diff options
Diffstat (limited to '3785')
71 files changed, 1217 insertions, 0 deletions
diff --git a/3785/CH1/EX1.1/Ex1_1.sce b/3785/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..cadc32351 --- /dev/null +++ b/3785/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,10 @@ +// Example 1_1
+clc;funcprot(0);
+// Given data
+h_avg=3800;// The ocean's average depth in m
+deltaT=1;// The increase in ocean temperature in K
+alpha=1.6*10^-4;// The average thermal expansion coefficient in K^-1
+
+// Calculation
+deltah=alpha*deltaT*h_avg;// The rise in sea level in m
+printf('\nThe rise in sea level is %0.3f m',deltah);
diff --git a/3785/CH1/EX1.2/Ex1_2.sce b/3785/CH1/EX1.2/Ex1_2.sce new file mode 100644 index 000000000..a38ac94ba --- /dev/null +++ b/3785/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,13 @@ +// Example 1_2
+clc;funcprot(0);
+// Given data
+T=98.6;// Temperature in °F
+p=1.0133*10^5;// Pressure in N/m^2
+M=32;// The molecular weight of oxygen
+R=8.3143*10^3;// Universal gas constant in J/kg.K
+O=20/100;// The maximum oxygen concentration in oxygenated blood in %
+
+// Calculation
+rho=(p*M)/(R*(273.15+((5/9)*(T-32))));// Density in kg/m^3
+rho_O2=O*rho;// The partial density of blood oxygen in kg/m^3
+printf("\nThe partial density of oxygen in blood at this concentration is %0.4f kg/m^3",rho_O2);
diff --git a/3785/CH1/EX1.3/Ex1_3.sce b/3785/CH1/EX1.3/Ex1_3.sce new file mode 100644 index 000000000..4d544e7ea --- /dev/null +++ b/3785/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,11 @@ +// Example 1_3
+clc;funcprot(0);
+// Given data
+// From table 1.6
+// Conversion between EES and SI units
+ft=3.048*10^-1;// Conversion of EES uint of foot to SI unit value(m)
+acre=4.048*10^3;// Conversion of EES uint of acre to SI unit value(m^2)
+
+// Calculation
+acreft=1*acre*ft;// m^3
+printf("\n%0.0f cubic meters of water are there in one acre-foot",acreft);
diff --git a/3785/CH10/EX10.3/Ex10_3.sce b/3785/CH10/EX10.3/Ex10_3.sce new file mode 100644 index 000000000..25dfe6663 --- /dev/null +++ b/3785/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,32 @@ +// Example 10_3
+clc;funcprot(0);
+// Given data
+L_p=100;// Length in m
+L_m=2;// The model length in m
+v_m=5*10^-2;// Displaced volume in m^3
+A_wm=0.9*1;// Wetted area in m^2
+rho_m=1*10^3;// kg/m^3
+V_m=1.1;// m/s
+D_m=2.66;// The drag force in N
+rho_p=1.03*10^3;// kg/m^3
+SR=1/50;
+nu=1*10^-6;// m^2/s
+
+// Calculation
+// (a)
+V_p=V_m/(sqrt(SR));
+V_pn=V_p*(3600/(1.852*10^3));// naut.mi/h
+// (b)
+Re_Lm=(V_m*L_m)/nu;// Reynolds number
+C_Dm=0.455/(log10(Re_Lm))^2.58;// The drag coefficient
+D_fm=((1/2)*rho_m*V_m^2*A_wm)*C_Dm;// Drag force in N
+D_wm=D_m-D_fm;// N
+D_wp=D_wm*(rho_p/rho_m)*(1/SR)^3;// N
+A_wp=A_wm*(1/SR)^2;// m^2
+Re_Lp=(V_p*L_p)/nu;// Reynolds number
+C_Dp=0.455/(log10(Re_Lp))^2.58;// The drag coefficient
+D_fp=((1/2)*rho_p*V_p^2*A_wp)*C_Dp;// Drag force in N
+D_p=D_wp+D_fp;// Drag force in N
+// (c)
+P_p=(D_p*V_p)/10^6;// The power in kW
+printf("\n(a)The corresponding speed V_p=%2.2f naut.mi/h \n(b)The drag force V_p in ocean water,D_p=%1.3e N \n(c)The propulsive power,P=%1.3f MW",V_pn,D_p,P_p);
diff --git a/3785/CH10/EX10.4/Ex10_4.sce b/3785/CH10/EX10.4/Ex10_4.sce new file mode 100644 index 000000000..7dd7f7c5d --- /dev/null +++ b/3785/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,14 @@ +// Example 10_4
+clc;funcprot(0);
+// Given data
+V_wbyomegaD_m=0.1;
+C_pm=0.50;// The power coefficient
+V_w=10;// The wind speed in m/s
+P_wtp=100;// kW
+rho=1.2;// The density of air in kg/m^3
+
+// Calculation
+omega_p=sqrt((%pi*C_pm*rho*V_w^5)/(8*P_wtp*10^3*V_wbyomegaD_m^2));// s^-2
+omega=omega_p*(60/(2*%pi));// RPM
+D_p=(1/V_wbyomegaD_m)*(V_w/omega_p);// m
+printf("\nThe turbine diameter,D_p=%2.1f m",D_p);
diff --git a/3785/CH10/EX10.5/Ex10_5.sce b/3785/CH10/EX10.5/Ex10_5.sce new file mode 100644 index 000000000..a5d0bb4e9 --- /dev/null +++ b/3785/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,19 @@ +// Example 10_5
+clc;funcprot(0);
+// Given data
+Q=1000;// GPM
+h=100;// Head in m
+g=9.807;// The acceleration due to gravity in m/s^2
+// Reading values from figure 10.5
+C_Q=7*10^-3;
+C_h=0.116;
+C_p=1.16*10^-3;
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+Q=Q*((3.785*10^-3)/60);// m^3/s
+omega=((g*h)^(3/4)*(C_Q)^(1/2))/(Q^(1/2)*(C_h)^(3/4));// s^-1
+omega_rpm=omega*(60/(2*%pi));// rpm
+D=(Q/(omega*C_Q));// The diameter D in m
+P=(rho*omega^3*D^5*C_p);// The power in kW
+printf("\nThe pump speed=%4.0f \nDiameter,D=%0.4f m \nThe power=%2.2f kW",omega_rpm,D,P);
diff --git a/3785/CH10/EX10.6/Ex10_6.sce b/3785/CH10/EX10.6/Ex10_6.sce new file mode 100644 index 000000000..476476205 --- /dev/null +++ b/3785/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,9 @@ +// Example 10_6
+clc;funcprot(0);
+// Given data
+C_Du=1.4;// Drag coefficient of the upwind facing cup
+C_Dd=0.4;// Drag coefficient of the downwind facing cup
+
+// Calculation
+omegaRbyV=sqrt((C_Du-C_Dd)/(C_Du+(4*C_Dd)));// The dimensionless angular speed
+printf("\nThe dimensionless angular speed at which the anemometer rotates is %0.4f.",omegaRbyV)
diff --git a/3785/CH12/EX12.1/Ex12_1.sce b/3785/CH12/EX12.1/Ex12_1.sce new file mode 100644 index 000000000..88e59ce10 --- /dev/null +++ b/3785/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,16 @@ +// Example 12_1
+clc;funcprot(0);
+// Given data
+T=300;// Temperature in K
+R_a=287.0;// Gas constant for air in J/kg.K
+C_pbyR_a=3.5;
+R_h=2077;// Gas constant for helium in J/kg.K
+C_pbyR_h=2.5;
+
+// Calculation
+// (a)
+a=sqrt((C_pbyR_a/(C_pbyR_a-1))*R_a*T);// m/s
+printf("\n(a)The speed of sound in air,a=%3.1f m/s",a);
+// (b)
+a=sqrt((C_pbyR_h/(C_pbyR_h-1))*R_h*T);// m/s
+printf("\n(b)The speed of sound in helium,a=%4.0f m/s",a);
diff --git a/3785/CH12/EX12.10/Ex12_10.sce b/3785/CH12/EX12.10/Ex12_10.sce new file mode 100644 index 000000000..284179459 --- /dev/null +++ b/3785/CH12/EX12.10/Ex12_10.sce @@ -0,0 +1,14 @@ +// Example 12_10
+clc;funcprot(0);
+// Given data
+u_0=100;// The average speed in m/s
+r=1.31;// The specific heat ratio
+a_0=446.1;// m/s
+
+// Calculation
+// (a)
+u_out_1=((2/(r+1))*a_0)+(((r-1)/(r+1))*u_0);// The outflow speed of section 1 in m/s
+mfr_ratio_1=(u_out_1/a_0)^(2/(r-1))*(u_out_1/u_0);// The mass flow rate ratio of section 1
+u_out_2=-(((2/(r+1))*a_0)-(((r-1)/(r+1))*u_0));// The outflow speed of section 2 in m/s
+mfr_ratio_2=(-u_out_2/a_0)^(2/(r-1))*(-u_out_2/u_0);// The mass flow rate ratio of section 2
+printf("\n(a)The outflow speed of section 1,u_out=%3.1f m/s \n The outflow speed of section 1,u_out=%3.1f m/s \n(b)The mass flow rate ratio of section 1 is %1.3f \n The mass flow rate ratio of section 2 is %1.3f",u_out_1,u_out_2,mfr_ratio_1,mfr_ratio_2);
diff --git a/3785/CH12/EX12.2/Ex12_2.sce b/3785/CH12/EX12.2/Ex12_2.sce new file mode 100644 index 000000000..4f059fa2a --- /dev/null +++ b/3785/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,19 @@ +// Example 12_2
+clc;funcprot(0);
+// Given data
+T=20;// °C
+SPL=20;// Sound Pressure level in dB
+// From table 1.1,
+rho_0=1.204;// kg/m^3
+gamma=3.5/2.5;// Specific heat ratio
+
+// Calculation
+// (a) Inverting equation 12.18,
+Pa=2*10^-5*(1*10^(20/10));// Pa
+// (b) From equation 12.17,
+a=(gamma*1.013*10^5*rho_0)^(1/2);// m/s
+va=Pa/(rho_0*a);
+//(c) From equation 12.17,
+P_sw=(Pa)^2/(rho_0*a);
+printf('\n(a)The pressure amplitude is %1.0e Pa \n(b)The velocity amplitude is %1.2e m/s \n The power per unit area,P_sw=%1.2e W/m^2',Pa,va,P_sw);
+// The answer provided in the book is wrong
diff --git a/3785/CH12/EX12.3/Ex12_3.sce b/3785/CH12/EX12.3/Ex12_3.sce new file mode 100644 index 000000000..f80edcbc5 --- /dev/null +++ b/3785/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,10 @@ +// Example 12_3
+clc;funcprot(0);
+// Given data
+r=1.4;// The specific heat ratio
+p_s=6*10^5;// The pressure in the large tank in Pa
+p_t=5*10^4;// The pressure in the test section in Pa
+
+// Calculation
+M_t=sqrt((2/(r-1))*((p_t/p_s)^(-(r-1)/r)-1));// Mach number
+printf("\nThe Mach number M_t in the test section is %1.3f",M_t);
diff --git a/3785/CH12/EX12.4/Ex12_4.sce b/3785/CH12/EX12.4/Ex12_4.sce new file mode 100644 index 000000000..dd3e876fc --- /dev/null +++ b/3785/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,17 @@ +// Example 12_4
+clc;funcprot(0);
+// Given data
+p_s=4*10^5;// Pressure in Pa
+a_s=347.2;// Sound speed in m/s
+A_c=1*10^-4;// The flow area in m^2
+p_a=1*10^5;// The atmospheric pressure in Pa
+r=1.4;// The specific heat ratio
+V_c=0.5787;
+
+// Calculation
+rho_c=(r*p_s)/a_s;// kg/m^3
+m_c=rho_c*V_c*A_c;// kg/s
+V_c=a_s/(sqrt(1+(r-1)/2));// m/s
+p_c=((2/(r+1))^(r/(r-1)))*p_s;// N
+F=(m_c*V_c)+((p_c-p_a)*A_c);// N
+printf('\nThe mass flow rate of air from the tank=%1.2e kg/s \nThe external force F required to restrain the tank from moving is %2.2f N',m_c,F);
diff --git a/3785/CH12/EX12.5/Ex12_5.sce b/3785/CH12/EX12.5/Ex12_5.sce new file mode 100644 index 000000000..1563e54cc --- /dev/null +++ b/3785/CH12/EX12.5/Ex12_5.sce @@ -0,0 +1,16 @@ +// Example 12_5
+clc;funcprot(0);
+// Given data
+r=1.3;// The specific heat ratio
+V_e=0.90;// Exit velocity in % of maximum possible velocity
+
+// Calculation
+V_mebya_s=sqrt(2/(r-1));// The maximum possible exit velocity
+function[X]=machnumber(y)
+ X(1)=(V_e*V_mebya_s)-(y(1)*(1+((0.3*y(1)^2)/2))^(-1/2));
+endfunction
+y=[1];
+M_e=fsolve(y,machnumber);// The Mach number of the exit flow
+A_ebyA_c=M_e*((2/(r+1))*((1+(((r-1)/2)*(M_e)^2))))^((r+1)/(2*(r-1)));// The area ratio
+D_r=sqrt(A_ebyA_c);// Corresponding diameter ratio
+printf("\nThe exit Mach number,M_e=%1.3f \nThe area ratio=%1.3e \nThe diameter ratio=%2.2f",M_e,A_ebyA_c,D_r)
diff --git a/3785/CH12/EX12.6/Ex12_6.sce b/3785/CH12/EX12.6/Ex12_6.sce new file mode 100644 index 000000000..71c4b7641 --- /dev/null +++ b/3785/CH12/EX12.6/Ex12_6.sce @@ -0,0 +1,18 @@ +// Example 12_6
+clc;funcprot(0);
+// Given data
+a_1=347.2;// m/s
+p_1=1*10^5;// Pa
+r=1.4;// The specific heat ratio
+V_p=100;// The velocity of piston in m/s
+
+// Calculation
+C=((r+1)*V_p)/(2*a_1);
+function[X]=machnumber(y)
+ X(1)=y(1)^2-(C*y(1))-1;
+endfunction
+y=[1];
+M_1=fsolve(y,machnumber);// The shock Mach number
+V_s=M_1*a_1;// The shock speed in m/s
+p_2=(((2*r*M_1^2)-(r-1))/(r+1))*p_1;// The pressure on the piston face in Pa
+printf("\nThe shock Mach number,M_1=%1.3f \nThe shock speed,V_s=%3.1f m/s \nThe pressure on the piston face,p_2=%1.2e Pa",M_1,V_s,p_2);
diff --git a/3785/CH12/EX12.7/Ex12_7.sce b/3785/CH12/EX12.7/Ex12_7.sce new file mode 100644 index 000000000..6c237a46b --- /dev/null +++ b/3785/CH12/EX12.7/Ex12_7.sce @@ -0,0 +1,27 @@ +// Example 12_7
+clc;funcprot(0);
+// Given data
+D=1;// The diameter of gas pipeline in m
+epsilon=5*10^-5;// Surface roughness in m
+p_1=2*10^6;// Pressure in Pa
+T_1=20;// Temperature in °C
+a_1=446.1;// The natural gas sound speed in m/s
+mu_1=9*10^-6;// Viscosity in Pa s
+r=1.31;// The specific heat ratio
+R=518.3;// The gas constant in J/kg.K
+V_1=10;// The pipe flow speed in m/s
+
+// Calculation
+rho_1=(p_1)/(R*(T_1+273.15));// The density in kg/m^3
+Re_D=(rho_1*V_1*D)/mu_1;// Reynolds number
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);// Friction factor
+M_1=V_1/a_1;// Mach number
+L_max=((D/f)*(((1-M_1^2)/(r*M_1^2))+(((r+1)/(2*r))*log(((r+1)*M_1^2)/(2*(1+((r-1)*M_1^2/2)))))))/1000;// The pipe length at which the flow would be choked in m
+V_c=((V_1/M_1)*(sqrt((2+((r-1)*M_1^2))/(r+1))));// The flow speed in m/s
+p_c=(p_1*V_c*M_1^2)/V_1;// The pressure at the point of choked flow in Pa
+printf("\nThe pipe length at which the flow would be choked,L_max=%3.1f km \nThe flow speed at the point of choked flow,V_c=%3.0f m/s \nThe pressure at the point of choked flow,p_c=%1.3e Pa",L_max,V_c,p_c);
diff --git a/3785/CH12/EX12.8/Ex12_8.sce b/3785/CH12/EX12.8/Ex12_8.sce new file mode 100644 index 000000000..c2fae6951 --- /dev/null +++ b/3785/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,10 @@ +// Example 12_8
+clc;funcprot(0);
+// Given data
+theta=10;// The half angle of two dimensional wedge in degree
+beta=20;// The attached shock wave angle in degree
+r=1.4;// The specific heat ratio
+
+// Calculation
+M=sqrt(((2*tand(theta))+(2*cotd(beta)))/((2*cosd(beta)*sind(beta))-((r+(cosd(beta)*cosd(beta)))*tand(theta))));// The Mach number M of the flow in the wind tunnel
+printf("The Mach number M of the flow in the wind tunnel,M=%1.3f",M);
diff --git a/3785/CH2/EX2.1/Ex2_1.sce b/3785/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..1f3772062 --- /dev/null +++ b/3785/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,13 @@ +// Example 2_1
+clc;funcprot(0);
+// Given data
+p_0=1.0133*10^5;// The sea level pressure in Pa
+alpha=1.2*10^-4;// m^-1
+
+// Calculation
+z=0;// km
+minusdelp=alpha*p_0*exp(-alpha*z);// The pressure force per unit volume in N/m^3
+printf("\nAt z=0,The pressure force per unit volume -delp=(%0.2f N/m^3)i_z",minusdelp);
+z=5;// km
+minusdelp=alpha*p_0*exp(-alpha*z*10^3);// The pressure force per unit volume in N/m^3
+printf("\nAt z=5,The pressure force per unit volume -delp=(%0.3f N/m^3)i_z",minusdelp);
diff --git a/3785/CH2/EX2.10/Ex2_10.sce b/3785/CH2/EX2.10/Ex2_10.sce new file mode 100644 index 000000000..54a5602bc --- /dev/null +++ b/3785/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,10 @@ +// Example 2_10
+clc;funcprot(0);
+// Given data
+rho_w=1*10^3;// The density of water in kg/m^3
+r=7.3*10^-2;// The air/water interfacial tension in N/m
+g=9.8066;// The acceleration due to gravity in m/s^2
+
+// Calculation
+d=sqrt((6*r)/(rho_w*g))*10^3;// The approximate maximum diameter d of a bubble of air in water in mm
+printf("\nThe approximate maximum diameter d of a bubble of air in water is %1.1f m",d);
diff --git a/3785/CH2/EX2.3/Ex2_3.sce b/3785/CH2/EX2.3/Ex2_3.sce new file mode 100644 index 000000000..83aa1c965 --- /dev/null +++ b/3785/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,16 @@ +// Example 2_3
+clc;funcprot(0);
+// Given data
+D=1;// The diameter of a circular flat plate in m
+h=3;// Distance in m
+theta=45;// Angle in degrees
+rho=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+p_c=rho*g*h;// The gage pressure at the plate centroid in Pa
+A=(%pi*D^2)/4;// Area in m^2
+F=p_c*A;// The total force exerted on the plate by the water in N
+printf("\nThe total force exerted on the plate by the water,F=%1.3e N",F);
+y_cp=-(rho*g*D^2)/(16*sqrt(2)*p_c);// The distance between the center of pressure cp and the centroid of the circular plate in m
+printf("\nThe distance between the center of pressure cp and the centroid of the circular plate,y_cp=%1.3e m",y_cp);
diff --git a/3785/CH2/EX2.7/Ex2_7.sce b/3785/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..5c4a9e20b --- /dev/null +++ b/3785/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,12 @@ +// Example 2_7
+clc;funcprot(0);
+// Given data
+//The block has a width W, a height H and a specific gravity SG.
+// GM=0;
+
+//Solution
+SG=(3-sqrt(3))/6;
+printf("\nSG<=%0.4f",SG)
+SG=(3+sqrt(3))/6;
+printf("\nSG>=%0.4f",SG)
+printf("\nIce cubes and styrofoam cubes will float upright, but not soap cubes !");
diff --git a/3785/CH2/EX2.8/Ex2_8.sce b/3785/CH2/EX2.8/Ex2_8.sce new file mode 100644 index 000000000..e992c6e45 --- /dev/null +++ b/3785/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,15 @@ +// Example 2_8
+clc;funcprot(0);
+// Given data
+d=3;// The internal diameter of a horizontal cylindrical fuel oil storage tank in m
+SG=0.87;// Specific gravity of water oil
+t=0.2;// Thickness in m
+z_0=0;// The initial height in m
+z_1=-1.3;// The height of the water-oil interface in m
+z_2=-1.5;// The height of the bottom of the tank in m
+rho_w=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+p_bminusp_0=rho_w*-g*((SG*(z_1-z_0))+(z_2-z_1));// The gage pressure at the bottom of the tank in Pa
+printf("The gage pressure at the bottom of the tank is %0.4e Pa",p_bminusp_0);
\ No newline at end of file diff --git a/3785/CH2/EX2.9/Ex2_9.sce b/3785/CH2/EX2.9/Ex2_9.sce new file mode 100644 index 000000000..fa92c6172 --- /dev/null +++ b/3785/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,18 @@ +// Example 2_9
+clc;funcprot(0);
+// Given data
+// From table 2.1
+z=5;// Altitude in km
+z_i=0;// The initial height in km
+dTbydz=-6.5;// The temperature gradient from 0 to 5 km in K/km
+T_i=288.15;// Temperature in K
+p_i=1.0133*10^5;// Pressure in Pa
+R=287;// Gas constant in J/kg.K
+
+//Calculation
+// Using equation 2.41,
+T=T_i+((dTbydz)*(z-z_i));// Temperature in K
+// Using equation 2.42,
+p=p_i*(T/T_i)^(-1/((dTbydz*10^-3)*29.26));// The pressure in Pa
+rho=p/(R*T);// The density in kg/m^3
+printf("\nT=%0.1f K \np=%1.4e Pa \nrho=%0.4f kg/m^3",T,p,rho);
diff --git a/3785/CH3/EX3.5/Ex3_5.sce b/3785/CH3/EX3.5/Ex3_5.sce new file mode 100644 index 000000000..6e92cd493 --- /dev/null +++ b/3785/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,9 @@ +// Example 3_5
+clc;funcprot(0);
+// Given data
+R=1;// The radius of a cylindrical tank in m
+V_w=1;// The velocity in mm/s
+
+// Calculation
+Q=%pi*R^2*V_w*10^-3;// The volume flow rate of water through the pump in m^3/s
+printf("\nThe volume flow rate Q of water through the pump is %1.3e m^3/s",Q);
diff --git a/3785/CH3/EX3.6/Ex3_6.sce b/3785/CH3/EX3.6/Ex3_6.sce new file mode 100644 index 000000000..6f3578b76 --- /dev/null +++ b/3785/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,10 @@ +// Example 3_6
+clc;funcprot(0);
+// Given data
+Q=1*10^3;// The water volume flow rate in m^3/s
+D=2;// The diameter of the fire hose at exit nozzle in inch
+
+// Calculation
+V=(4*(Q/60)*3.785*10^-3)/(%pi*(D*2.54*10^-2)^2);// The velocity of the water leaving the nozzle in m/s
+// We have used table 1.6 to convert gallons to cubic meters.
+printf("\nThe velocity of the water leaving the nozzle,v=%2.2f m/s",V);
diff --git a/3785/CH3/EX3.9/Ex3_9.sce b/3785/CH3/EX3.9/Ex3_9.sce new file mode 100644 index 000000000..2a6ddc467 --- /dev/null +++ b/3785/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,13 @@ +// Example 3_9
+clc;funcprot(0);
+// Given data
+v=10;// The volume of the tank in m^3
+rho_s0=3.0;// The initial salt density in kg/m^3
+t=0;// Time in s
+Q=0.01;// The volume flow rate in m^3/s
+
+// Calculation
+// (b)
+// V=Q*t;
+V=v*log(2);//
+printf("\nThe volume of fresh water,V=%0.3f m^3",V);
diff --git a/3785/CH4/EX4.4/Ex4_4.sce b/3785/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..38bd31ff9 --- /dev/null +++ b/3785/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,11 @@ +// Example 4_4
+clc;funcprot(0);
+// Given data
+V_1=50;// Velocity in m/s
+alpha=45;// Angle in degree
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+w_1=V_1*sind(alpha);// m/s
+h=(w_1)^2/(2*g);// Height in m
+printf("\nThe maximum value of h is %2.2f m",h)
diff --git a/3785/CH4/EX4.5/Ex4_5.sce b/3785/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..4d1a14b44 --- /dev/null +++ b/3785/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,10 @@ +// Example 4_5
+clc;funcprot(0);
+// Given data
+rho=1.225;// The density of air in kg/m^3
+A_0=1.0;// Area of orifice in cm^2
+A_pV=2.5*10^-4;// The volumetric flow rate in m^3/s
+
+// Calculation
+P=(rho*(A_pV)^3)/(2*(A_0*10^-4)^2);// The power expended in inhaling in W
+printf("\nThe power P expended in inhaling (or exhaling) is %1.2e W",P)
diff --git a/3785/CH4/EX4.6/Ex4_6.sce b/3785/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..c6e8180ff --- /dev/null +++ b/3785/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,12 @@ +// Example 4_6
+clc;funcprot(0);
+// Given data
+D_t=30;// The diameter of an oil storage tank in m
+H=5;// The depth of the oil in m
+D_p=5;// The inside diameter of pipe in cm
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+t=(D_t/(D_p/100))^2*sqrt((2*H)/g);// Time in s
+t=t/3600;// Time in hours
+printf("\nIt will take %3.0f hr for the oil to drain completely from the tank.",t)
diff --git a/3785/CH4/EX4.8/Ex4_8.sce b/3785/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..d9b7d7120 --- /dev/null +++ b/3785/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,11 @@ +// Example 4_8
+clc;funcprot(0);
+// Given data
+rho=8.6*10^2;// The density of gasoline in kg/m^3
+L=1.0;// The tank length in m
+H=0.6;// The tank height in m
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+p=rho*g*(H+(2*L));// Pa
+printf("\nThe maximum pressure in the tank,p=p_a+%1.3e Pa",p)
diff --git a/3785/CH4/EX4.9/Ex4_9.sce b/3785/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..3142398d7 --- /dev/null +++ b/3785/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,11 @@ +// Example 4_9
+clc;funcprot(0);
+// Given data
+R=5;// The radius of a jar in cm
+n=33;// tThe turntable has been revolving at a steady speed in rpm
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+omega=(2*%pi*n)/60;// Acceleration
+h=(omega*R*10^-2)^2/(2*g);// The height h in m
+printf("\nThe height,h=%1.3e m",h);
diff --git a/3785/CH5/EX5.10/Ex5_10.sce b/3785/CH5/EX5.10/Ex5_10.sce new file mode 100644 index 000000000..015f2b725 --- /dev/null +++ b/3785/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,13 @@ +// Example 5_10
+clc;funcprot(0);
+// Given data
+A_w=100;// The wake area in m^2
+x=100;// m
+// From example 5.7,
+rho_w=0.4;// The density of air in kg/m^3
+V_f=250;// The speed of flight in m/s
+F=2.6*10^4;// The restraining force in N
+
+// Calculation
+V_w=V_f+(F/(rho_w*A_w*V_f));// m/s
+printf("\nThe wake speed,V_w=%3.1f m/s",V_w);
diff --git a/3785/CH5/EX5.11/Ex5_11.sce b/3785/CH5/EX5.11/Ex5_11.sce new file mode 100644 index 000000000..e8072f038 --- /dev/null +++ b/3785/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,11 @@ +// Example 5_11
+clc;funcprot(0);
+// Given data
+V_s=1;// The speed of water jet in m/s
+D_s=3;// The diameter of a hole in cm
+D_j=10;// The jet diameter in cm
+x=1;// Distance from the source in m
+
+// Calculation
+V_j=V_s*(D_s/D_j);// m/s
+printf("\nThe value of the jet speed V_j at that point is %0.1f m/s.",V_j);
diff --git a/3785/CH5/EX5.12/Ex5_12.sce b/3785/CH5/EX5.12/Ex5_12.sce new file mode 100644 index 000000000..f6d099d2c --- /dev/null +++ b/3785/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,19 @@ +// Example 5_12
+clc;funcprot(0);
+// Given data
+D_s=1;// The diameter of jet in inch
+D=3;// The inside diameter of a pipe in inch
+Q_s=100;// The jet volumetric flow rate in GPM (gallons per minute)
+Q_1=500;// The volumetric flow rate in GPM
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+A_s=(%pi/4)*(D_s*2.54*10^-2)^2;// m^2
+A=9*A_s;// m^2
+Q_s=(Q_s*3.785*10^-3)/60;// m^3/s
+Q_1=5*Q_s;// m^3/s
+V_1=Q_1/(A-A_s);// m/s
+V_s=Q_s/A_s;// m/s
+// Assume dp=p_2-p_1;
+dp=(A_s/A)*(1-(A_s/A))*rho*(V_s-V_1)^2;// The pressure rise in the jet pump in Pa
+printf("\nThe pressure rise in the jet pump,p_2-p_1=%1.3e Pa",dp);
diff --git a/3785/CH5/EX5.13/Ex5_13.sce b/3785/CH5/EX5.13/Ex5_13.sce new file mode 100644 index 000000000..e77b1d298 --- /dev/null +++ b/3785/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,20 @@ +// Example 5_13
+clc;funcprot(0);
+// Given data
+h=100;// Height in m
+A_n=1.0;// The area of the turbine jet stream in in^2
+alpha=20;// The blade angle in degree
+g=9.807;// The acceleration due to gravity in m/s^2
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+// (a)
+V_n=sqrt(2*g*h);// The nozzle velocity V_n in m/s
+printf("\n(a)The nozzle velocity V_n=%2.2f m/s",V_n);
+// (b)
+maxP_b=((1+cosd(alpha))/2)*(rho*(A_n*2.54*10^-2)^2*(V_n^3/2))/1000;// The maximum power P, of the turbine in kW
+printf("\n(b)The maximum power P_t of the turbine is %2.2f kW.",maxP_b);
+// (c)
+V_b=V_n/2;// The blade speed in m/s
+F_b=rho*(A_n*2.54*10^-2)^2*(V_n-V_b)^2*(1+cosd(alpha));// The force in N
+printf("\n(c)The blade speed,V_b=%2.2f m/s \n The force when maximum power is being produced,F_b=%3.1f N",V_b,F_b);
diff --git a/3785/CH5/EX5.16/Ex5_16.sce b/3785/CH5/EX5.16/Ex5_16.sce new file mode 100644 index 000000000..09c9cb345 --- /dev/null +++ b/3785/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,14 @@ +// Example 5_16
+clc;funcprot(0);
+// Given data
+A=10;// The internal area of the rotating tube in mm^2
+V=5;// The speed of water flow in m/s
+alpha=30;// Angle in degree
+R=10;// The tip radial dimension in mm
+T=2*10^-2;// Torque in Nm
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+omega=(V/(R*10^-2)*cosd(alpha))-((T)/(2*rho*(A*10^-6)*(R/100)^2*V));// The angular speed of the sprinkler rotor in s^-1
+V=[(V*sind(alpha)),((V*cosd(alpha))-(omega*R*10^-2))];// The velocity V of the fluid stream relative to the ground in m/s
+printf("\n(a)The angular speed of the sprinkler rotor,omega=%2.2f s^-1 \n(b)The velocity V in the ground reference frame is:V=(%1.1f m/s)i_r+(%1.1f m/s)i_theta",omega,V(1),V(2));
diff --git a/3785/CH5/EX5.4/Ex5_4.sce b/3785/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..368374171 --- /dev/null +++ b/3785/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,13 @@ +// Example 5_4
+clc;funcprot(0);
+// Given data
+Q=150;// The water stream volume flow rate in gal/min
+D=1;// The nozzle exit diameter in inch
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+Q=(Q*3.785*10^-3)/60;// The water stream volume flow rate in m^3/s
+V_out=(4*Q)/(%pi*(D*2.54*10^-2)^2);// The velocity in m/s
+F_e=rho*Q*V_out;// The force in N
+F_e=F_e/4.448;// The force in lbf
+printf("\nThe force,F_e=%2.2f lbf",F_e);
diff --git a/3785/CH5/EX5.5/Ex5_5.sce b/3785/CH5/EX5.5/Ex5_5.sce new file mode 100644 index 000000000..46f27d1ef --- /dev/null +++ b/3785/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,13 @@ +// Example 5_5
+clc;funcprot(0);
+// Given data
+D=1;// Diameter of hose at inlet in inch
+d=2;// Diameter of hose at outlet in inch
+// From example 5.4,F_e=rho*Q*V_out
+F_e=176.8;// The force in N
+
+// Calculation
+// F_c=rho*Q*V_out*[1/2*((A_in/A_out)+(A_out/A_in)-1];
+// A_in=4*A_out
+F_c=F_e*((1/2)*(4+(1/4))-1);// The force exerted on the nozzle by the coupling in N
+printf("\n The force exerted on the nozzle by the coupling,F_c=%3.1f N",F_c);
diff --git a/3785/CH5/EX5.6/Ex5_6.sce b/3785/CH5/EX5.6/Ex5_6.sce new file mode 100644 index 000000000..a9342e564 --- /dev/null +++ b/3785/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,9 @@ +// Example 5_6
+clc;funcprot(0);
+// Given data
+m=2;// The mass flow rate in kg/s
+V_e=200;// The rocket exhaust velocity in m/s
+
+// Calculation
+F=m*V_e;// The restraining force required to hold the rocket in place in N
+printf("\nThe restraining force required to hold the rocket in place,F_c=%0.0f N",F);
diff --git a/3785/CH5/EX5.7/Ex5_7.sce b/3785/CH5/EX5.7/Ex5_7.sce new file mode 100644 index 000000000..e9fe0a268 --- /dev/null +++ b/3785/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,16 @@ +// Example 5_7
+clc;funcprot(0);
+// Given data
+V_f=250;// The speed of flight in m/s
+rho_a=0.4;// The density of air in kg/m^3
+A_in=1;// The inlet area in m^2
+m_f=2;// The mass flow rate of fuel in kg/s
+V_e=500;// The speed of exhaust jet in m/s
+
+// Calculation
+m_in=rho_a*V_f*A_in;// The mass flow rate of air at inlet in kg/s
+m_out=m_in+m_f;// The mass flow rate of air at outlet in kg/s
+F=(m_out*V_e)-(m_in*V_f);// The force exerted on the engine by the airframe in N
+printf("\nThe value of the force F exerted on the engine by the airframe is %1.1e N",F);
+
+
diff --git a/3785/CH5/EX5.8/Ex5_8.sce b/3785/CH5/EX5.8/Ex5_8.sce new file mode 100644 index 000000000..4ffb4feab --- /dev/null +++ b/3785/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,22 @@ +// Example 5_8
+clc;funcprot(0);
+// Given data
+V_f=200;// The speed of flying air plane in km/h
+rho=1.2;// The density of air in kg/m^3
+F=3*10^3;// The propulsive force in N
+D_p=2;// The diameter of the propeller in m
+
+// Calculation
+// (a)
+V_f=(V_f*10^3)/3600;// The speed of flying air plane in m/s
+A_p=(%pi*D_p^2)/4;// Area of propeller in m^2
+V_w=sqrt((V_f^2)+((2*F)/(rho*A_p)));// The wake speed in m/s
+printf("\nThe wake speed,V_w=%2.2f m/s",V_w);
+
+// (b)
+n_prop=(2*V_f)/(V_w+V_f)*100;// The propulsive efficiency in %
+printf("\nThe propulsive efficiency is %2.2f percentage",n_prop);
+// (c)
+
+P_p=(F*(V_w+V_f))/(2*10^3);// The engine powerin kW
+printf("\nThe engine power for this air craft is %3.1f kW",P_p);
diff --git a/3785/CH5/EX5.9/Ex5_9.sce b/3785/CH5/EX5.9/Ex5_9.sce new file mode 100644 index 000000000..681055ee6 --- /dev/null +++ b/3785/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,11 @@ +// Example 5_9
+clc;funcprot(0);
+// Given data
+D=6;// The diameter of wind turbine in m
+V_w=20;// The wind speed in m/s
+rho=1.2;// The density of air in kg/m^3
+
+// Calculation
+A_p=((%pi/4)*(6)^2);// m^2
+maxP_wt=((8/27)*(rho)*A_p*(V_w*0.447)^3)/1000;// The maximum power that can be generated by a wind turbine in kW
+printf("\nThe maximum power that can be generated by a wind turbine is %1.3f kW",maxP_wt);
diff --git a/3785/CH6/EX6.1/Ex6_1.sce b/3785/CH6/EX6.1/Ex6_1.sce new file mode 100644 index 000000000..7cc96c70d --- /dev/null +++ b/3785/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,15 @@ +// Example 6_1
+clc;funcprot(0);
+// Given data
+a=1.0;// s^-1
+b=0.1;// s^-1
+c=2.0;// s^-1,where a,b,c are constants
+z=1;// m
+mu=1.82*10^-5;// Pa s
+
+// Calculation
+tau_xz=mu*(a-(2*b*z));// The non-zero viscous stress component in Pa
+tau_zx=tau_xz;// The non-zero viscous stress component in Pa
+tau_yz=mu*c;// The non-zero viscous stress component in Pa
+tau_zy=tau_yz;// The non-zero viscous stress component in Pa
+printf("The numerical values of all the viscous stress components,tau_xz=tau_zx=%1.3e Pa & tau_yz=tau_zy=%1.2e Pa",tau_xz,tau_yz);
diff --git a/3785/CH6/EX6.10/Ex6_10.sce b/3785/CH6/EX6.10/Ex6_10.sce new file mode 100644 index 000000000..cb8de56f8 --- /dev/null +++ b/3785/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,15 @@ +// Example 6_10
+clc;funcprot(0);
+// Given data
+D=1.0*10^-6;// Diameter of solid particle in m
+rho_p=2*10^3;// The density of particle in kg/m^3
+rho_f=1.206;// The density of air in kg/m^3
+mu=1.80*10^-5;// Viscosity in Pa s
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+// (a)
+V_f=(2*(rho_p-rho_f)*g*D^2)/(9*mu);// The free fall velocity in m/s
+// (b)
+Re_D=(rho_f*V_f*D)/mu;// The Reynolds number
+printf("\n(a)The free fall velocity,V_f=%1.3e m/s \n(b)The Reynolds number,Re_D=%1.3e",V_f,Re_D);
diff --git a/3785/CH6/EX6.11/Ex6_11.sce b/3785/CH6/EX6.11/Ex6_11.sce new file mode 100644 index 000000000..9a6d98202 --- /dev/null +++ b/3785/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,14 @@ +// Example 6_11
+clc;funcprot(0);
+// Given data
+D_i=3;// The inner diameter of hollow cylinder in cm
+D_o=10;// The outer diameter of hollow cylinder in cm
+L=20;// Length in cm
+Q=1;// The fuel flow rate in l/min
+mu=2*10^-6;// The fuel viscosity in Pa s
+k=1*10^-6;// The fuel filter permeability in m^2
+
+// Calculation
+// Assume dp=p_in-p_out
+dp=((mu*Q/60)/(2*%pi*k*L/100))*log(D_o/D_i);// The pressure drop in Pa
+printf("\n The pressure drop,p_in-p_out=%1.3e Pa",dp);
diff --git a/3785/CH6/EX6.12/Ex6_12.sce b/3785/CH6/EX6.12/Ex6_12.sce new file mode 100644 index 000000000..8e540bd87 --- /dev/null +++ b/3785/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,12 @@ +// Example 6_12
+clc;funcprot(0);
+// Given data
+// From Example 6_4
+h=0.1;// The gap betwen the shaft and the bearing in mm
+mu=6.7*10^-5;// Viscosity in Pa/s
+rho=8.0*10^2;// kg/m^3
+
+//Calculation
+// (b)
+t=(rho*(h*10^-3)^2)/mu;// s
+printf("\nThe numerical value of t is %0.4f s",t);
diff --git a/3785/CH6/EX6.13/Ex6_13.sce b/3785/CH6/EX6.13/Ex6_13.sce new file mode 100644 index 000000000..a5a7c25b5 --- /dev/null +++ b/3785/CH6/EX6.13/Ex6_13.sce @@ -0,0 +1,12 @@ +// Example 6_13
+clc;funcprot(0);
+// Given data
+D=0.5;// The diameter of cirrcular disk in m
+mu=1.0;// The viscosity of oil in Pa s
+rho=9.0*10^2;// Density in kg/m^3
+omega=1*10^3;// The angular frequency in s^-1
+phi=1*10^-3;// The angular amplitude
+
+// Calculation
+P=(%pi/32)*mu*(omega*phi)^2*((omega*rho)/(2*mu))^(1/2)*D^4;// W
+printf("\nThe power absorbed by the vibration damper,P=%1.3f W",P);
diff --git a/3785/CH6/EX6.14/Ex6_14.sce b/3785/CH6/EX6.14/Ex6_14.sce new file mode 100644 index 000000000..9d2de4083 --- /dev/null +++ b/3785/CH6/EX6.14/Ex6_14.sce @@ -0,0 +1,11 @@ +// Example 6_14
+clc;funcprot(0);
+// Given data
+// L=10h;
+Lbyh=10;
+
+// Calculation
+// Re=(V*h)/nu;
+Re=Lbyh*(12/1.328)^2;// Reynolds number
+printf("For flow velocities having Vh/v «%3.1f. the pressure drop would be given by (a),while for Vh/v »%3.1f it would be given by (b).",Re,Re);
+// The answer provided in the text book is wrong
\ No newline at end of file diff --git a/3785/CH6/EX6.2/Ex6_2.sce b/3785/CH6/EX6.2/Ex6_2.sce new file mode 100644 index 000000000..b8d854ee6 --- /dev/null +++ b/3785/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,12 @@ +// Example 6_2
+clc;funcprot(0);
+// Given data
+a=1.0;// s^-1
+b=0.1;// s^-1
+c=2.0;// s^-1 where a,b,c are constants
+z=1;// m
+mu=1.82*10^-5;// Pa s
+
+// Calculation
+delp=mu*(2*b);// Pa/m
+printf("[delp=%1.2e Pa/m]i_x",delp)
diff --git a/3785/CH6/EX6.4/Ex6_4.sce b/3785/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..14c8919ac --- /dev/null +++ b/3785/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,14 @@ +// Example 6_4
+clc;funcprot(0);
+// Given data
+D=10;// The diameter of circular shaft in cm
+L=10;// The bearing length in cm
+h=0.1;// The gap betwen the shaft and the bearing in mm
+mu=6.7*10^-5;// Viscosity in Pa/s
+n=3600;// rpm
+
+// Calculation
+omega=(2*%pi*n)/60;// s^-1
+T=(%pi*mu*omega*(L/100)*(D/100)^3)/(4*(h/1000));// The torque applied to the shaft in Nm
+P=T*omega;// The power consumed in the bearing by friction in W
+printf("\nThe torque applied to the shaft,T=%1.3e Nm \nThe power consumed in the bearing by friction,P=%1.3f W",T,P);
diff --git a/3785/CH6/EX6.5/Ex6_5.sce b/3785/CH6/EX6.5/Ex6_5.sce new file mode 100644 index 000000000..91d1add8d --- /dev/null +++ b/3785/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,13 @@ +// Example 6_5
+clc;funcprot(0);
+// Given data
+W=1.0;// Width of concrete slabs in m
+L=0.1;// Depth in m
+h=1.0;// Width of a crack in mm
+mu=1.13*10^-3;// Pa s
+rho=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+Q=(rho*g*(h*10^-3)^3*W)/(12*mu);// m^3/s (or) l/s
+printf("\nThe volume flow rate of rainwater through the crack,Q=%1.3e m^3/s (or) %0.4f l/s",Q,Q*1000);
diff --git a/3785/CH6/EX6.6/Ex6_6.sce b/3785/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..8bd416b17 --- /dev/null +++ b/3785/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,10 @@ +// Example 6_6
+clc;funcprot(0);
+// Given data
+V=0.1;// The speed of coating liquid in m/s
+nu=1.0*10^-6;// The liquid kinematic viscosity in m^2/s
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+h=sqrt((2*nu*V)/g);// m
+printf("\nThe film thickness h=%1.3e m",h);
diff --git a/3785/CH6/EX6.7/Ex6_7.sce b/3785/CH6/EX6.7/Ex6_7.sce new file mode 100644 index 000000000..36a6aa120 --- /dev/null +++ b/3785/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,8 @@ +// Example 6_7
+clc;funcprot(0);
+// Solution
+// P_out=(3*%pi*mu*W*omega^2*D^3)/(16*h);
+// P_in=(5*%pi*mu*W*omega^2*D^3)/(8*h);
+// n_p=P_out/P_in;
+n_p=(((3*%pi)/16)/((5*%pi)/8))*100;//The pump efficiency in %
+printf("\nThe pump efficiency,n_p=%0.0f percentage",n_p);
diff --git a/3785/CH6/EX6.8/Ex6_8.sce b/3785/CH6/EX6.8/Ex6_8.sce new file mode 100644 index 000000000..dbe396d20 --- /dev/null +++ b/3785/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,15 @@ +// Example 6_8
+clc;funcprot(0);
+// Given data
+H=3;// Distance in m
+L=30;// Length in cm
+D=3;// Diameter in mm
+V=100;// cm^3
+t=152;// s
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+Q=(V*10^-6)/t;// The flow rate in m^3/s
+nu=((%pi*((D*10^-3)^4)*g)/(128*Q))*(1+(H/L));// The kinematic viscosity of the oil mixture in m/s^2
+printf("\nThe kinematic viscosity of the oil mixture is %1.3e m^2/s",nu);
+// The answer provided in the text book is wrong
diff --git a/3785/CH6/EX6.9/Ex6_9.sce b/3785/CH6/EX6.9/Ex6_9.sce new file mode 100644 index 000000000..9bf257ee8 --- /dev/null +++ b/3785/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,23 @@ +// Example 6_9
+clc;funcprot(0);
+// Given data
+a=1.5;// Radius in cm
+W=3;// Length in cm
+hbar=5*10^-5;// Clearance in m
+mu=2*10^-2;// Viscosity of lubricating oil in Pa s
+rho=9*10^2;// Density of lubricating oil in kg/m^3
+N=3600;// rpm
+n=0.5;// The eccentricity
+
+// Calculation
+// (a)
+omega=(2*%pi*N)/60;// s^-1
+L=(12*%pi*mu*omega*W*10^-2)*((a*10^-2)^3/(hbar)^2)*(n/((sqrt(1-n^2))*(2+n^2)));// The load force in N
+// (b)
+T=(4*%pi*mu*omega*W*10^-2)*((a*10^-2)^3/(hbar))*((1+(2*n^2))/((sqrt(1-n^2))*(2+n^2)));// The torque in Nm
+P=omega*T;// Power in W
+// (c)
+Re_h=(rho*omega*a*10^-2*hbar*(1-n^2))/(mu*(2+n^2));// Reynolds number
+printf("\n(a)The maximum load F=%1.3e N \n(b)The torque,T=%0.4f Nm \n The frictional power of the bearing,P=%2.2f W \n(c)The reynolds number,Re_h=%2.2f",L,T,P,Re_h);
+Re_h=((a*10^-2)/hbar)
+// The answer provided in the text book is wrong
diff --git a/3785/CH7/EX7.1/Ex7_1.sce b/3785/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..d65b2e7a8 --- /dev/null +++ b/3785/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,27 @@ +// Example 7_1
+clc;funcprot(0);
+// Given data
+D=6;// The diameter of a steel pipe in inch
+Q=2000;// Volume flow rate in gpm
+L=1.0;// Length in km
+nu=1.0*10^-6;// Kinematic viscosity in m^2/s
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+// (a)
+D=D*2.54*10^-2;// m
+Q=(Q*3.782*10^-3)/60;// m^3/s
+Vbar=(4*Q)/(%pi*D^2);// m/s
+Re_D=(Vbar*D)/nu;// Reynolds number
+// (b)
+epsilon=5*10^-5;// physical height in m
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+dp=f*((1/2)*rho*Vbar^2)*((L*10^3)/D);// The pressure drop in Pa
+P=dp*Q;// The power required to maintain the flow in W
+printf("\n(a)Re_D=%1.3e.The How is turbulent since the Reynolds number exceeds the transition value of 2300. \n(b)The pressure drop,deltap=%1.3e Pa \n(c)The power required to maintain the flow,P=%1.3e W",Re_D,dp,P);
+// The answer is varied due to round off error
diff --git a/3785/CH7/EX7.3/Ex7_3.sce b/3785/CH7/EX7.3/Ex7_3.sce new file mode 100644 index 000000000..c59def7cf --- /dev/null +++ b/3785/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,19 @@ +// Example 7_3
+clc;funcprot(0);
+// Given data
+L=100;// The length of the ship in m
+A=3*10^3;// Surface area in m^2
+rho=1.03*10^3;// The density of sea water in kg/m^3
+V=8;// Speed in m/s
+epsilon=1*10^-4;// The surface roughness in m
+nu=1*10^-6;// The kinematic viscosity in m^2/s
+
+// Calculation
+Re_L=(V*L)/nu;// The length Reynolds number Re_L
+// If the ship surface were smooth,
+C_D_fp=0.455/(log10(Re_L))^2.58;// The drag coefficient
+// For a rough surface,
+C_D_fp=0.30/(log10(14.7*(L/epsilon))^2.5);// The drag coefficient for a rough surface
+D=((1/2)*rho*V^2)*A*C_D_fp;// The ship's frictional drag force in N
+P=D*V;// The power in MW
+printf("\nThe ships frictional drag force,D=%1.4e N \nThe power required to overcome drag force,DV=%1.3f MW",D,P/10^6);
diff --git a/3785/CH7/EX7.4/Ex7_4.sce b/3785/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..7702c9253 --- /dev/null +++ b/3785/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,14 @@ +// Example 7_4
+clc;funcprot(0);
+// Given data
+z_1=1;// m
+z_2=10;// m
+k=0.4;// The von Karman constant
+ubar_1=6;// m/s
+ubar_2=9;// m/s
+
+// Calculation
+ustar=(ubar_2-ubar_1)/(2.5*log(10));// m/s
+y_0=10/exp(ubar_2/(2.5*ustar));// m
+C_f=(2*ustar^2)/ubar_2^2;// The friction coefficient
+printf("\nu_*=%0.3f m/s \ny_0=%1.2e m \nThe friction coefficient,C_f=%1.2e",ustar,y_0,C_f);
diff --git a/3785/CH7/EX7.6/Ex7_6.sce b/3785/CH7/EX7.6/Ex7_6.sce new file mode 100644 index 000000000..04609301b --- /dev/null +++ b/3785/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,13 @@ +// Example 7_6
+clc;funcprot(0);
+// Given data
+x=40;// Fixed distance in m
+V_v=100;// Vehicular speed in m/s
+C_D=1.0;// The truck drag coefficient
+A=9;// The trucks frontal area in m^2
+alpha_w=0.05;
+
+// Calculation
+V_w=V_v*((C_D*A)/(%pi*(2*alpha_w*x)^2));// km/h
+dV=V_v-V_w;// The relative air speed in km/h
+printf("\nThe relative air speed,V_v-V_w=%2.1f km/h",dV)
diff --git a/3785/CH8/EX8.2/Ex8_2.sce b/3785/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..0aa710cc4 --- /dev/null +++ b/3785/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,14 @@ +// Example 8_2
+clc;funcprot(0);
+// Given data
+D=2;// The diameter of the pipe in inch
+h_in=10;// Elevation in m
+Q=425;// The volumetric flow rate in gal/min
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+D=D*2.54*10^-2;// m
+Q=(Q*3.785*10^-3)/60;// The volumetric flow rate in m^3/s
+V=(4*Q)/(%pi*D^2);// m/s
+deltah=h_in-(V^2/(2*g));// m
+printf("The reduction in head,h_in-h_out=%1.3f m",deltah);
diff --git a/3785/CH8/EX8.3/Ex8_3.sce b/3785/CH8/EX8.3/Ex8_3.sce new file mode 100644 index 000000000..66878ce60 --- /dev/null +++ b/3785/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,12 @@ +// Example 8_3
+clc;funcprot(0);
+// Given data
+P=8*10^6;// The mechanical power delivered to an electric generator in MW
+deltah=10;// The change in head between the turbine inlet and outlet in m
+Q=100;// The volumetric flow rate in m^3/s
+rho=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+n_t=(P/(rho*g*Q*deltah))*100;// The turbine efficiency in %
+printf(" The turbine efficiency n_t=%2.2f percentage",n_t);
diff --git a/3785/CH8/EX8.4/Ex8_4.sce b/3785/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..f14162dd1 --- /dev/null +++ b/3785/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,15 @@ +// Example 8_4
+clc;funcprot(0);
+// Given data
+lambda=4;// W/mK
+// From example 6.4
+mu=6.7*10^-5;// Pa s
+V=18.85;// m/s
+h=1*10^-4;// m
+
+// Calculation
+// (a)
+q_w=-(mu)*((V^2)/h);// The heat flux to the wall (y =0) for the bearing in W/m^2
+// (b)
+deltaT=(mu/lambda)*((V^2)/(2*h));// The temperature difference T_h-T_o across the oil gap in K
+printf("\n(a)The heat flux to the wall (y =0) for the bearing,q_w=%1.3e W/m^2 \n(b)The temperature difference T_h-T_o across the oil gap is %2.2f K",q_w,deltaT);
diff --git a/3785/CH8/EX8.5/Ex8_5.sce b/3785/CH8/EX8.5/Ex8_5.sce new file mode 100644 index 000000000..49dda9167 --- /dev/null +++ b/3785/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,12 @@ +// Example 8_5
+clc;funcprot(0);
+// Given data
+Q=5;// The flow rate of water through a pipe in gal/min
+q=10*10^3;// kW
+c_p=4.18;// The specific heat in J/kg.K
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+Q=(Q*3.785*10^-3)/60;// The flow rate of water through a pipe in m^3/s
+deltaT=q/(rho*Q*c_p*10^3);// The temperature rise in the water in K
+printf("The temperature rise in the water,T_out-T_in=%1.3f K",deltaT);
diff --git a/3785/CH9/EX9.1/Ex9_1.sce b/3785/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..45e115cf9 --- /dev/null +++ b/3785/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,32 @@ +// Example 9_1
+clc;funcprot(0);
+// Given data
+D=8;// The diameter of the steel pipe in inch
+z_in=100;// Elevation in m
+z_out=22;// Elevation in m
+L=2.2;// The distance in km
+Q=1000;// The flow rate in m^3/s
+g=9.807;// The acceleration due to gravity in m/s^2
+nu=1.0*10^-6;// The kinematic viscosity in m/s^2
+rho=1*10^3;// The density of water in kg/m^3
+
+// Calculation
+// (a)
+D=D*2.54*10^-2;// m
+Q=Q*(3.782*10^-3)/60;// m^3/s
+V=(4*Q)/(%pi*D^2);// m/s
+Re_D=(V*D)/nu;// Reynolds number
+epsilon=5*10^-5;// physical height
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+
+K_f=f*((L*10^3)/D);// The head loss coefficient
+// (b)
+deltah_f=K_f*((V^2)/(2*g));// The head loss in m
+// (c)
+dp=(deltah_f-(z_in-z_out))*rho*g;// The static pressure change between the pipe inlet and outlet
+printf("\n(a)The head loss coefficient,K_f=%1.3e \n(b)The head loss,deltah_f=%2.2f \n(c)The static pressure change between the pipe inlet and outlet,p_in-p_out=%1.3e Pa",K_f,deltah_f,dp);
diff --git a/3785/CH9/EX9.2/Ex9_2.sce b/3785/CH9/EX9.2/Ex9_2.sce new file mode 100644 index 000000000..5f170284f --- /dev/null +++ b/3785/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,23 @@ +// Example 9_2
+clc;funcprot(0);
+// Given data
+// From Example 9_1
+D=8;// The diameter of the steel pipe in inch
+z_in=100;// Elevation in m
+z_out=22;// Elevation in m
+L=2.2;// The distance in km
+g=9.807;// The acceleration due to gravity in m/s^2
+nu=1.0*10^-6;// The kinematic viscosity in m/s^2
+rho=1*10^3;// The density of water in kg/m^3
+dp=0;// The static pressure in Pa
+
+// Calculation
+D=D*2.54*10^-2;// m
+deltah_f=(dp/(rho*g))+(z_in-z_out);// m
+// From equation 9.9
+sqrtoffintoRe_D=((2*g*deltah_f*D^3)/(((nu)^2)*L*10^3))^(1/2);
+epsilon=5*10^-5;// physical height in m
+Re_D=-2*sqrtoffintoRe_D*log10(((epsilon/D)/3.7)+(2.51/(sqrtoffintoRe_D)));// Reynolds number
+Q=(%pi*D*nu*Re_D)/4;// The volume flow rate in m^3/s
+Q=(Q*60)/(3.782*10^-3)// The volume flow rate in gal/min
+printf("The volume flow rate,Q=%4.0f gal/min",Q);
diff --git a/3785/CH9/EX9.3/Ex9_3.sce b/3785/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..d39b27e27 --- /dev/null +++ b/3785/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,48 @@ +// Example 9_3
+clc;funcprot(0);
+// Given data
+dp=100;// The pressure drop in psi
+rho=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+Q=2000;// The flow rate of water in gal/min
+D=4;// The next pipe size in inch
+L=100;// Length in m
+nu=1*10^-6;// m^2/s
+
+// Calculation
+deltah=(dp*6.895*10^3)/(rho*g);// m
+printf("\nh_in-h_out=%2.2f m",deltah);
+D=D*2.54*10^-2;// m
+Q=Q*(3.782*10^-3)/60;// m^3/s
+V=(4*Q)/(%pi*D^2);// m/s
+Re_D=(V*D)/nu;// Reynolds number
+epsilon=5*10^-5;// physical height
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+deltah_f=K_f*((V^2)/(2*g));// The head loss in m
+printf("\nD=%0.4f m \nQ=%1.3e m^3/s \nV=%2.2f m/s \nRe_D=%1.3e \nf=%1.3e \nK_f=%2.2f \nh_in-h_out=%3.1f m",D,Q,V,Re_D,f,K_f,deltah_f)
+printf("\nThe head loss of 205.9 m is greater than the allowable los s of 70.31 m.");
+
+// If we try the next size pipe, D = 6 in,
+D=6;// inch
+D=D*2.54*10^-2;// m
+Q=2000;// The flow rate of water in gal/min
+Q=Q*(3.782*10^-3)/60;// m^3/s
+V=(4*Q)/(%pi*D^2);// m/s
+Re_D=(V*D)/nu;// Reynolds number
+epsilon=5*10^-5;// physical height
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+deltah_f=K_f*((V^2)/(2*g));// The head loss in m
+printf("\nD=%0.4f m \nQ=%1.3e m^3/s \nV=%1.3f m/s \nRe_D=%1.3e \nf=%1.3e \nK_f=%2.2f \nh_in-h_out=%2.2f m",D,Q,V,Re_D,f,K_f,deltah_f)
+printf("\nThis is smaller than the allowable head loss so that a 6 in diameter pipe is acceptable.")
diff --git a/3785/CH9/EX9.4/Ex9_4.sce b/3785/CH9/EX9.4/Ex9_4.sce new file mode 100644 index 000000000..1ad0ec16d --- /dev/null +++ b/3785/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,31 @@ +// Example 9_4
+clc;funcprot(0);
+// Given data
+l=6;// in
+b=12;// in
+A=6*12;// in^2
+L=20;// Length in ft
+Q=1000;// ft^3/min
+epsilon=1*10^-5;// The duct roughness in m
+nu=1.51*10^-5;// m/s
+rho=1.204;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+
+// Calculation
+D=(4*(l*b))/(2*(l+b));// m
+D=D*2.54*10^-2;// m
+Q=Q*(2.832*10^-2)/60;// m^3/s
+A=A*(2.54*10^-2)^2;// m^2
+V=Q/A;// m/s
+L=L*0.3048;// m
+Re_D=(V*D)/nu;// Reynolds number
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+deltah=K_f*((V^2)/(2*g));// The head loss in m
+dp=rho*g*deltah;// The pressure drop in Pa
+printf("\nThe head loss,h_in-h_out=%1.3f m \nThe pressure drop,p*_in-p*_out=%2.2f Pa",deltah,dp);
diff --git a/3785/CH9/EX9.5/Ex9_5.sce b/3785/CH9/EX9.5/Ex9_5.sce new file mode 100644 index 000000000..d470102ba --- /dev/null +++ b/3785/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,26 @@ +// Example 9_5
+clc;funcprot(0);
+// Given data
+D=1;// Diameter in cm
+L=22;// The lngth of a copper tube in m
+Q=4;// The water flow rate in the circuit in l/min
+g=9.807;// The acceleration due to gravity in m/s^2
+nu=1.0*10^-6;// The kinematic viscosity in m/s^2
+
+// Calculation
+D=1*10^-2;// m
+Q=(Q*1*10^-3)/60;// m^3/s
+A=(%pi*(D)^2)/4;// m^2
+V=Q/A;// m/s
+Re_D=(V*D)/nu;// Reynolds number
+epsilon=1*10^-6;// Roughness in m
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+SigmaK=11.4;
+deltah=(K_f+SigmaK)*((V^2)/(2*g));// The total head loss in m
+printf("The total head loss,deltah=%1.2f m",deltah);
diff --git a/3785/CH9/EX9.6/Ex9_6.sce b/3785/CH9/EX9.6/Ex9_6.sce new file mode 100644 index 000000000..7e77b6b01 --- /dev/null +++ b/3785/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,33 @@ +// Example 9_6
+clc;funcprot(0);
+// Given data
+L=1.5;// The length in km
+D=6;// Diameter in inch
+h=80;// m
+// Assume
+deltah_l=20;// m
+g=9.807;// The acceleration due to gravity in m/s^2
+nu=1*10^-6;// m/s^2
+epsilon=5*10^-5;// roughness in m
+
+// Calculation
+D=D*2.54*10^-2;// m
+sqrtoffintoRe_D=((2*g*deltah_l*D^3)/(((nu)^2)*L*10^3))^(1/2);
+Re_D=-2*sqrtoffintoRe_D*log10(((epsilon/D)/3.7)+(2.51/(sqrtoffintoRe_D)));// Reynolds number
+Q=(%pi*D*nu*Re_D)/4;// The volume flow rate in m^3/s
+Q_20=(Q*60)/(3.782*10^-3)// The volume flow rate in gal/min
+deltah=150*(1-(Q_20/1000)^2);// m
+dh_20=deltah-(h+deltah_l);// m
+deltah_l=40;// m
+sqrtoffintoRe_D=((2*g*deltah_l*D^3)/(((nu)^2)*L*10^3))^(1/2);
+Re_D=-2*sqrtoffintoRe_D*log10(((epsilon/D)/3.7)+(2.51/(sqrtoffintoRe_D)));// Reynolds number
+Q=(%pi*D*nu*Re_D)/4;// The volume flow rate in m^3/s
+Q_40=(Q*60)/(3.782*10^-3)// The volume flow rate in gal/min
+deltah=150*(1-(Q_40/1000)^2);// m
+dh_40=deltah-(h+deltah_l);// m
+Q=((((dh_20)/(dh_20-dh_40))*(Q_40-Q_20))+Q_20);// GPM
+deltah=150*(1-(Q/1000)^2);// m
+deltah_l=deltah-h;// m
+printf("\nThe flow rate through the system,Q=%3.1f GPM \ndeltah=%3.1f m \ndeltah_l=%2.2f m",Q,deltah,deltah_l);
+printf("\nContinuing this process to improve our estimate of Q and Ah we finally arrive at:Q=527.7(GPM);deltah=108.3 m")
+
diff --git a/3785/CH9/EX9.7/Ex9_7.sce b/3785/CH9/EX9.7/Ex9_7.sce new file mode 100644 index 000000000..b9a0cb27b --- /dev/null +++ b/3785/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,29 @@ +// Example 9_7
+clc;funcprot(0);
+// Given data
+h=100;
+Q=10;
+n_t=.85;
+D=1.5;
+L=300;
+delta_t=93.99;
+epsilon=1*10^-4;
+nu=1.0*10^-6;// The kinematic viscosity in m/s^2
+rho=1*10^3;// The density of water in kg/m^3
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+V=(4*Q)/(%pi*D^2);// m/s
+Re_D=(V*D)/nu;// Reynolds number
+function[X]=frictionfactor(y)
+ X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+SigmaK=3.681;
+deltah_1=SigmaK*((V^2)/(2*g));// The head loss in m
+P=n_t*(rho*Q)*g*deltah_1;
+P=P/10^3;
+printf("\nThe head loss in the piping,deltah_1=%1.3f m \nThe power produced by the turbine,P=%3.0f kW",deltah_1,P);
diff --git a/3785/CH9/EX9.8/Ex9_8.sce b/3785/CH9/EX9.8/Ex9_8.sce new file mode 100644 index 000000000..c39ef76ed --- /dev/null +++ b/3785/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,46 @@ +// Example 9_8
+clc;funcprot(0);
+// Given data
+L=50;// Lengths of garden hose in ft
+D_A=3/4;// Diameter of hose A in inch
+D_B=1/2;// Diameter of hose B in inch
+p=40;// Pressure in the tank in psig
+nu=1.0*10^-6;// The kinematic viscosity in m/s^2
+rho=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+epsilon=0;
+
+// Calculation
+D_A=D_A*2.54*10^-2;// m
+D_B=D_B*2.54*10^-2;// m
+L=L*0.3048;// m
+deltah_l1=(p*6.895*10^3)/(rho*g);// m
+deltah_A1=10;// m
+deltah_B1=18.12;// m
+sqrtoffintoRe_D_A=((2*g*deltah_A1*D_A^3)/(((nu)^2)*L))^(1/2);
+Re_D_A=-2*sqrtoffintoRe_D_A*log10(2.51/(sqrtoffintoRe_D_A));// Reynolds number
+Q_A1=(%pi*D_A*nu*Re_D_A)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_B=((2*g*deltah_B1*D_B^3)/(((nu)^2)*L))^(1/2);
+Re_D_B=-2*sqrtoffintoRe_D_B*log10((2.51/(sqrtoffintoRe_D_B)));// Reynolds number
+Q_B1=(%pi*D_B*nu*Re_D_B)/4;// The volume flow rate in m^3/s
+V_A=(4*Q_A1)/(%pi*D_A^2);// m/s
+V_B=(4*Q_B1)/(%pi*D_B^2);// m/s
+// Assume deltah=SigmaK*((V^2)/(2*g))
+deltah=((0.4*V_A^2)+(0.4*V_B^2))/(2*g);// m
+deltah_f=deltah_l1-deltah;// m
+// We decide to allocate this total to
+deltah_A2=2;// m
+deltah_B2=25.43;// m
+sqrtoffintoRe_D_A=((2*g*deltah_A2*D_A^3)/(((nu)^2)*L))^(1/2);
+Re_D_A=-2*sqrtoffintoRe_D_A*log10((2.51/(sqrtoffintoRe_D_A)));// Reynolds number
+Q_A2=(%pi*D_A*nu*Re_D_A)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_B=((2*g*deltah_B2*D_B^3)/(((nu)^2)*L))^(1/2);
+Re_D_B=-2*sqrtoffintoRe_D_B*log10((2.51/(sqrtoffintoRe_D_B)));// Reynolds number
+Q_B2=(%pi*D_B*nu*Re_D_B)/4;// The volume flow rate in m^3/s
+V_A=(4*Q_A2)/(%pi*D_A^2);// m/s
+V_B=(4*Q_B2)/(%pi*D_B^2);// m/s
+deltah_l2=((0.4*V_A^2)+(0.4*V_B^2))/(2*g);// m
+//Indicating the first and second guesses by '1' and '2' we find a third guess to be:
+deltah=deltah_A2-((Q_A2-Q_B2)*((deltah_A1-deltah_A2)/((Q_A1-Q_B1)-(Q_A2-Q_B2))));// m
+printf('\nThe flow rate through the hoses Q_A=%1.3e m^3/s;Q_B=%1.3e m^3/s;SigmaK(V^2/2g)=%0.4f m',Q_A2,Q_B2,deltah_l2);
+// The answer is vary due to roundoff error
diff --git a/3785/CH9/EX9.9/Ex9_9.sce b/3785/CH9/EX9.9/Ex9_9.sce new file mode 100644 index 000000000..f99179a03 --- /dev/null +++ b/3785/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,62 @@ +// Example 9_9
+clc;funcprot(0);
+// Given data
+D=1*10^-2;// m
+h_1=10;// m
+h_4=0;// m
+L_12=3;// m
+L_13=4;// m
+L_14=5;// m
+g=9.807;// The acceleration due to gravity in m/s^2
+nu=1.0*10^-6;// The kinematic viscosity in m/s^2
+
+// Calculation
+// Because of the symmetry of the network. h1- h2 = h3-h4,Q12 = Q34 and Q13 = Q24.
+// Assume
+h_2a=5;// m
+h_3a=5;// m
+deltah_12=h_1-h_2a;// m
+deltah_13=h_1-h_3a;// m
+sqrtoffintoRe_D_12=((2*g*deltah_12*D^3)/(((nu)^2)*L_12))^(1/2);
+Re_D_12=-2*sqrtoffintoRe_D_12*log10((2.51/(sqrtoffintoRe_D_12)));// Reynolds number
+Q_12=(%pi*D*nu*Re_D_12)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_13=((2*g*deltah_13*D^3)/(((nu)^2)*L_13))^(1/2);
+Re_D_13=-2*sqrtoffintoRe_D_13*log10((2.51/(sqrtoffintoRe_D_13)));// Reynolds number
+Q_13=(%pi*D*nu*Re_D_13)/4;// The volume flow rate in m^3/s
+Q_23=0;// The volume flow rate in m^3/s
+Q_24=Q_13;// The volume flow rate in m^3/s
+deltaQ_2a=Q_12-Q_23-Q_24;// m^3/s
+// Assume
+h_2b=6;// m
+h_3b=4;// m
+deltah_12=4;// m
+deltah_13=6;// m
+deltah_23=2;// m
+sqrtoffintoRe_D_12=((2*g*deltah_12*D^3)/(((nu)^2)*L_12))^(1/2);
+Re_D_12=-2*sqrtoffintoRe_D_12*log10((2.51/(sqrtoffintoRe_D_12)));// Reynolds number
+Q_12=(%pi*D*nu*Re_D_12)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_13=((2*g*deltah_13*D^3)/(((nu)^2)*L_13))^(1/2);
+Re_D_13=-2*sqrtoffintoRe_D_13*log10((2.51/(sqrtoffintoRe_D_13)));// Reynolds number
+Q_13=(%pi*D*nu*Re_D_13)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_23=((2*g*deltah_23*D^3)/(((nu)^2)*1))^(1/2);
+Re_D_23=-2*sqrtoffintoRe_D_23*log10((2.51/(sqrtoffintoRe_D_23)));// Reynolds number
+Q_23=(%pi*D*nu*Re_D_23)/4;// The volume flow rate in m^3/s
+deltaQ_2b=Q_12-Q_23-Q_24;// m
+h_2=h_2a-(((h_2b-h_2a)/(deltaQ_2b-deltaQ_2a))*deltaQ_2b);// m
+// Proceeding in this manner for two more iterations, we converge to the solution:
+h_2=5.11;// m
+h_3=4.89;// m
+deltah_12=h_1-h_2;// m
+deltah_13=h_1-h_3;// m
+deltah_23=h_2-h_3;// m
+sqrtoffintoRe_D_12=((2*g*deltah_12*D^3)/(((nu)^2)*L_12))^(1/2);
+Re_D_12=-2*sqrtoffintoRe_D_12*log10((2.51/(sqrtoffintoRe_D_12)));// Reynolds number
+Q_12=(%pi*D*nu*Re_D_12)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_13=((2*g*deltah_13*D^3)/(((nu)^2)*L_13))^(1/2);
+Re_D_13=-2*sqrtoffintoRe_D_13*log10((2.51/(sqrtoffintoRe_D_13)));// Reynolds number
+Q_13=(%pi*D*nu*Re_D_13)/4;// The volume flow rate in m^3/s
+sqrtoffintoRe_D_23=((2*g*deltah_23*D^3)/(((nu)^2)*1))^(1/2);
+Re_D_23=-2*sqrtoffintoRe_D_23*log10((2.51/(sqrtoffintoRe_D_23)));// Reynolds number
+Q_23=(%pi*D*nu*Re_D_23)/4;// The volume flow rate in m^3/s
+Q_2=Q_13+Q_12;// m^3/s
+printf("\nThe flow rate,Q=%1.3e m^3/s",Q_2);
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