diff options
Diffstat (limited to '3831/CH5')
| -rw-r--r-- | 3831/CH5/EX5.1/Ex5_1.sce | 35 | ||||
| -rw-r--r-- | 3831/CH5/EX5.2/Ex5_2.sce | 15 | ||||
| -rw-r--r-- | 3831/CH5/EX5.3/Ex5_3.sce | 14 | ||||
| -rw-r--r-- | 3831/CH5/EX5.4/Ex5_4.sce | 18 | ||||
| -rw-r--r-- | 3831/CH5/EX5.5/Ex5_5.sce | 16 | ||||
| -rw-r--r-- | 3831/CH5/EX5.6/Ex5_6.sce | 37 | ||||
| -rw-r--r-- | 3831/CH5/EX5.7/Ex5_7.sce | 20 | ||||
| -rw-r--r-- | 3831/CH5/EX5.8/Ex5_8.sce | 18 |
8 files changed, 173 insertions, 0 deletions
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);
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