diff options
Diffstat (limited to '3831/CH10')
| -rw-r--r-- | 3831/CH10/EX10.1/Ex10_1.sce | 25 | ||||
| -rw-r--r-- | 3831/CH10/EX10.12/Ex10_12.sce | 21 | ||||
| -rw-r--r-- | 3831/CH10/EX10.13/Ex10_13.sce | 18 | ||||
| -rw-r--r-- | 3831/CH10/EX10.14/Ex10_14.sce | 12 | ||||
| -rw-r--r-- | 3831/CH10/EX10.15/Ex10_15.sce | 31 | ||||
| -rw-r--r-- | 3831/CH10/EX10.16/Ex10_16.sce | 23 | ||||
| -rw-r--r-- | 3831/CH10/EX10.2/Ex10_2.sce | 20 | ||||
| -rw-r--r-- | 3831/CH10/EX10.3/Ex10_3.sce | 41 | ||||
| -rw-r--r-- | 3831/CH10/EX10.4/Ex10_4.sce | 38 | ||||
| -rw-r--r-- | 3831/CH10/EX10.5/Ex10_5.sce | 14 | ||||
| -rw-r--r-- | 3831/CH10/EX10.6/Ex10_6.sce | 18 | ||||
| -rw-r--r-- | 3831/CH10/EX10.7/Ex10_7.sce | 31 | ||||
| -rw-r--r-- | 3831/CH10/EX10.8/Ex10_8.sce | 43 | ||||
| -rw-r--r-- | 3831/CH10/EX10.9/Ex10_9.sce | 32 |
14 files changed, 367 insertions, 0 deletions
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
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