgmole.K
+
+// Calculation
+m_oct=m_f/M_octane;// kgmole
+// The reaction equation for 50.0% excess pure oxygen is C8H18+1.5(12.5)O2--->8(CO2)+9(H2O)+6.25(O2)
+n_CO2=8;// The stoichiometric coefficient of the reaction
+n_H2O=9;// The stoichiometric coefficient of the reaction
+n_O2=6.25;// The stoichiometric coefficient of the reaction
+m_oy=m_oct*(n_CO2+n_H2O+n_O2);// kgmole of product
+n_p=m_oy*2.2046;// lbmole of product
+h_f_C8H18=-249.952;// MJ/kgmole
+h_f_CO2=-393.522;// MJ/kgmole
+h_f_H2O_g=-241.827;// MJ/kgmole
+h_f_N2=0;// MJ/kgmole
+h_f_O2=0;// MJ/kgmole
+N=h_f_C8H18+(0-(1.5*12.5*R_u*T))-(n_CO2*(h_f_CO2-(R_u*T)))-(n_H2O*(h_f_H2O_g-(R_u*T)))-(n_O2*(h_f_O2-(R_u*T)));// The numerator in MJ
+c_v_CO2=0.04987;// MJ/kgmole.K
+c_v_H2O=0.03419;// MJ/kgmole.K
+c_v_O2=0.02468;// MJ/kgmole.K
+D=(n_CO2*c_v_CO2)+(n_H2O*c_v_H2O)+(n_O2*c_v_O2);// The denominator in MJ/K
+T_A_bc=(T-273.15)+(N/D);// °C
+T_A_bc=T_A_bc+273.15;// K
+T_A_bc=T_A_bc*1.8;// R
+P_max=(n_p*R*T_A_bc)/(V_p*144);// psi
+printf("\nThe maximum possible explosion pressure inside the bomb,P_max=%5.0f psi",P_max);
+// The answer vary due to round off error
diff --git a/3831/CH15/EX15.12/Ex15_12.sce b/3831/CH15/EX15.12/Ex15_12.sce new file mode 100644 index 000000000..e197e3af2 --- /dev/null +++ b/3831/CH15/EX15.12/Ex15_12.sce @@ -0,0 +1,44 @@ +// Example 15_12
+clc;funcprot(0);
+// Given data
+T=25.0+273;// K
+p_m=0.100;// MPa
+T_b=200+273;// K
+q_r=-134.158;// MJ
+R=8.3143;// kJ/(kgmole.K)
+
+// Calculation
+// The reaction equation for 100.% theoretical air is CH4+2O2+3.76N2-->CO2+2(H2O)+7.52(N2)
+n_CH4=1;// The stoichiometric coefficient of the reaction
+n_O2=2;// The stoichiometric coefficient of the reaction
+n_N2=7.52;// The stoichiometric coefficient of the reaction
+n_R=(n_CH4+n_O2+n_N2);// The stoichiometric coefficient of the reaction
+p_CH4=(n_CH4/n_R)*p_m;// kPa
+p_O2=(n_O2/n_R)*p_m;// kPa
+p_N2=(n_N2/n_R)*p_m;// kPa
+n_CO2=1;// The stoichiometric coefficient of the reaction
+n_H2O=2;// The stoichiometric coefficient of the reaction
+n_N2=7.52;// The stoichiometric coefficient of the reaction
+n_P=(n_CO2+n_H2O+n_N2);// The stoichiometric coefficient of the reaction
+p_CO2=(n_CO2/n_P)*p_m;// kPa
+p_H2O=(n_H2O/n_P)*p_m;// kPa
+p_N2=(n_N2/n_P)*p_m;// kPa
+s0_CH4=186.256;// kJ/(kgmole.K)
+s0_O2=205.138;// kJ/(kgmole.K)
+s0_N2=191.610;// kJ/(kgmole.K)
+sbar_CH4=s0_CH4-(R*log(p_CH4/p_m));// kJ/(kgmole.K)
+sbar_O2=s0_O2-(R*log(p_O2/p_m));// kJ/(kgmole.K)
+sbar_N2=s0_N2-(R*log(p_N2/p_m));// kJ/(kgmole.K)
+sbar_iR=(n_CH4*sbar_CH4)+(n_O2*sbar_O2)+(n_N2*sbar_N2);// kJ/(kgmole.K)
+s0_CO2=213.795;// kJ/(kgmole.K)
+s0_H2O=188.833;// kJ/(kgmole.K)
+s0_N2=191.610;// kJ/(kgmole.K)
+c_p_CO2=37.19;// kJ/(kgmole.K)
+c_p_H2O=33.64;// kJ/(kgmole.K)
+c_p_N2=29.08;// kJ/(kgmole.K)
+sbar_CO2=s0_CH4+(c_p_CO2*log(T_b/T))-(R*log(p_CO2/p_m));// kJ/(kgmole.K)
+sbar_H2O=s0_O2++(c_p_H2O*log(T_b/T))-(R*log(p_H2O/p_m));// kJ/(kgmole.K)
+sbar_N2=s0_N2+(c_p_N2*log(T_b/T))-(R*log(p_N2/p_m));// kJ/(kgmole.K)
+sbar_iP=(n_CO2*sbar_CO2)+(n_H2O*sbar_H2O)+(n_N2*sbar_N2);// kJ/(kgmole.K)
+sbar_p_r=sbar_iP-sbar_iR-((q_r*10^3)/T_b);// kJ/(kgmole.K)
+printf("\nThe entropy produced per mole of fuel,(sbar_p)_r=%3.0f kJ/(kgmole.K)",sbar_p_r);
diff --git a/3831/CH15/EX15.13/Ex15_13.sce b/3831/CH15/EX15.13/Ex15_13.sce new file mode 100644 index 000000000..5fba06f20 --- /dev/null +++ b/3831/CH15/EX15.13/Ex15_13.sce @@ -0,0 +1,16 @@ +// Example 15_13
+clc;funcprot(0);
+// Given data
+T=25+273.15;// K
+n_C=1;// The stoichiometric coefficient of the reaction
+n_H2=2;// The stoichiometric coefficient of the reaction
+n_CH4=1;// The stoichiometric coefficient of the reaction
+sbar0_CH4=186.256;// kJ/kgmole.K
+sbar0_C=5.740;// kJ/kgmole.K
+sbar0_H2=130.684;// kJ/kgmole.K
+h_f_CH4=-74.873;// MJ/kgmole.K
+
+// Calculation
+sbar0_f_CH4=sbar0_CH4-[((n_C/n_CH4)*sbar0_C)+((n_H2/n_CH4)*sbar0_H2)];// kJ/kgmole.K
+gbar0_f_CH4=h_f_CH4-(T*sbar0_f_CH4*1/1000);// The specific molar Gibbs function of formation of methane in MJ/kgmole
+printf("\nThe molar specific entropy of formation,(sbar0_f)_CH4=%2.3f kJ/kgmole.K \nThe specific molar Gibbs function of formation of methane,(gbar0_f)_CH4=%2.3f MJ/kgmole",sbar0_f_CH4,gbar0_f_CH4);
diff --git a/3831/CH15/EX15.14/Ex15_14.sce b/3831/CH15/EX15.14/Ex15_14.sce new file mode 100644 index 000000000..6dbd4c652 --- /dev/null +++ b/3831/CH15/EX15.14/Ex15_14.sce @@ -0,0 +1,37 @@ +// Example 15_14
+clc;funcprot(0);
+// Given data
+p=0.100;// MPa
+T_a=298;// K
+T_b=2000;// K
+R=0.0083143;// MJ/kgmole.K
+
+// Calculation
+// (a)
+gbar0_f_H2O=-228.583;// kJ/kgmole
+// since H2 and O2 are elements, their molar specific Gibbs function of formation is zero. Then, from Table 15.7,
+gbar0_f_H2=0;// kJ/kgmole
+gbar0_f_O2=0;// kJ/kgmole
+K_e=exp(gbar0_f_H2O/(R*T_a));// The equilibrium constant
+printf("\n(a)The equilibrium constant,K_e=%1.2e",K_e);
+// (b)
+T_b_R=T_b*1.8;// R
+// Eq. (15.34) with Tables 15.7 and C.16c in Thermodynamic Tables to accompany Modern Engineering Thermodynamics give
+h_a_H2O=4258.3;// Btu/lbmole
+h_b_H2O=35540.1;// Btu/lbmole
+h_a_H2=3640.3;// Btu/lbmole
+h_b_H2=26398.5;// Btu/lbmole
+h_a_O2=3725.1;// Btu/lbmole
+h_b_O2=29173.5;// Btu/lbmole
+s_a_H2O=188.833;// kJ/(kgmole.K)
+s_b_H2O=63.221;// Btu/(lbmole.R)
+s_a_H2=130.684;// kJ/(kgmole.K)
+s_b_H2=44.978;// Btu/(lbmole.R)
+s_a_O2=205.138;// kJ/(kgmole.K)
+s_b_O2=64.168;// Btu/(lbmole.R)
+// Note: The multipliers 2.3258 and 4.1865 in these equations are necessary to convert the Btu/lbmole and Btu/(lbmole.R) values in Table C.16c into kJ/kgmole and kJ/(kgmole.K), respectively.
+gbar_f_H2O=(gbar0_f_H2O*10^3)+((h_b_H2O-h_a_H2O)*2.3258)-[((T_b*s_b_H2O)*4.1865)-(T_a*s_a_H2O)];// kJ/kgmole
+gbar_f_H2=gbar0_f_H2+((h_b_H2-h_a_H2)*2.3258)-[((T_b*s_b_H2)*4.1865)-(T_a*s_a_H2)];// kJ/kgmole
+gbar_f_O2=gbar0_f_O2+((h_b_O2-h_a_O2)*2.3258)-[((T_b*s_b_O2)*4.1865)-(T_a*s_a_O2)];// kJ/kgmole
+K_e=exp([gbar_f_H2O-gbar_f_H2-((1/2)*gbar_f_O2)]/(R*10^3*T_b));// The equilibrium constant
+printf("\n(b)The equilibrium constant,K_e=%1.2e",K_e);
diff --git a/3831/CH15/EX15.17/Ex15_17.sce b/3831/CH15/EX15.17/Ex15_17.sce new file mode 100644 index 000000000..996a09a91 --- /dev/null +++ b/3831/CH15/EX15.17/Ex15_17.sce @@ -0,0 +1,20 @@ +// Example 15_17
+clc;funcprot(0);
+// Given data
+T=5000;// K
+
+// Calculation
+// (a)
+K_e1=10^0.450;// The equilibrium constant for the reaction
+K_e2=1/K_e1;// The equilibrium constant for a second reaction
+printf("\n(a)The equilibrium constant for the first reaction,K_e1=%1.2f \n The equilibrium constant for a second reaction,K_e2=%0.3f",K_e1,K_e2);
+// (b)
+K_e1=10^-0.298;// The equilibrium constant for the reaction
+printf("\n(b)The equilibrium constant for the reaction,K_e1=%0.3f",K_e1);
+// (c)
+alpha=1;// Constant
+beta=3.76;// Constant
+K_e1=10^(1.719);// The equilibrium constant for the first reaction
+K_e2=10^-0.570;// The equilibrium constant for a second reaction
+K_e3=(K_e1^alpha)*(K_e2^beta);// The equilibrium constant for a third reaction
+printf("\n(c)The equilibrium constant for the first reaction,K_e1=%2.1f \n The equilibrium constant for a second reaction,K_e2=%0.3f \n The equilibrium constant for the combined reaction,K_e3=%0.3f",K_e1,K_e2,K_e3);
diff --git a/3831/CH15/EX15.18/Ex15_18.sce b/3831/CH15/EX15.18/Ex15_18.sce new file mode 100644 index 000000000..d8c0d2368 --- /dev/null +++ b/3831/CH15/EX15.18/Ex15_18.sce @@ -0,0 +1,17 @@ +// Example 15_18
+clc;funcprot(0);
+// Given data
+T=25.0;// °C
+p=0.100;// MPa
+g_f_H2O=-237.178;// MJ/kgmole
+h_f_H2O=-285.838;// MJ/kgmole
+j=2;// kgmole of electrons per kgmole of H2
+F=96487;// kJ/(V.kgmole electrons)
+
+// Calculation
+n_H2=1;// The stoichiometric coefficient of the reaction
+n_H2O=1;// The stoichiometric coefficient of the reaction
+n_r_max=g_f_H2O/h_f_H2O;// The maximum reaction efficiency
+phi_0=([-(n_H2O/n_H2)*(h_f_H2O*10^3)]*[n_r_max])/(j*F);// The theoretical open circuit voltage in V
+W_maxbyn_fuel=phi_0*j*F;// kJ/kgmoleH2
+printf("\nThe maximum theoretical reaction efficiency,(n_r)max=%2.1f percentage \nThe theoretical open circuit voltage,V=%1.2f V \nThe maximum theoretical work output,W_max/n_fuel=%6.0f kJ/kgmole H2",n_r_max*100,phi_0,W_maxbyn_fuel);
diff --git a/3831/CH15/EX15.19/Ex15_19.sce b/3831/CH15/EX15.19/Ex15_19.sce new file mode 100644 index 000000000..4a8fd672b --- /dev/null +++ b/3831/CH15/EX15.19/Ex15_19.sce @@ -0,0 +1,15 @@ +// Example 15_19
+clc;funcprot(0);
+// Given data
+T=25;// °C
+p=0.1;// MPa
+
+// Calculation
+n_H2=1;// The stoichiometric coefficient of the reaction
+n_O2=0.5;// The stoichiometric coefficient of the reaction
+n_H2O=1;// The stoichiometric coefficient of the reaction
+g_f_H2O=-237.178;// MJ/kgmole
+// [(abar_f)_i]_chemical=gbar0_i+RT ln[1].
+abar_H2O=g_f_H2O;// MJ/kgmole
+adot_fc=0+0-[(n_H2O/n_H2)*abar_H2O];// The net molar specific flow availability in MJ/kgmoleH2
+printf("\nThe net molar specific flow availability of the hydrogen–oxygen fuel cell,(a_flow chemical)_net=%3.3f MJ/kgmoleH2",adot_fc);
diff --git a/3831/CH15/EX15.2/Ex15_2.sce b/3831/CH15/EX15.2/Ex15_2.sce new file mode 100644 index 000000000..8daa72593 --- /dev/null +++ b/3831/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,10 @@ +// Example 15_2
+clc;funcprot(0);
+// Given data
+CH_4=1.00;// kgmole of methane
+C_3H_8=3.00;// kgmoles of propane
+
+// Solution
+n=1+(3*(3));// Carbon balance
+m=4+(3*(8));// Hydrogen balance
+printf("\nThe hydrocarbon fuel model for this mixture is C_%2.0fH_%2.0f.",n,m);
diff --git a/3831/CH15/EX15.3/Ex15_3.sce b/3831/CH15/EX15.3/Ex15_3.sce new file mode 100644 index 000000000..4d5c1036a --- /dev/null +++ b/3831/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,37 @@ +// Example 15_3
+clc;funcprot(0);
+// Given data
+T=20.0;// °C
+CO_2=7.10;// %
+CO=0.800;// %
+O_2=9.90;// %
+N_2=82.2;// %
+M_air=28.97;// lbmdry air/lbmole dry air
+
+// Solution
+// (a)
+n=7.10+0.800;// Carbon (C) balance
+// m=2*b
+a=82.2/3.76;// Nitrogen (N2) balance
+b=2*(a-(7.10+(0.800/2)+9.90));// Oxygen (O2) balance
+m=2*b;// Hydrogen (H) balance
+printf("\n(a)The hydrocarbon model (CnHm) of the fuel is C_%1.2fH_%2.0f",n,m);
+// (b)
+M_fuel=(7.90*(12))+(18.0*(1));// lbm/lbmole
+Fc_C=7.90*12.0*113;// lbm C/lbm fuel
+Fc_H=((9.00)*(2.016))/113;// lbmH/lbmfuel
+printf("\n(b)The molecular mass of the fuel in this model,M_fuel=%3.0f lbm/lbmole \n The fuel’s composition on a mass basis is %0.3f lbmC/lbmfuel and %0.3f lbmH/lbm fuel",M_fuel,Fc_C,Fc_H);
+// (c)
+n_air=21.9*(1+3.76);// The stoichiometric coefficient of the reaction
+n_fuel=1;// The stoichiometric coefficient of the reaction
+AF_molar=n_air/n_fuel;// moles air/mole fuel
+AF_mass=AF_molar*(28.97/(M_fuel));// lbm air/lbm fuel
+printf("\n(c)The air-fuel ratio on a molar and a mass basis,(A/F)_molar=%3.0f moles air/molefuel and (A/F)_mass=%2.1f lbm air/lbm fuel",AF_molar,AF_mass);
+// (d)
+b=7.90;// Carbon (C) balance
+c=18.0;// Hydrogen (H) balance
+a=b+(c/2);// Oxygen (O2) balance
+d=3.76*a;// Nitrogen (N2) balance
+AF_mt=(12.4*(1+3.76))/1;// mole air/mole fuel
+per_ta=(AF_molar/AF_mt)*100;// The percent of theoretical air used in the actual combustion process (%)
+printf("\n(d)The percentage of theoretical air used in the combustion process,Percentage of theoritical air=%3.0f percentage or %2.0f percentage excess air",per_ta,(per_ta-100));
diff --git a/3831/CH15/EX15.4/Ex15_4.sce b/3831/CH15/EX15.4/Ex15_4.sce new file mode 100644 index 000000000..f0030907e --- /dev/null +++ b/3831/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,14 @@ +// Example 15_4
+clc;funcprot(0);
+// Given data
+m_H2O=9.00;// moles
+m_m=109;// moles
+p_t=14.7;// The total pressure of the mixture in psia
+
+// Calculation
+X_H2O=m_H2O/m_m;// The mole fraction
+p_H2O=X_H2O*p_t;// psia
+// By interpolation in Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that
+T_DP=108;// °F
+T_DP=(108-32)/1.8;// °C
+printf("\nThus, the exhaust products must be cooled to %3.0f°F(%2.1f°C)or below to condense the water of combustion and have an essentially dry exhaust gas.",(T_DP*1.8+32),T_DP);
diff --git a/3831/CH15/EX15.5/Ex15_5.sce b/3831/CH15/EX15.5/Ex15_5.sce new file mode 100644 index 000000000..4f885f12f --- /dev/null +++ b/3831/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,24 @@ +// Example 15_5
+clc;funcprot(0);
+// Given data
+T_DB=90.0;// °F
+T_WB=75.0;// °F
+phi=50;// The relative humidity in %
+w=105*1/7000;// lbm H2O/lbm dry air
+M_da=28.97;// lbmdry air/lbmole dry air
+M_H2O=18.016;// lbmH2O/lbmoleH2O
+
+// Calculation
+w=w*(M_da/M_H2O);// lbmole H2O/lbmole dry air
+// From the balanced reaction equation of part a of Example 15.3, we find that the amount of dry air used per mole of fuel is
+a_da=21.9*(1+3.76);// moles
+a_w=w*a_da;// moles of water
+n_H2O=9.00+a_w;// moles per mole of fuel
+n_total=111.5;// moles per mole of fuel
+X_H2O=n_H2O/n_total;// The mole fraction of water vapor in the exhaust
+p_H20=X_H2O*14.7;// psia
+// Again, interpolating in Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find
+T_DP=116.0;//°F
+T_DP=(T_DP-32)/1.8;// °C
+T_sat=T_DP;// °C
+printf("\n(a)The amount of water carried into the engine in the form of inlet humidity,w=%0.4f lbmole H2O/lbmole dry air \n(b)The new dew point temperature of the exhaust products,T_DP=%2.1f°C",w,T_DP);
diff --git a/3831/CH15/EX15.6/Ex15_6.sce b/3831/CH15/EX15.6/Ex15_6.sce new file mode 100644 index 000000000..9dc50fc5c --- /dev/null +++ b/3831/CH15/EX15.6/Ex15_6.sce @@ -0,0 +1,13 @@ +// Example 15_6
+clc;funcprot(0);
+// Given data
+h_f_CO2=393.522;// MJ/kgmole
+h_f_H2O_g=241.827;//MJ/kgmole
+h_f_H2O_l=285.838;// MJ/kgmole
+HHV_CH4=-890.4;// MJ/kgmole
+
+// Calculation
+n=1;// The stoichiometric coefficient for the reaction
+m=4;// The stoichiometric coefficient for the reaction
+q_f=-[(n*h_f_CO2)+((m/2)*h_f_H2O_l)+HHV_CH4];// MJ/kgmole of CH_4
+printf("\nThe heat of formation of methane gas CH4(g) at the standard reference state,(qbar_f)_CH4=%2.1f MJ/kgmole of CH_4",q_f);
diff --git a/3831/CH15/EX15.7/Ex15_7.sce b/3831/CH15/EX15.7/Ex15_7.sce new file mode 100644 index 000000000..808af3edc --- /dev/null +++ b/3831/CH15/EX15.7/Ex15_7.sce @@ -0,0 +1,14 @@ +// Example 15_7
+clc;funcprot(0);
+// Given data
+m=0.160;// kg of liquid water
+T=25.0;// °C
+p=0.100;// MPa
+
+// Calculation
+h_f_H2O=285.838;// MJ/kg mole
+q_f_H2O=285.838;// MJ/kg mole
+q_r=q_f_H2O;// MJ/kg mole
+M=18.016;// kg/kgmole
+Q_r=m*(-q_r/M);// MJ
+printf("\nThe total heat transfer required,Q_r=%1.2f MJ",Q_r);
diff --git a/3831/CH15/EX15.8/Ex15_8.sce b/3831/CH15/EX15.8/Ex15_8.sce new file mode 100644 index 000000000..45370e823 --- /dev/null +++ b/3831/CH15/EX15.8/Ex15_8.sce @@ -0,0 +1,22 @@ +// Example 15_8
+clc;funcprot(0);
+// Given data
+// For 100.% theoretical air, the combustion equation for methane is,CH_4+2.00[O_2+3.76 N_2]-->CO_2+2.00(H_2O)+7.52(N_2)
+// From Table 15.1, we find that
+h_f_CH4=-74.873;// MJ/kgmoleCH4
+h_R=-74.873;// MJ/kgmoleCH4
+h_f_N_2=0;// MJ/kgmole N2
+h_f_CO2=-393.522;// MJ/kgmole CO2
+h_f_H2O_g=-241.827;// MJ/kgmole H2O_g
+h_f_H2O_l=-285.838;// MJ/kgmole H2O_l
+
+// Calculation
+h_p_LHV=h_f_CO2+(2*h_f_H2O_g)+(7.52*h_f_N_2);// MJ/kgmole CH4
+h_p_HHV=h_f_CO2+(2*h_f_H2O_l)+(7.52*h_f_N_2);// MJ/kgmole CH4
+LHV=h_p_LHV-h_R;// MJ/kgmole CH4
+HHV=h_p_HHV-h_R;// MJ/kgmole CH4
+h_fg_H2O=44.00;// MJ/kgmole CH4
+n_H2O=2.00;// The stoichiometric coefficient for the reaction
+n_fuel=1.00;// The stoichiometric coefficient for the reaction
+HHV=LHV-((n_H2O/n_fuel)*h_fg_H2O);// MJ/kgmole CH4
+printf("\nThe higher heating value of methane,LHV=%3.2f MJ/kgmole CH4 \nThe lower heating value of methane,HHV=%3.2f MJ/kgmole CH4",HHV,LHV);
diff --git a/3831/CH15/EX15.9/Ex15_9.sce b/3831/CH15/EX15.9/Ex15_9.sce new file mode 100644 index 000000000..4d312c855 --- /dev/null +++ b/3831/CH15/EX15.9/Ex15_9.sce @@ -0,0 +1,18 @@ +// Example 15_9
+clc;funcprot(0);
+// Given data
+h_R=-74.873;// MJ/kgmole CH4
+h_f_N_2=0;// MJ/kgmole
+h_f_CO2=-393.522;// MJ/kgmole
+h_f_H2O_g=-241.827;// MJ/kgmole
+h_f_H2O_l=-285.838;// MJ/kgmole
+c_p_CO2=0.03719;// MJ/kgmole.K
+c_p_H2O=0.03364;// MJ/kgmole.K
+c_p_N2=0.02908;// MJ/kgmole.K
+T=500;// °C
+T_0=25;// °C
+
+// Calculation
+h_P=h_f_CO2+(2*h_f_H2O_g)+(7.52*h_f_N_2)+([c_p_CO2+(2.00*c_p_H2O)+(7.52*c_p_N2)]*(T-T_0));// MJ/kgmole CH4
+q_r=h_P-h_R;// MJ/kgmole CH4
+printf("\nThe heat of reaction of methane,qbar_r=%3.3f MJ/kgmole CH4",q_r);
diff --git a/3831/CH16/EX16.1/Ex16_1.sce b/3831/CH16/EX16.1/Ex16_1.sce new file mode 100644 index 000000000..0ba509586 --- /dev/null +++ b/3831/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,12 @@ +// Example 16_1
+clc;funcprot(0);
+// Given data
+T=20+273.15;// K
+V=90.0;// km/h
+g_c=1;// The gravitational constant
+c_p=1.004;// kJ/kg.K
+
+// Solution
+T_0=T*(1+(((V*10^3/(3600*1000))^2)/(2*g_c*c_p*T)));// K
+T_0=T_0-273.15;// °C
+printf("\nThe stagnation temperature,T_0=%2.1f°C",T_0)
diff --git a/3831/CH16/EX16.10/Ex16_10.sce b/3831/CH16/EX16.10/Ex16_10.sce new file mode 100644 index 000000000..2d64a15c2 --- /dev/null +++ b/3831/CH16/EX16.10/Ex16_10.sce @@ -0,0 +1,16 @@ +// Example 16_10
+clc;funcprot(0);
+// Given data
+m=5.00*10^-3;// kg
+T=20.0+273.15;// K
+p=101.3*10^3;// kg/(m.s^2)
+R=286;// m^2/(s^2.K)
+D=3.00*10^-3;// m
+g=9.81;// m/s^2
+g_c=1;// The gravitational constant
+
+// Calculation
+W=(m*g)/g_c;// N
+rho=p/(R*T);// kg/m^3
+V_in=((4*g_c*W)/(rho*%pi*D^2))^(1/2);// m/s
+printf("\nThe velocity of the jet,V_in=%2.1f m/s",V_in);
diff --git a/3831/CH16/EX16.11/Ex16_11.sce b/3831/CH16/EX16.11/Ex16_11.sce new file mode 100644 index 000000000..d6b365224 --- /dev/null +++ b/3831/CH16/EX16.11/Ex16_11.sce @@ -0,0 +1,16 @@ +// Example 16_11
+clc;funcprot(0);
+// Given data
+M_x=5.50;// The Mach number
+p_x=14.7;// lbf/in^2
+T_x=70.0+459.67;// °F
+k=1.4;// The specific heat ratio
+R=53.34;// ft.lbf/lbm.R
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+M_y=((((k-1)*M_x^2)+2)/((2*k*M_x^2)+1-k))^(1/2);// The Mach number
+T_y=T_x*[(1+(((k-1)/2)*M_x^2))/(1+(((k-1)/2)*M_y^2))];// R
+p_y=p_x*(M_x/M_y)*(T_y/T_x)^(1/2);// lbf/in^2
+V_wind=(M_x*sqrt(k*g_c*R*T_x))-(M_y*sqrt(k*g_c*R*T_y));// ft/s
+printf("\nThe pressure directly behind the shock wave,p_y=%3.0f lbf/in^2 \nThe temperature directly behind the shock wave,T_y=%4.0f R \nThe wind velocity directly behind the shock wave,V_wind=%1.0e ft/s",p_y,T_y,V_wind);
diff --git a/3831/CH16/EX16.12/Ex16_12.sce b/3831/CH16/EX16.12/Ex16_12.sce new file mode 100644 index 000000000..07c18c308 --- /dev/null +++ b/3831/CH16/EX16.12/Ex16_12.sce @@ -0,0 +1,45 @@ +// Example 16_12
+clc;funcprot(0);
+// Given data
+p_os=7.00;// MPa
+T_os=2000;// °C
+D_t=0.0200;// m
+D_e=0.100;// m
+k=1.40;// The specific heat ratio
+R=286;// m^2/(s^2.K)
+g_c=1;// The gravitational constant
+
+// Calculation
+// (a)
+A_t=(%pi*D_t^2)/4;// m^2
+mdot=(0.0404*(p_os*10^6)*A_t)/sqrt(T_os+273.15);// kg/s
+// (b)
+A_r=(D_e/D_t)^2;// (A_r=A_exit/A*)
+M_e=5.00;// Mach number at exit
+// Assume p_exit/p_os=p_r
+p_r=1.89*10^-3;// Pressure ratio
+// Assume T_exit/T_os=T_r
+T_r=0.16667;// Temperature ratio
+p_e=p_r*p_os*10^3;// The exit pressure in kN/m^2
+T_exit=T_r*(T_os+273.15);// K
+c_e=sqrt(k*g_c*R*T_exit);// The velocity of sound at the exit in m/s
+V_exit=c_e*M_e;// m/s
+// (c)
+M_x=5.0;// The Mach number
+p_x=13.23;// kN/m^2
+T_x=378.8;// K
+// Table C.19 is a tabular version of these equations, and at Mx = 5.0, we again have a direct entry
+M_y=0.415;// The Mach number
+// Assume p_osy/p_osx=p_ros
+p_ros=0.06172;
+// Assume p_y/p_x=p_rxy
+p_rxy=29.00;
+// Assume p_osy/p_x=p_rosyx
+p_rosyx=32.654;
+// Assume T_y/T_x=T_yx
+T_yx=5.800;
+p_osx=p_os*10^3;// kN/m^2
+p_B=p_ros*p_osx;// The required back pressure in kN/m^2
+// Alternatively
+p_B=p_rosyx*p_x;// The required back pressure in kN/m^2
+printf("\n(a)The mass flow rate required for supersonic flow in the diverging section,mdot=%1.2f kg/s \n(b)The Mach number, pressure,temperature and velocity at the exit of the diverging section with this massflow rate,M_exit=%1.2f,p_exit=%2.1f kN/m^2,T_exit=%3.1f K,V_exit=%4.0f m/s \n(c)The outside back pressure required to produce a standing normal shock wave at the exit of the diverging section,p_B=%3.0f kN/m^2",mdot,M_e,p_e,T_exit,V_exit,p_B);
diff --git a/3831/CH16/EX16.13/Ex16_13.sce b/3831/CH16/EX16.13/Ex16_13.sce new file mode 100644 index 000000000..59fc65793 --- /dev/null +++ b/3831/CH16/EX16.13/Ex16_13.sce @@ -0,0 +1,28 @@ +// Example 16_13
+clc;funcprot(0);
+// Given data
+p_os=3.00;// atm
+T_os=20.0;// °C
+p_B=1.00;// atm
+A_r=2.0;// The exit to throat area ratio fo r the nozzle
+k=1.4;// The specific heat ratio
+R=286;// m^2/(s^2.K)
+g_c=1;// The gravitational constant
+
+// Calculation
+p_a=p_os*(2/(k+1))^(k/(k-1));// atm
+// Since we are given Aexit/A* = A_E/A*= 2.00, we can find ME by inverting Eq. (16.23b).However, in this case, it is again much easier to use Table C.18 for this area ratio and read (approximately),
+M_E=2.20;// The Mach number at exit
+// Assume p_rEos=p_E/p_os
+p_rEos=0.09352;
+p_E=p_rEos*p_os;// atm
+// Assume p_r=p_osy/p_osx
+p_r=1.00/3.00;
+// From Table C.19 at p_osy/p_osx=0.333
+M_x=2.98;// The Mach number
+M_y=0.476;// The Mach number
+T_e=0.50813*(T_os+273.15);// K
+c_exit=sqrt(k*g_c*R*T_e);// m/s
+M_exit=M_E;// The Mach number at exit
+V_exit=M_exit*c_exit;// m/s
+printf("\nThe exit pressure,p_E=%0.3f atm\nThe exit temperature,T_exit=%3.2f K \nThe exit velocity,V_exit=%3.0f m/s",p_E,T_e,V_exit);
diff --git a/3831/CH16/EX16.14/Ex16_14.sce b/3831/CH16/EX16.14/Ex16_14.sce new file mode 100644 index 000000000..69adc41ae --- /dev/null +++ b/3831/CH16/EX16.14/Ex16_14.sce @@ -0,0 +1,30 @@ +// Example 16_14
+clc;funcprot(0);
+// Given data
+p_inlet=456.2;// kN/m^2
+T_inlet=283.7;// K
+p_exit=370.4;// kN/m^2
+T_exit=260.1;// K
+V_exit=474.8;// m/s
+k=1.67;// The specific heat ratio for helium
+R=2077.0;// m^2/(s^2.K)
+g_c=1;// The gravitational constant
+
+// Calculation
+// (a)
+c_osi=sqrt(k*g_c*R*T_inlet);// m/s
+c_inlet=c_osi;// m/s
+n_N=((((k-1)/2)*(V_exit/c_inlet)^2)/(1-((p_exit/p_inlet)^((k-1)/k))));// The nozzle’s efficiency
+// (b)
+C_v=sqrt(n_N);// The nozzle’s velocity coefficient
+// (c)
+R=2.077;// kJ/kg.K
+rho_e=p_exit/(R*T_exit);// kg/m^3
+M_exit=1.0;// The exit Mach number
+T_os=T_inlet;// K
+p_os=p_inlet;// kN/m^2
+T_es=T_os*(2/(k+1));// K
+rho_es=(p_os/(R*T_os))*[2/(k+1)]^(1/(k-1));// kg/m^3
+V_es=sqrt(k*g_c*R*10^3*T_es);// m/s
+C_d=(rho_e*V_exit)/(rho_es*V_es);// The nozzle’s discharge coefficient
+printf("\n(a)The nozzle’s efficiency,n_N=%0.3f \n(b)The nozzle’s velocity coefficient,C_v=%0.3f \n(c)The nozzle’s discharge coefficient,C_d=%0.3f",n_N,C_v,C_d);
diff --git a/3831/CH16/EX16.15/Ex16_15.sce b/3831/CH16/EX16.15/Ex16_15.sce new file mode 100644 index 000000000..6872faec3 --- /dev/null +++ b/3831/CH16/EX16.15/Ex16_15.sce @@ -0,0 +1,15 @@ +// Example 16_15
+clc;funcprot(0);
+// Given data
+M_in=0.890;// The inlet Mach number
+p_osi=314.7;// kPa
+p_ose=249.3;// kPa
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+n_D=(((((1+((((k-1)/2)*M_in^2)))*(p_ose/p_osi)^((k-1)/k)))-1)/(((k-1)*M_in^2)/2))*100;// %
+// (b)
+p_i=p_osi/((1+(((k-1)/2)*M_in^2))^(k/(k-1)));// kPa
+C_p=(p_ose-p_i)/(p_osi-p_i);// The diffuser’s pressure recovery coefficient
+printf("\n(a)The diffuser’s efficiency,n_D=%2.1f percentage \n(b)The diffuser’s pressure recovery coefficient,C_p=%0.3f",n_D,C_p);
diff --git a/3831/CH16/EX16.2/Ex16_2.sce b/3831/CH16/EX16.2/Ex16_2.sce new file mode 100644 index 000000000..7f40f256f --- /dev/null +++ b/3831/CH16/EX16.2/Ex16_2.sce @@ -0,0 +1,16 @@ +// Example 16_2
+clc;funcprot(0);
+// Given data
+T=20+273.15;// K
+V=25.0;// m/s
+k=1.40;// The specific heat ratio
+p=0.101;// MPa
+g_c=1;// The gravitational constant
+c_p=1.004;// kJ/kg.K
+R=0.286;// kJ/kg.K
+
+// Solution
+p_os=p*(1+((V^2/1000)/(2*g_c*c_p*T)))^(k/(k-1));// The isentropic stagnation pressure in MPa
+rho=(p*10^3)/(R*T);// kg/m^3
+rho_os=rho*(1+((V^2/1000)/(2*g_c*c_p*T)))^(1/(k-1));// The isentropic stagnation density in kg/m^3
+printf("\nThe isentropic stagnation pressure,p_os=%0.4f MPa \nThe isentropic stagnation density,rho_os=%1.4f kg/m^3",p_os,rho_os);
diff --git a/3831/CH16/EX16.3/Ex16_3.sce b/3831/CH16/EX16.3/Ex16_3.sce new file mode 100644 index 000000000..c18a212f8 --- /dev/null +++ b/3831/CH16/EX16.3/Ex16_3.sce @@ -0,0 +1,23 @@ +// Example 16_3
+clc;funcprot(0);
+// Given data
+p_1=14.7;// psia
+T_1=1000;// °F
+V_1=1612;// ft/s
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+// Station 1
+p_1=14.7;// psia
+T_1=1000;// °F
+h_1=1534.4;// Btu/lbm
+s_1=2.1332;// Btu/lbm.R
+// Station os
+s_os=s_1;// Btu/lbm.R
+h_os=h_1+(V_1^2/(2*g_c));// Btu/lbm
+//Table C.3a, in Thermodynamic Tables to accompany Modern Engineering Thermodynamics a Mollier diagram for steam
+p_os=20.0;// psia
+T_os=1100;// °F
+v_os=46.4;// ft^3/lbm
+rho_os=1/v_os;// lbm/ft^3;
+printf("\nThe isentropic stagnation temperature,T_0=%4.0f°F \nThe isentropic stagnation pressure,p_os=%2.1f psia \nThe isentropic stagnation density,rho_os=%0.3f lbm/ft^3",T_os,p_os,rho_os);
diff --git a/3831/CH16/EX16.4/Ex16_4.sce b/3831/CH16/EX16.4/Ex16_4.sce new file mode 100644 index 000000000..3c15fa3f2 --- /dev/null +++ b/3831/CH16/EX16.4/Ex16_4.sce @@ -0,0 +1,14 @@ +// Example 16_4
+clc;funcprot(0);
+// Given data
+T=35+273.15;// K
+V=300;// m/s
+
+// Solution
+// Using Table C.13b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics for the values of the specific heat ratio and the gas constant for methane, we get
+k_methane=1.30;// The specific heat ratio
+g_c=1;// The gravitational constant
+R_methane=518;// J/kg.K
+c_methane=sqrt(k_methane*g_c*R_methane*T);// m/s
+M_methane=V/c_methane;//The Mach number
+printf("\nThe Mach number of the methane,M_methane=%0.3f",M_methane);
diff --git a/3831/CH16/EX16.5/Ex16_5.sce b/3831/CH16/EX16.5/Ex16_5.sce new file mode 100644 index 000000000..4f5044cec --- /dev/null +++ b/3831/CH16/EX16.5/Ex16_5.sce @@ -0,0 +1,17 @@ +// Example 16_5
+clc;funcprot(0);
+// Given data
+T=-20.0+273.15;// K
+p=0.500;// atm
+M=0.850;// The Mach number
+k=1.40;// The specific heat ratio
+R=286;// J/kg.K
+g_c=1;// The gravitational constant
+
+// Solution
+V=M*sqrt(k*g_c*R*T);// m/s
+T_os=T*(1+(((k-1)*M^2)/2));// K
+T_os=T_os-273.15;// °C
+p_os=p*(1+(((k-1)*M^2)/2))^(k/(k-1));// atm
+p_os=p_os*1.013*10^2;// kPa
+printf("\nThe aircraft’s velocity,V=%3.0f m/s \nThe isentropic stagnation temperature,T_os=%2.1f°C \nThe isentropic stagnation pressure,p_os=%2.1f KPa",V,T_os,p_os);
diff --git a/3831/CH16/EX16.6/Ex16_6.sce b/3831/CH16/EX16.6/Ex16_6.sce new file mode 100644 index 000000000..771b7dc2b --- /dev/null +++ b/3831/CH16/EX16.6/Ex16_6.sce @@ -0,0 +1,24 @@ +// Example 16_6
+clc;funcprot(0);
+// Given data
+p_os=1.00;// MPa
+T_os=20.0+273.15;// K
+k=1.40;// The specific heat ratio
+p=0.1013;// MPa
+g_c=1;// The gravitational constant
+R=286;// J/kg.K
+
+// Solution
+// (a)
+p_r=p/p_os;// The pressure ratio
+M=((2/(k-1))*(((p_os/p)^((k-1)/k))-1))^(1/2);// The exit Mach number
+// (b)
+T=(T_os/(1+(((k-1)*M^2)/2)))-273.15;// The exit temperature in °C
+// (c)
+V=M*sqrt(k*g_c*R*(T+273.15));// The exit velocity in m/s
+// (d)
+p_throat=p_os*[2/(k+1)]^(k/(k-1));// The pressure at the throat of the nozzle in MPa
+// (e)
+T_throat=T_os*[2/(k+1)];// The temperature at the throat of the nozzle in K
+T_throat=T_throat-273.15;// The temperature at the throat of the nozzle in °C
+printf("\n(a)The exit Mach number,M=%1.2f \n(b)The exit temperature,T=%3.0f°C \n(c)The exit velocity,V=%3.0f m/s \n(d)The pressure at the throat of the nozzle,p_throat=%0.3f MPa \n(e)The temperature at the throat of the nozzle,T_throat=%2.1f°C",M,T,V,p_throat,T_throat);
diff --git a/3831/CH16/EX16.7/Ex16_7.sce b/3831/CH16/EX16.7/Ex16_7.sce new file mode 100644 index 000000000..5d16670ca --- /dev/null +++ b/3831/CH16/EX16.7/Ex16_7.sce @@ -0,0 +1,19 @@ +// Example 16_7
+clc;funcprot(0);
+// Given data
+D_bag=3.00;// ft
+t_fill=30;// milliseconds
+p_air=15.00;// psia
+p_os=1500;// psia
+T_os=70.0+459.67;// R
+k=1.40;// The specific heat ratio
+R_air=53.34;// ft.lbf/lbm.R
+
+// Solution
+V_bag=(%pi*D_bag^3)/6;// ft^3
+T_air=T_os*(2/(k+1));// R
+rho_air=(p_air*144)/(R_air*T_air);// lbm/ft^3
+m_avg=(rho_air*V_bag)/(t_fill*10^-3);// lbm/s
+D_tube=[(4*m_avg*sqrt(T_os+459.67))/(0.532*%pi*p_os)]^(1/2);// in
+printf("\nThe minimum tube diameter,D_tube=%1.2f in",D_tube);
+// The answer vary due to round off error
diff --git a/3831/CH16/EX16.8/Ex16_8.sce b/3831/CH16/EX16.8/Ex16_8.sce new file mode 100644 index 000000000..3cf90b793 --- /dev/null +++ b/3831/CH16/EX16.8/Ex16_8.sce @@ -0,0 +1,21 @@ +// Example 16_8
+clc;funcprot(0);
+// Given data
+D_exit=0.0938;// in
+T_os=70.0;// °F
+p_osi=50.0;// psia
+V_T=1.00;// ft^3
+k=1.40;// The specific heat ratio
+
+// Calculation
+// (a)
+p_r1=(2/(k+1))^(k/(k-1));// The pressure ratio
+p_exit=14.7;// psia
+p_exitbyp_os=p_exit/p_osi;// The pressure ratio
+// (b)
+p_os=p_exit/p_r1;// psia
+p_os=p_os*0.472;// psig
+// (c)
+A_a=(%pi*D_exit^2)/(4*144);// ft^2
+tau=31.95*log(p_osi/(p_os/0.472));// s
+printf("\n(a)p_exit/p_os=%0.3f,which is <0.528 therefore, initially, the flow is choked.\n(b)The flow remains choked until the tire deflates to a pressure of p_os=%2.1f psig \n(c)The valve stem unchokes at time,tau=%2.1f s",p_exitbyp_os,p_os,tau);
diff --git a/3831/CH17/EX17.1/Ex17_1.sce b/3831/CH17/EX17.1/Ex17_1.sce new file mode 100644 index 000000000..0fa0f278a --- /dev/null +++ b/3831/CH17/EX17.1/Ex17_1.sce @@ -0,0 +1,17 @@ +// Example 17_1
+clc;funcprot(0);
+// Given data
+T=37.0;// °C
+// From table 17.2
+c_Na_c=14.0;// osmoles/cm^3
+c_Na_o=144;// osmoles/cm^3
+c_K_c=140;// osmoles/cm^3
+c_K_o=4.1;// osmoles/cm^3
+c_Cl_c=4.00;// osmoles/cm^3
+c_Cl_o=107;// osmoles/cm^3
+
+// Solution
+E_Na=(26.7/1)*log(c_Na_o/c_Na_c);// mV
+E_K=(26.7/1)*log(c_K_o/c_K_c);// mV
+E_Cl=(26.7/-1)*log(c_Cl_o/c_Cl_c);// mV
+printf("\nThe membrane potential of sodium in a human cell,E_Na+=%2.1f mV \nThe membrane potential of potassium in a human cell,E_K+=%2.1f mV \nThe membrane potential of chlorine in a human cell,E_Cl-=%2.1f mV",E_Na,E_K,E_Cl);
diff --git a/3831/CH17/EX17.10/Ex17_10.sce b/3831/CH17/EX17.10/Ex17_10.sce new file mode 100644 index 000000000..bc0464b58 --- /dev/null +++ b/3831/CH17/EX17.10/Ex17_10.sce @@ -0,0 +1,9 @@ +// Example 17_10
+clc;funcprot(0);
+// Given data
+T=27+273;
+k_d=0.0350;
+
+// Calculation
+alpha=k_d/(T*exp((9.62*10^4*((T-330)/(330*T)))-33.2));
+disp(alpha)
diff --git a/3831/CH17/EX17.2/Ex17_2.sce b/3831/CH17/EX17.2/Ex17_2.sce new file mode 100644 index 000000000..93dcbee13 --- /dev/null +++ b/3831/CH17/EX17.2/Ex17_2.sce @@ -0,0 +1,18 @@ +// Example 17_2
+clc;funcprot(0);
+// Given data
+n_ech=20.0;// The energy conversion efficiency of the plants eaten by grazing herbivores in %
+n_ecc=5.0;// The energy conversion efficiency of the carnivores in %
+n_o=(0.100*0.200*0.0500)*100;// %
+E_avg=15.3;// The average daily solar energy reaching the surface of the Earth MJ/d.m^2
+E_c=10.0;// MJ/d
+
+// Calculation
+// car-carnivore,her-herbivore,ec-energy conversion efficiency
+E_car=E_c/(n_ecc/100);// MJ/d
+E_her=E_car/(n_ech/100);// MJ/d
+n_ec=1/100;// Energy conversion rate
+E_hreq=E_her/(n_ec);// MJ/d
+A=E_hreq/E_avg;// Area in m^2
+A_acre=A*(1/4047);// acres
+printf("\n%1.2f acres of land is required to grow the plants needed to feed the herbivores eaten by a large carnivore that requires 10.0 MJ/d to stay alive.",A_acre);
diff --git a/3831/CH17/EX17.3/Ex17_3.sce b/3831/CH17/EX17.3/Ex17_3.sce new file mode 100644 index 000000000..82198db42 --- /dev/null +++ b/3831/CH17/EX17.3/Ex17_3.sce @@ -0,0 +1,10 @@ +// Example 17_3
+clc;funcprot(0);
+// Given data
+m_h=80.0;// kg
+m_m=0.008;// kg
+
+// Solution
+BMRbym_human=293*(m_h^-0.25);// kJ/kg.d
+BMRbym_mouse=293*(m_m^-0.25);// kJ/kg.d
+printf("\nThe BMR per unit mass of an 80.0 kg human,(BMR/m)_human=%2.0f kJ/kg.d \nThe BMR per unit mass of an 8.00 gram mouse,(BMR/m)_mouse=%3.0f kJ/kg.d",BMRbym_human,BMRbym_mouse);
diff --git a/3831/CH17/EX17.4/Ex17_4.sce b/3831/CH17/EX17.4/Ex17_4.sce new file mode 100644 index 000000000..bbc133197 --- /dev/null +++ b/3831/CH17/EX17.4/Ex17_4.sce @@ -0,0 +1,17 @@ +// Example 17_4
+clc;funcprot(0);
+// Given data
+e=10.5;// MJ
+C=45/100;// MJ/kg
+P=15.0/100;// MJ/kg
+F=40.0/100;// MJ/kg
+
+// Calculation
+// (a)
+e_C=4.20;// MJ/kg meal
+e_P=8.40;// MJ/kg meal
+e_F=33.1;// MJ/kg meal
+e_avgMeal=(C*e_C)+(P*e_P)+(F*e_F);// MJ/kg meal
+// (b)
+mdot_avgMeal=(e/e_avgMeal)*2.187;// lbm of average meal/day
+printf("\n(a)The specific energy content of an average meal with natural state foods,e_avg meal=%2.1f MJ/kg meal \n(b)The total mass of an average meal,mdot_avg meal=%1.1f lbm of average meal/day",e_avgMeal,mdot_avgMeal);
diff --git a/3831/CH17/EX17.5/Ex17_5.sce b/3831/CH17/EX17.5/Ex17_5.sce new file mode 100644 index 000000000..482423719 --- /dev/null +++ b/3831/CH17/EX17.5/Ex17_5.sce @@ -0,0 +1,14 @@ +// Example 17_5
+clc;funcprot(0);
+// Given data
+m_h=1.00;// kg
+E_me=33.1;// MJ
+E_na=10.5;// MJ
+m_fat=10.0;// kg
+
+// Calculation
+// (a)
+mdot_fat=E_na/E_me;// The mass of body fat consumed per day in kg of body/d
+// (b)
+t=m_fat/mdot_fat;// d
+printf("\n(a)The mass of body fat consumed per day,mdot_fat=%0.3f kg of body/d \n(b)The number of fasting days required to lose (consume) 10.0 kg of body fat,t=%2.1f d",mdot_fat,t);
diff --git a/3831/CH17/EX17.6/Ex17_6.sce b/3831/CH17/EX17.6/Ex17_6.sce new file mode 100644 index 000000000..38cc267a8 --- /dev/null +++ b/3831/CH17/EX17.6/Ex17_6.sce @@ -0,0 +1,18 @@ +// Example 17_6
+clc;funcprot(0);
+// Given data
+mg=490;// N
+Z=1.00;// m
+g_c=1;// The gravitational constant
+delt=1.00;// s
+
+// Calculation
+E=(mg*Z)/g_c;// J
+W=E/delt;// J/s
+n_T_muscle=25/100;// The energy conversion efficiency
+U_body=-W/n_T_muscle;// J/s
+Q=U_body+W;// J/s
+delU=-(1)*(2.51);// MJ
+tau=delU/(U_body*10^-6);// s
+tau=tau/60;// min
+printf("\nThe time required to produce a change in the total internal energy of the system that equals the energy content of one pint of ice cream,tau=%2.1f min",tau);
diff --git a/3831/CH17/EX17.7/Ex17_7.sce b/3831/CH17/EX17.7/Ex17_7.sce new file mode 100644 index 000000000..ffe32aff4 --- /dev/null +++ b/3831/CH17/EX17.7/Ex17_7.sce @@ -0,0 +1,15 @@ +// Example 17_7
+clc;funcprot(0);
+// Given data
+m_m=0.0300;// kg
+m_h=70.0;// kg
+m_e=4000;// kg
+
+// Calculation
+Hr_m=241*(m_m^(-0.25));// Beats/min
+Hr_h=241*(m_h^(-0.25));// Beats/min
+Hr_e=241*(m_e^(-0.25));// Beats/min
+Br_m=54*(m_m^(-0.25));// Beats/min
+Br_h=54*(m_h^(-0.25));// Beats/min
+Br_e=54*(m_e^(-0.25));// Beats/min
+printf("\nThe heartbeat rates of the mouse, human, and elephant are\n(Heartbeat rate)_mouse=%3.0f Beats/min \n(Heartbeat rate)_house=%2.1f Beats/min \n(Heartbeat rate)_elephant=%2.1f Beats/min \nThe breathing rates of the mouse, human, and elephant are \n(Breathing rate)_mouse=%3.0f Breaths/min \n(Breathing rate)_human=%2.1f Breaths/min \n(Breathing rate)_elephant=%1.2f Breaths/min",Hr_m,Hr_h,Hr_e,Br_m,Br_h,Br_e);
diff --git a/3831/CH17/EX17.8/Ex17_8.sce b/3831/CH17/EX17.8/Ex17_8.sce new file mode 100644 index 000000000..589492dd3 --- /dev/null +++ b/3831/CH17/EX17.8/Ex17_8.sce @@ -0,0 +1,8 @@ +// Example 17_8
+clc;funcprot(0);
+// Given data
+d=5.00*10^-3;// The base diameter of the tree in m
+
+// Calculation
+h_critical=68.0*(d^(2/3));// The critical buckling height of a small tree in m
+printf("\nThe critical buckling height of a small tree,h_critical=%1.2f m",h_critical);
diff --git a/3831/CH17/EX17.9/Ex17_9.sce b/3831/CH17/EX17.9/Ex17_9.sce new file mode 100644 index 000000000..7f4b4f067 --- /dev/null +++ b/3831/CH17/EX17.9/Ex17_9.sce @@ -0,0 +1,14 @@ +// Example 17_9
+clc;funcprot(0);
+// Given data
+m=60.0;// kg
+m_bc=15.0;// kg
+P=400;// W
+V=15.0;// miles/h
+g=9.81;// m/s^2
+
+// Calculation
+w=(m+m_bc)*9.81;// N
+V=(V*1.609)*1000;// m/h
+T=(P*3600)/(w*V);// The locomotion transport number
+printf("\nThe locomotion transport number,T=%0.4f",T);
diff --git a/3831/CH18/EX18.1/Ex18_1.sce b/3831/CH18/EX18.1/Ex18_1.sce new file mode 100644 index 000000000..b2beb8143 --- /dev/null +++ b/3831/CH18/EX18.1/Ex18_1.sce @@ -0,0 +1,18 @@ +// Example 18_1
+clc;funcprot(0);
+// Given data
+T=20+273.15;// K
+m=1.00;// kg
+R=296;// J/kg.K
+M=28.0;// kg/kgmole
+N_o=6.022*10^26;// molecules/kgmole
+k=1.380*10^-23;// J/molecule.K
+
+// Calculation
+// (a)
+V_rms=sqrt(3*R*T);// The kinetic theory root mean square molecular velocity in m/s
+// (b)
+m_molecule=M/N_o;// kg/molecule
+N=m/m_molecule;// molecules
+U_trans=(3/2)*(N*k*T)/1000;// The total translational internal energy in kJ
+printf("\n(a)The kinetic theory root mean square molecular velocity,V_rms=%3.0f m/s \n(b)The total translational internal energy,U=%3.0f kJ",V_rms,U_trans);
diff --git a/3831/CH18/EX18.10/Ex18_10.sce b/3831/CH18/EX18.10/Ex18_10.sce new file mode 100644 index 000000000..ad0bf0273 --- /dev/null +++ b/3831/CH18/EX18.10/Ex18_10.sce @@ -0,0 +1,34 @@ +// Example 18_10
+clc;funcprot(0);
+// Given data
+theta_r=0.562;// K
+theta_v1=1932;// K
+theta_v3=960;// K
+theta_v2=theta_v3;// K
+theta_v4=3380;// K
+p=101325;// Pa
+T=1000;// K
+R_u=8.314;// kJ/kg.K
+M=44.01;// The molecular mass of Carbon dioxide
+h_c=6.626*10^-34;// Planck's constant
+N_o=6.023*10^26;// molecules/kgmole
+k=1.38*10^-23;// J/molecule.K
+
+
+// Calculation
+m=M/N_o;// kg/molecule
+R=R_u/M;// kJ/kg.K
+u_o_vib=R*((theta_v1+theta_v2+theta_v3+theta_v4)/2);// kJ/kg
+u_vib=u_o_vib+(R*((theta_v1*exp(theta_v1-1)^-1)+(theta_v2*exp(theta_v2-1)^-1)+(theta_v3*exp(theta_v3-1)^-1)+(theta_v4*exp(theta_v4-1)^-1)));// kJ/kg
+u_trans=(3/2)*R*T;// kJ/kg
+u_rot=R*T;// kJ/kg
+u=u_trans+u_rot+u_vib;// kJ/kg
+h=u+(R*T);// kJ/kg
+Sigma=2;// molecules/m^3
+d=((((2*%pi*m)/(h_c^2))^(3/2))*(k*T)^(5/2))/p;// per molecule
+s_trans=R*(log(d)+(5/2));// kJ/kg.K
+s_rot=R*(log(T/(Sigma*theta_r))+1);// kJ/kg.K
+s_vib=R*[(((log(1-exp(-theta_v1/T))^-1)+((theta_v1/T)*[exp(theta_v1/T)-1]^-1))+((log(1-exp(-theta_v2/T))^-1)+((theta_v2/T)*[exp(theta_v2/T)-1]^-1))+((log(1-exp(-theta_v3/T))^-1)+((theta_v3/T)*[exp(theta_v3/T)-1]^-1))+((log(1-exp(-theta_v4/T))^-1)+((theta_v4/T)*[exp(theta_v4/T)-1]^-1)))];// kJ/kg.K
+s=s_trans+s_rot+s_vib;// kJ/kg.K
+printf("\nThe specific internal energy of CO2,u=%4.0f kJ/kg \nThe specific enthalpy of CO2,h=%4.0f kJ/kg \nThe specific entropy of CO2=%1.3f kJ/kg.K",u,h,s);
+// The answer provided in the text book is wrong
diff --git a/3831/CH18/EX18.2/Ex18_2.sce b/3831/CH18/EX18.2/Ex18_2.sce new file mode 100644 index 000000000..0749538a1 --- /dev/null +++ b/3831/CH18/EX18.2/Ex18_2.sce @@ -0,0 +1,19 @@ +// Example 18_2
+clc;funcprot(0);
+// Given data
+T=273;// K
+p=0.113;// MPa
+M=20.183;// kg/kg mole
+N_o=6.022*10^26;// molecules/kgmole
+k=1.380*10^-23;// J/(molecules.K)
+
+
+// Calculation
+m=M/N_o;// kg/molecule
+V_rms=((3*k*T)/m)^(1/2);// m/s
+r=1.3*10^-10;// The radius of the neon molecule in m
+sigma=4*%pi*r^2;// The collision cross-section in m^2
+NbyV=(p*10^6)/(k*T);// molecules/m^3
+F=sigma*V_rms*NbyV*(8/(3*%pi))^(1/2);// The collision frequency in collisions/s
+lambda=1/(NbyV*sigma);// The molecular mean free path in m
+printf("\nThe collision frequency,F=%1.2e collisions/s \nThe molecular mean free path,lambda=%1.2e m",F,lambda);
diff --git a/3831/CH18/EX18.3/Ex18_3.sce b/3831/CH18/EX18.3/Ex18_3.sce new file mode 100644 index 000000000..dd48dc042 --- /dev/null +++ b/3831/CH18/EX18.3/Ex18_3.sce @@ -0,0 +1,26 @@ +// Example 18_3
+clc;funcprot(0);
+// Given data
+T=273;// K
+m=3.35*10^-26;// kg
+k=1.38*10^-23;// J/(molecule.K)
+
+// Calculation
+// (a) The fraction having velocities greater than Vmp is given by Eq. (18.26) with x = Vmp/Vmp = 1.0
+x=1.00;// The velocity ratio
+NV_mpbyN=1-erf(x)+((2/sqrt(%pi))*x*exp(-(x^2)));// The fraction of molecules whose velocities lie in the range from V to infinity
+// (b)
+x=sqrt(8/(2*%pi));// The velocity ratio
+NV_avgbyN=1-erf(x)+((2/sqrt(%pi))*x*exp(-(x^2)));// The fraction of molecules whose velocities lie in the range from V to infinity
+// (c)
+// x=V_rms/V_mp;
+x=sqrt(3/2);// The velocity ratio
+NV_rmsbyN=1-erf(x)+((2/sqrt(%pi))*x*exp(-(x^2)));// The fraction of molecules whose velocities lie in the range from V to infinity
+// (d)
+x=10.0;// The velocity ratio
+NVbyN=((2/sqrt(%pi))*x*exp(-(x^2)));// The fraction of molecules whose velocities lie in the range from V to infinity
+c=3.00*10^8;// m/s
+V_mp=sqrt((2*k*T)/m);// m/s
+x=c/V_mp;// The velocity ratio
+NcbyN=((2/sqrt(%pi))*x*exp(-(x^2)));// The fraction of molecules whose velocities lie in the range from c to infinity
+printf("\n(a)%2.2f percentage of the molecules have velocities faster than V_mp. \n(b)%2.2f percentage of the molecules have velocities faster than V_avg. \n(c)%2.2f percentage of the molecules have velocities faster than V_rms. \n(d)The fraction of molecules whose velocities lie in the range from c to infinity is %0.0f.",NV_mpbyN*100,NV_avgbyN*100,NV_rmsbyN*100,NcbyN*100);
diff --git a/3831/CH18/EX18.4/Ex18_4.sce b/3831/CH18/EX18.4/Ex18_4.sce new file mode 100644 index 000000000..1250d6df0 --- /dev/null +++ b/3831/CH18/EX18.4/Ex18_4.sce @@ -0,0 +1,17 @@ +// Example 18_4
+clc;funcprot(0);
+// Given data
+T_in=500;// K
+T_out=1200;// K
+mdot=1.00;// kg/min
+R_u=8.314;// kJ/(kgmole.K)
+
+// Calculation
+// For CC1_4,
+b=5;// The number of atoms in the molecule
+F=3*b;// The degrees of freedom per molecule
+M=12.0+(4*(35.5));// kg/kgmole
+R=R_u/M;// kJ/(kg.K)
+c_p=(1+(F/2))*R;// kJ/(kg.K)
+Qdot=mdot*c_p*(T_out-T_in);// kJ/min
+printf("\nThe heat transfer rate required to heat low-pressure gaseous carbon tetrachloride,Qdot=%3.0f kJ/min",Qdot);
diff --git a/3831/CH18/EX18.6/Ex18_6.sce b/3831/CH18/EX18.6/Ex18_6.sce new file mode 100644 index 000000000..eb8c7e1d6 --- /dev/null +++ b/3831/CH18/EX18.6/Ex18_6.sce @@ -0,0 +1,10 @@ +// Example 18_6
+clc;funcprot(0);
+// Given data
+P_ace=4/52;// The probability of getting ace
+P_spade=13/52;// The probability of getting spade
+P_aceofspades=1/52;// The probability of getting ace of spades
+
+// Calculation
+P=(P_ace+P_spade-P_aceofspades)*100;// The probability that it will be an ace or a spade in %
+printf("\nThe probability that it will be an ace or a spade,P=%2.1f percentage",P);
diff --git a/3831/CH18/EX18.7/Ex18_7.sce b/3831/CH18/EX18.7/Ex18_7.sce new file mode 100644 index 000000000..ec18ef7bc --- /dev/null +++ b/3831/CH18/EX18.7/Ex18_7.sce @@ -0,0 +1,20 @@ +// Example 18_7
+clc;funcprot(0);
+// Given data
+N=10;// The number of available students
+R=5;// The number of ordered groups
+
+// Calculation
+// (a)
+P_a=(factorial(N))/(factorial(N-R));// P_using each student only once
+// (b)
+P_b=N^R;// P_using each student more than once
+// (c)
+P_c=(factorial(N))/(factorial(N-R)*factorial(R));// C_using each student only once
+// (d)
+P_d=(factorial(N+R-1))/(factorial(N-1)*factorial(R));// C_using each student more than once
+// (e)
+R_1=4;
+R_2=6
+P_e=(factorial(N))/((factorial(R_1))*(factorial(R_2)));
+printf("\n(a)P_using each student only once=%5.0f groups \n(b)P_using each student more than once=%5.0f groups \n(c)C_using each student only once=%3.0f groups \n(d)C_using each student more than once=%4.0f groups \n(e)P_4,6=%3.0f groups",P_a,P_b,P_c,P_d,P_e);
diff --git a/3831/CH18/EX18.8/Ex18_8.sce b/3831/CH18/EX18.8/Ex18_8.sce new file mode 100644 index 000000000..499697759 --- /dev/null +++ b/3831/CH18/EX18.8/Ex18_8.sce @@ -0,0 +1,20 @@ +// Example 18_8
+clc;funcprot(0);
+// Given data
+m=3.50;// kg
+T_1=20.0+273.15;// K
+p_1=0.101325;// MPa
+p_2=10.0;// MPa
+R_u=8.3143;// kJ/kg.K
+W_12=-100;// kJ
+
+// Calculation
+// (a)
+M_krypton=83.80;
+R_krypton=R_u/M_krypton;// kJ/kg.K
+Q_12=0;// kJ
+T_2=T_1-((W_12/(3*m*R_krypton/2)));// K
+// (b)
+S_p12=m*R_krypton*log(((T_2/T_1)^(5/2))*(p_1/p_2));// kJ/kg.K
+printf("\n(a)The final temperature of the krypton gas after compression,T_2=%3.0f K \n(b)The entropy production of the compression process,1(S_p)2=%1.2f kJ/kg.K",T_2,S_p12);
+// The answer provided in the textbook is wrong
diff --git a/3831/CH18/EX18.9/Ex18_9.sce b/3831/CH18/EX18.9/Ex18_9.sce new file mode 100644 index 000000000..161812b5e --- /dev/null +++ b/3831/CH18/EX18.9/Ex18_9.sce @@ -0,0 +1,14 @@ +// Example 18_9
+clc;funcprot(0);
+// Given data
+T=20.0+273.15;// K
+
+// Calculation
+theta_v=2740;// K
+c_vbyR=(5/2)+((((theta_v/T)^2)*exp((theta_v/T)))/(exp(theta_v/T)-1)^2);
+Y=8.3143;// kJ/kg.K
+M_NO=30.01;// The molecular mass of nitrous oxide
+R_NO=Y/M_NO;// kJ/kg.K
+c_v_NO=R_NO*c_vbyR;// kJ/kg.K
+printf("\nThe value of c_v/R for nitrous oxide is %1.2f.",c_vbyR);
+
diff --git a/3831/CH19/EX19.1/Ex19_1.sce b/3831/CH19/EX19.1/Ex19_1.sce new file mode 100644 index 000000000..b73f02a26 --- /dev/null +++ b/3831/CH19/EX19.1/Ex19_1.sce @@ -0,0 +1,14 @@ +// Example 19_1
+clc;funcprot(0);
+// Given data
+T=20.0+273.16;// K
+d=0.0100;// m
+alpha_cu=3.50*10^-6;// V/K
+rho_e=5.00*10^-9;// ohm m
+dphibydx=1.00;// Voltage gradient in V/m
+
+// Solution
+A=(%pi/4)*d^2;// m^2
+I=(A/rho_e)*dphibydx;// A
+Q_P=alpha_cu*T*I;// W
+printf('\nThe Peltier heat flow,Q_P=%2.1f W',Q_P);
diff --git a/3831/CH19/EX19.2/Ex19_2.sce b/3831/CH19/EX19.2/Ex19_2.sce new file mode 100644 index 000000000..2b57c3e9e --- /dev/null +++ b/3831/CH19/EX19.2/Ex19_2.sce @@ -0,0 +1,11 @@ +// Example 19_2
+clc;funcprot(0);
+// Given data
+T=100.0;// °C
+
+// Solution
+// (a)
+alpha_fecu=-(-13.4+(0.028*T)+(0.00039*T^2))*10^-6;// V/K
+// (b)
+pi_fecu=(T+273.16)*alpha_fecu;// V
+printf('\n(a)The relative Seebeck coefficient,alpha_fecu=%1.2e V/K \n(b)The relative Peltier coefficient,pi_fecu=%1.2e V',alpha_fecu,pi_fecu);
diff --git a/3831/CH19/EX19.3/Ex19_3.sce b/3831/CH19/EX19.3/Ex19_3.sce new file mode 100644 index 000000000..2897ca7f9 --- /dev/null +++ b/3831/CH19/EX19.3/Ex19_3.sce @@ -0,0 +1,20 @@ +// Example 19_3
+clc;funcprot(0);
+// Given data
+T_H=100;// °C
+T_C=0;// °C
+alpha_ch=23.0*10^-6;// V/K
+alpha_al=-18.0*10^-6;// V/K
+
+// Solution
+// (a)
+alpha_chal=alpha_ch-alpha_al;// V/K
+phi_alch=alpha_chal*(T_H-T_C);// V
+// (b)
+pi_ch1=alpha_ch*(T_C+273.15);// V
+pi_al1=alpha_al*(T_C+273.15);// V
+pi_chal1=pi_ch1-pi_al1;// V
+pi_ch2=alpha_ch*(T_H+273.15);// V
+pi_al2=alpha_al*(T_H+273.15);// V
+pi_chal2=pi_ch2-pi_al2;// V
+printf('\n(a)alpha_ch-al=%2.0e V/K \n phi_al-ch=%1.1e V \n(b)At the 0.00°C = 273.15 K junction, \npi_ch=%1.2e V \npi_al=%1.2e V \npi_ch-al=%2.1e V \nAt the 100.°C = 373.15 K junction,\npi_ch=%1.2e V \npi_al=%1.2e V \npi_ch-al=%2.1e V ',alpha_chal,phi_alch,pi_ch1,pi_al1,pi_chal1,pi_ch2,pi_al2,pi_chal2);
diff --git a/3831/CH19/EX19.4/Ex19_4.sce b/3831/CH19/EX19.4/Ex19_4.sce new file mode 100644 index 000000000..7b33c8310 --- /dev/null +++ b/3831/CH19/EX19.4/Ex19_4.sce @@ -0,0 +1,12 @@ +// Example 19_4
+clc;funcprot(0);
+// Given data
+mu=1.50*10^-5;// The viscosity of the CO_2 in kg/(m.s)
+T_1=300;// K
+T_2=305;// K
+k_p=1.00*10^-6;// m^2
+k_o=2.00*10^4;// The osmotic heat conductivity in m^2/s
+
+// Solution
+dp=-((mu*k_o)/k_p)*log(T_2/T_1);// N/m^2
+printf('\nThe steady state thermomolecular pressure difference across the membrane,p_2-p_1=%4.0f N/m^2',dp);
diff --git a/3831/CH19/EX19.5/Ex19_5.sce b/3831/CH19/EX19.5/Ex19_5.sce new file mode 100644 index 000000000..d0febe114 --- /dev/null +++ b/3831/CH19/EX19.5/Ex19_5.sce @@ -0,0 +1,22 @@ +// Example 19_5
+clc;funcprot(0);
+// Given data
+T_1=30+273.15;// K
+T_2=T_1;// K
+dp=10.0;// kPa
+d=0.0100;// m
+rho=996;// kg/m^3
+k_p=1.00*10^-12;// m^2
+mu=891*10^-6;// kg/(s.m)
+dx=0.100;// m
+Q=15.0;// The isothermal energy transport rate in this system in J/s
+
+// Solution
+// (a)
+A=(%pi/4)*d^2;// m^2
+m=-((rho*A*k_p)/mu)*((dp*10^3)/dx);// kg/s
+// (b)
+k_o=-(Q/A)/((-dp*10^3)/dx);// m^2/s
+// (c)
+S_i=Q/T_1;// J/(s.K)
+printf('\n(a)The thermomechanical mass flow rate between the vessels,m=%1.2e kg/s \n(b)The osmotic heat conductivity coefficient,k_o=%1.2f m^2/s \n(c)The isothermal entropy transport rate induced by the thermomechanical mass flow rate,S_i=%0.4f J/(s.K)',m,k_o,S_i);
diff --git a/3831/CH19/EX19.6/Ex19_6.sce b/3831/CH19/EX19.6/Ex19_6.sce new file mode 100644 index 000000000..56b8330e9 --- /dev/null +++ b/3831/CH19/EX19.6/Ex19_6.sce @@ -0,0 +1,17 @@ +// Example 19_6
+clc;funcprot(0);
+// Given data
+rho=996;// kg/m^3
+Q=8.70;// J/s
+T=30+273;// K
+k_t=0.610;// J/(s.K.m)
+k_o=1.91;// m^2/s
+k_p=1.00*10^-12;// m^2
+mu=891*10^-6;// kg/(s.m)
+dx=0.100;// m
+
+// Solution
+m=(rho*Q)/((T*(k_t/k_o))+(mu*(k_o/k_p)));// kg/s
+dTbydx=-(T*m)/(rho*k_o);// K/m
+dT=dTbydx*dx;// K
+printf('\nThe induced isobaric mass flow rate,m=%1.2e kg/s \nThe resulting temperature difference between the vessels,dT=%1.2e K',m,dT);
diff --git a/3831/CH2/EX2.1/Ex2_1.sce b/3831/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..d1bda3137 --- /dev/null +++ b/3831/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,13 @@ +// Example 2_1
+clc;funcprot(0);
+// Given data
+C_c=120000;// The number of chips per day to its customers in chips/day
+C_s=100000;// The number of chips receives per day from its suppliers in chips/day
+C_m=30000;// The number of chips manufactures of its own in chips/day
+C_r=3000;// The number of chips are rejected as defective and are destroyed in chips/day
+
+// Solution
+X_T=C_s-C_c;// The net transport of chips into the facility in chips/day
+X_P=C_m-C_r;// The net production of chips in chips/day
+X_G=X_T+X_P;// The net gain in computer chips at the end of each day in in chips/day
+printf('\nThe net gain in computer chips at the end of each day,X_G=%4.0f chips per day.',X_G);
diff --git a/3831/CH2/EX2.4/Ex2_4.sce b/3831/CH2/EX2.4/Ex2_4.sce new file mode 100644 index 000000000..25008096c --- /dev/null +++ b/3831/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,12 @@ +// Example 2_4
+clc;funcprot(0);
+// Given data
+m_pendulum=5.0;// The mass of the pendulum in kg
+m_projectile=0.01;// The mass of the projectile in kg
+g=9.81;// The acceleration due to gravity in m/s^2
+R=1.5;// The length of the pendulum support cable in m
+theta=15;// degree
+
+// Solution
+V_projectile=(1+(m_pendulum/m_projectile))*(2*g*R*[1-cosd(theta)])^(1/2);// The muzzle velocity in m/s
+printf('\nThe muzzle velocity,V_projectile=%1.0e m/s',V_projectile);
diff --git a/3831/CH3/EX3.2/Ex3_2.sce b/3831/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..2d2904ef7 --- /dev/null +++ b/3831/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,14 @@ +// Example 3_2
+clc;funcprot(0);
+// Given data
+T_1=250;// K
+T_2=800;// K
+beta_1=48.0*10^-6;// K^-1
+beta_2=60.7*10^-6;// K^-1
+V_1=1.00;// cm^3
+
+// Solution
+beta_avg=(beta_2+beta_1)/2;// K^-1
+beta=beta_avg;// K^-1
+V_2=V_1*exp(beta*(T_2-T_1));// The final volume in cm^3
+printf('\nThe volume of the block,V_2=%1.2f cm^3',V_2);
diff --git a/3831/CH3/EX3.3/Ex3_3.sce b/3831/CH3/EX3.3/Ex3_3.sce new file mode 100644 index 000000000..ec925fed7 --- /dev/null +++ b/3831/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,11 @@ +// Example 3_3
+clc;funcprot(0);
+// Given data
+u=82.77;// The specific internal energy in kJ/kg
+v=0.0009928;// The specific volume of liquid water in m^3/kg
+T=20.0;// °C
+P=20.0;// MPa
+
+// Solution
+h=u+(P*10^3*v);// The specific enthalpy of the water in kJ/kg
+printf('\n The specific enthalpy of the water,h=%3.0f kJ/kg',h);
diff --git a/3831/CH3/EX3.4/Ex3_4.sce b/3831/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..e8cbc2f17 --- /dev/null +++ b/3831/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,31 @@ +// Example 3_4
+clc;funcprot(0);
+// Given data
+T=212;// °F
+V=3.00;// The total volume in ft^3
+m=0.200;// lbm
+p=14.696;// psia
+v_f=0.01672;// ft^3/lbm
+v_g=26.80;// ft^3/lbm
+u_f=180.1;// Btu/lbm
+u_g=1077.6;// Btu/lbm
+h_f=180.1;// Btu/lbm
+h_g=1150.5;// Btu/lbm
+
+// Solution
+// (a)
+v=V/m;// The specific volume in ft^3/lbm
+// (b)
+v_fg=v_g-v_f;// ft^3/lbm
+x=(v-v_f)/v_fg;// The quality
+x_m=1-x;// The amount of moisture present
+// (c)
+u_fg=u_g-u_f;// Btu/lbm
+u=u_f+(x*u_fg);// The specific internal energy in Btu/lbm
+// (d)
+h_fg=h_g-h_f;// Btu/lbm
+h=h_f+(x*h_fg);// The specific enthalpy in Btu/lbm
+// (e)
+m_g=x*m;// The mass of water in the vapor phase in lbm
+m_f=m-m_g;// The mass of water in the liquid phase in lbm
+printf('\n(a)The specific volume,v=%2.0f ft^3/lbm \n(b)The quality,x=%0.3f (or) %2.1f percentage \n The amount of moisture present,1-x=%0.3f (or) %2.1f percentage \n(c)The specific internal energy,u=%3.0f Btu/lbm \n(d)The specific enthalpy,h=%3.0f Btu/lbm \n(e)The mass of water in the liquid and vapor phases,m_f=%0.3f lbm & m_g=%0.3f lbm',v,x,x*100,x_m,x_m*100,u,h,m_f,m_g);
diff --git a/3831/CH3/EX3.5/Ex3_5.sce b/3831/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..77b853daf --- /dev/null +++ b/3831/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,16 @@ +// Example 3_5
+clc;funcprot(0);
+// Given data
+V=0.500;// ft^3
+p_c=3203.8;// psia
+T_c=1165.1;// R
+v_c=0.05053;// ft^3/lbm
+p_1=14.696;// psia
+T_1=212;// °F
+v_f1=0.01672;// ft^3/lbm
+v_g1=26.8;// ft^3/lbm
+
+// Solution
+m=V/v_c;// lbm
+x_1=((v_c-v_f1)/(v_g1-v_f1))*100;// % percentage
+printf('\nThe initial quality in the vessel,x_1=%0.3f percentage vapor',x_1);
diff --git a/3831/CH3/EX3.6/Ex3_6.sce b/3831/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..a427ae48b --- /dev/null +++ b/3831/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,15 @@ +// Example 3_6
+clc;funcprot(0);
+// Given data
+T_1=20;// °C
+T_2=100;// °C
+p_1=0.100;// MPa
+p_2=1.00;// MPa
+rho=515;// kg/m^3
+c=1.76;// kJ/kg.K.
+
+// Solution
+deltau=c*((T_2+273.15)-(T_1+273.15));// The change in specific internal energy in kJ/kg
+v=1/rho;// The specific volume in m^3/kg
+deltah=deltau+(v*((p_2*10^3)-(p_1*10^3)));// The change in specific enthalpy in kJ/kg
+printf('\nThe change in specific internal energy,u_2-u_1=%3.0f kJ/kg \nThe change in specific enthalpy,h_2-h_1=%3.0f kJ/kg',deltau,deltah);
diff --git a/3831/CH3/EX3.7/Ex3_7.sce b/3831/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..b60011599 --- /dev/null +++ b/3831/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,26 @@ +// Example 3_7
+clc;funcprot(0);
+// Given data
+T_1=240;// °F
+T_2=80;// °F
+p_1=150;// psia
+p_2=14.7;// psia
+c_p=0.240;// Btu/lbm · R
+c_v=0.172;// Btu/lbm · R
+
+// Solution
+// (a)
+deltau=c_v*((T_2+459.67)-(T_1+459.67));// Btu/lbm
+deltah=c_p*(T_2-T_1);// Btu/lbm
+printf('\n(a)The change in specific internal energy,u_2-u_1=%2.1f Btu/lbm \n The change in specific enthalpy,h_2-h_1=%2.1f kJ/kg',deltau,deltah);
+// (b)
+// Values for u and h for variable specific heat air can be found in Table C.16.
+T_1=T_1+459.67;// R
+h_1=167.56;// Btu/lbm
+u_1=119.58;// Btu/lbm
+T_2=T_2+459.67;// R
+h_2=129.06;// Btu/lbm
+u_2=92.04;// Btu/lbm
+deltau=u_2-u_1;// Btu/lbm
+deltah=h_2-h_1;// Btu/lbm
+printf('\n(b)The change in specific internal energy,u_2-u_1=%2.1f Btu/lbm \n The change in specific enthalpy,h_2-h_1=%2.1f kJ/kg',deltau,deltah);
diff --git a/3831/CH3/EX3.8/Ex3_8.sce b/3831/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..b61d7d16b --- /dev/null +++ b/3831/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,14 @@ +// Example 3_8
+clc;funcprot(0);
+// Given data
+T_max=2830;// The maximum temperature in °C
+rho=200;// The density of the propellant gases in kg/m^3
+R=8314.3;// N.m/(kgmole.K)
+M=23.26;// The molecular mass of the propellant gases in kg/kgmole
+b=0.960*10^-3;// The volume occupied by the molecules of the propellant gases in m^3/kg
+
+// Solution
+v=1/rho;// m^3/kg
+p_max=(R*(T_max+273.15))/(M*(v-b));// N/m^2
+p_max=p_max/6894.76;// lbf/ in^2 absolute
+printf('\nThe maximum pressure in the breech as the cannon fires,p_max=%5.0f psia',p_max);
diff --git a/3831/CH3/EX3.9/Ex3_9.sce b/3831/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..f44902c43 --- /dev/null +++ b/3831/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,18 @@ +// Example 3_9
+clc;funcprot(0);
+// Given data
+T=100;// °F
+p=95.0;// psia
+
+// Calculation
+// From Table C.7a,C.8a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics,
+v_1=0.5751;// ft^3/lbm (100°F,90.0 psia)
+v_2=0.5086;// ft^3/lbm (100°F,100.0 psia)
+p_i1=100;// psia (Pressure used for interpolation)
+p_i2=90;// psia
+v=v_1+(((p-p_i2)/(p_i1-p_i2))*(v_2-v_1));// ft^3/lbm (100°F,95.0 psia)
+h_1=118.39;// Btu/lbm (100°F,90 psia)
+h_2=117.73;// Btu/lbm (100°F,100 psia)
+h=h_1+(((p-p_i2)/(p_i1-p_i2))*(h_2-h_1));// Btu/lbm (100°F,95.0 psia)
+printf("\nThe specific volume of Refrigerant-134a,v (100°F,95.0 psia)=%0.5f ft^3/lbm \nThe specific enthalpy of Refrigerant-134a,h (100°F,95.0 psia)=%3.2f Btu/lbm",v,h);
+
diff --git a/3831/CH4/EX4.1/Ex4_1.sce b/3831/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..eae71bdc4 --- /dev/null +++ b/3831/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,20 @@ +// Example 4_1
+clc;funcprot(0);
+// Given data
+p_1=10.0;// psia
+x_1=1.00;// The quality of saturated vapor
+V_1=25000;// mph
+Z_1=200;// miles
+v_1=38.42;// ft^3/lbm
+m=3.0;// lbm
+u_2=950.0;// The final specific internal energy in Btu/lbm
+v_2=v_1;;// ft^3/lbm
+g=32.174;// The acceleration due to gravity in m/s^2
+
+// Solution
+// Table C.2a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics gives
+u_1=1072.2;// Btu/lbm
+U_1=m*u_1;// Btu
+U_2=m*u_2;// Btu
+E_T=(U_2-U_1)-([((m/2)*(V_1*(5280/3600))^2)]*((1/(g*778.16))))-[((m*g)/g)*Z_1*5280/778.16];// Btu
+printf('\nThe energy transport is required to decelerate the water to zero velocity and bring it down to the surface of the Earth,E_T=%5.0f Btu',E_T);
diff --git a/3831/CH4/EX4.10/Ex4_10.sce b/3831/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..33b054473 --- /dev/null +++ b/3831/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,17 @@ +// Example 4_10
+clc;funcprot(0);
+// Given data
+T=20;// °C
+mu_0=4*%pi*10^-7;// V.s/A
+Shi_m=-2.20*10^-5;// The electric susceptibility
+H_2=1.00*10^3;// A/m
+V=5.00*10^-6;// m^3
+
+// Solution
+// (a)
+H_1=0;// A/m
+W_12=-mu_0*V*(1+Shi_m)*((H_2^2-H_1^2)/2);// J
+printf('\n(a)The total magnetic work required,(W_12)magnetic=%1.2e J',W_12);
+// (b)
+W_12=-mu_0*V*Shi_m*((H_2^2-H_1^2)/2);// J
+printf('\n(b)The magnetic work required to change the magnetic field strength,(W)_magnetic=%1.2e J',W_12);
diff --git a/3831/CH4/EX4.11/Ex4_11.sce b/3831/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..67a05560e --- /dev/null +++ b/3831/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,15 @@ +// Example 4_11
+clc;funcprot(0);
+// Given data
+W_actual=150;// hp
+W_reversible=233;// hp
+m_in=1.10;// lbm/min
+E=20.0*10^3;// Btu/lbm
+
+// Solution
+W_in=(E*m_in*60)/2545;// hp
+// (a)
+n_c=(W_actual/W_in)*100;// The energy conversion efficiency of the engine in %
+// (b)
+n_W=(W_actual/W_reversible)*100;// The work efficiency of the engine.
+printf('\n(a)The energy conversion efficiency of the engine,n_c=%2.1f percentage \n(b)The work efficiency of the engine,n_W=%2.1f percentage',n_c,n_W);
diff --git a/3831/CH4/EX4.2/Ex4_2.sce b/3831/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..73b8513f9 --- /dev/null +++ b/3831/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,16 @@ +// Example 4_2
+clc;funcprot(0);
+// Given data
+E_fuel=15000;// Btu/min
+E_exhaust=500;// Btu/min
+W_1=200;// hp
+W_2=50;// hp
+E_thl=180000;// Top heat loss in Btu/h
+E_Bhl=54000;// Bottom heat loss in Btu/h
+
+// Solution
+Q=-E_thl-E_Bhl;// The net heat transfer into the system in Btu/h
+W=W_1+W_2;// The net work rate out of the system in hp
+E_massflow=E_fuel-E_exhaust;// The net mass flow of energy into the system in Btu/min
+E_T=(Q/60)-(W*42.4)+E_massflow;// The total energy transport rate in Btu/min
+printf('\nThe total energy transport rate,E_T=%1.2f Btu/min',E_T);
diff --git a/3831/CH4/EX4.4/Ex4_4.sce b/3831/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..22d6af91d --- /dev/null +++ b/3831/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,10 @@ +// Example 4_4
+clc;funcprot(0);
+// Given data
+p=20.0;// Pressure in psia
+D_1=1.00;// Initial diameter in ft
+D_2=10.0;// Final diameter in ft
+
+// Solution
+W_12=p*144*(%pi/6)*(D_2^3-D_1^3);// ft.lbf
+printf('\nThe moving system boundary work,W_12=%1.2e ft.lbf',W_12);
diff --git a/3831/CH4/EX4.5/Ex4_5.sce b/3831/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..84f10efb6 --- /dev/null +++ b/3831/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,16 @@ +// Example 4_5
+clc;funcprot(0);
+// Given data
+T_1=20.0;// °C
+n=1.35;// The polytropic index
+m=0.0100;// kg
+p_1=0.100;// MPa
+m_2=0.0100;// kg
+p_2=10.0;// MPa
+
+// Solution
+T_2=((T_1+273.15)*(p_2/p_1)^((n-1)/n))-273.15;// °C
+// Using Table C.13b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics to find the value of the gas constant for methane,
+R_methane=0.518;// kJ/kg.K
+W_12=(m*R_methane*((T_2+273.15)-(T_1+273.15)))/(1-n);// kJ
+printf('\nThe moving boundary work required,W_12=%1.2f kJ',W_12);
diff --git a/3831/CH4/EX4.7/Ex4_7.sce b/3831/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..da941b99c --- /dev/null +++ b/3831/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,14 @@ +// Example 4_7
+clc;funcprot(0);
+// Given data
+D_1=0;// m
+D_2=0.0500;// m
+Sigma_s=0.0400;// N/m (constant)
+
+// Solution
+A_1=0;// m^2
+R_2=D_2/2;// m
+A_2=2*(4*%pi*R_2^2);// m^2
+W_12=-Sigma_s*(A_2-A_1);// J
+W_12=W_12/1055;// Btu
+printf('\nThe amount of surface tension work required to inflate the soap bubble,(W_12)_surface tension=%1.2e Btu',W_12);
diff --git a/3831/CH4/EX4.8/Ex4_8.sce b/3831/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..cbaed2db6 --- /dev/null +++ b/3831/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,14 @@ +// Example 4_8
+clc;funcprot(0);
+// Given data
+phi_e=120;// V
+R=144;// ohm
+t=1.50;// h
+
+// Solution
+// (a)
+i_e=phi_e/R;// A
+W_12=-phi_e*i_e*t;// The electrical current work in W.h
+// (b)
+W_ec=-phi_e*i_e;// W
+printf('\n(a)The electrical current work,W_12=%3.0f W.h \n(b)The electrical power consumption,W_electrical current=%3.0f W',W_12,W_ec);
diff --git a/3831/CH4/EX4.9/Ex4_9.sce b/3831/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..c9d62348b --- /dev/null +++ b/3831/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,16 @@ +// Example 4_9
+clc;funcprot(0);
+// Given data
+deltaphi=120;// volts
+L=0.0100;// The distance between two plates in m
+d=0.100;// The length of the plate on square side in m
+epsilon_0=8.85419*10^-12;// The electric permittivity of vacuum in N/V^2
+
+// Solution
+E_1=0;// V/m
+A=0.100*0.100;// m^2
+V=A*L;// m^3
+E_2=deltaphi/L;// V/m
+Shi_e=77.5;// The electric susceptibility
+W_12=-(epsilon_0*Shi_e*V*(E_2^2-E_1^2))/2;// The polarization work required in the charging of the capacitor in J
+printf('\nThe polarization work required in the charging of the capacitor,W_12=%1.2e N.m',W_12);
diff --git a/3831/CH5/EX5.1/Ex5_1.sce b/3831/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..7cda0334c --- /dev/null +++ b/3831/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,35 @@ +// Example 5_1
+clc;funcprot(0);
+// Given data
+V=1.00;// m^3
+m=2.00;// kg
+T_1=20.0;// °C
+T_2=95.0;// °C
+
+// Calculation
+v_1=V/m;// m^3/kg
+v_2=v_1;// m^3/kg
+// Step 7
+// From Table C.1b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that
+// At 20.0°C
+v_f1=0.001002;// m^3/kg
+v_g1=57.79;// m^3/kg
+v_fg1=v_g1-v_f1;// m^3/kg
+u_f1=83.9;// kJ/kg
+u_g1=2402.9;// kJ/kg
+u_fg1=u_g1-u_f1;// kJ/kg
+// At 95.0°C
+v_f2=0.00104;// m^3/kg
+v_g2=1.982;// m^3/kg
+v_fg2=v_g2-v_f2;// m^3/kg
+u_f2=397.9;// kJ/kg
+u_g2=2500.6;// kJ/kg
+u_fg2=u_g2-u_f2;// kJ/kg
+x_1=(v_1-v_f1)/v_fg1;// The quality in the container when the contents are at 20.0°C
+x_1p=x_1*100;// %
+x_2=(v_2-v_f2)/v_fg2;// The quality in the container when the contents are at 95.0°C.
+x_2p=x_2*100;// %
+u_1=u_f1+(x_1*u_fg1);// kJ/kg
+u_2=u_f2+(x_2*u_fg2);// kJ/kg
+Q_12=m*(u_2-u_1);// kJ
+printf('\n(a)The quality in the container when the contents are at 20.0°C,x_1=%0.3f percentage \n(b)The quality in the container when the contents are at 95.0°C,x_2=%2.1f percentage \n(c)The heat transport of energy required to raise the temperature of the contents from 20.0 to 95.0°C,Q_12=%4.0f kJ/kg',x_1p,x_2p,Q_12);
diff --git a/3831/CH5/EX5.2/Ex5_2.sce b/3831/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..e441e2a80 --- /dev/null +++ b/3831/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,15 @@ +// Example 5_2
+clc;funcprot(0);
+// Given data
+W=100;// W
+
+// Calculation
+// (a)
+// Since we are assuming a constant bulb temperature in part a, U=constant and
+U=0;// W
+Q=U-W;// kW
+printf("\n(a)The heat transfer rate of an illuminated 100 W incandescent lightbulb in a room,Q=%3.0f W",Q);
+// (b)
+Q=0;
+Udot=W;// W
+printf("\n(b)The rate of change of its internal energy,Udot=%3.0f W",Udot);
diff --git a/3831/CH5/EX5.3/Ex5_3.sce b/3831/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..116898d56 --- /dev/null +++ b/3831/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,14 @@ +// Example 5_3
+clc;funcprot(0);
+// Given data
+Q_B=950*10^5;// kJ/h
+W_p=23.0;// kW
+Q_c=-600*10^5;// kJ/h
+
+// Calculation
+Q_net=(Q_B+Q_c);// kJ/h
+W_T_net=Q_net/3600;// kJ/h
+W_T_net=W_T_net/1000;// MW
+W_T_total=(W_T_net*10^3)+W_p;// kW
+printf("\nThe net power of the turbine,(W_T)_total=%4.0f kW(round off error)",W_T_total);
+// The answer vary due to round off error
diff --git a/3831/CH5/EX5.4/Ex5_4.sce b/3831/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..dd031c017 --- /dev/null +++ b/3831/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,18 @@ +// Example 5_4
+clc;funcprot(0);
+// Given data
+W=0.250;// hp
+V=1.00;// quart of water
+p_1=14.7;// psia
+T_1=60.0;// °F
+p_2=p_1;// psia
+t=10;// min
+c=1.00;// Btu/(lbm.R)
+
+// Calculation
+V=V*(1/4)*0.13368;// ft^3
+v=0.01603;// ft^3/lbm
+m=V/v;// lbm
+Q_12bymc=0;
+T_2=T_1+Q_12bymc-((-W*t*(1/60)*(2545))/(m*c));// °F
+printf('\nThe temperature of the water when the machine is turned off,T_2=%3.0f°F',T_2)
diff --git a/3831/CH5/EX5.5/Ex5_5.sce b/3831/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..02aa4388a --- /dev/null +++ b/3831/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,16 @@ +// Example 5_5
+clc;funcprot(0);
+// Given data
+V_2=0.0400;// m^3
+T_1=20.0;// °C
+p_1=0.0100;// MPa
+Q_12=0.100;// kJ
+V_1=0.0100;// m^3
+R=0.208;// kJ/kg.K
+c_v=0.315;// kJ/kg.K
+
+// Calculation
+m=((p_1*10^3)*V_1)/(R*(T_1+273.15));// kg
+T_2=T_1+(Q_12/(m*c_v));// K
+p_2=(m*R*(T_2+273.15))/V_2;// kPa
+printf('\nThe pressure and temperature inside the box after the balloon bursts p_2=%1.2f kPa and T_2=%3.0f°C',p_2,T_2);
diff --git a/3831/CH5/EX5.6/Ex5_6.sce b/3831/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..8aff52e26 --- /dev/null +++ b/3831/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,37 @@ +// Example 5_6
+clc;funcprot(0);
+// Given data
+// State 1
+m=0.100;// lbm
+p_1=100;// psia
+T_1=180;// °F
+// State 2
+p_2=30.0;// psia
+T_2=120;// °F
+// State 3
+p_3=p_2;// psia
+
+// Calculation
+// (a)
+// From Table C.7e of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that at p1 = 100 psia and T1 = 180°F,
+v_1=0.6210;// ft^3/lbm
+u_1=125.99;// Btu/lbm
+// At p2= 30 psia and T2 = 120°F,
+v_2=1.966;// ft^3/lbm
+u_2=115.47;// Btu/lbm
+W_12=-m*(u_2-u_1);// Btu
+// (b)
+v_3=v_1/2;// ft^3/lbm
+// At p2= 30 psia
+v_f3=0.01209;// ft^3/lbm
+v_g3=1.5408;// ft^3/lbm
+u_f3=16.24;// Btu/lbm
+u_g3=95.40;// Btu/lbm
+x_3=(v_3-v_f3)/(v_g3-v_f3);// The quality of steam
+x_3p=x_3*100;// %
+u_3=u_f3+(x_3*(u_g3-u_f3));// Btu/lbm
+Q_23=(m*(u_3-u_2))+(m*(p_3*144)*((v_3-v_2)*(1/778.17)));// Btu
+// (c)
+// From Table C.7b
+T_3=15.38;// °F
+printf('\n(a)The work transport of energy during the adiabatic expansion,W_12=%1.2f Btu \n(b)The heat transport of energy during the isobaric compression,Q_23=%1.2f Btu \n(c)Since state 3 is saturated (a mixture of liquid and vapor), T3 must be equal to the saturation temperature at 30.0 psia,which, from Table C.7b, is T_3 =%2.2f°F',W_12,Q_23,T_3);
diff --git a/3831/CH5/EX5.7/Ex5_7.sce b/3831/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..cd61fe86a --- /dev/null +++ b/3831/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,20 @@ +// Example 5_7
+clc;funcprot(0);
+// Given data
+D=0.100;// m
+T_1=200;// °C
+p_1=0.140;// MPa
+h=3.50;// W/(m^2.K)
+T_infinitive=15.0;// °C
+c_v=3.123;// kJ/kg.K
+R=2.077;// kJ/kg.K
+t=5.00;// seconds
+
+// Calculation
+V=(%pi/6)*D^3;// m^3
+A=%pi*D^2;// m^2
+m=((p_1*10^3)*V)/(R*(T_1+273.15));// kg
+hAbymc_v=(h*A)/(m*c_v*1000);// s^-1
+T_2=((T_1-T_infinitive)*exp((-(h*A)/(m*c_v*1000))*t))+T_infinitive;// °C
+delU=m*c_v*(T_2-T_1);// kJ
+printf('\n(a)The final temperature of the helium,T_2=%2.1f°C \n(b)The change in total internal energy of the helium,U_2-U_1=%0.3f kJ',T_2,delU);
diff --git a/3831/CH5/EX5.8/Ex5_8.sce b/3831/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..927fe8697 --- /dev/null +++ b/3831/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,18 @@ +// Example 5_8
+clc;funcprot(0);
+// Given data
+P_1=600;// psia
+T_1=800;// ºF
+V=250;// ft^3
+gamma_TNT=1400;// Btu/lbm
+
+// Calculation
+// From the superheated steam table, Table C.3a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics,we find that at 600. psia and 800.ºF,
+u_1=1275.4;// Btu/lbm
+v_1=1.190;// ft^3/lbm
+u_f2=38.1;// Btu/lbm
+u_2=u_f2;// Btu/lbm
+gamma=(u_1-u_2)/v_1;// Btu/ft^3
+Ee=gamma*V;// Btu
+n=Ee/gamma_TNT;// The number of one-pound sticks of TNT to match the boiler explosion
+printf('\n(a)The explosive energy per unit volume of superheated steam,gamma=%4.1f Btu/ft^3 \n(b)%3.0f one-pound sticks of TNT to match the boiler explosion',gamma,n);
diff --git a/3831/CH6/EX6.1/Ex6_1.sce b/3831/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..b8fa4fa12 --- /dev/null +++ b/3831/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,22 @@ +// Example 6_1
+clc;funcprot(0);
+// Given data
+V=300;// ft/s
+D=6/12;// ft
+R=D/2;// ft
+Z=15;// ft
+g=32.174;// ft/s^2
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+// From the superheated steam table, Table C.3a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at 100. psia and 500.°F,
+v=5.587;// ft^3/lbm
+h=1279.1;// Btu/lbm
+A=%pi*(3/12)^2;// ft^2
+mdot=(A*V)/v;// lbm/s
+ke=(V^2)/(2*g_c);// ft.lbf/lbm
+ke=ke*(1/778.16);// Btu/lbm
+pe=(g*Z)/g_c;// // ft.lbf/lbm
+pe=pe*(1/778.16);// Btu/lbm
+E_mf=-[mdot*(h+ke+pe)];// Btu/s
+printf("\nThe mass flow energy transport rate of steam,E_mass flow=%1.2e Btu/s",E_mf);
diff --git a/3831/CH6/EX6.10/Ex6_10.sce b/3831/CH6/EX6.10/Ex6_10.sce new file mode 100644 index 000000000..7831668a3 --- /dev/null +++ b/3831/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,15 @@ +// Example 6_10
+clc;funcprot(0);
+// Given data
+p_1=2000;// psig
+T_1=200+459.67;// R
+T_T=70.0+459.67;// R
+m_R=0.500;// lbm/s
+W_c=-3.00;// hp
+k=1.4;// The specific heat ratio of nitrogen
+
+// Calculation
+m_Rbym_D=(k-1)/[(k*(T_1/T_T))-1];// The ratio of recycled mass flow rate to discharge mass flow rate
+c_p=0.248;// Btu/(lbm.R)
+Q_H=(m_R*c_p*(T_1-T_T))+[(W_c)*550*(1/778)];// Btu/s
+printf("\nThe rate of recycle heat transfer required,Q_H=%2.1f Btu/s",Q_H);
diff --git a/3831/CH6/EX6.2/Ex6_2.sce b/3831/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..8fecb1a65 --- /dev/null +++ b/3831/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,18 @@ +// Example 6_2
+clc;funcprot(0);
+// Given data
+D=1.00;// inch
+T=60.0;// °F
+p=80.0;// psig
+mdot=0.800;// lbm/s
+v=0.01603;// ft^3/lbm
+g_c=32.174;// lbm.ft/lbf.s^2
+g=32.174;// ft/s^2
+
+// Calculation
+V_in=(4*mdot*v)/(%pi*D^2*(1/12)^2);// ft/s
+p_in=94.7;// psia
+p_out=14.7;// psia
+V_out=[(V_in^2)+(2*g_c*v*(p_in-p_out)*144)]^(1/2);// ft/s
+Z_out=V_out^2/(2*g);// ft
+printf("\n(a)The outlet velocity from the nozzle,(V_out)_a=%3.0f ft/s \n(b)The height to which the stream of water rises above the nozzle outlet when the nozzle is pointed straight up,(Z_out)_b=%3.0f ft.",V_out,Z_out)
diff --git a/3831/CH6/EX6.3/Ex6_3.sce b/3831/CH6/EX6.3/Ex6_3.sce new file mode 100644 index 000000000..b0ddc7201 --- /dev/null +++ b/3831/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,18 @@ +// Example 6_3
+clc;funcprot(0);
+// Given data
+p_1=2.00;// MPa
+p_2=0.100;// MPa
+T_2=150;// °C
+h_1=2776.4;// kJ/kg
+h_2=2776.4;// kJ/kg
+
+// Calculation
+h_f1=908.8;// kJ/kg
+h_fg1=1890.7;// kJ/kg
+h_g1=2799.5;// kJ/kg
+x_1=(h_1-h_f1)/h_fg1;// The quality of steam
+x_1=x_1*100;// The quality of steam in %
+T_1=212.4;// °C
+mu_J=(T_1-T_2)/(p_1-p_2);// °C/MPa
+printf("\nThe quality of the wet steam in the pipe,x=%2.1f percentage \nJoule-Thomson coefficient,mu_J=%2.1f°C/MPa",x_1,mu_J);
diff --git a/3831/CH6/EX6.4/Ex6_4.sce b/3831/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..3686f743a --- /dev/null +++ b/3831/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,17 @@ +// Example 6_4
+clc;funcprot(0);
+// Given data
+Q=0;// kW
+W=0;// kW
+m_s=12.0;// kg/min
+p_1=1.00;// MPa
+T_1=500;// °C
+T_3=15;// °C
+T_4=20;// °C
+
+// Calculation
+h_1=3478.4;// kJ/kg
+h_2=762.8;// kJ/kg
+c_w=4.2;// kJ/kg.K
+m_w=m_s*(h_1-h_2)/[c_w*(T_4-T_3)];// kg/min
+printf("\nThe flow rate of cooling water taken from a local river,m_w=%4.0f kg/min",m_w);
diff --git a/3831/CH6/EX6.5/Ex6_5.sce b/3831/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..0e152bf5d --- /dev/null +++ b/3831/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,16 @@ +// Example 6_5
+clc;funcprot(0);
+// Given data
+p_1=85.0;// psig
+p_2=10.0;// psig
+t=8.00;// hour
+m=20.0;// gal
+
+// Calculation
+mv=20.0/8.00;// gal/h
+mv=mv*0.13368*(1/3600);// ft^3/s
+W_shaft=mv*(p_1-p_2)*144;// ft.lbf/s
+W_shaft=W_shaft*(1/550);// hp
+W_shaft=W_shaft*746;// W
+W_shaft_ins=W_shaft*5*60*(1/2.50);// W
+printf("\nThe hydraulic power produced,(W_shaft)_instantaneous=%3.0f W",W_shaft_ins);
diff --git a/3831/CH6/EX6.6/Ex6_6.sce b/3831/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..f546890e8 --- /dev/null +++ b/3831/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,17 @@ +// Example 6_6
+clc;funcprot(0);
+// Given data
+p_1=2.00;// MPa
+T_1=800;// °C
+p_2=1.00;// MPa
+Wbymdot=2000;// kJ/kg
+
+// Calculation
+h_1=4150.4;// kJ/kg
+h_f2=29.30;// kJ/kg
+h_fg2=2484.9;// kJ/kg
+h_g2=2514.2;// kJ/kg
+h_2=h_1-Wbymdot;// kJ/kg
+x_2=(h_2-h_f2)/h_fg2;// The quality of steam
+x_2=x_2*100;// % vapor at the turbine’s outlet
+printf("\nThe quality of the steam at the outlet of an insulated steam turbine,x_2=%2.1f percentage.",x_2);
diff --git a/3831/CH6/EX6.7/Ex6_7.sce b/3831/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..78e861d2b --- /dev/null +++ b/3831/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,12 @@ +// Example 6_7
+clc;funcprot(0);
+// Given data
+T_in=20.0;// °C
+p_in=50.0;// MPa
+c=4.126;// kN.m/kg.K
+
+// Calculation
+v_f=0.001002;// m^3/kg
+v=0.0009804;// m^3/kg
+T_finalfilled=T_in+((v*(p_in*10^3))/c);// °C
+printf("\nThe final temperature of the water in the tank,T_final filled=%2.1f°C",T_finalfilled);
diff --git a/3831/CH6/EX6.8/Ex6_8.sce b/3831/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..7118053eb --- /dev/null +++ b/3831/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,11 @@ +// Example 6_8
+clc;funcprot(0);
+// Given data
+T_in=20.0;// °C
+p_in=1.40;// MPa
+k=1.40;// The specific heat ratio
+
+// Calculation
+T_finalfilling=k*(T_in+273.15);// K
+T_finalfilling=T_finalfilling-273.15;// °C
+printf("\nThe final temperature of the air in the tank,T_final filling=%3.0f°C",T_finalfilling);
diff --git a/3831/CH6/EX6.9/Ex6_9.sce b/3831/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..721e20bdb --- /dev/null +++ b/3831/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,11 @@ +// Example 6_9
+clc;funcprot(0);
+// Given data
+// From Example 6_8
+T_initial=137+273.15;// K
+k=1.4;// The specific heat ratio
+
+// Calculation
+T_finalemptying=T_initial*((2/k)-1);// K
+T_finalemptying=T_finalemptying-273.15;// °C
+printf("\nThe final temperature inside the tank immediately after the tank is empty,T_final emptying=%2.1f°C.",T_finalemptying);
diff --git a/3831/CH7/EX7.1/Ex7_1.sce b/3831/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..8fc6040e4 --- /dev/null +++ b/3831/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,9 @@ +// Example 7_1
+clc;funcprot(0);
+// Given data
+T_L=70.0;// °F
+T_H=4000.0;// °F
+
+// Solution
+n_T_max=(1-((T_L+459.67)/(T_H+459.67)))*100;// The maximum possible thermal efficiency of this engine in %
+printf('\nThe maximum possible thermal efficiency of this engine,(n_T)_max=%2.1f percentage',n_T_max);
diff --git a/3831/CH7/EX7.10/Ex7_10.sce b/3831/CH7/EX7.10/Ex7_10.sce new file mode 100644 index 000000000..b3c5b0586 --- /dev/null +++ b/3831/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,19 @@ +// Example 7_10
+clc;funcprot(0);
+// Given data
+m_1=1.00;// lbm
+p_1=14.7;// psia
+T_1=70.0;// °F
+p_2=50.0;// psia
+T_2=T_1;// °F
+W_act=-42.0*10^3;// ft.lbf
+R=53.34;// ft.lbf
+
+// Solution
+P_1=p_1*144;// lbf/ft^2
+V_1=(m_1*R*(T_1+459.67))/P_1;// ft^3
+W_rev=P_1*V_1*log(p_1/p_2);// ft.lbf
+W_in=W_rev-W_act;// ft.lbf
+S_pW=W_in/(T_1+459.67);// ft.lbf/R
+S_pW=S_pW/778.16;// Btu/R
+printf('\nThe work mode entropy production,(S_p)_w=%0.4f Btu/R',S_pW);
diff --git a/3831/CH7/EX7.11/Ex7_11.sce b/3831/CH7/EX7.11/Ex7_11.sce new file mode 100644 index 000000000..b73055d44 --- /dev/null +++ b/3831/CH7/EX7.11/Ex7_11.sce @@ -0,0 +1,11 @@ +// Example 7_11
+clc;funcprot(0);
+// Given data
+T=30;// °C
+mu=0.10;// N.s/m^2
+dVbydx=1000;// s^-1
+
+// Calculation
+Sigma_w=(mu*dVbydx^2)/(T+273.15);// N/m^2.s.K
+Sigma_w=Sigma_w/10^3;// kJ/(m^3.s.K)
+printf('\nThe entropy production rate per unit volume,Sigma_w-vis=%0.2f kJ/(m^3.s.K)',Sigma_w);
diff --git a/3831/CH7/EX7.12/Ex7_12.sce b/3831/CH7/EX7.12/Ex7_12.sce new file mode 100644 index 000000000..cb68055f7 --- /dev/null +++ b/3831/CH7/EX7.12/Ex7_12.sce @@ -0,0 +1,15 @@ +// Example 7_12
+clc;funcprot(0);
+// Given data
+T=600;// K
+I=0.10;// amp
+L=10.0*10^-3;// m
+b=5.00*10^-3;// m
+w=1.00*10^-3;// m
+rho_e=0.10;// ohm.m
+
+// Calculation
+A=b*w;// m^2
+R_e=rho_e*(L/A);// W/A^2
+S_pW=(I^2*R_e)/T;// W/K
+printf('\nThe entropy production rate of the chip,(S_p)_W=%0.4f W/K',S_pW);
diff --git a/3831/CH7/EX7.2/Ex7_2.sce b/3831/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..b27fff3da --- /dev/null +++ b/3831/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,17 @@ +// Example 7_2
+clc;funcprot(0);
+// Given data
+T_L=10;// °C
+W_E=5.00;// MW
+W_P=100;// kW
+Q_L=8.00;// MW
+
+// Solution
+// (a)
+Q_H=abs(-Q_L)+(W_E-abs(-W_P/10^3));// MW
+n_T=((W_E-abs(-W_P/10^3))/Q_H);// The actual thermal efficiency of the power plant
+printf('\nThe actual thermal efficiency of the power plant,n_T=%2.1f percentage',n_T*100);
+// (b)
+T_H=(T_L+273.15)/(1-n_T);// K
+T_H=T_H-273.15;// °C
+printf('\nThe equivalent heat source temperature,T_H=%3.0f°C',T_H);
diff --git a/3831/CH7/EX7.3/Ex7_3.sce b/3831/CH7/EX7.3/Ex7_3.sce new file mode 100644 index 000000000..f91547ec8 --- /dev/null +++ b/3831/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,9 @@ +// Example 7_3
+clc;funcprot(0);
+// Given data
+T_H=95;// °F
+T_L=70;// °F
+
+// Solution
+COP=(T_L+459.67)/((T_H+459.67)-(T_L+459.67));// Coefficient of performance
+printf('\nThe Coefficient of performance,COP_Carnot air conditioner=%2.0f',COP);
diff --git a/3831/CH7/EX7.4/Ex7_4.sce b/3831/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..e68d41c10 --- /dev/null +++ b/3831/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,14 @@ +// Example 7_4
+clc;funcprot(0);
+// Given data
+m=1.5;// kg
+x_1=0;// The dryness fraction
+T_1=20.0;// °C
+p_1=0.10;// MPa
+p_2=0.10;// MPa
+c=4.19;// kJ/kg.°C
+
+// Solution
+T_2=T_1;// °C
+deltaS=c*log(T_2/T_1);// kJ/kg.K
+printf('\nThe change in specific entropy of the water,s_2-s_1=%0.0f.Consequently, the entropy of an incompressible material is not altered by changing its pressure.',deltaS);
diff --git a/3831/CH7/EX7.5/Ex7_5.sce b/3831/CH7/EX7.5/Ex7_5.sce new file mode 100644 index 000000000..5dabbe103 --- /dev/null +++ b/3831/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,15 @@ +// Example 7_5
+clc;funcprot(0);
+// Given data
+m=0.035;// kg
+p_1=0.100;// MPa
+T_1=20.0;// °C
+p_2=5.00;// MPa
+k=1.4;// The specific heat ratio for air
+R_air=0.286;// kJ/kg.K
+
+// Solution
+T_2=((T_1+273.15)*(p_2/p_1)^((k-1)/k))-273.15;// °C
+v_1=(m*R_air*(T_1+273.15))/(p_1*10^3);// m^3/kg
+v_2=v_1*((T_2+273.15)/(T_1+273.15))^(1/(1-k));// m^3/kg
+printf('\nThe final temperature,T_2=%3.0f°C \nThe specific volume of the air,v_2=%0.5f m^3/kg',T_2,v_2);
diff --git a/3831/CH7/EX7.6/Ex7_6.sce b/3831/CH7/EX7.6/Ex7_6.sce new file mode 100644 index 000000000..2e1c65538 --- /dev/null +++ b/3831/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,16 @@ +// Example 7_6
+clc;funcprot(0);
+// Given dataS
+m=3.00;// lbm
+T_1=100.0;// °F
+x_1=80.0/100;// Quality of steam
+p_2=200;// psia
+T_2=800.0;// °F
+s_f1=0.1296;// Btu/lbm.R
+s_fg1=1.8528;// Btu/lbm.R
+s_2=1.7662;// Btu/lbm.R
+
+// Solution
+s_1=s_f1+(x_1*s_fg1);// Btu/lbm.R
+deltaS=m*(s_2-s_1);// Btu/R
+printf('\nThe change in total entropy,S_2-S_1=%0.3f Btu/R',deltaS);
diff --git a/3831/CH7/EX7.7/Ex7_7.sce b/3831/CH7/EX7.7/Ex7_7.sce new file mode 100644 index 000000000..6a03cc870 --- /dev/null +++ b/3831/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,13 @@ +// Example 7_7
+clc;funcprot(0);
+// Given data
+mdot=3.00;// kg/min
+x_in=0;// The quality of steam at inlet
+x_out=75;// The quality of steam at outlet
+T_in=100;// °C
+h_fg=2257;// kJ/kg
+
+// Solution
+Qdot=mdot*(x_out/100)*h_fg;// kJ/min
+S_T_Q=Qdot/(T_in+273.15);// kJ/min.K
+printf('\nThe heat transport rate of entropy for this process,(S_T)_Q=%2.1f kJ/min.K',S_T_Q);
diff --git a/3831/CH7/EX7.8/Ex7_8.sce b/3831/CH7/EX7.8/Ex7_8.sce new file mode 100644 index 000000000..71fc465b8 --- /dev/null +++ b/3831/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,10 @@ +// Example 7_8
+clc;funcprot(0);
+// Given data
+V=2.50*10^-3;// m^3
+Sigma_Q=53.7;// W/k.m^3
+tau=30.0;// min
+
+// Solution
+S_pQ=Sigma_Q*V*tau*60;// J/K
+printf('\nThe heat production of entropy inside this motor,(S_p)_Q=%3.0f J/K',S_pQ);
diff --git a/3831/CH8/EX8.1/Ex8_1.sce b/3831/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..cc16a821d --- /dev/null +++ b/3831/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,20 @@ +// Example 8_1
+clc;funcprot(0);
+// Given data
+m=2.00;// kg
+// State 1
+T_1=50.0;// °C
+x_1=0;// The quality of steam
+// State 2
+T_2=50.0;// °C
+p_2=5.00;// kPa
+
+// Calculation
+s_1=0.7036;// kJ/(kg.K)
+s_2=8.4982;// kJ/(kg.K)
+u_1=209.3;// kJ/kg
+u_2=2444.7;// kJ/kg
+T_b=T_1;// °C
+Q_12=m*(T_b+273.15)*(s_2-s_1);// kJ
+W_12=(m*(u_1-u_2))+Q_12;// kJ
+printf("\nThe heat and work transports of energy for this process,Q_12=%4.0f kJ & W_12=%3.0f kJ",Q_12,W_12);
diff --git a/3831/CH8/EX8.10/Ex8_10.sce b/3831/CH8/EX8.10/Ex8_10.sce new file mode 100644 index 000000000..fca8c6bb8 --- /dev/null +++ b/3831/CH8/EX8.10/Ex8_10.sce @@ -0,0 +1,54 @@ +// Example 8_10
+clc;funcprot(0);
+// Given data
+// State 1
+m=0.100;// lbm
+p_1=100;// psia
+T_1=180;// °F
+v_1=0.6210;// ft^3/lbm
+h_1=137.49;// Btu/lbm
+s_1=0.2595;// Btu/(lbm.R)
+// State 2
+p_2=30.0;// psia
+T_2=120;// °F
+v_2=1.9662;// ft^3/lbm
+h_2=126.39;// Btu/lbm
+s_2=0.2635;// Btu/(lbm.R)
+// State 3
+p_3=p_2;// psia
+v_3=v_1/2;// ft^3/lbm
+x_3=0.1952;// The quality of steam
+s_3=0.07241;// Btu/(lbm.R)
+K=5.00;// Btu/R
+
+// Calculation
+// (a)
+// From Table C.7e of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that at p1 = 100 psia and T1 = 180°F,
+v_1=0.6210;// ft^3/lbm
+u_1=125.99;// Btu/lbm
+// At p2= 30 psia and T2 = 120°F,
+v_2=1.966;// ft^3/lbm
+u_2=115.47;// Btu/lbm
+W_12=-m*(u_2-u_1);// Btu
+// (b)
+v_3=v_1/2;// ft^3/lbm
+// At p2= 30 psia
+v_f3=0.01209;// ft^3/lbm
+v_g3=1.5408;// ft^3/lbm
+u_f3=16.24;// Btu/lbm
+u_g3=95.40;// Btu/lbm
+x_3=(v_3-v_f3)/(v_g3-v_f3);// The quality of steam
+x_3p=x_3*100;// %
+u_3=u_f3+(x_3*(u_g3-u_f3));// Btu/lbm
+Q_23=(m*(u_3-u_2))+(m*(p_3*144)*((v_3-v_2)*(1/778.17)));// Btu
+// (c)
+// From Table C.7b
+T_3=15.38;// °F
+dQ=0;// Btu
+S_p12=m*(s_1-s_2)-0;// Btu/R
+s_f3=0.0364;// Btu/(lbm.R)
+s_fg3=0.2209;// Btu/(lbm.R)
+s_3=s_f3+(x_3*(s_fg3-s_f3));// Btu/(lbm.R)
+S_p23=(m*[s_3-s_2])-(K*log((T_3+459.67)/(T_2+459.67)));// Btu/R
+S_p13=S_p12+S_p23;// Btu/R
+printf('\n(a)The work transport of energy during the adiabatic expansion,W_12=%1.2f Btu \n(b)The heat transport of energy during the isobaric compression,Q_23=%1.2f Btu \n(c)Since state 3 is saturated (a mixture of liquid and vapor), T3 must be equal to the saturation temperature at 30.0 psia,which, from Table C.7b, is T_3 =%2.2f°F \n(d)The total entropy production for both processes,1(S_p)3=%0.3f Btu/R',W_12,Q_23,T_3,S_p13);
diff --git a/3831/CH8/EX8.11/Ex8_11.sce b/3831/CH8/EX8.11/Ex8_11.sce new file mode 100644 index 000000000..5e9cebdb6 --- /dev/null +++ b/3831/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,29 @@ +// Example 8_11
+clc;funcprot(0);
+// Given data
+// from Example 5.7
+D=0.100;// m
+T_1=200;// K
+p_1=0.140;// MPa
+h=3.50;// W/(m^2.K)
+T_infinitive=15.0;// K
+c_v=3.123;// kJ/kg.K
+R=2.077;// kJ/kg.K
+t=5.00;// seconds
+
+// Calculation
+// (a)
+V=(%pi/6)*D^3;// m^3
+A=%pi*D^2;// m^2
+m=((p_1*10^3)*V)/(R*(T_1+273.15));// kg
+hAbymc_v=(h*A)/(m*c_v*1000);// s^-1
+T_2=((T_1-T_infinitive)*exp((-(h*A)/(m*c_v*1000))*t))+T_infinitive;// °C
+// (b)
+delU=m*c_v*(T_2-T_1);// kJ
+// (c)
+// Let s_2-s_1=ds
+ds=(c_v*log((T_2+273.15)/(T_1+273.15)))+0;// kJ/(kg.K)
+dQbyT_b=-1.35*10^-4;// kJ/K
+S_P=((m*ds)-(dQbyT_b));// kJ/K
+S_P=S_P*10^3;// J/K
+printf("\n(a)The final temperature of the helium,T_2=%2.1f°C \n(b)The change in total internal energy of the helium,U_2-U_1=%0.3f kJ \n(c)The total entropy production in the helium,S_P=%0.4f J/K",T_2,delU,S_P);
diff --git a/3831/CH8/EX8.12/Ex8_12.sce b/3831/CH8/EX8.12/Ex8_12.sce new file mode 100644 index 000000000..e9bebd2c4 --- /dev/null +++ b/3831/CH8/EX8.12/Ex8_12.sce @@ -0,0 +1,13 @@ +// Example 8_12
+clc;funcprot(0);
+// Given data
+h=3.50;// W/(m^2.K)
+A=1.00*10^-4;// m^2
+P=0.0400;// m
+T_infinitive=20.0+273.15;// K
+k_t=204;// W/(m.K)
+T_f=95.0+273.15;// K
+
+// Solution
+S_P_Q=sqrt(h*P*k_t*A)*((log(T_f/T_infinitive))+(T_infinitive/T_f)-1);// W/K
+printf("\nThe entropy production rate for the fin,S_P=%0.5f W/K",S_P_Q);
diff --git a/3831/CH8/EX8.13/Ex8_13.sce b/3831/CH8/EX8.13/Ex8_13.sce new file mode 100644 index 000000000..d19e2aa1c --- /dev/null +++ b/3831/CH8/EX8.13/Ex8_13.sce @@ -0,0 +1,14 @@ +// Example 8_13
+clc;funcprot(0);
+// Given data
+T=20.0+273.15;// K
+mu=0.700;// N.s/m^2
+L=0.100;// m
+R_1=0.0500;// m
+R_2=0.0510;// m
+n=1000;// rev/min
+
+// Solution
+omega=(2*%pi*n)/60;// rad/s
+S_P_W=((2*%pi*L*omega^2*R_1^4*mu)/((R_2^2-R_1^2)^2*T))*((2*R_2^2*(log(R_2/R_1)))+((R_2^4)/(2*R_1^2))-(R_1^2/2));// W/K
+printf("\nThe rate of entropy production due to laminar viscous losses,(S_P)w=%1.2f W/K",S_P_W);
diff --git a/3831/CH8/EX8.14/Ex8_14.sce b/3831/CH8/EX8.14/Ex8_14.sce new file mode 100644 index 000000000..16d02a06f --- /dev/null +++ b/3831/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,10 @@ +// Example 8_14
+clc;funcprot(0);
+// Given data
+T=30.0;// °C
+phi=5.00;// V
+I=10.0;// mA
+
+// Solution
+S_P_W=(phi*I*10^-3)/(T+273.15);// W/K
+printf("\nThe entropy production rate of the circuit board,(S_P)_W=%1.2e W/K",S_P_W);
diff --git a/3831/CH8/EX8.15/Ex8_15.sce b/3831/CH8/EX8.15/Ex8_15.sce new file mode 100644 index 000000000..306df8d1f --- /dev/null +++ b/3831/CH8/EX8.15/Ex8_15.sce @@ -0,0 +1,14 @@ +// Example 8_15
+clc;funcprot(0);
+// Given data
+m_a=3.00;// g
+T_a=10.0;// °C
+m_b=200;// g
+T_b=80.0;// °C
+c=4186;// J/kg.K
+
+// Solution
+// Let a=cream,b=coffee
+r=m_a/(m_a+m_b);// The mass ratio
+S_p12=((m_a+m_b)/1000)*c*log([1+((r*(((T_a+273.15)/(T_b+273.15))-1)))]*((T_b+273.15)/(T_a+273.15))^(r));// J/K
+printf("\nThe entropy produced,1(S_P)2=%0.3f J/K",S_p12);
diff --git a/3831/CH8/EX8.2/Ex8_2.sce b/3831/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..1cdd31feb --- /dev/null +++ b/3831/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,10 @@ +// Example 8_2
+clc;funcprot(0);
+// Given data
+Q_solar=100*10^3;// Btu/h
+T_river=40+459.67;// R
+T_collector=200+459.67;// R
+
+// Calculation
+W_e_rev=(Q_solar*(1-(T_river/T_collector)))/3412;// kW
+printf("\nThe maximum steady state electrical power (in kW) that can be produced by this power plant,(W_electrical)_rev=%1.2f kW",W_e_rev);
diff --git a/3831/CH8/EX8.5/Ex8_5.sce b/3831/CH8/EX8.5/Ex8_5.sce new file mode 100644 index 000000000..c0f831eee --- /dev/null +++ b/3831/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,48 @@ +// Example 8_5
+clc;funcprot(0);
+// Given data
+V=1.00;// m^3
+m=2.00;// kg
+T_1=20.0;// °C
+T_2=95.0;// °C
+T_b=100.0;// °C
+
+// Calculation
+// (a)
+v_1=V/m;// m^3/kg
+v_2=v_1;// m^3/kg
+// From Table C.1b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that
+// At 20.0°C
+v_f1=0.001002;// m^3/kg
+v_g1=57.79;// m^3/kg
+v_fg1=v_g1-v_f1;// m^3/kg
+u_f1=83.9;// kJ/kg
+u_g1=2402.9;// kJ/kg
+u_fg1=u_g1-u_f1;// kJ/kg
+// At 95.0°C
+v_f2=0.00104;// m^3/kg
+v_g2=1.982;// m^3/kg
+v_fg2=v_g2-v_f2;// m^3/kg
+u_f2=397.9;// kJ/kg
+u_g2=2500.6;// kJ/kg
+u_fg2=u_g2-u_f2;// kJ/kg
+x_1=(v_1-v_f1)/v_fg1;// The quality in the container when the contents are at 20.0°C
+x_1p=x_1*100;// %
+// (b)
+x_2=(v_2-v_f2)/v_fg2;// The quality in the container when the contents are at 95.0°C.
+x_2p=x_2*100;// %
+// (c)
+u_1=u_f1+(x_1*u_fg1);// kJ/kg
+u_2=u_f2+(x_2*u_fg2);// kJ/kg
+Q_12=m*(u_2-u_1);// kJ
+// (d)
+s_f1=0.2965;// kJ/kg.K
+s_fg1=8.3715;// kJ/kg.K
+s_f2=1.2503;// kJ/kg.K
+s_fg2=6.1664;// kJ/kg.K
+s_1=s_f1+((x_1)*s_fg1);// kJ/kg.K
+s_2=s_f2+((x_2)*s_fg2);// kJ/kg.K
+S_p_12=((m*(s_2-s_1))-(Q_12/(T_b+273.15)))*1000;// J/K
+T_b_minimum=Q_12/(m*(s_2-s_1));// K
+T_b_minimum=T_b_minimum-273.15;// °C
+printf('\n(a)The quality in the container when the contents are at 20.0°C,x_1=%0.3f percentage \n(b)The quality in the container when the contents are at 95.0°C,x_2=%2.1f percentage \n(c)The heat transport of energy required to raise the temperature of the contents from 20.0 to 95.0°C,Q_12=%4.0f kJ/kg \n(d)The entropy production,S_P=%3.0f J/K \n The minimum boundary temperature,(T_b)minimum=%2.1f°C',x_1p,x_2p,Q_12,S_p_12,T_b_minimum);
diff --git a/3831/CH8/EX8.6/Ex8_6.sce b/3831/CH8/EX8.6/Ex8_6.sce new file mode 100644 index 000000000..c817f7de0 --- /dev/null +++ b/3831/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,19 @@ +// Example 8_6
+clc;funcprot(0);
+// Given data
+W=100;// W
+T_b=110.0+273.15;// K
+
+// Calculation
+// (a)
+// Since we are assuming a constant bulb temperature in part a, U=constant and
+U=0;// W
+Q=U-W;// kW
+printf("\n(a)The heat transfer rate of an illuminated 100. W incandescent lightbulb in a room,Q=%3.0f W",Q);
+// (b)
+Q_dot=0;
+Udot=W;// W
+printf("\n(b)The rate of change of its internal energy,Udot=%3.0f W",Udot);
+Sdot=0;// W/K
+S_p=Sdot-(Q/(T_b));// W/K
+printf("\n(c)The value of the entropy production rate,S_p=%0.3f W/K",S_p);
diff --git a/3831/CH8/EX8.7/Ex8_7.sce b/3831/CH8/EX8.7/Ex8_7.sce new file mode 100644 index 000000000..61ebfe489 --- /dev/null +++ b/3831/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,20 @@ +// Example 8_7
+clc;funcprot(0);
+// Given data
+Q_b=950*10^5;// kJ/h
+T_b=500;// K
+W_p=-23.0;// kW
+Q_c=-600*10^5;// kJ/h
+T_c=10.0;// °C
+
+// Calculation
+Q_net=(Q_b+Q_c);// kJ/h
+W_T_net=Q_net/3600;// kJ/h
+W_T_net=W_T_net/1000;// MW
+W_T_total=(W_T_net*10^3)+W_p;// kW
+S_p=-((Q_b/(T_b+273.15))+(Q_c/(T_c+273.15)));// kJ/(h.K)
+Q_in=Q_b;// kJ/h
+Q_out=Q_c;// kJ/h
+n_T_act=(1-((abs(Q_out))/Q_in))*100;// The actual thermal efficiency of this power plant in %
+n_T_rev=(1-((T_c+273.15)/(T_b+273.15)))*100;// The theoretical reversible (Carnot) efficiency in %
+printf("\nThe net power of the turbine,(W_T)_total=%4.0f kW(round off error) \nThe rate of entropy production,S_p=%2.1e kJ/(h.K)",W_T_total,S_p);
diff --git a/3831/CH8/EX8.8/Ex8_8.sce b/3831/CH8/EX8.8/Ex8_8.sce new file mode 100644 index 000000000..a03bbd874 --- /dev/null +++ b/3831/CH8/EX8.8/Ex8_8.sce @@ -0,0 +1,19 @@ +// Example 8_8
+clc;funcprot(0);
+// Given data
+W=0.250;// hp
+V=1.00;// quart of water
+p_1=14.7;// psia
+T_1=60.0;// °F
+p_2=p_1;// psia
+t=10;// min
+c=1.00;// Btu/(lbm.R)
+
+// Calculation
+V=V*(1/4)*0.13368;// ft^3
+v=0.01603;// ft^3/lbm
+m=V/v;// lbm
+Q_12bymc=0;// Btu/lbm
+T_2=T_1+Q_12bymc-((-W*t*(1/60)*(2545))/(m*c));// °F
+S_p12=m*c*log((T_2+459.67)/(T_1+459.67));// Btu/R
+printf("\nThe temperature of the water when the machine is turned off,T_2=%3.0f°F \nThe amount of entropy produced,1(S_p)2=%0.3f Btu/R",T_2,S_p12);
diff --git a/3831/CH8/EX8.9/Ex8_9.sce b/3831/CH8/EX8.9/Ex8_9.sce new file mode 100644 index 000000000..82ca4a7ae --- /dev/null +++ b/3831/CH8/EX8.9/Ex8_9.sce @@ -0,0 +1,20 @@ +// Example 8_9
+clc;funcprot(0);
+// Given data
+V_2=0.0400;// m^3
+T_1=20.0;// °C
+p_1=0.0100;// MPa
+Q_12=0.100;// kJ
+V_1=0.0100;// m^3
+R=0.208;// kJ/kg.K
+T_w=400;// K
+c_p=0.523;// kJ/kg.K
+c_v=0.315;// kJ/kg.K
+
+// Calculation
+m=((p_1*10^3)*V_1)/(R*(T_1+273.15));// kg
+T_2=T_1+(Q_12/(m*c_v));// K
+p_2=(m*R*(T_2+273.15))/V_2;// kPa
+S_p12=(m*[(c_p*log((T_2+273.15)/(T_1+273.15)))-(R*log(p_2/(p_1*10^3)))])-(Q_12/T_w);// kJ/K
+S_p12=S_p12*10^3;// J/K
+printf('\nThe pressure and temperature inside the box after the balloon bursts p_2=%1.2f kPa and T_2=%3.0f°C \nThe entropy produced,1(S_P)2=%0.3f J/K',p_2,T_2,S_p12);
diff --git a/3831/CH9/EX9.1/Ex9_1.sce b/3831/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..d1e1a8c6d --- /dev/null +++ b/3831/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,18 @@ +// Example 9_1
+clc;funcprot(0);
+// Given data
+T_1=15+273.15;// K
+T_2=50+273.15;// K
+Q=0.100;// The electrical energy in W
+c=4.186;// kJ/kg.K
+T_b=20+273.15;// K
+
+// Calculation
+m=Q/(c*(T_2-T_1));// The expected water flow rate in kg/s
+// Assume ds=s_out-s_in
+ds=c*log(T_2/T_1);// kJ/kg.K
+S_p=(m*ds)-(Q/T_b);// kJ/s.K
+printf("\nThe entropy production rate,S_p=%1.2e kJ/s.K ",S_p);
+if(S_p<0)
+ printf("\nSince the entropy production rate is negative, this water heater cannot possibly meet the claims of the inventor, so we should reject the patent application.")
+ end
diff --git a/3831/CH9/EX9.10/Ex9_10.sce b/3831/CH9/EX9.10/Ex9_10.sce new file mode 100644 index 000000000..c81a6d189 --- /dev/null +++ b/3831/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,18 @@ +// Example 9_10
+clc;funcprot(0);
+// Given data
+m=500;// lbm/s
+T=50.0;// °F
+y_1=1.00;// The inlet height in ft
+y_2=1.80;// The exit height in ft
+v_1=8.00;// The inlet velocity ft/s
+v_2=5.14;// The exit velocity in ft/s
+g=32.174;// ft/s^2
+g_c=32.174;// lbm.ft/(lbf.s^2)
+c=1.00; // Btu/(lbm.R)
+
+// Solution
+h_L12=(y_2-y_1)^3/(4*y_1*y_2);// ft
+E_dr=(m*(g/g_c)*h_L12)/778.17;// The energy dissipation rate in Btu/s
+S_p=m*c*log(1+(g*[(h_L12)]/(c*g_c*(T+459.67))));// The entropy production rate in Btu/(s.R)
+printf('\nThe energy dissipation rate=%0.4f Btu/s \nThe entropy production rate,S_p=%0.4f Btu/(s.R)',E_dr,S_p);
diff --git a/3831/CH9/EX9.11/Ex9_11.sce b/3831/CH9/EX9.11/Ex9_11.sce new file mode 100644 index 000000000..86c027d95 --- /dev/null +++ b/3831/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,11 @@ +// Example 9_11
+clc;funcprot(0);
+// Given data
+mu=10.1*10^-3;// The viscosity of the water in kg/(m.s)
+L=10.0;// The length of the pipe in m
+V_m=0.500;// The maximum velocity of the fluid in m/s
+T=20.0;// °C
+
+// Solution
+S_pW=(2*%pi*mu*L*V_m^2)/(T+273.15);// The entropy production rate in W/K
+printf('\nThe entropy production rate,(S_p)_W=%1.3e W/K',S_pW);
diff --git a/3831/CH9/EX9.2/Ex9_2.sce b/3831/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..4b97626b2 --- /dev/null +++ b/3831/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,23 @@ +// Example 9_2
+clc;funcprot(0);
+// Given data
+m=0.2000;// lbm/s
+// Station 1
+p_1=14.7;// psia
+T_1=50.00;// °F
+// Station 2
+p_2=95.00;// psia
+D_1=1.000;// The inlet diameter of the nozzle in m
+D_2=0.2500;// The outlet diameter of the nozzle in m
+c=1.0;// Btu/lbm.R
+g_c=32.174;// lbm.ft/(lbf.s^2)
+
+// Calculation
+v_f=0.01602;// ft^3/lbm
+v=v_f;// ft^3/lbm
+V_1=(4*m*v*144)/(%pi*D_1^2);// ft/s
+V_2=V_1*(D_1/D_2)^2;// ft/s
+T_2=(T_1+459.67)+(v*(((p_2-p_1)*144)/(c*778.17)))-((V_2^2-V_1^2)/(2*c*g_c*778.17));// R
+S_p=m*c*log(T_2/(T_1+459.7));// Btu/(s.R)
+S_p=S_p*778.17;// ft.lbf/(s.R)
+printf("\nThe rate of entropy production,S_p=%0.4f ft.lbf/(s.R)",S_p);
diff --git a/3831/CH9/EX9.3/Ex9_3.sce b/3831/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..36301b55d --- /dev/null +++ b/3831/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,18 @@ +// Example 9_3
+clc;funcprot(0);
+// Given data
+m=0.800;// kg/s
+V_1=93.0;// m/s
+// Station 1
+p_1=97.0;// kPa
+T_1=80.0;// °C
+// Station 2
+p_2=101.3;// kPa
+g_c=1;// The gravitational constant
+c_p=523;// J/(kg.K)
+R=208;// J/(kg.K)
+
+// Calculation
+T_2=(T_1+273.15)+((V_1^2)/(2*g_c*c_p));// K
+S_p=m*((c_p*log(T_2/(T_1+273.15)))-(R*log(p_2/p_1)));// The rate of entropy production within the diffuser in W/K
+printf("\nThe rate of entropy production within the diffuser,S_p=%1.2f W/K",S_p);
diff --git a/3831/CH9/EX9.4/Ex9_4.sce b/3831/CH9/EX9.4/Ex9_4.sce new file mode 100644 index 000000000..f26f376cd --- /dev/null +++ b/3831/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,41 @@ +// Example 9_4
+clc;funcprot(0);
+// Given data
+m=0.100;// lbm/s
+// Station 1
+x_1=0.00;// The quality of steam at inlet
+T_1=100;// °F
+// Station 2
+x_2=0.530;// The quality of steam at exit
+T_2=20;// °F
+T_b=60.0;// °F
+
+// Calculation
+// (a)
+// From Table C.7a for R-134a, we find
+h_f1=44.23;// Btu/lbm
+h_1=h_f1;// Btu/lbm
+s_f1=0.0898;// Btu/(lbm.R)
+s_1=s_f1;// Btu/(lbm.R)
+h_f2=17.74;// Btu/lbm
+h_fg2=86.87;// Btu/lbm
+s_f2=0.0393;// Btu/(lbm.R)
+s_fg2=0.2206-s_f2;// Btu/(lbm.R)
+h_2=h_f2+(x_2*h_fg2);// Btu/lbm
+s_2=s_f2+(x_2*s_fg2);// Btu/(lbm.R)
+Q=m*(h_2-h_1);// Btu/s
+S_pa=((m*(s_2-s_1))-(Q/(T_b+459.67)));// The entropy production rate inside the valve in Btu/(s.R)
+S_p=S_pa*778.17;// ft.lbf/(s.R)
+printf("\n(a)The entropy production rate inside the valve if the valve is not insulated and has an isothermal external surface temperature of 60.0°F,S_p=%0.4f ft.lbf/(s.R)",S_p);
+// (b)
+h_2=h_1;// Btu/lbm
+x_2=(h_2-h_f2)/h_fg2;// The quality of steam
+x_2p=x_2*100;// % (in x_2p,p refers the quality of steam in percentage)
+s_2=s_f2+(x_2*s_fg2);// Btu/(lbm.R)
+Q=0;// W
+S_pb=m*(s_2-s_1)-(Q/T_b);// Btu/(s.R)
+S_p=S_pb*778.17;// lbf/(s.R)
+printf("\n(b)The entropy production rate inside the valve if it is insulated and assuming it has the same inlet conditions and exit temperature,S_p=%0.3f ft.lbf/(s.R)",S_p);
+//(c)
+S_p_pd=((S_pa-S_pb)/S_pa)*100;// The percentage decrease in S_p brought about by adding the insulation in %
+printf("\n(c)The percentage decrease in S_p brought about by adding the insulation is %2.1f percentage.",S_p_pd);
diff --git a/3831/CH9/EX9.5/Ex9_5.sce b/3831/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..feb1c574c --- /dev/null +++ b/3831/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,23 @@ +// Example 9_5
+clc;funcprot(0);
+// Given data
+m_a=0.200;// kg/s
+T_ain=90.0;// °C
+T_aout=75.0;// °C
+T_win=20.0;// °C
+T_wout=40.0;// °C
+U=140;// W/(m^2.K)
+c_pa=1.004;// The specific heat of air in kJ/kg.K
+c_pw=4.186;// The specific heat of water in kJ/kg.K
+
+// Calculation
+// (a) Parallel flow
+delT_LMTDa=((T_aout-T_wout)-(T_ain-T_win))/(log((T_aout-T_wout)/(T_ain-T_win)));// K
+//(b) Counter flow
+delT_LMTDb=((T_aout-T_win)-(T_ain-T_wout))/(log((T_aout-T_win)/(T_ain-T_wout)));// K
+Q=abs(m_a*c_pa*10^3*(T_aout-T_ain));// J/s
+A_pf=Q/(U*delT_LMTDa);// m^2
+A_cf=Q/(U*delT_LMTDb);// m^2
+m_w=m_a*(c_pa/c_pw)*((T_ain-T_aout)/(T_wout-T_win));// kg/s
+S_p=(m_a*c_pa*10^3*log((T_aout+273.15)/(T_ain+273.15)))+(m_w*c_pw*10^3*log((T_wout+273.15)/(T_win+273.15)));// W/K
+printf("\nThe corresponding heat exchanger area for parallel flow,A_parallel flow=%0.3f m^2 \nThe corresponding heat exchanger area for counter flow,A_counter flow=%0.3f m^2 \nThe entropy production rate,S_p=%1.2f W/K",A_pf,A_cf,S_p);
diff --git a/3831/CH9/EX9.6/Ex9_6.sce b/3831/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..166f942d5 --- /dev/null +++ b/3831/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,26 @@ +// Example 9_6
+clc;funcprot(0);
+// Given data
+m_H=0.300;// lbm/s
+T_H=140.0;// °F
+m_C=0.300;// lbm/s
+T_C=50.0;// °F
+c=1.00;// Btu/(lbm.R)
+
+// Calculation
+// (a)
+m_M=m_H+m_C;// lbm/s
+gamma=m_H/m_M;// The mass flow rate ratio
+T_1=T_H;// °F
+T_2=T_C;// °F
+T_1byT_2=(T_H+459.67)/(T_C+459.67);// The temperature ratio
+T_3=T_C+(gamma*(T_H-T_C));// °F
+m_3=m_M;// lbm/s
+S_p_mixing=m_3*c*log((1+(gamma*(T_1byT_2-1)))*(T_1byT_2)^(-gamma));// Btu/(s.R)
+S_p_mixing=S_p_mixing*778.17;// ft.lbf/(s.R)
+printf("\n(a)The shower mixture temperature,T_3=%2.0f°F \n The entropy production rate,(S_p)_mixing=%1.2f lbf/(s.R)",T_3,S_p_mixing);
+// (b)
+gamma_c=((1-T_1byT_2)+log(T_1byT_2))/((1-T_1byT_2)*log(T_1byT_2));// The critical mass fraction
+S_p_mixing=m_3*c*log((1+(gamma_c*(T_1byT_2-1)))*(T_1byT_2)^(-gamma_c));// // Btu/(s.R)
+S_p_mixing=S_p_mixing*778.17;// ft.lbf/(s.R)
+printf("\n(b)The critical mass fraction,gamma_c=%0.3f \n The value of the maximum entropy production rate,(S_p)_mixing=%1.2f ft.lbf/(s.R)",gamma_c,S_p_mixing);
diff --git a/3831/CH9/EX9.7/Ex9_7.sce b/3831/CH9/EX9.7/Ex9_7.sce new file mode 100644 index 000000000..91cfa862f --- /dev/null +++ b/3831/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,17 @@ +// Example 9_7
+clc;funcprot(0);
+// Given data
+mdot=0.500;// kg/s
+p_1=8.00;// MPa
+T_1=300;// °C
+T_2=100;// °C
+x_2=1.00;// The quality of steam at station 2
+T_b=20.0;// °C
+h_1=2785.0;// kJ/kg
+h_2=2676.0;// kJ/kg
+s_1=5.7914;// kJ/kg.K
+s_2=7.3557;// kJ/kg.K
+
+// Calculation
+W_max=mdot*[(h_1-((T_b+273.15)*s_1))-(h_2-((T_b+273.15)*s_2))];// kW
+printf("\nThe maximum (reversible) power,W_max=%3.0f kW",W_max);
diff --git a/3831/CH9/EX9.8/Ex9_8.sce b/3831/CH9/EX9.8/Ex9_8.sce new file mode 100644 index 000000000..61c745009 --- /dev/null +++ b/3831/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,21 @@ +// Example 9_8
+clc;funcprot(0);
+// Given data
+V_2=3.00;// ft^3
+T_in=70+459.67;// °F
+p_2=2000;// psia
+
+// Calculation
+// From Table C.13a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find for oxygen
+c_p=0.219;// Btu/(lbm.R)
+R=48.29;// ft.lbf/(lbm.R)
+k=1.39;// The specific heat ratio
+T_2_af=k*T_in;// R
+T_2_if=T_in;// R
+m_2_af=(p_2*144*V_2)/(R*T_2_af);// lbm
+m_2_if=(p_2*144*V_2)/(R*T_2_if);// lbm
+// (a)
+S_p_12_af=m_2_af*c_p*2.303*log10(k);// Btu/R
+// (b)
+S_p_12_if=m_2_if*R/778.16;// Btu/R
+printf("\n(a)The amount of entropy produced when the container is filled adiabatically by insulating it,[1(S_P)2]adiabatic filling=%1.2f Btu/R \n(b)The amount of entropy produced when the container is filled isothermally,[1(S_P)2]isothermal filling=%1.2f Btu/R",S_p_12_af,S_p_12_if)
diff --git a/3831/CH9/EX9.9/Ex9_9.sce b/3831/CH9/EX9.9/Ex9_9.sce new file mode 100644 index 000000000..37c9f8bdb --- /dev/null +++ b/3831/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,36 @@ +// Example 9_9
+clc;funcprot(0);
+// Given data
+gamma=0.500;// The specific heat ratio for air
+T_in=70.0;// °F
+p_in_psig=[0.000,20.00,40.00,60.00,80.00,100.00,120.00,140.00];// psig
+p_in=[14.7,34.7,54.7,74.7,94.7,114.7,134.7,154.7];// psia
+T_hot=[70.0,119.0,141.0,150.0,156.0,161.0,164.0,166.0];// °F
+T_cold=[70.0,19.5,-3.00,-14.0,-22.0,-29.0,-34.0,-39.0];// °F
+T_r=[1.000,1.209,1.315,1.368,1.406,1.441,1.465,1.487];// Note:T_r=(T_hot+460)/(T_cold+460)
+p_e=14.7;// The exit pressure in psia
+R=0.0685;// Btu/(lbm.R)
+c_p=0.240;// Btu/(lbm.R)
+
+// Calculation
+Sdot_pbymdot_3_1=((c_p*log(((T_r(1)^gamma)/(1+(gamma*(T_r(1)-1))))))+(R*log(p_in(1)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_2=((c_p*log(((T_r(2)^gamma)/(1+(gamma*(T_r(2)-1))))))+(R*log(p_in(2)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_3=((c_p*log(((T_r(3)^gamma)/(1+(gamma*(T_r(3)-1))))))+(R*log(p_in(3)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_4=((c_p*log(((T_r(4)^gamma)/(1+(gamma*(T_r(4)-1))))))+(R*log(p_in(4)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_5=((c_p*log(((T_r(5)^gamma)/(1+(gamma*(T_r(5)-1))))))+(R*log(p_in(5)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_6=((c_p*log(((T_r(6)^gamma)/(1+(gamma*(T_r(6)-1))))))+(R*log(p_in(6)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_7=((c_p*log(((T_r(7)^gamma)/(1+(gamma*(T_r(7)-1))))))+(R*log(p_in(7)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3_8=((c_p*log(((T_r(8)^gamma)/(1+(gamma*(T_r(8)-1))))))+(R*log(p_in(8)/p_e)));// Btu/(lbm.R)
+Sdot_pbymdot_3=[Sdot_pbymdot_3_1,Sdot_pbymdot_3_2,Sdot_pbymdot_3_3,Sdot_pbymdot_3_4,Sdot_pbymdot_3_5,Sdot_pbymdot_3_6,Sdot_pbymdot_3_7,Sdot_pbymdot_3_8];// Btu/(lbm.R)
+plot(p_in_psig,Sdot_pbymdot_3);
+xlabel('Inlet pressure(psig)');
+ylabel('Sdot_p/mdot_3(Btu/lbm.R)');
+xtitle('Sdot_p/mdot_3 vs. inlet pressure for a vortex tube');
+disp('Remaining Results for Example 9.9');
+disp('The entropy production rate per unit mass flow rate for each pressure shown');
+disp('Inlet pressure psig');
+disp(p_in_psig);
+disp('T_1/T_2');
+disp(T_r);
+disp('Sdot_P/mdot_3 Btu/(lbm⋅R)');
+disp(Sdot_pbymdot_3);
diff --git a/3831/CH9/EX9.9/Figure9_20.pdf b/3831/CH9/EX9.9/Figure9_20.pdf Binary files differnew file mode 100644 index 000000000..ad20ac117 --- /dev/null +++ b/3831/CH9/EX9.9/Figure9_20.pdf |