diff options
Diffstat (limited to '3720/CH5')
| -rw-r--r-- | 3720/CH5/EX5.1/Ex5_1.sce | 20 | ||||
| -rw-r--r-- | 3720/CH5/EX5.12/Ex5_12.sce | 23 | ||||
| -rw-r--r-- | 3720/CH5/EX5.13/Ex5_13.sce | 16 | ||||
| -rw-r--r-- | 3720/CH5/EX5.14/Ex5_14.sce | 24 | ||||
| -rw-r--r-- | 3720/CH5/EX5.15/Ex5_15.sce | 16 | ||||
| -rw-r--r-- | 3720/CH5/EX5.2/Ex5_2.sce | 13 | ||||
| -rw-r--r-- | 3720/CH5/EX5.3/Ex5_3.sce | 24 | ||||
| -rw-r--r-- | 3720/CH5/EX5.5/Ex5_5.sce | 10 | ||||
| -rw-r--r-- | 3720/CH5/EX5.6/Ex5_6.sce | 9 | ||||
| -rw-r--r-- | 3720/CH5/EX5.7/Ex5_7.sce | 21 | ||||
| -rw-r--r-- | 3720/CH5/EX5.8/Ex5_8.sce | 11 | ||||
| -rw-r--r-- | 3720/CH5/EX5.9/Ex5_9.sce | 23 |
12 files changed, 210 insertions, 0 deletions
diff --git a/3720/CH5/EX5.1/Ex5_1.sce b/3720/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..ea27ff59f --- /dev/null +++ b/3720/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,20 @@ +//Example 5_1
+clc;clear;funcprot(0);
+// Given values
+V=10; // Volume of water in gallon
+dt=50;// Time in seconds
+rho=1;//The density of water in kg/L
+r_e=0.4;// Radius of nozzle at exit in cm
+
+// Calculation (a)
+v=V/dt; // Volume flow rate in L/s
+v=v*3.7854;// Convert gal into L
+printf('(a)The volume flow rate of water,v=%0.3f L/s\n',v);
+m=rho*v;// Mass flow rate of water in kg/s
+printf('The mass flow rate of water,m=%0.3f kg/s\n',m);
+
+// Calculation (b)
+A_e=%pi*r_e^2;// The cross sectional area of nozzle at exit in cm^2
+V_e=v/A_e;
+V_e=(V_e*10000/1000);// Convert to m/s
+printf('(b)The average velocity of water at the nozzle exit,V_e=%0.1f m/s\n',V_e);
diff --git a/3720/CH5/EX5.12/Ex5_12.sce b/3720/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..4b84d95ee --- /dev/null +++ b/3720/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,23 @@ +//Example 5_12
+clc;clear;funcprot(0);
+// Given values
+rho=1;//The density of water to be 1 kg/L = 1000 kg/m^3
+v=50;// The water flow rate through the pump in L/s
+n_m=.90;//The efficiency of electric motor
+W_e=15;//Power in kW;
+P_1=100;// The pressure at the inlet of the pump in kPa
+P_2=300;// The pressure at the outlet of the pump in kPa
+rho_1=1000;//The density of water in kg/m^3
+c=4.18;// The specific heat in kJ/kg °C.
+
+// Calculation
+// (a)
+m=rho*v;//The mass flow rate of water through the pump in kg/s
+W_p=n_m*W_e;//The mechanical (shaft) power delivers to the pump kW
+dE_m=m*((P_2-P_1)/rho_1);//The increase in the mechanical energy of the fluid in kW
+n_p=dE_m/W_p;// The mechanical efficiency of the pump
+printf('(a)The mechanical efficiency of the pump,n_pump=%0.3f (or)%0.1f percentage \n',n_p,n_p*100);
+//(b)
+E_mloss=W_p-dE_m;// “lost” mechanical energy in kW
+dT=E_mloss/(m*c);// °C
+printf('(b)The temperature rise of water due to the mechanical inefficiency,dT=%0.3f degree Celsius\n',dT);
diff --git a/3720/CH5/EX5.13/Ex5_13.sce b/3720/CH5/EX5.13/Ex5_13.sce new file mode 100644 index 000000000..6ace2d325 --- /dev/null +++ b/3720/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,16 @@ +//Example 5_13
+clc;clear;
+// Given values
+rho=1000;// the density of water in kg/m^3
+v=100;//Flow rate of water in kg/m^3
+z_1=120;// m
+h_l=35;// m
+n_t=0.8;
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+m=rho*v;//The mass flow rate of water in kg/s
+h_t=z_1-h_l;// m
+W_t=(m*g*h_t)/1000;// kW
+W_e=(n_t*(W_t/1000));// MW
+printf('The electric power generated by the actual unit=%0.1f MW\n',W_e);
diff --git a/3720/CH5/EX5.14/Ex5_14.sce b/3720/CH5/EX5.14/Ex5_14.sce new file mode 100644 index 000000000..4894d34b0 --- /dev/null +++ b/3720/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,24 @@ +//Example 5_14
+clc;clear;funcprot(0);
+// Given values
+rho=1.20;//The density of air in kg/m^3
+alpha_2=1.10;
+dt=1;// s
+D=0.05;// Diameter in m
+n_f=0.3;// Efficiency of fan motor
+
+// Calculation
+//(a)
+V=0.5*(12*40*40);//The air volume in the computer case in cm^3
+V=V/10^6;// cm^3 to m^3
+v=V/dt;//The volume flow rate of air through the case in m^3/s
+m=rho*V;//The mass flow rate of air through the case in kg/s
+A=(%pi*D^2)/4;// m^2
+V_1=v/A;//m/s
+W_fan=m*alpha_2*(V_1^2/2);
+W_e=W_fan/n_f;//Electric power input to the fan in W
+printf('(a)Electric power input to the fan,W_elect=%0.3f W\n',W_e);
+//(b)
+dP=(rho*W_fan)/m;//dp=P_4-P_3
+printf('(b)The pressure rise across the fan is %0.1f Pa.\n',dP);
+//The answer is bit different due to round off error in the book
diff --git a/3720/CH5/EX5.15/Ex5_15.sce b/3720/CH5/EX5.15/Ex5_15.sce new file mode 100644 index 000000000..4b0356227 --- /dev/null +++ b/3720/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,16 @@ +//Example 5_15
+clc;clear;funcprot(0);
+// Properties
+rho=1000;//The density of water in kg/m^3
+// Given values
+v=0.03;//The flow rate of water in m^3/s
+W_p=20;// kW
+g=9.81;//The acceleration due to gravity in m/s^2
+z_2=45;// m
+
+// Calculation
+m=rho*v;//The mass flow rate of water through the system in kg/s
+E_ml=(W_p-(m*g*z_2)/1000);
+printf('The lost mechanical power,E_mechloss=%0.2f kW\n',E_ml);
+h_l=E_ml*1000/(m*g);
+printf('The irreversible head loss,h_L=%0.1f m\n',h_l);
diff --git a/3720/CH5/EX5.2/Ex5_2.sce b/3720/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..b9acd38c6 --- /dev/null +++ b/3720/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,13 @@ +//Example 5_2
+clc;clear;funcprot(0);
+// Given values
+h_0=4;
+h_2=2;// Corresponding heights in ft
+D_tank=3*12;
+D_jet=0.5;// Corresponding diameters in inch
+g=32.2;// The acceleration due to gravity in ft/s^2
+
+// Calculation
+t=((sqrt(h_0)-sqrt(h_2))/sqrt(g/2))*((D_tank/D_jet)^2);
+t=t/60;// Convert seconds to minutes
+printf('The time of discharge,t=%0.1f min\n',t);
diff --git a/3720/CH5/EX5.3/Ex5_3.sce b/3720/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..6a23fea28 --- /dev/null +++ b/3720/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,24 @@ +//Example 5_3
+clc;clear;funcprot(0);
+// Given values
+m=5000;// Mass flow rate of water in kg/s
+W_eout=1862;//The electric power generated is measured in kW
+rho=1000;// The density of water in kg/m^3
+h=50;// The depth of the water in m
+g=9.81;// m/s^2
+e_min=g*h;// kJ/kg
+e_mout=0;// kJ/kg
+n_gen=0.95;// The generator efficiency
+
+// Calculation
+//(a)
+dE_mech=(m*(e_min-e_mout))/1000;//kW
+n_o=(W_eout/dE_mech);// The over all efficiency
+printf('(a)The over all efficiency,n_o=%0.2f\n',n_o);
+//(b)
+n_t=n_o/n_gen;// )The mechanical efficiency of the turbine
+printf('(b)The mechanical efficiency of the turbine,n_t=%0.2f\n',n_t);
+//(c)
+W_sout=n_t*dE_mech;// kW
+printf('(c)The shaft power output,W_shaft,out=%0.0f kW\n',W_sout);
+//The answer is a bit different due to rounding off error in textbook
diff --git a/3720/CH5/EX5.5/Ex5_5.sce b/3720/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..0d23a597e --- /dev/null +++ b/3720/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,10 @@ +// Example 5_5
+clc;clear;
+// given values
+P_gage=400;// kPa
+rho=1000;// the density of water in kg/m^3
+g=9.81;// the accleration due to gravity in m/s^2
+
+// Calculation
+z_2=P_gage*1000/(rho*g);// m
+printf('The water jet can rise as high,z_2=%0.1f m\n',z_2);
diff --git a/3720/CH5/EX5.6/Ex5_6.sce b/3720/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..17c9147d2 --- /dev/null +++ b/3720/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,9 @@ +// Example 5_6
+clc;clear;funcprot(0);
+// Given values
+z_1=5;// m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+V_2=sqrt(2*g*z_1);// Toricelli equation
+printf('The water leaves the tank with an initial velocity,V_2=%0.1f m/s\n',V_2);
diff --git a/3720/CH5/EX5.7/Ex5_7.sce b/3720/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..caed91990 --- /dev/null +++ b/3720/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,21 @@ +// Example 5_7
+clc;clear;funcprot(0);
+// Given values
+P_atm=101.3;// The atmospheric pressure in kPa;
+rho=750;//The density of gasoline in kg/m^3
+g=9.81;//m/s^2
+z_1=0.75;// m
+z_3=2.75;// m
+D=(5/1000);// m
+
+// Calculation
+//(a)
+V_2=sqrt(2*g*z_1);
+A=(%pi*D^2)/4;//The cross-sectional area of the tube in m^2
+v=V_2*A*1000;//The flow rate of gasoline in L/s
+V=4;// Volume of gasoline in litre
+gradt=V/v;
+printf('(a)The time needed to siphon 4 L of gasoline from the tank,gradt=%0.1f s\n',gradt);
+//(b)
+P_3=P_atm-((rho*g*z_3)/1000);// kPa
+printf('(b)The pressure at point 3,P_3=%0.1f kPa\n',P_3);
diff --git a/3720/CH5/EX5.8/Ex5_8.sce b/3720/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..0aa769523 --- /dev/null +++ b/3720/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,11 @@ +//Example 5_8
+clc;clear;funcprot(0);
+// Given values
+h_1=0.03;// m
+h_2=0.07;// m
+h_3=0.12;// m
+g=9.81;//m/s^2
+
+//Calculation
+V_1=sqrt(2*g*h_3);// m/s
+printf('The velocity at the center of the pipe,V_1=%0.2f m/s\n',V_1);
diff --git a/3720/CH5/EX5.9/Ex5_9.sce b/3720/CH5/EX5.9/Ex5_9.sce new file mode 100644 index 000000000..d759e75bd --- /dev/null +++ b/3720/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,23 @@ +//Example 5_9
+clc;clear;funcprot(0);
+// Given values
+rho_hg=848;//The density of mercury in lbm/ft^3
+rho_sw=64;//The density of seawater in lbm/ft^3
+rho_atm=0.076;//The density of atmosphereic air in lbf/ft^3
+H_hg=(30-22);// inch
+V_a=155;//mph
+V_a=155*1.4667;// convert mph into ft/s
+P_air=22;// The hurricane atmospheric pressure at the eye of the storm is in Hg
+P_atm=30;// in hg
+g=32.2;// ft/s^2
+
+// Calculation
+//(a)
+h_1=((rho_hg/rho_sw)*H_hg)/12;
+printf('(a)The pressure difference between points 1 and 3 in terms of the seawater column height,h_1=%0.2f ft\n',h_1);
+//(b)
+H_air=((V_a^2)/(2*g));//ft
+rho_air=(P_air/P_atm)*rho_atm;//the density of air in the hurricane in lbm/ft^3
+h_dynamic=(rho_air/rho_sw)*H_air;//ft
+h_2=h_1+h_dynamic;//ft
+printf('(b)The total storm surge at point 2,h_2=%0.2f ft\n',h_2);
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