diff options
Diffstat (limited to '629')
243 files changed, 2422 insertions, 0 deletions
diff --git a/629/CH1/EX1.4/ex1_4.txt b/629/CH1/EX1.4/ex1_4.txt new file mode 100644 index 000000000..99a1f7dbd --- /dev/null +++ b/629/CH1/EX1.4/ex1_4.txt @@ -0,0 +1,4 @@ +(a)The thrust force in newtons = 270 N
+
+(b)The thrust force in units of pounds-force = 60.5 lbf.
+
\ No newline at end of file diff --git a/629/CH1/EX1.4/example1_4.sce b/629/CH1/EX1.4/example1_4.sce new file mode 100644 index 000000000..69fa3d928 --- /dev/null +++ b/629/CH1/EX1.4/example1_4.sce @@ -0,0 +1,15 @@ +clear
+clc
+//Example 1.4 GRID METHOD APPLIED TO A ROCKET
+//Part(a)
+m=9; //mass flow rate[Kg/s]
+V=30; //velocity[m/s]
+//Thrust force
+T=m*V //[N]
+printf("\n(a)The thrust force in newtons = %.f N\n",T)
+//Part(b)
+m=19.8; //[lbm/s]
+V=98.4; //[ft/s]
+//1lbf.s^2=32.2(lbm.ft)
+T=m*V/32.2 //[lbf]
+printf("\n(b)The thrust force in units of pounds-force = %.1f lbf.\n",T)
\ No newline at end of file diff --git a/629/CH10/EX10.1/ex10_1.txt b/629/CH10/EX10.1/ex10_1.txt new file mode 100644 index 000000000..d754d835f --- /dev/null +++ b/629/CH10/EX10.1/ex10_1.txt @@ -0,0 +1,9 @@ +
+(a)Since Re= 3352 > 3000, the flow is turbulent.
+
+The entrance length for air = 0.25 m.
+
+(b)Since, Re= 1787 < 2000, the flow is laminar.
+
+The entrance length for water = 0.447 m.
+
\ No newline at end of file diff --git a/629/CH10/EX10.1/example10_1.sce b/629/CH10/EX10.1/example10_1.sce new file mode 100644 index 000000000..1ad20a100 --- /dev/null +++ b/629/CH10/EX10.1/example10_1.sce @@ -0,0 +1,21 @@ +clear
+clc
+//Example 10.1 CLASSIFYING FLOW IN CONDUITS
+D=0.005; //[m]
+//For Air
+Va=12; //velocity[m/s]
+va=1.79*10^-5; //viscosity[m^2/s]
+//Reynolds number
+Re_a=Va*D/va
+printf("\n(a)Since Re= %.f > 3000, the flow is turbulent.\n",Re_a)
+//Entrance length
+Le_a=50*D //[m]
+printf("\nThe entrance length for air = %.2f m.\n",Le_a)
+
+//For Water
+m=0.008; //mass flow rate[kg/s]
+mu=1.14*10^-3; //[N.s/m^2]
+Re_w=4*m/(%pi*D*mu)
+printf("\n(b)Since, Re= %.f < 2000, the flow is laminar.\n",Re_w)
+Le_w=0.05*Re_w*D //[m]
+printf("\nThe entrance length for water = %.3f m.\n",Le_w)
\ No newline at end of file diff --git a/629/CH10/EX10.2/ex10_2.txt b/629/CH10/EX10.2/ex10_2.txt new file mode 100644 index 000000000..190dfc8f5 --- /dev/null +++ b/629/CH10/EX10.2/ex10_2.txt @@ -0,0 +1,2 @@ +
+The head loss per 100m length of the pipe = 9.84 m.
\ No newline at end of file diff --git a/629/CH10/EX10.2/example10_2.sce b/629/CH10/EX10.2/example10_2.sce new file mode 100644 index 000000000..1b824367f --- /dev/null +++ b/629/CH10/EX10.2/example10_2.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 10.2 HEAD LOSS FOR LAMINAR FLOW
+g=9.81; //[m/s^2]
+L=100; //[m]
+D=0.15; //diameter[m]
+A=%pi*D^2/4 //area[m^2]
+Q=0.02 ;//[m^3/s]
+v=6*10^-4; //[m^2/s]
+V=Q/A //[m/s]
+//Reynolds number
+Re=V*D/v
+//Re<2000, the flow is laminar.
+//Head loss(laminar flow)
+hf=32*v*L*V/(g*D^2) //[m]
+printf("\nThe head loss per 100m length of the pipe = %.2f m.\n",hf)
\ No newline at end of file diff --git a/629/CH10/EX10.3/ex10_3.txt b/629/CH10/EX10.3/ex10_3.txt new file mode 100644 index 000000000..263bb6186 --- /dev/null +++ b/629/CH10/EX10.3/ex10_3.txt @@ -0,0 +1,2 @@ +
+The head loss per kilometer length of the pipe = 12.3 m.
\ No newline at end of file diff --git a/629/CH10/EX10.3/example10_3.sce b/629/CH10/EX10.3/example10_3.sce new file mode 100644 index 000000000..ca97b526f --- /dev/null +++ b/629/CH10/EX10.3/example10_3.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 10.3 HEAD LOSS IN A PIPE (CASE 1)
+g=9.81; //[m/s^2]
+L=1000; //[m]
+D=0.20; //diameter[m]
+A=%pi*D^2/4 //area[m^2]
+Q=0.05 ;//[m^3/s]
+v=10^-6; //[m^2/s]
+V=Q/A //[m/s]
+//Reynolds number
+Re=V*D/v
+ks=0.12*10^-3; //[m]
+//Relative roughness
+Rr=ks/D
+//From Moody diagram for Re and Rr,
+f=0.019;
+//Darcy-Weisbach equation
+hf=f*(L/D)*(V^2/(2*g)) //[m]
+printf("\nThe head loss per kilometer length of the pipe = %.1f m.\n",hf)
\ No newline at end of file diff --git a/629/CH10/EX10.4/ex10_4.txt b/629/CH10/EX10.4/ex10_4.txt new file mode 100644 index 000000000..a0e3a5eb8 --- /dev/null +++ b/629/CH10/EX10.4/ex10_4.txt @@ -0,0 +1,2 @@ +
+The flow rate of water through the pipe = 0.05 m^3/s.
\ No newline at end of file diff --git a/629/CH10/EX10.4/example10_4.sce b/629/CH10/EX10.4/example10_4.sce new file mode 100644 index 000000000..f8c2217d4 --- /dev/null +++ b/629/CH10/EX10.4/example10_4.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 10.4 FLOW RATE IN A PIPE (CASE 2)
+g=9.81; //[m/s^2]
+L=1000; //[m]
+D=0.20; //diameter[m]
+A=%pi*D^2/4 //area[m^2]
+v=10^-6; //viscosity[m^2/s]
+hf=12.2; //[m]
+P=D^(3/2)*sqrt(2*g*hf/L)/v
+ks=0.12; //[mm]
+//Relative roughness
+Rr=ks/D
+//From Moody diagram for P and Rr
+f=0.019; //friction factor
+//Darcy-Weisbach equation
+V=sqrt(hf*2*g*(D/L)/f) //[m/s]
+//Discharge
+Q=V*A //[m^3/s]
+printf("\nThe flow rate of water through the pipe = %.2f m^3/s.\n",Q)
\ No newline at end of file diff --git a/629/CH10/EX10.7/ex10_7.txt b/629/CH10/EX10.7/ex10_7.txt new file mode 100644 index 000000000..1e18b6b30 --- /dev/null +++ b/629/CH10/EX10.7/ex10_7.txt @@ -0,0 +1,3 @@ +
+The elevation of the oil surface in the upper reservoir = 136 m.
+
\ No newline at end of file diff --git a/629/CH10/EX10.7/example10_7.sce b/629/CH10/EX10.7/example10_7.sce new file mode 100644 index 000000000..d3a64612b --- /dev/null +++ b/629/CH10/EX10.7/example10_7.sce @@ -0,0 +1,25 @@ +clear
+clc
+//Example 10.7 PIPE SYSTEM WITH COMBINED HEAD LOSS
+//From energy equation, z1=z2+hL
+z2=130; //[m]
+g=9.81; //[m/s^2]
+L=197; //[m]
+D=0.15; //diameter[m]
+A=%pi*D^2/4; //area[m^2]
+Q=0.028; //[m^3/s]
+V=Q/A //[m/s]
+v=4*10^-5; //[m^2/s]
+//Reynolds number
+Re=V*D/v
+//Re>3000, flow is turbulent
+//Minor head coefficients
+Ke=0.5; //entrance
+Kb=0.19; //bend
+KE=1.0; //outlet
+//Swamee-Jain equation, ks/D is neglected
+f=0.25/(log10(5.74/Re^0.9))^2
+//head loss
+hL=(V^2/(2*g))*(f*L/D+2*Kb+Ke+KE)
+z1=z2+hL //[m]
+printf("\nThe elevation of the oil surface in the upper reservoir = %.f m.\n",z1)
diff --git a/629/CH10/EX10.8/ex10_8.txt b/629/CH10/EX10.8/ex10_8.txt new file mode 100644 index 000000000..95b2dabbc --- /dev/null +++ b/629/CH10/EX10.8/ex10_8.txt @@ -0,0 +1,2 @@ +
+The pressure drop in inches of water per 50 m of duct = 0.869 inch H2O.
\ No newline at end of file diff --git a/629/CH10/EX10.8/example10_8.sce b/629/CH10/EX10.8/example10_8.sce new file mode 100644 index 000000000..7a055baa3 --- /dev/null +++ b/629/CH10/EX10.8/example10_8.sce @@ -0,0 +1,27 @@ +clear
+clc
+//Example 10.8 PRESSURE DROP IN AN HVAC DUCT
+g=9.81; //[m/s^2]
+rho=1.2; //[kg/m^3]
+L=50; //[m]
+b=0.6; //[m]
+h=0.3; //[m]
+A=b*h //area of cross section [m^2]
+Q=2.5; //[m^3/s]
+V=Q/A //[m/s]
+v=15.1*10^-6; //[m^2/s]
+//Hydraulic perimeter
+Dh=4*A/(2*(b+h)) //[m]
+//Reynolds number
+Re=V*Dh/v
+//Thus, flow is turbulent
+ks=0.000046;
+//Relative roughness
+Rr=ks/Dh
+//From Moody diagram for Re and Rr,
+f=0.015;
+//Darcy-Weisbach equation
+hf=f*(L/Dh)*(V^2/(2*g)) //[m]
+//1 inch H20=249.7 Pa
+delp=rho*g*hf/249.7 //pressure drop in inch H2O
+printf("\nThe pressure drop in inches of water per 50 m of duct = %.3f inch H2O.\n",delp)
\ No newline at end of file diff --git a/629/CH11/EX11.1/ex11_1.txt b/629/CH11/EX11.1/ex11_1.txt new file mode 100644 index 000000000..e81c0d9dc --- /dev/null +++ b/629/CH11/EX11.1/ex11_1.txt @@ -0,0 +1,4 @@ +
+The drag force acting on the cylinder = 1323 N.
+
+The bending moment at the base of the cylinder = 19845 N.m
\ No newline at end of file diff --git a/629/CH11/EX11.1/example11_1.sce b/629/CH11/EX11.1/example11_1.sce new file mode 100644 index 000000000..f03cac6b8 --- /dev/null +++ b/629/CH11/EX11.1/example11_1.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 11.1 DRAG FORCE ON A CYLINDER
+Vo=35; //speed of air [m/s]
+d=0.3; //[m]
+L=30; //[m]
+Ap=d*L //area[m^2]
+rho=1.2; //density[Kg/m^3]
+mu=1.81*10^-5; //[N.s/m^2]
+Re=Vo*d*rho/mu //Reynolds number
+//From fig. 11.4
+Cd=0.2; //coefficient of drag
+//Drag force
+Fd=(Cd*Ap*rho*Vo^2)/2 //[N]
+printf("\nThe drag force acting on the cylinder = %.f N.\n",Fd)
+//Moment at the base
+M=Fd*(L/2) //[N.m]
+printf("\nThe bending moment at the base of the cylinder = %.f N.m\n",M)
\ No newline at end of file diff --git a/629/CH11/EX11.2/ex11_2.txt b/629/CH11/EX11.2/ex11_2.txt new file mode 100644 index 000000000..7e3b7e9b9 --- /dev/null +++ b/629/CH11/EX11.2/ex11_2.txt @@ -0,0 +1,2 @@ +
+The drag of a 12 mm sphere = 0.00163 N.
\ No newline at end of file diff --git a/629/CH11/EX11.2/example11_2.sce b/629/CH11/EX11.2/example11_2.sce new file mode 100644 index 000000000..096067235 --- /dev/null +++ b/629/CH11/EX11.2/example11_2.sce @@ -0,0 +1,14 @@ +clear
+clc
+//Example 11.2 DRAG ON A SPHERE
+V=0.08; //speed[m/s]
+d=0.012; //[m]
+Ap=%pi*d^2/4 //area[m^2]
+rho=850; //density[Kg/m^3]
+mu=10^-1; //[N.s/m^2]
+Re=V*d*rho/mu //Reynolds number
+//From fig. 11.4
+Cd=5.3; //coefficient of drag
+//Drag force
+Fd=(Cd*Ap*rho*V^2)/2 //[N]
+printf("\nThe drag of a 12 mm sphere = %.5f N.\n",Fd)
\ No newline at end of file diff --git a/629/CH11/EX11.3/ex11_3.txt b/629/CH11/EX11.3/ex11_3.txt new file mode 100644 index 000000000..08447e511 --- /dev/null +++ b/629/CH11/EX11.3/ex11_3.txt @@ -0,0 +1,2 @@ +
+The speed of the cyclist, Vc = 9.12 m/s(= 20.4 mph).
\ No newline at end of file diff --git a/629/CH11/EX11.3/example11_3.sce b/629/CH11/EX11.3/example11_3.sce new file mode 100644 index 000000000..53b7cb536 --- /dev/null +++ b/629/CH11/EX11.3/example11_3.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 11.3 SPEED OF A BICYCLE RIDER
+m=70; //mass[Kg]
+g=9.81; //[m/s^2]
+Cr=0.007; //coefficient of rolling resistance
+//Rolling resistance
+Fr=Cr*m*g //[N]
+Cd=0.88;
+A=0.362; //[m^2]
+rho=1.2; //density [Kg/m^3]
+P=300; //power supply [W]
+//Drag force, Fd=Cd*A*rho*Vo^2/2, Vo=Vc+3
+//P=(Fd+Fr)*Vc
+q=[Cd*A*rho/2 3*Cd*A*rho (9*Cd*A*rho/2)+Fr -P]; //cubic polynomial in Vc
+R=roots(q); //roots of poly.q
+//R(3), real root of q
+Vc=R(3) //speed of cyclist [m/s]
+//1m/s=3600/(1.61*1000)mph
+printf("\nThe speed of the cyclist, Vc = %.2f m/s(= %.1f mph).\n",Vc,Vc*2.236)
\ No newline at end of file diff --git a/629/CH11/EX11.4/ex11_4.txt b/629/CH11/EX11.4/ex11_4.txt new file mode 100644 index 000000000..377ac0d81 --- /dev/null +++ b/629/CH11/EX11.4/ex11_4.txt @@ -0,0 +1,3 @@ +
+The terminal velocity Vo = 0.436 m/s.
+
\ No newline at end of file diff --git a/629/CH11/EX11.4/example11_4.sce b/629/CH11/EX11.4/example11_4.sce new file mode 100644 index 000000000..f1a09a482 --- /dev/null +++ b/629/CH11/EX11.4/example11_4.sce @@ -0,0 +1,35 @@ +clear
+clc
+//Example 11.4 TERMINAL VELOCITY OF A SPHERE IN WATER
+//To find Approx Value
+function [A]= approx (V,n)
+ A= round(V*10^n)/10^n; //V-Value, n-to what place
+ funcprot (0)
+endfunction
+d=0.02; //diameter [m]
+A=%pi*(d^2)/4 //area [m^2]
+Vol=%pi*(d^3)/6 //volume [m^3]
+v=10^-6; //viscosity [m^2/s]
+//Specific weights
+g_sphere=12.7*10^3; //[N/m^3]
+g_water=9.79*10^3; //[N/m^3]
+rho=998; //density [kg/m^3]
+//Force equilibrium, F_drag+F_buoyancy=W
+//F_drag=CD*A*rho*Vo^2/2
+W=g_sphere*Vol //weight [N]
+F_b=g_water*Vol //buoyant force [N]
+V(1)=0;
+//Assume initial value of Vo=1
+V(2)=1;
+//Iterate until Vo reaches a constant value
+for i=2:1:7 //say 6 iterations
+if(V(i)~=V(i-1))
+ Re=V(i)*d/v;
+ CD=24*(1+0.15*(Re^0.687))/Re +0.42/(1+4.25*10^4*Re^(-1.16));
+ V(i+1)=approx((2*(W-F_b)/(CD*rho*A))^0.5,3);
+else
+ Vo=V(i)
+ break;
+end
+end
+printf("\nThe terminal velocity Vo = %.3f m/s.\n",Vo)
\ No newline at end of file diff --git a/629/CH11/EX11.5/ex11_5.txt b/629/CH11/EX11.5/ex11_5.txt new file mode 100644 index 000000000..0b8b76cb8 --- /dev/null +++ b/629/CH11/EX11.5/ex11_5.txt @@ -0,0 +1,2 @@ +
+The ratio of drag forces for streamlined shape to cylinder shape = 0.17.
\ No newline at end of file diff --git a/629/CH11/EX11.5/example11_5.sce b/629/CH11/EX11.5/example11_5.sce new file mode 100644 index 000000000..bf12077b9 --- /dev/null +++ b/629/CH11/EX11.5/example11_5.sce @@ -0,0 +1,10 @@ +clear +clc +//Example 11.5 COMPARING DRAG ON BLUFF AND STREAMLINED SHAPES +Re=7*10^5; //Reynolds number +Cdc=0.2; //Cd for cylinder, from Ex.11.1 +//Interpolating Re for streamlined shape +Cds=0.034; +//Drag force ratio, from Eq.11.4 +Dfr=Cds/Cdc //(Fds/Fdc) +printf("\nThe ratio of drag forces for streamlined shape to cylinder shape = %.2f.\n",Dfr)
\ No newline at end of file diff --git a/629/CH11/EX11.6/ex11_6.txt b/629/CH11/EX11.6/ex11_6.txt new file mode 100644 index 000000000..1d9c72be8 --- /dev/null +++ b/629/CH11/EX11.6/ex11_6.txt @@ -0,0 +1,5 @@ +
+The lift force on the ball = 0.011 N and the direction of lift is downward.
+
+The drag force on the ball = 0.0271 N.
+
\ No newline at end of file diff --git a/629/CH11/EX11.6/example11_6.sce b/629/CH11/EX11.6/example11_6.sce new file mode 100644 index 000000000..260346ef6 --- /dev/null +++ b/629/CH11/EX11.6/example11_6.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 11.6 LIFT ON A ROTATING SPHERE
+rho=1.2; //density[Kg/m^3]
+d=0.03; //diameter [m]
+r=d/2; //radius [m]
+A=%pi*d^2/4 //area[m^2]
+Vo=10; //velocity [m/s]
+w=100*2*%pi //angular speed [rad/s]
+p=w*r/Vo //rotational parameter
+//From fig.11.17
+CL=0.26; //lift coefficient
+CD=0.64; //drag coefficient
+//Lift force
+FL=(rho*Vo^2*CL*A)/2 //[N]
+printf("\nThe lift force on the ball = %.3f N and the direction of lift is downward.\n",FL)
+//Drag force
+FD=(rho*Vo^2*CD*A)/2 //[N]
+printf("\nThe drag force on the ball = %.4f N.\n",FD)
\ No newline at end of file diff --git a/629/CH11/EX11.7/ex11_7.txt b/629/CH11/EX11.7/ex11_7.txt new file mode 100644 index 000000000..c2146dbad --- /dev/null +++ b/629/CH11/EX11.7/ex11_7.txt @@ -0,0 +1,2 @@ +
+The wing area = 394 ft^2 and the minimum induced drag = 86.1 lbf.
\ No newline at end of file diff --git a/629/CH11/EX11.7/example11_7.sce b/629/CH11/EX11.7/example11_7.sce new file mode 100644 index 000000000..781e79bdf --- /dev/null +++ b/629/CH11/EX11.7/example11_7.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 11.7 WING AREA FOR AN AIRPLANE
+T=-67+460; //temperature [°R]
+R=1716; //[ft.lbf/slug.°R]
+//1ft^2=144in^2
+p=3.3*144; //pressure[lbf/ft^2]
+rho=p/(R*T) //[slug/ft^3]
+CL=0.2;
+Vo=600; //[ft/s]
+W=10000; //weight [lbf]
+//for steady flight,
+FL=W
+//Wing area
+S=2*W/(rho*Vo^2*CL) //[ft^2]
+b=54; //span of wing [ft]
+CDi=CL^2/(%pi*(b^2/S)) //min.induced drag coefficient
+//Induced drag
+Di=(rho*Vo^2*CDi*S)/2 //[lbf]
+printf("\nThe wing area = %.f ft^2 and the minimum induced drag = %.1f lbf.\n",S,Di)
\ No newline at end of file diff --git a/629/CH11/EX11.8/ex11_8.txt b/629/CH11/EX11.8/ex11_8.txt new file mode 100644 index 000000000..38104743b --- /dev/null +++ b/629/CH11/EX11.8/ex11_8.txt @@ -0,0 +1,5 @@ +
+The angle of attack for a take the given take off speed = 7 degrees.
+
+The stall speed is 119 km/h.
+
\ No newline at end of file diff --git a/629/CH11/EX11.8/example11_8.sce b/629/CH11/EX11.8/example11_8.sce new file mode 100644 index 000000000..1675d17a7 --- /dev/null +++ b/629/CH11/EX11.8/example11_8.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 11.8 TAKEOFF CHARACTERISTICS OF AN AIRPLANE
+Vo=140000/3600; //velocity [m/s]
+rho=1.2; //density [Kg/m^3]
+b=10; //wing span[m]
+c=1.5; //chord length[m]
+S=b*c //area [m^2]
+FL=11600; //lift force[N]
+CL=FL/(S*rho*Vo^2/2) //lift coefficient
+A=b/c //aspect ratio
+//Interpolating for A from fig.11.23,
+alpha=7; //angle of attack in degrees
+printf("\nThe angle of attack for a take the given take off speed = %.f degrees.\n",alpha)
+//stall occurs at CL=1.18, from fig.11.23
+Cl=1.18;
+Vstall=sqrt(2*FL/(Cl*S*rho))*(3600/1000) //stall speed [Km/hr]
+printf("\nThe stall speed is %.f km/h.\n",Vstall)
\ No newline at end of file diff --git a/629/CH11/EX11.9/ex11_9.txt b/629/CH11/EX11.9/ex11_9.txt new file mode 100644 index 000000000..7df69f9cc --- /dev/null +++ b/629/CH11/EX11.9/ex11_9.txt @@ -0,0 +1,4 @@ +
+The downward thrust from the vane = 1148 N.
+
+The drag from the vane = 86.4 N.
\ No newline at end of file diff --git a/629/CH11/EX11.9/example11_9.sce b/629/CH11/EX11.9/example11_9.sce new file mode 100644 index 000000000..ec3e7fdcd --- /dev/null +++ b/629/CH11/EX11.9/example11_9.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 11.9 NEGATIVE LIFT ON A RACE CAR
+l=1.5; //[m]
+c=0.25; //[m]
+A=l/c //aspect ratio
+//Interpolating for A, from fig 11.23
+CL=0.93; //lift coefficient
+CD=0.07; //drag coefficient
+S=l*c //area [m^2]
+Vo=75; //velocity [m/s]
+rho=1.17; //[Kg/m^3]
+//Lift force
+FL=CL*S*rho*Vo^2/2 //[N]
+printf("\nThe downward thrust from the vane = %.f N.\n",FL)
+//Drag force
+FD=CD*S*rho*Vo^2/2
+printf("\nThe drag from the vane = %.1f N.\n",FD)
\ No newline at end of file diff --git a/629/CH12/EX12.1/ex12_1.txt b/629/CH12/EX12.1/ex12_1.txt new file mode 100644 index 000000000..50cf5ff23 --- /dev/null +++ b/629/CH12/EX12.1/ex12_1.txt @@ -0,0 +1,3 @@ +
+ The speed of sound in air at 15°C, c = 340 m/s.
+
\ No newline at end of file diff --git a/629/CH12/EX12.1/example12_1.sce b/629/CH12/EX12.1/example12_1.sce new file mode 100644 index 000000000..80ad3445a --- /dev/null +++ b/629/CH12/EX12.1/example12_1.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 12.1 SPEED OF SOUND CALCULATION
+R=287; //gas constant [J/Kg.K]
+k=1.4;
+T=15+273; //temperature [K]
+//Speed of sound
+c=sqrt(k*R*T) //[m/s]
+printf("\n The speed of sound in air at 15°C, c = %.f m/s.\n",c)
\ No newline at end of file diff --git a/629/CH12/EX12.10/ex12_10.txt b/629/CH12/EX12.10/ex12_10.txt new file mode 100644 index 000000000..b661e3a0f --- /dev/null +++ b/629/CH12/EX12.10/ex12_10.txt @@ -0,0 +1,2 @@ +
+ Because (pe=38.7) < (pb=100), the nozzle is overexpanded.
\ No newline at end of file diff --git a/629/CH12/EX12.10/example12_10.sce b/629/CH12/EX12.10/example12_10.sce new file mode 100644 index 000000000..77d9d7ce2 --- /dev/null +++ b/629/CH12/EX12.10/example12_10.sce @@ -0,0 +1,10 @@ +clear
+clc
+//Example 12.10 NOZZLE EXIT CONDITION
+k=1.4;
+//From table A.1, interpolating for A/Ao=4,
+M=2.94; //Mach number
+pb=100; //back pressure[kPa]
+pt=1300; //total pressure[kPa]
+pe=pt/((1+[(k-1)/2]*M^2)^(k/(k-1))) //[kPa]
+printf("\n Because (pe=%.1f) < (pb=%.f), the nozzle is overexpanded.\n",pe,pb)
\ No newline at end of file diff --git a/629/CH12/EX12.11/ex12_11.txt b/629/CH12/EX12.11/ex12_11.txt new file mode 100644 index 000000000..8530424f3 --- /dev/null +++ b/629/CH12/EX12.11/ex12_11.txt @@ -0,0 +1,3 @@ +
+ The static pressure at the exit = 603 kPa.
+
\ No newline at end of file diff --git a/629/CH12/EX12.11/example12_11.sce b/629/CH12/EX12.11/example12_11.sce new file mode 100644 index 000000000..7fff6f466 --- /dev/null +++ b/629/CH12/EX12.11/example12_11.sce @@ -0,0 +1,24 @@ +clear
+clc
+//Example 12.11 SHOCK WAVE IN LAVAL NOZZLE
+k=1.4;
+AoA=1/2 ;//AoA=(Ao/A)
+AeAo=4; //AeAo=(Ae/Ao)
+//From table A.1, interpolation for A/Ao=2,
+//for supersonic flow
+M1=2.2;
+//for normal shock
+M2=0.547;
+pt21=0.6281; //pt21=(pt2/pt1)
+pt1=1000; //[kPa]
+pt2=pt21*pt1 //[kPa]
+//for subsonic part at M=M2,
+AAv=1.26; //AAv=(A/Av)
+//Exit area ratio
+//Ae/Av=(Ae/Ao)*(Ao/A)*(A/Av)
+AeAv=AeAo*AoA*AAv //AeAv=(Ae/Av)
+//for subsonic flow at A/Ao=(Ae/Av)
+M=0.24;
+//Exit pressure
+pe=pt2/((1+(k-1)*M^2/2)^(k/(k-1))) //[kPa]
+printf("\n The static pressure at the exit = %.f kPa.\n",pe)
\ No newline at end of file diff --git a/629/CH12/EX12.12/ex12_12.txt b/629/CH12/EX12.12/ex12_12.txt new file mode 100644 index 000000000..ad41c1119 --- /dev/null +++ b/629/CH12/EX12.12/ex12_12.txt @@ -0,0 +1,2 @@ +
+ The mass flow rate = 0.238 kg/s.
\ No newline at end of file diff --git a/629/CH12/EX12.12/example12_12.sce b/629/CH12/EX12.12/example12_12.sce new file mode 100644 index 000000000..df847404a --- /dev/null +++ b/629/CH12/EX12.12/example12_12.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 12.12 MASS FLOW IN TRUNCATED NOZZLE
+k=1.4;
+d=0.03; //diameter[m]
+A=%pi*d^2/4 //area[m^2]
+pt=160; //[kPa]
+pb=100; //[kPa]
+Tt=273+80; //total temp.[K]
+//Mach number at exit
+Me=sqrt((2/(k-1))*[(pt/pb)^((k-1)/k)-1])
+//Static temperature at exit
+Te=Tt/(1+((k-1)/2)*Me^2) //[K]
+R=287; //[J/kg.K]
+//Static density at exit
+rho=pb*10^3/(R*Te) //[kg/m^3]
+c=sqrt(k*R*Te) //speed of sound[m/s]
+//Mass flow rate
+m=rho*A*Me*c //[kg/s]
+printf("\n The mass flow rate = %.3f kg/s.\n",m)
\ No newline at end of file diff --git a/629/CH12/EX12.2/ex12_2.txt b/629/CH12/EX12.2/ex12_2.txt new file mode 100644 index 000000000..dc9b178d2 --- /dev/null +++ b/629/CH12/EX12.2/ex12_2.txt @@ -0,0 +1,3 @@ +
+ The Mach number of the aircraft, M = 1.58.
+
\ No newline at end of file diff --git a/629/CH12/EX12.2/example12_2.sce b/629/CH12/EX12.2/example12_2.sce new file mode 100644 index 000000000..d693c926e --- /dev/null +++ b/629/CH12/EX12.2/example12_2.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 12.2 MACH-NUMBER CALCULATION
+z=13; //[km]
+alpha=5.87; //[K/km]
+To=296; //[K]
+//Temperature at z
+T=To-alpha*z //[K]
+R=287; //[J/Kg.K]
+k=1.4;
+//Speed of sound
+c=sqrt(k*R*T) //[m/s]
+V=470; //speed of fighter[m/s]
+//Mach number
+M=V/c
+printf("\n The Mach number of the aircraft, M = %.2f.\n",M)
\ No newline at end of file diff --git a/629/CH12/EX12.3/ex12_3.txt b/629/CH12/EX12.3/ex12_3.txt new file mode 100644 index 000000000..b9f026826 --- /dev/null +++ b/629/CH12/EX12.3/ex12_3.txt @@ -0,0 +1,2 @@ +
+ The surface temperature of the aircraft = 337 K.
\ No newline at end of file diff --git a/629/CH12/EX12.3/example12_3.sce b/629/CH12/EX12.3/example12_3.sce new file mode 100644 index 000000000..d9e9cb90b --- /dev/null +++ b/629/CH12/EX12.3/example12_3.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 12.3 TOTAL TEMPERATURE CALCULATION
+T=273+(-50) //static temperature [K]
+k=1.4;
+M=1.6; //Mach number
+//Total temperature
+Tt=T*(1+(k-1)*M^2/2) //[K]
+printf("\n The surface temperature of the aircraft = %.f K.\n",Tt)
\ No newline at end of file diff --git a/629/CH12/EX12.4/ex12_4.txt b/629/CH12/EX12.4/ex12_4.txt new file mode 100644 index 000000000..7c7ae6ac9 --- /dev/null +++ b/629/CH12/EX12.4/ex12_4.txt @@ -0,0 +1,2 @@ +
+ The drag force on the sphere = 2.58 N.
\ No newline at end of file diff --git a/629/CH12/EX12.4/example12_4.sce b/629/CH12/EX12.4/example12_4.sce new file mode 100644 index 000000000..3cf8712ef --- /dev/null +++ b/629/CH12/EX12.4/example12_4.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 12.4 DRAG FORCE ON A SPHERE
+k=1.4;
+p=101; //pressure[kPa]
+M=0.7; //Mach number
+q=k*p*M^2/2 //kinetic pressure[kPa]
+CD=0.95;
+D=0.01; //[m]
+A=%pi*D^2/4 //area[m^2]
+//Drag force
+FD=CD*q*A*10^3 //[N]
+printf("\n The drag force on the sphere = %.2f N.\n",FD)
\ No newline at end of file diff --git a/629/CH12/EX12.5/ex12_5.txt b/629/CH12/EX12.5/ex12_5.txt new file mode 100644 index 000000000..fd0cafe91 --- /dev/null +++ b/629/CH12/EX12.5/ex12_5.txt @@ -0,0 +1,9 @@ +
+The Mach number for downstream of the shock wave, M = 0.668.
+
+
+The pressure for downstream of the shock wave, P = 282 kPa, absolute.
+
+
+The temperature for downstream of the shock wave, T = 400 K or 127°C.
+
\ No newline at end of file diff --git a/629/CH12/EX12.5/example12_5.sce b/629/CH12/EX12.5/example12_5.sce new file mode 100644 index 000000000..1848c77e1 --- /dev/null +++ b/629/CH12/EX12.5/example12_5.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 12.5 PROPERTY CHANGES ACROSS NORMAL SHOCK WAVE
+k=1.4;
+M1=1.6; //Upstream Mach number
+p1=100; //pressure[kPa]
+T1=15+273; //[K]
+//Downstream Mach number
+M2=sqrt(((k-1)*M1^2+2)/(2*k*M1^2-(k-1)))
+printf("\nThe Mach number for downstream of the shock wave, M = %.3f.\n\n",M2)
+//Downstream pressure
+p2=p1*[(1+k*M1^2)/(1+k*M2^2)] //[kPa]
+printf("\nThe pressure for downstream of the shock wave, P = %.f kPa, absolute.\n\n",p2)
+//Downstream temperature
+T2=T1*{(1+[(k-1)/2]*M1^2)/(1+[(k-1)/2]*M2^2)} //[K]
+printf("\nThe temperature for downstream of the shock wave, T = %.f K or %.f°C.\n",T2,T2-273)
\ No newline at end of file diff --git a/629/CH12/EX12.6/ex12_6.txt b/629/CH12/EX12.6/ex12_6.txt new file mode 100644 index 000000000..51179500f --- /dev/null +++ b/629/CH12/EX12.6/ex12_6.txt @@ -0,0 +1,3 @@ +
+ The change in entropy across the wave = 20.5 J/kg.K.
+
\ No newline at end of file diff --git a/629/CH12/EX12.6/example12_6.sce b/629/CH12/EX12.6/example12_6.sce new file mode 100644 index 000000000..f50a6ce45 --- /dev/null +++ b/629/CH12/EX12.6/example12_6.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 12.6 ENTROPY INCREASE ACROSS SHOCK WAVE
+//To find Approx Value
+function [A]= approx (V,n)
+ A= round(V*10^n)/10^n; //V-Value, n-to what place
+ funcprot (0)
+endfunction
+k=1.4;
+M1=1.5;
+//Downstream Mach number
+M2=approx(sqrt(((k-1)*M1^2+2)/(2*k*M1^2-(k-1))),3)
+//Pressure ratio, (p21=p2/p1)
+p21=approx([(1+k*M1^2)/(1+k*M2^2)],2)
+//Temperature ratio, (T21=T2/T1)
+T21=approx((1+[(k-1)/2]*M1^2)/(1+[(k-1)/2]*M2^2),2)
+R=287; //[J/kg.K]
+//Entropy change
+del_s=R*log((T21^(k/(k-1))/p21)) //[J/kg.K]
+printf("\n The change in entropy across the wave = %.1f J/kg.K.\n",del_s)
\ No newline at end of file diff --git a/629/CH12/EX12.7/ex12_7.txt b/629/CH12/EX12.7/ex12_7.txt new file mode 100644 index 000000000..2721bfa5a --- /dev/null +++ b/629/CH12/EX12.7/ex12_7.txt @@ -0,0 +1,2 @@ +
+ The cross-sectional area of the test section, A = 42.3 cm^2.
\ No newline at end of file diff --git a/629/CH12/EX12.7/example12_7.sce b/629/CH12/EX12.7/example12_7.sce new file mode 100644 index 000000000..7c0f8a7b0 --- /dev/null +++ b/629/CH12/EX12.7/example12_7.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 12.7 TEST SECTION SIZE IN SUPERSONIC WIND TUNNEL
+k=1.4;
+M=3; //Mach number
+Ao=10; //area[cm^2]
+//Cross-sectional area
+A=Ao*(1/M)*{(1+[(k-1)/2]*M^2)/((k+1)/2)}^((k+1)/(2*(k-1))) //[cm^2]
+printf("\n The cross-sectional area of the test section, A = %.1f cm^2.\n",A)
\ No newline at end of file diff --git a/629/CH12/EX12.8/ex12_8.txt b/629/CH12/EX12.8/ex12_8.txt new file mode 100644 index 000000000..d2264ac13 --- /dev/null +++ b/629/CH12/EX12.8/ex12_8.txt @@ -0,0 +1,9 @@ +
+ The Mach number at the test section, M = 3.91.
+
+
+ At the test section, pressure p = 29.7 kPa and temperature T = 86 K.
+
+
+ The velocity at test section, V = 727 m/s.
+
\ No newline at end of file diff --git a/629/CH12/EX12.8/example12_8.sce b/629/CH12/EX12.8/example12_8.sce new file mode 100644 index 000000000..fa95625a4 --- /dev/null +++ b/629/CH12/EX12.8/example12_8.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 12.8 FLOW PROPERTIES IN SUPERSONIC WIND TUNNEL
+k=1.4;
+R=287; //[J/kg.K]
+//From table A.1, interpolating for A/Ao=10,
+M=3.91; //Mach number
+printf("\n The Mach number at the test section, M = %.2f.\n\n",M)
+pt=4000; //pressure[kPa]
+Tt=350; //temp.[K]
+//In test section
+p=0.00743*pt //[kPa]
+T=0.246*Tt //[K]
+printf("\n At the test section, pressure p = %.1f kPa and temperature T = %.f K.\n\n",p,T)
+//Speed of sound
+c=sqrt(k*R*T) //[m/s]
+//Velocity
+V=M*c //[m/s]
+printf("\n The velocity at test section, V = %.f m/s.\n",V)
\ No newline at end of file diff --git a/629/CH12/EX12.9/ex12_9.txt b/629/CH12/EX12.9/ex12_9.txt new file mode 100644 index 000000000..6259f24c3 --- /dev/null +++ b/629/CH12/EX12.9/ex12_9.txt @@ -0,0 +1,3 @@ +
+ The mass flow rate = 14.8 kg/s.
+
\ No newline at end of file diff --git a/629/CH12/EX12.9/example12_9.sce b/629/CH12/EX12.9/example12_9.sce new file mode 100644 index 000000000..c7d9e299b --- /dev/null +++ b/629/CH12/EX12.9/example12_9.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 12.9 MASS FLOW RATE IN SUPERSONIC WIND TUNNEL
+k=1.4;
+R=287; //[J/kg.K]
+M=3; //Mach number
+A=225*10^-4; //[m^2]
+//Throat area
+Ao=A/((1/M)*{(1+[(k-1)/2]*M^2)/((k+1)/2)}^((k+1)/(2*(k-1)))) //[m^2]
+//Static T,p
+T=273+(-20) //[K]
+p=50; //[kPa]
+//Total temperature
+Tt=T*(1+(k-1)*M^2/2) //[K]
+//Total pressure
+pt=p*(1+(k-1)*M^2/2)^(k/(k-1)) //[kPa]
+//Mass flow rate
+m=k^(1/2)*[(2/(k+1))^((k+1)/(2*(k-1)))]*pt*10^3*Ao/((R*Tt)^(1/2)) //[kg]
+printf("\n The mass flow rate = %.1f kg/s.\n",m)
\ No newline at end of file diff --git a/629/CH13/EX13.1/ex13_1.txt b/629/CH13/EX13.1/ex13_1.txt new file mode 100644 index 000000000..98ed2bc53 --- /dev/null +++ b/629/CH13/EX13.1/ex13_1.txt @@ -0,0 +1,3 @@ +
+The volume rate of flow = 29.7 m^3/s.
+
\ No newline at end of file diff --git a/629/CH13/EX13.1/example13_1.sce b/629/CH13/EX13.1/example13_1.sce new file mode 100644 index 000000000..b23f5d559 --- /dev/null +++ b/629/CH13/EX13.1/example13_1.sce @@ -0,0 +1,22 @@ +clear
+clc
+//Example 13.1 DISCHARGE FROM VELOCITY DATA
+r=[0 5 10 15 20 25 30 35 40 45 47.5 50]; //[cm]
+V=[50 49.5 49 48 46.5 45 43 40.5 37.5 34 25 0]; //velocity[m/s]
+
+//dA=2*pi*r*dr, Q=V*dA
+del_A(1)=0; //dA
+q(1)=0;
+for i=2:1:11
+ del_r(i)=(r(i+1)-r(i-1))/2;//dr
+ del_A(i)=2*%pi*r(i)*del_r(i)*10^-4; //[m^2]
+ q(i)=V(i)*del_A(i); //[m^3/s]
+end
+//for i=12,
+ del_r(12)=(r(12)-r(11)/2);
+ del_A(12)=2*%pi*r(12)*del_r(12)*10^-4; //[m^2]
+ q(12)=V(12)*del_A(12); //[m^3/s]
+
+//Discharge
+Q=sum(q) //m^3/s
+printf("\nThe volume rate of flow = %.1f m^3/s.\n",Q)
\ No newline at end of file diff --git a/629/CH13/EX13.10/ex13_10.txt b/629/CH13/EX13.10/ex13_10.txt new file mode 100644 index 000000000..360821622 --- /dev/null +++ b/629/CH13/EX13.10/ex13_10.txt @@ -0,0 +1,3 @@ +
+The uncertainty of the calculated discharge = 0.0046 m^3/s.
+
\ No newline at end of file diff --git a/629/CH13/EX13.10/example13_10.sce b/629/CH13/EX13.10/example13_10.sce new file mode 100644 index 000000000..71455724e --- /dev/null +++ b/629/CH13/EX13.10/example13_10.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 13.10 UNCERTAINTY ESTIMATE FOR AN ORIFICE METER
+K=0.66;
+d=0.15; //[m]
+A=%pi*d^2/4 //area[m^2]
+D=0.24; //[m]
+g=9.81; //[m/s^2]
+h=0.25; //[m]
+S=13.6; //specific gravity of Hg
+del_h=h*(S-1) //piezometric head[m]
+Q=K*A*sqrt(2*g*del_h)
+//From equation of uncertainty, (UQ/Q)^2=(UK/K)^2+(2Ud/d)^2+(Uh/2h)^2
+//from example 13.2
+UK=0.03;
+Ud=0.00015;//[m]
+Uh=0.01;//[m]
+//Uncertainty in Q
+UQ=Q*sqrt((UK/K)^2+(2*Ud/d)^2+(Uh/(2*h))^2)
+printf("\nThe uncertainty of the calculated discharge = %.4f m^3/s.\n",UQ)
\ No newline at end of file diff --git a/629/CH13/EX13.3/example13_3.sce b/629/CH13/EX13.3/example13_3.sce new file mode 100644 index 000000000..3a23e8846 --- /dev/null +++ b/629/CH13/EX13.3/example13_3.sce @@ -0,0 +1,28 @@ +clear
+clc
+//Example 13.3 ANALYSIS OF AN ORIFICE METER
+g=9.81; //[m/s^2]
+l=0.25; //deflection[m]
+S=13.6; //specific gravity of Hg
+h=l*(S-1) //piezometric head[m]
+d=0.15; //[m]
+D=0.24; //[m]
+Ao=%pi*d^2/4 //[m^2]
+A1=%pi*D^2/4 //[m^2]
+v=10^-6; //[m^2/s]
+//Flow coefficient
+ReK=(d/v)*sqrt(2*g*h) //ReK=Re/K
+//From fig.13.14, for Re/K, d/D=0.625
+K=0.66;
+Q=K*Ao*sqrt(2*g*h) //[m^3/s]
+printf("\nThe discharge in the system = %.3f m^3/s.\n",Q)
+Cv=0.98;
+//K=Cv*Cc/(1-(Cc*Ao/A1)^2)
+P=[K*(Ao/A1)^2 Cv -K] //polynomial in Cc
+q=roots(P) //roots of P
+Cc=q(2) //positive root
+V1=Q/A1 //[m/s]
+V2=Q/(Cc*Ao) //[m/s]
+//Head loss
+hL=(V2-V1)^2/(2*g) //[m]
+printf("\nThe head loss produced by the orifice = %.2f m.\n",hL)
\ No newline at end of file diff --git a/629/CH13/EX13.4/ex13_4.txt b/629/CH13/EX13.4/ex13_4.txt new file mode 100644 index 000000000..8568fc440 --- /dev/null +++ b/629/CH13/EX13.4/ex13_4.txt @@ -0,0 +1,3 @@ +
+The deflection on the manometer = 1.1 ft.
+
\ No newline at end of file diff --git a/629/CH13/EX13.4/example13_4.sce b/629/CH13/EX13.4/example13_4.sce new file mode 100644 index 000000000..61ee6759b --- /dev/null +++ b/629/CH13/EX13.4/example13_4.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 13.4 MANOMETER DEFLECTION FOR AN ORIFICE METER
+d=8/12; //[ft]
+Ao=%pi*d^2/4 //area [ft^2]
+v=1.22*10^-5; //[ft^2/s]
+Q=2; //flow rate [ft^3/s]
+g=32.2; //[ft/s^2]
+Re=4*Q/(%pi*d*v) //Reynolds number
+//interpolate for d/D=8/12 from fig. 13.14
+K=0.68
+del_h=Q^2/(2*g*K^2*Ao^2) //change in piezometric head [ft]
+//del_h=del_l(g_w-g_air)/g_w
+//g_w>>g_a
+del_l=del_h //manometer deflection [ft]
+printf("\nThe deflection on the manometer = %.1f ft.\n",del_l)
\ No newline at end of file diff --git a/629/CH13/EX13.5/ex13_5.txt b/629/CH13/EX13.5/ex13_5.txt new file mode 100644 index 000000000..732f57a6b --- /dev/null +++ b/629/CH13/EX13.5/ex13_5.txt @@ -0,0 +1,2 @@ +
+The mass flow rate of natural gas = 0.302 kg/s.
\ No newline at end of file diff --git a/629/CH13/EX13.5/example13_5.sce b/629/CH13/EX13.5/example13_5.sce new file mode 100644 index 000000000..59606cf2c --- /dev/null +++ b/629/CH13/EX13.5/example13_5.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 13.5 MASS FLOW RATE OF NATURAL GAS
+d0=0.07; //[m]
+d1=0.1; //[m]
+A0=%pi*d0^2/4 //area[m^2]
+A1=%pi*d1^2/4 //area[m^2]
+rho=0.678; //density of methane[kg/m^3]
+k=1.31;
+v=1.59*10^-5; //[m^2/s]
+p1=101000; //pressure[Pa]
+delp=10000; //difference in pressure[Pa]
+p2=p1-delp //[Pa]
+ReK=(d0/v)*sqrt(2*delp/rho) //ReK=Re/K
+//From fig 13.14,
+K=0.7; //flow coefficient
+Y=1-((1/k)*(1-p2/p1)*(0.41+0.35*(A0/A1)^2)) //Compressibility factor
+m=Y*A0*K*sqrt(2*rho*(p1-p2)) //mass flow rate[kg/s]
+printf("\nThe mass flow rate of natural gas = %.3f kg/s.\n",m)
\ No newline at end of file diff --git a/629/CH13/EX13.6/ex13_6.txt b/629/CH13/EX13.6/ex13_6.txt new file mode 100644 index 000000000..747059965 --- /dev/null +++ b/629/CH13/EX13.6/ex13_6.txt @@ -0,0 +1,2 @@ +
+The discharge of water at 10°C is 0.268 m^3/s.
\ No newline at end of file diff --git a/629/CH13/EX13.6/example13_6.sce b/629/CH13/EX13.6/example13_6.sce new file mode 100644 index 000000000..aa5acb49b --- /dev/null +++ b/629/CH13/EX13.6/example13_6.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 13.6 FLOW RATE USING A VENTURI METER
+del_p=35000; //pressure difference [Pa]
+Gamma=9810; //specific weight[N/m^3]
+del_h=del_p/Gamma //change in piezometric head [m]
+d=0.2; //[m]
+v=1.31*10^-6; //[m^2/s]
+g=9.81; //[m/s^2]
+ReK=(d/v)*sqrt(2*g*del_h) //(ReK=Re/K)
+//Interpolating from fig.13.14
+K=1.02; //flow coefficient
+A=%pi*d^2/4 //area [m^2]
+//Discharge
+Q=1.02*A*sqrt(2*g*del_h) //[m^3/s]
+printf("\nThe discharge of water at 10°C is %.3f m^3/s.\n",Q)
\ No newline at end of file diff --git a/629/CH13/EX13.7/ex13_7.txt b/629/CH13/EX13.7/ex13_7.txt new file mode 100644 index 000000000..b5e8e9b96 --- /dev/null +++ b/629/CH13/EX13.7/ex13_7.txt @@ -0,0 +1,3 @@ +
+The discharge of water over the weir = 0.23 m^3/s.
+
\ No newline at end of file diff --git a/629/CH13/EX13.7/example13_7.sce b/629/CH13/EX13.7/example13_7.sce new file mode 100644 index 000000000..8d600774b --- /dev/null +++ b/629/CH13/EX13.7/example13_7.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 13.7 FLOW RATE FOR A RECTANGULAR WEIR
+H=0.21; //head on weir[m]
+P=0.6; //height of weir[m]
+L=1.3; //width[m]
+g=9.81; //acceleration due to gravity[m/s^2]
+//Flow coefficient
+K=0.4+0.05*(H/P)
+//Discharge
+Q=K*L*sqrt(2*g*H^3) //[m^3/s]
+printf("\nThe discharge of water over the weir = %.2f m^3/s.\n",Q)
\ No newline at end of file diff --git a/629/CH13/EX13.8/ex13_8.txt b/629/CH13/EX13.8/ex13_8.txt new file mode 100644 index 000000000..3a0a2fe2e --- /dev/null +++ b/629/CH13/EX13.8/ex13_8.txt @@ -0,0 +1,2 @@ +
+ The flow of water over the weir = 0.096 m^3/s.
\ No newline at end of file diff --git a/629/CH13/EX13.8/example13_8.sce b/629/CH13/EX13.8/example13_8.sce new file mode 100644 index 000000000..86af6e81b --- /dev/null +++ b/629/CH13/EX13.8/example13_8.sce @@ -0,0 +1,8 @@ +clear
+clc
+//Example 13.8 FLOW RATE FOR A TRIANGULAR WEIR
+H=0.43; //head on weir[m]
+g=9.81; //[m/s^2]
+//Discharge
+Q=0.179*sqrt(2*g*(H^5)) //[m^3/s]
+printf("\n The flow of water over the weir = %.3f m^3/s.\n",Q)
\ No newline at end of file diff --git a/629/CH13/EX13.9/ex13_9.txt b/629/CH13/EX13.9/ex13_9.txt new file mode 100644 index 000000000..a5d3e2441 --- /dev/null +++ b/629/CH13/EX13.9/ex13_9.txt @@ -0,0 +1,3 @@ +
+The mass flow rate of air flowing through a venturi meter = 0.0264 kg/s.
+
\ No newline at end of file diff --git a/629/CH13/EX13.9/example13_9.sce b/629/CH13/EX13.9/example13_9.sce new file mode 100644 index 000000000..d169454c0 --- /dev/null +++ b/629/CH13/EX13.9/example13_9.sce @@ -0,0 +1,17 @@ +clear
+clc
+//Example 13.9 COMPRESSIBLE FLOW
+k=1.4;
+p1=150000; //upstream pressure[Pa]
+p2=100000; //throat pressure[Pa]
+T1=300; //temperature[K]
+R=287; //[J/Kg.K]
+//Ideal gas law
+rho1=p1/(R*T1) //[Kg/m^3]
+D1=0.03; //[m]
+D2=0.01; //[m]
+A2=%pi*D2^2/4 //area[m^2]
+Cd=1;
+//Mass flow rate
+m=Cd*A2*((p2/p1)^(1/k))*{([2*k/(k-1)]*p1*rho1*[1-(p2/p1)^((k-1)/k)])/(1-(p2/p1)^(2/k)*(D2/D1)^4)}^(1/2) //[Kg/s]
+printf("\nThe mass flow rate of air flowing through a venturi meter = %.4f kg/s.\n",m)
\ No newline at end of file diff --git a/629/CH14/EX14.10/ex14_10.txt b/629/CH14/EX14.10/ex14_10.txt new file mode 100644 index 000000000..2140c89a6 --- /dev/null +++ b/629/CH14/EX14.10/ex14_10.txt @@ -0,0 +1,7 @@ +
+The power developed by the turbine = 14236 kW.
+
+Angular speed of wheel for maximum efficiency = 349 rpm.
+
+The torque on the turbine shaft = 390 kN.m
+
\ No newline at end of file diff --git a/629/CH14/EX14.10/example14_10.sce b/629/CH14/EX14.10/example14_10.sce new file mode 100644 index 000000000..448ee4aee --- /dev/null +++ b/629/CH14/EX14.10/example14_10.sce @@ -0,0 +1,33 @@ +clear
+clc
+//Example 14.10 IMPULSE TURBINE
+z1=1670; //[m]
+z2=1000; //[m]
+g=9.81; //[m/s^2]
+L=6000; //length[m]
+f=0.015;
+Dj=0.18; //diameter [m]
+Aj=%pi*Dj^2/4 //area[m^2]
+Dp=1; //diameter of penstock[m]
+Ap=%pi*Dp^2/4 //area[m^2
+//Energy equation, (p1/gamma)+(V1^2/2g)+z1=(p2/gamma)+(Vj^2/2g)+z2+hL
+//p1=0,p2=0 ,Vp=Vj*(Aj/Ap), hL=f*L*Vp^2/(D*2*g)
+Vj=sqrt(2*g*(z1-z2)/(1+f*L*(Aj/Ap)^2/Dp)) //jet velocity[m/s]
+Gamma=9810;
+Q=Aj*Vj //[m^3/s])
+P=Q*Gamma*Vj^2/(2*g*10^3) //gross power[kW]
+eta=0.85; //efficiency
+Pd=P*eta //power developed [kW]
+printf("\nThe power developed by the turbine = %.f kW.\n",Pd)
+
+//Angular speed of wheel
+r=1.5; //radius[m]
+Vb=Vj/2 //[m/s]
+w=Vb/r //[rad/s]
+//Wheel speed
+N=w*60/(2*%pi) //in rpm
+printf("\nAngular speed of wheel for maximum efficiency = %.f rpm.\n",N)
+
+//Torque
+T=Pd/w //[kN.m]
+printf("\nThe torque on the turbine shaft = %.f kN.m\n",T)
\ No newline at end of file diff --git a/629/CH14/EX14.11/ex14_11.txt b/629/CH14/EX14.11/ex14_11.txt new file mode 100644 index 000000000..04c62f864 --- /dev/null +++ b/629/CH14/EX14.11/ex14_11.txt @@ -0,0 +1,3 @@ +
+The guide vane angle for non seperating flow condition at runner entrance = 17.4°.
+
\ No newline at end of file diff --git a/629/CH14/EX14.11/example14_11.sce b/629/CH14/EX14.11/example14_11.sce new file mode 100644 index 000000000..e2b714b4a --- /dev/null +++ b/629/CH14/EX14.11/example14_11.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 14.11 FRANCIS TURBINE
+r1=0.6; //[m]
+beta1=110; //degrees
+w=600*(2*%pi)/60 //angular speed[rad/s]
+Q=4; //discharge[m^3/s]
+B=0.1; //blade height[m]
+//Radial velocity at inlet
+Vr1=Q/(2*%pi*r1*B) //[m/s]
+//Inlet guide vane angle
+alpha1=acotd((r1*w/Vr1)+cotd(beta1))
+printf("\nThe guide vane angle for non seperating flow condition at runner entrance = %.1f°.\n",alpha1)
\ No newline at end of file diff --git a/629/CH14/EX14.12/ex14_12.txt b/629/CH14/EX14.12/ex14_12.txt new file mode 100644 index 000000000..63fc78f65 --- /dev/null +++ b/629/CH14/EX14.12/ex14_12.txt @@ -0,0 +1,3 @@ +
+The minimum capture area of wind turbine = 8.20 m^2.
+
\ No newline at end of file diff --git a/629/CH14/EX14.12/example14_12.sce b/629/CH14/EX14.12/example14_12.sce new file mode 100644 index 000000000..fd42e36fc --- /dev/null +++ b/629/CH14/EX14.12/example14_12.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 14.12 CAPTURE AREA OF WIND TURBINE
+Pmax=5*100; //power produced by turbine[W]
+V=20*(1000/3600) //wind velocity [m/s]
+rho=1.2; //density [Kg/m^3]
+//Minimum capture area
+Amin=Pmax*(54/16)/(rho*V^3) //[m^2]
+printf("\nThe minimum capture area of wind turbine = %.2f m^2.\n",Amin)
\ No newline at end of file diff --git a/629/CH14/EX14.2/ex14_2.txt b/629/CH14/EX14.2/ex14_2.txt new file mode 100644 index 000000000..8fc466b9d --- /dev/null +++ b/629/CH14/EX14.2/ex14_2.txt @@ -0,0 +1,5 @@ +
+The discharge of water, Q = 0.180 m^3/s.
+
+The power required for these conditions, P = 4.12 kW.
+
\ No newline at end of file diff --git a/629/CH14/EX14.2/example14_2.sce b/629/CH14/EX14.2/example14_2.sce new file mode 100644 index 000000000..afe0f131a --- /dev/null +++ b/629/CH14/EX14.2/example14_2.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 14.2 DISCHARGE AND POWER FOR AXIAL-FLOW PUMP
+g=9.81; //[m/s^2]
+D=0.356; //[m]
+n=600/60 //speed[rps]
+rho=10^3; //density[Kg/m^3]
+H=2; //[m]
+CH=H/(D*n*g)
+//From fig.14.6
+CQ=0.4;
+CP=0.72;
+//Discharge
+Q=CQ*n*D^3 //[m^3/s]
+printf("\nThe discharge of water, Q = %.3f m^3/s.\n",Q)
+//Power
+P=CP*rho*D^5*n^3/10^3 //[kW]
+printf("\nThe power required for these conditions, P = %.2f kW.\n",P)
\ No newline at end of file diff --git a/629/CH14/EX14.3/ex14_3.txt b/629/CH14/EX14.3/ex14_3.txt new file mode 100644 index 000000000..72ba6c79a --- /dev/null +++ b/629/CH14/EX14.3/ex14_3.txt @@ -0,0 +1,5 @@ +
+The head developed is H = 2.76 m.
+
+The power required for the operation = 4.57 kW.
+
\ No newline at end of file diff --git a/629/CH14/EX14.3/example14_3.sce b/629/CH14/EX14.3/example14_3.sce new file mode 100644 index 000000000..52fea8cf6 --- /dev/null +++ b/629/CH14/EX14.3/example14_3.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 14.3 HEAD AND POWER FOR AXIAL FLOW PUMP
+g=9.81; //[m/s^2]
+Q=0.127; //[m^3/s]
+n=13.3; //[rps]
+D=0.3; //[m]
+rho=10^3; //density[Kg/m^3]
+//Discharge coefficient
+CQ=Q/(n*D^3)
+//From fig.14.6
+CH=1.7;
+CP=0.8;
+//Head produced
+H=CH*D^2*n^2/g //[m]
+printf("\nThe head developed is H = %.2f m.\n",H)
+//Power produced
+P=CP*rho*D^5*n^3/10^3 //[kW]
+printf("\nThe power required for the operation = %.2f kW.\n",P)
\ No newline at end of file diff --git a/629/CH14/EX14.4/ex14_4.txt b/629/CH14/EX14.4/ex14_4.txt new file mode 100644 index 000000000..dd9109e21 --- /dev/null +++ b/629/CH14/EX14.4/ex14_4.txt @@ -0,0 +1,5 @@ +
+The speed at which the pump should be operated = 1961 rpm.
+
+The discharge for the given conditions = 0.234 m^3/s.
+
\ No newline at end of file diff --git a/629/CH14/EX14.4/example14_4.sce b/629/CH14/EX14.4/example14_4.sce new file mode 100644 index 000000000..aa5223e69 --- /dev/null +++ b/629/CH14/EX14.4/example14_4.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 14.4 SPEED AND DISCHARGE OF CENTRIFUGAL PUMP
+N1=2133.5; //speed[rpm]
+H1=90; //[m]
+H2=76; //[m]
+//(g.H/n.D^2)_N1=(g.H/n.D^2)_N2
+N2=N1*(H2/H1)^(1/2) //[rpm]
+printf("\nThe speed at which the pump should be operated = %.f rpm.\n",N2)
+//(Q/n.D^3)_N1=(Q/n.D^3)_N2
+Q1=0.255; //discharge[m^3/s]
+Q2=Q1*N2/N1 //[m^3/s]
+printf("\nThe discharge for the given conditions = %.3f m^3/s.\n",Q2)
\ No newline at end of file diff --git a/629/CH14/EX14.5/ex14_5.txt b/629/CH14/EX14.5/ex14_5.txt new file mode 100644 index 000000000..4ea85768d --- /dev/null +++ b/629/CH14/EX14.5/ex14_5.txt @@ -0,0 +1,6 @@ +
+ At maximum efficiency, the expected values of
+ Head = 92.4 m
+ Discharge = 6.21 m^3/s
+ Power = 6222 kW
+
\ No newline at end of file diff --git a/629/CH14/EX14.5/example14_5.sce b/629/CH14/EX14.5/example14_5.sce new file mode 100644 index 000000000..3f27a3e4e --- /dev/null +++ b/629/CH14/EX14.5/example14_5.sce @@ -0,0 +1,15 @@ +clear
+clc
+//Example 14.5 HEAD, DISCHARGE, AND POWER OF CENTRIFUGAL PUMP
+g=9.81; //[ft/s^2]
+n=400/60 //speed[rps]
+D=1.98//diameter[m]
+//From fig.14.10
+CQ=0.12;
+CH=5.2;
+CP=0.69;
+H=CH*D^2*n^2/g //head[ft]
+Q=CQ*n*D^3 //discharge[m^3/s]
+rho=10^3; //density[Kg/m^3]
+P=CP*rho*D^5*n^3/10^3 //power[kW]
+printf("\n At maximum efficiency, the expected values of\n Head = %.1f m\n Discharge = %.2f m^3/s\n Power = %.f kW\n",H,Q,P)
\ No newline at end of file diff --git a/629/CH14/EX14.6/ex14_6.txt b/629/CH14/EX14.6/ex14_6.txt new file mode 100644 index 000000000..f07f37894 --- /dev/null +++ b/629/CH14/EX14.6/ex14_6.txt @@ -0,0 +1,2 @@ +
+For specific speed, ns= 0.035, radial flow pump is best choice.
\ No newline at end of file diff --git a/629/CH14/EX14.6/example14_6.sce b/629/CH14/EX14.6/example14_6.sce new file mode 100644 index 000000000..95b44ff58 --- /dev/null +++ b/629/CH14/EX14.6/example14_6.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 14.6 PUMP SELECTION USING SPECIFIC SPEED
+g=32.2; //[ft/s^2]
+n=1100/60; //in rps
+h=600; //[ft]
+Q=10; //[cfs]
+//Specific speed
+ns=n*Q^(1/2)/(g*h)^(3/4)
+//From fig 14.12,
+printf("\nFor specific speed, ns= %.3f, radial flow pump is best choice.\n",ns)
\ No newline at end of file diff --git a/629/CH14/EX14.7/ex14_7.txt b/629/CH14/EX14.7/ex14_7.txt new file mode 100644 index 000000000..064e963a7 --- /dev/null +++ b/629/CH14/EX14.7/ex14_7.txt @@ -0,0 +1,5 @@ +
+The net positive suction head = 26.1 ft.
+
+The traditional suction specific speed, Nss = 4538.
+
\ No newline at end of file diff --git a/629/CH14/EX14.7/example14_7.sce b/629/CH14/EX14.7/example14_7.sce new file mode 100644 index 000000000..566e74ba5 --- /dev/null +++ b/629/CH14/EX14.7/example14_7.sce @@ -0,0 +1,35 @@ +clear
+clc
+//Example 14.7 NET POSITIVE SUCTION HEAD
+//To find Approx Value
+function [A]= approx (V,n)
+ A= round(V*10^n)/10^n; //V-Value, n-to what place
+ funcprot (0)
+endfunction
+g=32.2;//[ft/s^2]
+Gamma=62.2; //[lbf/ft^3]
+d=8/12; //diameter[ft]
+A=%pi*d^2/4 //area[ft^2]
+Q=2; //discharge[cfs]
+V2=approx(Q/A,2) //[ft/s]
+p1=14.7; //[lbf/in^2]
+//Pressure head at reservoir,hr
+hr=approx(p1*144/Gamma,0) //[ft]
+//Energy equation, (p1/gamma)+(V1^2/2g)+z1=(p2/gamma)+(V2^2/2g)+z2+hL
+//V1=0,z1=0
+z2=6; //[m]
+Ce=0.1; //entrance loss coefficient
+Cb=0.2; //bend loss coefficient
+hL=(Ce+Cb)*V2^2/(2*g)
+//Head at pump entrance, he=p2/Gamma
+he=approx(hr-z2-V2^2/(2*g)-hL,1) //[ft]
+pvap=0.506; //vapor pressure[psi]
+hp=pvap*144/Gamma //[ft]
+//Net positive suction head
+NPSH=he-hp //[ft]
+printf("\nThe net positive suction head = %.1f ft.\n",NPSH)
+//Traditional suction specific speed
+N=1750; //in rpm
+//1cfs=449 gpm
+Nss=N*(Q*449)^(1/2)/NPSH^(3/4)
+printf("\nThe traditional suction specific speed, Nss = %.f.\n",Nss)
\ No newline at end of file diff --git a/629/CH14/EX14.8/ex14_8.txt b/629/CH14/EX14.8/ex14_8.txt new file mode 100644 index 000000000..9ab622b6c --- /dev/null +++ b/629/CH14/EX14.8/ex14_8.txt @@ -0,0 +1,3 @@ +
+The efficiency of an impeller of diameter 1.80 m = 0.89.
+
\ No newline at end of file diff --git a/629/CH14/EX14.8/example14_8.sce b/629/CH14/EX14.8/example14_8.sce new file mode 100644 index 000000000..d1b95e410 --- /dev/null +++ b/629/CH14/EX14.8/example14_8.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 14.8 VISCOUS EFFECTS ON PUMP EFFICIENCY
+D1=0.45; //diameter[m]
+D=1.8; //[m]
+eta1=0.85;
+//Efficiency
+eta=1-(1-eta1)/(D/D1)^(1/5)
+printf("\nThe efficiency of an impeller of diameter 1.80 m = %.2f.\n",eta)
\ No newline at end of file diff --git a/629/CH14/EX14.9/ex14_9.txt b/629/CH14/EX14.9/ex14_9.txt new file mode 100644 index 000000000..ca9201e50 --- /dev/null +++ b/629/CH14/EX14.9/ex14_9.txt @@ -0,0 +1,3 @@ +
+The shaft power required to operate the compressor = 118 kW.
+
\ No newline at end of file diff --git a/629/CH14/EX14.9/example14_9.sce b/629/CH14/EX14.9/example14_9.sce new file mode 100644 index 000000000..f5be518eb --- /dev/null +++ b/629/CH14/EX14.9/example14_9.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 14.9 CENTRIFUGAL COMPRESSOR
+p1=100; //pressure [kPa]
+p2=200; //[kPa]
+k=1.4;
+Q1=1;//discharge [m^3/s]
+eta=0.65; //efficiency
+//Theoretical power
+Ptheo=(k/(k-1))*Q1*p1*[(p2/p1)^((k-1)/k)-1] //[kW]
+//Shaft power
+Pshaft=Ptheo/eta //[kW]
+printf("\nThe shaft power required to operate the compressor = %.f kW.\n",Pshaft)
\ No newline at end of file diff --git a/629/CH15/EX15.1/ex15_1.txt b/629/CH15/EX15.1/ex15_1.txt new file mode 100644 index 000000000..c13fb31c8 --- /dev/null +++ b/629/CH15/EX15.1/ex15_1.txt @@ -0,0 +1,3 @@ +
+The maximum depth assured of having laminar flow = 0.062 ft.
+
\ No newline at end of file diff --git a/629/CH15/EX15.1/example15_1.sce b/629/CH15/EX15.1/example15_1.sce new file mode 100644 index 000000000..6e82e1160 --- /dev/null +++ b/629/CH15/EX15.1/example15_1.sce @@ -0,0 +1,15 @@ +clear
+clc
+//Example 15.1 CONDITIONS FOR LAMINAR OPEN-CHANNEL FLOW
+V=0.1; //velocity [ft/s]
+B=10; //width[ft]
+y=6; //depth[ft]
+Rh=B*y/(B+2*y) //[ft]
+v=1.22*10^-5; //[ft^2]
+Re=V*Rh/v //Reynolds number
+//Re>500, flow is turbulent
+//Depth, ymax for Re=500
+Re1=500;
+Rh1=Re1*v/V //[ft]
+ymax=B*Rh1/(B-2*Rh1) //[ft]
+printf("\nThe maximum depth assured of having laminar flow = %.3f ft.\n",ymax)
\ No newline at end of file diff --git a/629/CH15/EX15.2/ex15_2.txt b/629/CH15/EX15.2/ex15_2.txt new file mode 100644 index 000000000..e5941e9b4 --- /dev/null +++ b/629/CH15/EX15.2/ex15_2.txt @@ -0,0 +1,3 @@ +
+The discharge of water is 503 cfs.
+
\ No newline at end of file diff --git a/629/CH15/EX15.2/example15_2.sce b/629/CH15/EX15.2/example15_2.sce new file mode 100644 index 000000000..28ca05182 --- /dev/null +++ b/629/CH15/EX15.2/example15_2.sce @@ -0,0 +1,23 @@ +clear
+clc
+//Example 15.2 ESTIMATING Q FOR UNIFORM FLOW USING DARCY-WEISBACH EQUATION
+g=32.2; //[ft/s^2]
+l=10; //width[ft]
+y=6; //depth [ft]
+A=l*y //area[ft^2]
+So=0.0016; //slope of the channel
+v=1.2*10^-5; //[ft^2/s]
+Rh=l*y/(l+2*y); //[ft]
+ks=0.005; //[ft](Assume)
+Rr=ks/(4*Rh) //relative roughness
+//Estimating f from Rr, using Moody diagram
+f=0.016;
+//first iteration for V
+V=sqrt(8*g*Rh*So/f) //[ft/s]
+//Recalculate Re
+Re=V*4*Rh/v
+//Using Rr,new Re, read
+f1=0.016;
+//f1=f,meets reasonable convergence criterion for V
+Q=V*A //discharge [cfs]
+printf("\nThe discharge of water is %.f cfs.\n",Q)
\ No newline at end of file diff --git a/629/CH15/EX15.3/ex15_3.txt b/629/CH15/EX15.3/ex15_3.txt new file mode 100644 index 000000000..5ab75c576 --- /dev/null +++ b/629/CH15/EX15.3/ex15_3.txt @@ -0,0 +1,3 @@ +
+The value of resistance coefficient = 0.130
+
\ No newline at end of file diff --git a/629/CH15/EX15.3/example15_3.sce b/629/CH15/EX15.3/example15_3.sce new file mode 100644 index 000000000..957d7c74c --- /dev/null +++ b/629/CH15/EX15.3/example15_3.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 15.3 RESISTANCE COEFFICIENT FOR BOULDERS
+l=100; //width[ft]
+y=4.3; //depth[ft]
+d84=0.72; //[ft]
+//for wide channel, take Rh=b
+Rh=y
+//Resistance coefficient
+f=(1.2+(2.03*(log10(Rh/d84))))^-2
+printf("\nThe value of resistance coefficient = %.3f\n",f)
\ No newline at end of file diff --git a/629/CH15/EX15.4/ex15_4.txt b/629/CH15/EX15.4/ex15_4.txt new file mode 100644 index 000000000..a7aed98d8 --- /dev/null +++ b/629/CH15/EX15.4/ex15_4.txt @@ -0,0 +1,4 @@ +
+The discharge in the channel, Q = 2176 cfs.
+
+The numerical value of Mannings n for this channel = 0.0426
\ No newline at end of file diff --git a/629/CH15/EX15.4/example15_4.sce b/629/CH15/EX15.4/example15_4.sce new file mode 100644 index 000000000..3a7096919 --- /dev/null +++ b/629/CH15/EX15.4/example15_4.sce @@ -0,0 +1,22 @@ +clear
+clc
+//Example 15.4 CALCULATING DISCHARGE AND MANNING’S n USING CHEZY EQUATION
+//To find Approx Value
+function [A]= approx (V,n)
+ A= round(V*10^n)/10^n; //V-Value, n-to what place
+ funcprot (0)
+endfunction
+g=32.2; //[ft/s^2]
+l=100; //width[ft]
+y=4.3; //depth[ft]
+A=l*y //area[ft^2]
+//Estimate Rh to be y
+Rh=y
+f=0.13; //friction factor
+So=0.003; //slope
+V=approx(sqrt(8*g*Rh*So/f),2) //velocity[ft/s]
+Q=approx(V*A,0) //discharge[cfs]
+printf("\nThe discharge in the channel, Q = %.f cfs.\n",Q)
+//Manning's n
+n=1.49*A*Rh^(2/3)*So^(1/2)/Q
+printf("\nThe numerical value of Mannings n for this channel = %.4f\n",n)
\ No newline at end of file diff --git a/629/CH15/EX15.5/ex15_5.txt b/629/CH15/EX15.5/ex15_5.txt new file mode 100644 index 000000000..1f304f358 --- /dev/null +++ b/629/CH15/EX15.5/ex15_5.txt @@ -0,0 +1,3 @@ +
+The discharge in the concrete channel = 465 cfs.
+
\ No newline at end of file diff --git a/629/CH15/EX15.5/example15_5.sce b/629/CH15/EX15.5/example15_5.sce new file mode 100644 index 000000000..f9add62ed --- /dev/null +++ b/629/CH15/EX15.5/example15_5.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 15.5 DISCHARGE USING CHEZY EQUATION
+n=0.015;
+l=10; //width[ft]
+y=6; //depth[ft]
+Rh=l*y/(l+2*y)
+So=0.0016; //channel slope
+A=l*y //area[ft^2]
+//Discharge
+Q=1.49*A*Rh^(2/3)*So^(1/2)/n //[cfs]
+printf("\nThe discharge in the concrete channel = %.f cfs.\n",Q)
\ No newline at end of file diff --git a/629/CH2/EX2.1/ex2_1.txt b/629/CH2/EX2.1/ex2_1.txt new file mode 100644 index 000000000..e057e8c02 --- /dev/null +++ b/629/CH2/EX2.1/ex2_1.txt @@ -0,0 +1,2 @@ + The density of the air = 1.27 kg/m^3.
+
\ No newline at end of file diff --git a/629/CH2/EX2.1/example2_1.sce b/629/CH2/EX2.1/example2_1.sce new file mode 100644 index 000000000..321808ce2 --- /dev/null +++ b/629/CH2/EX2.1/example2_1.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 2.1 DENSITY OF AIR
+T=4+273; //[K]
+p=101*10^3; //[N/m^2]
+R=287; //[J/kg.K]
+//Density
+rho=p/(R*T) //[kg/m^3]
+printf("\n The density of the air = %.2f kg/m^3.\n",rho)
diff --git a/629/CH2/EX2.2/ex2_2.txt b/629/CH2/EX2.2/ex2_2.txt new file mode 100644 index 000000000..92efee69a --- /dev/null +++ b/629/CH2/EX2.2/ex2_2.txt @@ -0,0 +1,2 @@ +
+ The viscosity of water at 30°C = 8.02*10^(-4) N.s/m^2.
\ No newline at end of file diff --git a/629/CH2/EX2.2/example2_2.sce b/629/CH2/EX2.2/example2_2.sce new file mode 100644 index 000000000..961a4bf37 --- /dev/null +++ b/629/CH2/EX2.2/example2_2.sce @@ -0,0 +1,21 @@ +clear
+clc
+//Example 2.2 CALCULATING VISCOSITY OF LIQUID AS A FUNCTION OF TEMPERATURE
+T1=20+273; //[K]
+T2=40+273; //[K]
+mu1=10^-3; //[N.s/m^2]
+mu2=6.53*10^-4; //[N.s/m^2]
+
+//ln(mu)=ln(C)+b/T
+A=[1 1/T1;1 1/T2]
+B=[log(mu1);log(mu2)]
+//Az=B, z=[log(C);b]
+z=inv(A)*B
+C=exp(z(1))
+b=z(2) //[K]
+
+//At T=30°C,
+T=30+273; //[K]
+//Viscosity
+mu=C*exp(b/T)*10^4 //in 10^-4 N.s/m^2
+printf("\n The viscosity of water at 30°C = %.2f*10^(-4) N.s/m^2. \n",mu)
diff --git a/629/CH2/EX2.3/ex2_3.txt b/629/CH2/EX2.3/ex2_3.txt new file mode 100644 index 000000000..e0762f3d7 --- /dev/null +++ b/629/CH2/EX2.3/ex2_3.txt @@ -0,0 +1 @@ +The space between board and the tramp = 0.117 mm.
\ No newline at end of file diff --git a/629/CH2/EX2.3/example2_3.sce b/629/CH2/EX2.3/example2_3.sce new file mode 100644 index 000000000..14d04964a --- /dev/null +++ b/629/CH2/EX2.3/example2_3.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 2.3 MODELING A BOARD SLIDING ON A LIQUID LAYER
+mu=0.05; //[N.s/m^2]
+l=1; //[m]
+A=l^2 //area[m^2]
+delV=0.02; //ΔV[m]
+W=25; //[N]
+//Frebody analysis
+Ft=W*sind(20) //tangential force[N]
+Fs=Ft //shear force[N]
+dely=mu*delV*A*10^3/Fs //Δy[mm]
+printf("\n The space between board and the tramp = %.3f mm.\n",dely)
\ No newline at end of file diff --git a/629/CH2/EX2.4/ex2_4.txt b/629/CH2/EX2.4/ex2_4.txt new file mode 100644 index 000000000..87a897a14 --- /dev/null +++ b/629/CH2/EX2.4/ex2_4.txt @@ -0,0 +1 @@ + The height above the reservoir level, the water rises in a glass tube = 18.6 mm.
\ No newline at end of file diff --git a/629/CH2/EX2.4/example2_4.sce b/629/CH2/EX2.4/example2_4.sce new file mode 100644 index 000000000..95a035738 --- /dev/null +++ b/629/CH2/EX2.4/example2_4.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 2.4 CAPILLARY RISE IN A TUBE
+d=1.6*10^-3; //[m]
+A=%pi*d^2/4; //area[m^2]
+Gamma=9790; //[N/m^3]
+sigma=0.073; //[N/m]
+//Force balance, Ft-W=0, W=Gamma*Δh*A, theta=0°(assume)
+theta=0;
+Ft=sigma*%pi*d*cosd(theta) //surface tension force [N]
+delh=Ft*10^3/(Gamma*A) //Δh[mm]
+printf("\n The height above the reservoir level, the water rises in a glass tube = %.1f mm.\n",delh)
\ No newline at end of file diff --git a/629/CH3/EX3.1/ex3_1.txt b/629/CH3/EX3.1/ex3_1.txt new file mode 100644 index 000000000..5c3e5ed58 --- /dev/null +++ b/629/CH3/EX3.1/ex3_1.txt @@ -0,0 +1,2 @@ +
+The load that the jack can support = 12.2 kN.
\ No newline at end of file diff --git a/629/CH3/EX3.1/example3_1.sce b/629/CH3/EX3.1/example3_1.sce new file mode 100644 index 000000000..2f9d1bb63 --- /dev/null +++ b/629/CH3/EX3.1/example3_1.sce @@ -0,0 +1,20 @@ +clear
+clc
+//Example 3.1 LOAD LIFTED BY A HYDRAULIC JACK
+//Moment equilibrium at C, (F*l)-(F1*l1)=0
+l=0.33; //[m]
+F=100; //[N]
+l1=0.03; //[m]
+F1=l*F/l1 //[N]
+
+//Force equilibrium (small piston), (p1*A1)-F1=0
+d1=1.5*10^-2; //[m]
+A1=%pi*d1^2/4; //[m^2]
+p1=F1/A1 //[N/m^2]
+
+//Force equilibrium (lifter), F2-(p2*A2)=0
+d2=5*10^-2; //[m]
+A2=%pi*d2^2/4 //[m^2]
+p2=p1 //[N/m^2](as both are at same elevation)
+F2=p2*A2/10^3 //[kN]
+printf("\nThe load that the jack can support = %.1f kN.\n",F2)
\ No newline at end of file diff --git a/629/CH3/EX3.10/ex3_10.txt b/629/CH3/EX3.10/ex3_10.txt new file mode 100644 index 000000000..c5a5f05ea --- /dev/null +++ b/629/CH3/EX3.10/ex3_10.txt @@ -0,0 +1,3 @@ +
+The normal force F required to open the gate = 809 kN.
+
\ No newline at end of file diff --git a/629/CH3/EX3.10/example3_10.sce b/629/CH3/EX3.10/example3_10.sce new file mode 100644 index 000000000..854d46fde --- /dev/null +++ b/629/CH3/EX3.10/example3_10.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 3.10 FORCE TO OPEN AN ELLIPTICAL GATE
+g_water=9810; //specific weight of water[N/m^3]
+z_cen=10; //[m]
+p=(g_water*z_cen)/10^3 //pressure at the centroid level[kPa]
+a=2.5; //[m]
+b=2; //[m]
+A=%pi*a*b //area[m^2]
+Fp=p*A/10^3//resultant force[MN]
+I=%pi*a^3*b/4 //moment of inertia[m^4]
+y=12.5; //slant distance from the water surface to centroid[m]
+del_y=I/(y*A)//distance from center to center of pressure[m]
+//Moment equilibrium about the hinge
+F=Fp*10^3*(a+del_y)/(2*a) //force[kN]
+printf("\nThe normal force F required to open the gate = %.f kN.\n",F)
\ No newline at end of file diff --git a/629/CH3/EX3.11/ex3_11.txt b/629/CH3/EX3.11/ex3_11.txt new file mode 100644 index 000000000..331d992df --- /dev/null +++ b/629/CH3/EX3.11/ex3_11.txt @@ -0,0 +1,7 @@ +
+ The line of action for vertical force, xcp = 0.957m
+ and for horizontal force, ycp = 5.067m
+
+
+ Hydrostatic force on curved surface AB = 146.9 kN at 48 degrees to the horizontal.
+
\ No newline at end of file diff --git a/629/CH3/EX3.11/example3_11.sce b/629/CH3/EX3.11/example3_11.sce new file mode 100644 index 000000000..dd4d790df --- /dev/null +++ b/629/CH3/EX3.11/example3_11.sce @@ -0,0 +1,33 @@ +clear
+clc
+//Example 3.11 HYDROSTATIC FORCE ON A CURVED SURFACE
+//Equilibrium in horizontal direction
+g_w=9.81; //specific weight of water[kN/m^3]
+r=2; //[m]
+w=1; //[m]
+A=r*w //area[m^2]
+l=5; //[m]
+p=g_w*l //pressure[kN/m^3]
+Fx=p*A //horizontal force on side AC[kN]
+
+//Equilibrium in vertical direction
+h=4; //height[m]
+p0=g_w*h //[kN/m^2]
+Fv=p0*A //vertical force on side CB[kN]
+W=g_w*(%pi*r^2/4)*w //weight of water in ABC[kN]
+Fy=W+Fv //[kN]
+
+//Line of action (horizontal force)
+y=5; //[m]
+ycp=y+(w*r^3/12)/(y*A) //[m]
+
+//For line of action for vertical forces, sum moments about point C
+xW=4*r/(3*%pi) //distance of centroid from C[m]
+xcp=(Fv*r/2+W*xW)/Fy //[m]
+printf("\n The line of action for vertical force, xcp = %.3fm\n and for horizontal force, ycp = %.3fm\n\n",xcp,ycp)
+
+//tan(theta)=Fy/Fx
+theta=atand(Fy/Fx) //angle with the horizontal(degrees)
+F=sqrt(Fx^2+Fy^2) //resultant force[kN]
+printf("\n Hydrostatic force on curved surface AB = %.1f kN at %.f degrees to the horizontal.\n",F,theta)
+
diff --git a/629/CH3/EX3.12/ex3_12.txt b/629/CH3/EX3.12/ex3_12.txt new file mode 100644 index 000000000..c8aabf416 --- /dev/null +++ b/629/CH3/EX3.12/ex3_12.txt @@ -0,0 +1,4 @@ +
+ The tension in the cord = 0.110 N.
+
+ The mass of metal part = 17.8 grams.
\ No newline at end of file diff --git a/629/CH3/EX3.12/example3_12.sce b/629/CH3/EX3.12/example3_12.sce new file mode 100644 index 000000000..6649b7e1a --- /dev/null +++ b/629/CH3/EX3.12/example3_12.sce @@ -0,0 +1,24 @@ +clear
+clc
+//Example 3.12 BUOYANT FORCE ON A METAL PART
+//dimensions of wooden block
+l=50; //[mm]
+b=50; //[mm]
+h=10; //[mm]
+y=7.5; //submerged height[mm]
+V=l*b*y //volume of block submerged[mm^3]
+g_w=9800; //specific weight of water[N/m^3]
+Fb1=g_w*V*10^(-9) //buoyant force[N], (factor 10^(-9)m^3/mm^3)
+S1=0.3;
+V1=l*b*h //volume of block[mm^3]
+W1=g_w*S1*V1*10^(-9) //weight of block[N]
+T=Fb1-W1 //tension in cord[N]
+printf("\n The tension in the cord = %.3f N.\n",T)
+
+V2=6600; //volume of metal[mm^3]
+Fb2=g_w*V2*10^(-9) //buoyant force[N]
+//Equilibrium equation, W2-T-Fb2=0
+g=9.81; //[m/s^2]
+W2=T+Fb2 //weight of metal[N]
+m2=W2*10^3/g //mass of metal[gm](factor 10^3g/1kg)
+printf("\n The mass of metal part = %.1f grams.\n",m2)
\ No newline at end of file diff --git a/629/CH3/EX3.13/ex3_13.txt b/629/CH3/EX3.13/ex3_13.txt new file mode 100644 index 000000000..ef3405d34 --- /dev/null +++ b/629/CH3/EX3.13/ex3_13.txt @@ -0,0 +1,3 @@ +
+ Since the metacentric height is positive, the block will be stable in this slightly disturbed position.
+
\ No newline at end of file diff --git a/629/CH3/EX3.13/example3_13.sce b/629/CH3/EX3.13/example3_13.sce new file mode 100644 index 000000000..6f962ef19 --- /dev/null +++ b/629/CH3/EX3.13/example3_13.sce @@ -0,0 +1,31 @@ +clear
+clc
+//Example 3.13 STABILITY OF A FLOATING BLOCK
+//Dimensions of the block
+l=0.6; //[m]
+b=0.3; //[m]
+h=0.3; //[m]
+Vb=l*b*h //volume[m^3]
+W=318; //weight[N]
+g_w=9810; //[N/m^3]
+//For equilibrium(vertical direction), -weight+buoyant force=0
+d=W/(g_w*l*b) //[m]
+//Stability (longitudinal axis)
+Io1=((l*b^3)/12)//[m^4]
+CG1=0.06; //[m]
+V=d*l*b //[m^3]
+GM1=Io1/V-CG1 //metacentric height[m]
+//The metacentric height is negative hence the block is not stable about the longitudinal axis
+
+//For the block slightly disturbed
+VD=W/g_w //displaced volume[m^3]
+//(Displaced volume)=(Block volume)-(Volume above the waterline)
+//VD=Vb-(w^2*l/4)
+w=sqrt((Vb-VD)*4/l) //[m]
+
+//Moment of inertia of waterline
+Io2=(l*w^3)/12 //[m^4]
+CG2=0.0573; //[m]
+GM2=Io2/VD-CG2 //[m]
+printf("\n Since the metacentric height is positive, the block will be stable in this slightly disturbed position.\n")
+
diff --git a/629/CH3/EX3.2/ex3_2.txt b/629/CH3/EX3.2/ex3_2.txt new file mode 100644 index 000000000..d5dcfac5a --- /dev/null +++ b/629/CH3/EX3.2/ex3_2.txt @@ -0,0 +1,2 @@ +
+The water pressure at the depth of 35ft in the tank = 15.2 psig.
\ No newline at end of file diff --git a/629/CH3/EX3.2/example3_2.sce b/629/CH3/EX3.2/example3_2.sce new file mode 100644 index 000000000..a84ab7fbf --- /dev/null +++ b/629/CH3/EX3.2/example3_2.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 3.2 WATER PRESSURE IN A TANK
+//Hydrostatic equation, p1/Gamma +z1=p2/Gamma +z2
+p1=0; //[psig]
+z1=250; //[ft]
+z2=215; //[ft]
+Gamma=62.4; //specific weight of water[lbf/ft^3]
+//1psig=144psfg
+p2=p1+(z1-z2)*Gamma/144 //[psig]
+printf("\nThe water pressure at the depth of 35ft in the tank = %.1f psig.\n",p2)
\ No newline at end of file diff --git a/629/CH3/EX3.3/ex3_3.txt b/629/CH3/EX3.3/ex3_3.txt new file mode 100644 index 000000000..621dfd197 --- /dev/null +++ b/629/CH3/EX3.3/ex3_3.txt @@ -0,0 +1,2 @@ +
+The gage pressure at the bottom of the tank = 27.7 kPa gage.
\ No newline at end of file diff --git a/629/CH3/EX3.3/example3_3.sce b/629/CH3/EX3.3/example3_3.sce new file mode 100644 index 000000000..3e1d146f8 --- /dev/null +++ b/629/CH3/EX3.3/example3_3.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 3.3 PRESSURE IN TANK WITH TWO FLUIDS
+//Hydrostatic equation(oil), p1/g_oil +z1=p2/g_oil +z2
+p1=0; //[Pa]
+z1=3; //[m]
+z2=2.1; //[m]
+g_oil=0.8*9810; //specific weight of oil[N/m^3]
+p2=(p1+(z1-z2)*g_oil)/10^3//[kPa] (factor 1kPa/10^3Pa)
+
+//At oil-water interface, p2_oil=p2_water
+//Hydrostatic equation(water), p2/g_water +z2=p3/g_water +z3
+z3=0; //[m]
+g_water=9810; //specific weight of water[N/m^3]
+p3=p2+(z2-z3)*g_water/10^3 //[kPa]
+printf("\nThe gage pressure at the bottom of the tank = %.1f kPa gage.\n",p3)
\ No newline at end of file diff --git a/629/CH3/EX3.4/ex3_4.txt b/629/CH3/EX3.4/ex3_4.txt new file mode 100644 index 000000000..90496176e --- /dev/null +++ b/629/CH3/EX3.4/ex3_4.txt @@ -0,0 +1,3 @@ +
+The pressure at an elevation of 2000m = 80.0 kPa absolute.
+
\ No newline at end of file diff --git a/629/CH3/EX3.4/example3_4.sce b/629/CH3/EX3.4/example3_4.sce new file mode 100644 index 000000000..85e93b313 --- /dev/null +++ b/629/CH3/EX3.4/example3_4.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 3.4 PRESSURE IN THE TROPOSPHERE
+p0=101.3; //absolute pressure[kPa]
+T0=23+273; //absolute temp.[K]
+alpha=5.87*10^-3; //[K/m]
+delz=2000; //[m](delz=z-z0)
+k=5.823; //(k=g/alpha*R)
+
+//Pressure equation in troposphere
+p=p0*((T0-alpha*delz)/T0)^k //[kPa]
+printf("\nThe pressure at an elevation of 2000m = %.1f kPa absolute.\n",p)
\ No newline at end of file diff --git a/629/CH3/EX3.5/ex3_5.txt b/629/CH3/EX3.5/ex3_5.txt new file mode 100644 index 000000000..2a72701e2 --- /dev/null +++ b/629/CH3/EX3.5/ex3_5.txt @@ -0,0 +1,3 @@ +
+The pressure at an elevation of 55,000 ft, p = 1.43 psia (= 9.82 kPa absolute).
+
\ No newline at end of file diff --git a/629/CH3/EX3.5/example3_5.sce b/629/CH3/EX3.5/example3_5.sce new file mode 100644 index 000000000..0981b373e --- /dev/null +++ b/629/CH3/EX3.5/example3_5.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 3.5 PRESSURE IN THE LOWER STRATOSPHERE
+T=-71.5+460; //temperature[°R]
+z0=45000; //[ft]
+z=55000; //[ft]
+p0=2.31; //pressure at z0[psia]
+g=32.2; //[ft/s^2]
+R=1716; //[ft^2/R.s^2]
+p=p0*exp(-(z-z0)*(g/(R*T))) //pressure at z[psia]
+p1=p*(101.3/14.7); //in SI units[kPa]
+printf("\nThe pressure at an elevation of 55,000 ft, p = %.2f psia (= %.2f kPa absolute).\n",p,p1)
diff --git a/629/CH3/EX3.6/ex3_6.txt b/629/CH3/EX3.6/ex3_6.txt new file mode 100644 index 000000000..42bf141db --- /dev/null +++ b/629/CH3/EX3.6/ex3_6.txt @@ -0,0 +1,3 @@ +
+The gage pressure at the center of the pipe = 62.1 kPa gage.
+
\ No newline at end of file diff --git a/629/CH3/EX3.6/example3_6.sce b/629/CH3/EX3.6/example3_6.sce new file mode 100644 index 000000000..3f5fe4c9d --- /dev/null +++ b/629/CH3/EX3.6/example3_6.sce @@ -0,0 +1,15 @@ +clear
+clc
+//Example 3.6 PRESSURE MEASUREMENT (U-TUBE MANOMETER)
+p1=0; //[Pa]
+h12=0.6; //deflection[m]
+g_m=133000; //specific weight of mercury[N/m^3]
+p2=p1+g_m*h12 //[Pa]
+//From hydrostatic equation,as z3=z2
+p3=p2; //[Pa]
+
+//Pressure at the interface is constant, p3_mercury=p3_water=p3
+l=1.8; //[m]
+g_w=9810; //specific weight of water[N/m^3]
+p4=(p3-g_w*l)/10^3 //[kPa](factor 1kPa/10^3Pa)
+printf("\nThe gage pressure at the center of the pipe = %.1f kPa gage.\n",p4)
\ No newline at end of file diff --git a/629/CH3/EX3.7/ex3_7.txt b/629/CH3/EX3.7/ex3_7.txt new file mode 100644 index 000000000..ae1eb6e6f --- /dev/null +++ b/629/CH3/EX3.7/ex3_7.txt @@ -0,0 +1,3 @@ +
+The pressure of the air in the tank = 110 kPa gage.
+
\ No newline at end of file diff --git a/629/CH3/EX3.7/example3_7.sce b/629/CH3/EX3.7/example3_7.sce new file mode 100644 index 000000000..38f4cc6d4 --- /dev/null +++ b/629/CH3/EX3.7/example3_7.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 3.7 MANOMETER ANALYSIS
+l1=0.4; //[m]
+l2=1.0; //[m]
+l3=0.8; //[m]
+S_oil=0.8;
+//Specific weights
+g_water=9810; //[N/m^3]
+g_oil=S_oil*g_water //[N/m^3]
+g_air=0; //[N/m^3]
+g_mercury=133000; //[N/m^3]
+p1=0; //pressure[Pa]
+//From Manometer equation
+p2=(p1+g_mercury*l3-g_air*l2+g_oil*l1)/10^3 //[kPa],(factor 1kPa/10^3Pa)
+printf("\nThe pressure of the air in the tank = %.f kPa gage.\n",p2)
\ No newline at end of file diff --git a/629/CH3/EX3.8/ex3_8.txt b/629/CH3/EX3.8/ex3_8.txt new file mode 100644 index 000000000..0ddbbc80e --- /dev/null +++ b/629/CH3/EX3.8/ex3_8.txt @@ -0,0 +1,5 @@ +
+Between the points 1 and 2,
+the change in piezometric pressure = 65.4 psf
+and piezometric head = 1.05 ft.
+
\ No newline at end of file diff --git a/629/CH3/EX3.8/example3_8.sce b/629/CH3/EX3.8/example3_8.sce new file mode 100644 index 000000000..13fa5e921 --- /dev/null +++ b/629/CH3/EX3.8/example3_8.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 3.8 CHANGE IN PIEZOMETRIC HEAD FOR PIPE FLOW
+del_h=1/12; //deflection[ft]
+//Specific weights
+g_Hg=847; //[lbf/ft^3]
+g_water=62.4; //[lbf/ft^3]
+h=(del_h*((g_Hg/g_water)-1)) //[ft]
+//Piezometric pressure
+pz=h*g_water //[psf]
+printf("\nBetween the points 1 and 2,\nthe change in piezometric pressure = %.1f psf \nand piezometric head = %.2f ft.\n",pz,h)
\ No newline at end of file diff --git a/629/CH3/EX3.9/ex3_9.txt b/629/CH3/EX3.9/ex3_9.txt new file mode 100644 index 000000000..fd9f45811 --- /dev/null +++ b/629/CH3/EX3.9/ex3_9.txt @@ -0,0 +1,2 @@ +
+The force acting on one side of a concrete = 85.7 kN.
\ No newline at end of file diff --git a/629/CH3/EX3.9/example3_9.sce b/629/CH3/EX3.9/example3_9.sce new file mode 100644 index 000000000..bc474c6d2 --- /dev/null +++ b/629/CH3/EX3.9/example3_9.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 3.9 HYDROSTATIC FORCE DUE TO CONCRETE
+z=2.44; //height[m]
+w=1.22; //width[m]
+A=z*w //area of panel[m^2]
+g_concrete=23.6; //specific weight of concrete[kN/m^3]
+z_cen=z/2 //depth of centroid[m]
+p=g_concrete*z_cen //pressure at the centroid level[kPa]
+F=p*A //force[kN]
+printf("\nThe force acting on one side of a concrete = %.1f kN.\n",F)
\ No newline at end of file diff --git a/629/CH4/EX4.12/ex4_12.txt b/629/CH4/EX4.12/ex4_12.txt new file mode 100644 index 000000000..b067ae916 --- /dev/null +++ b/629/CH4/EX4.12/ex4_12.txt @@ -0,0 +1,5 @@ +
+The velocity at 10km from the center = 16 m/s.
+
+The pressure difference between two locations = 806 Pa.
+
\ No newline at end of file diff --git a/629/CH4/EX4.12/example4_12.sce b/629/CH4/EX4.12/example4_12.sce new file mode 100644 index 000000000..7639c1e58 --- /dev/null +++ b/629/CH4/EX4.12/example4_12.sce @@ -0,0 +1,14 @@ +clear
+clc
+//Example 4.12 VELOCITY AND PRESSURE DISTRIBUTION IN A FREE VORTEX
+rho=1.2; //density[kg/m^3]
+r1=4; //[km]
+r2=10; //[km]
+V1=40; //velocity[m/s]
+//Velocity distribution, V=C/r
+C=r1*V1
+V2=C/r2 //[m/s]
+printf("\nThe velocity at 10km from the center = %.f m/s.\n",V2)
+//Bernoulli equation, for horizontal plane
+delp=rho*(V1^2-V2^2)/2 //delp=p2-p1,[Pa]
+printf("\nThe pressure difference between two locations = %.f Pa.\n",delp)
\ No newline at end of file diff --git a/629/CH4/EX4.13/ex4_13.txt b/629/CH4/EX4.13/ex4_13.txt new file mode 100644 index 000000000..59d095eeb --- /dev/null +++ b/629/CH4/EX4.13/ex4_13.txt @@ -0,0 +1,2 @@ +
+The pressure difference between center and outer edge of mercury = -1.59 in Hg
\ No newline at end of file diff --git a/629/CH4/EX4.13/example4_13.sce b/629/CH4/EX4.13/example4_13.sce new file mode 100644 index 000000000..41b34a273 --- /dev/null +++ b/629/CH4/EX4.13/example4_13.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 4.13 PRESSURE DIFFERENCE IN TORNADO
+//1mi=5280ft, 1hr=3600s
+V=150*5280/3600 //[ft/s]
+//1slug=32.2lbm
+rho=0.075/32.2 //density[slug/ft^3]
+//Pressure difference, p1-p0=-rho*V^2
+//29.92 in Hg=2116 psf
+delp=-rho*V^2*(29.92/2116) //inches of Hg
+printf("\nThe pressure difference between center and outer edge of mercury = %.2f in Hg.\n",delp)
\ No newline at end of file diff --git a/629/CH4/EX4.2/ex4_2.txt b/629/CH4/EX4.2/ex4_2.txt new file mode 100644 index 000000000..f82815e54 --- /dev/null +++ b/629/CH4/EX4.2/ex4_2.txt @@ -0,0 +1,2 @@ +
+The gage pressure on the piston, p = 11.0 kPa, gage.
\ No newline at end of file diff --git a/629/CH4/EX4.2/example4_2.sce b/629/CH4/EX4.2/example4_2.sce new file mode 100644 index 000000000..d53b32650 --- /dev/null +++ b/629/CH4/EX4.2/example4_2.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 4.2 APPLICATION OF EULER’S EQUATION TO ACCELERATION OF A FLUID
+g=9.81; //[m/s^2]
+az=100; //acceleration in z direction [m/s^2]
+rho=10^3; //[kg/m^3]
+//Integrating, d(p+g*rho*z)/dz=-rho*az
+//(p2+rho*g*z2)-(p1+rho*g*z1)=-rho*az*(z2-z1)
+delz=0.1; //delz=z2-z1,[m]
+p2=0;
+p1=(p2+rho*(g+az)*delz)/10^3 //[kPa]
+printf("\nThe gage pressure on the piston, p = %.1f kPa, gage.\n",p1)
\ No newline at end of file diff --git a/629/CH4/EX4.3/ex4_3.txt b/629/CH4/EX4.3/ex4_3.txt new file mode 100644 index 000000000..8e3d17ada --- /dev/null +++ b/629/CH4/EX4.3/ex4_3.txt @@ -0,0 +1,4 @@ +
+(a)The pressure at the top front, p = 261 psfg (= 12.5 kPa,gage).
+
+(b)The maximum pressure in the tank, pmax = 513 psfg (= 24.6 kPa,gage)
\ No newline at end of file diff --git a/629/CH4/EX4.3/example4_3.sce b/629/CH4/EX4.3/example4_3.sce new file mode 100644 index 000000000..783c2a930 --- /dev/null +++ b/629/CH4/EX4.3/example4_3.sce @@ -0,0 +1,17 @@ +clear
+clc
+//Example 4.3 PRESSURE IN A DECELERATING TANK OF LIQUID
+p1=0;
+Gamma=42; //[lbf/ft^3]
+g=32.2; //[ft/s^2]
+al=-10; //[ft/s^2]
+l=20; //[ft]
+//Euler's equation along the top of tank, dp/dl=-Gamma*l/g
+p2=p1-Gamma*al*l/g //[psfg]
+//1kPa=20.88psfg
+printf("\n(a)The pressure at the top front, p = %.f psfg (= %.1f kPa,gage).\n",p2,p2/20.88)
+//Eulers equation in vertical direction,
+//d(p+Gamma*z)/dz=-rho*az, az=0
+delz=6; //delz=z2-z3,[ft]
+p3=p2+Gamma*delz //[psfg]
+printf("\n(b)The maximum pressure in the tank, pmax = %.f psfg (= %.1f kPa,gage).\n",p3,p3/20.88)
\ No newline at end of file diff --git a/629/CH4/EX4.4/ex4_4.txt b/629/CH4/EX4.4/ex4_4.txt new file mode 100644 index 000000000..b63c1ade1 --- /dev/null +++ b/629/CH4/EX4.4/ex4_4.txt @@ -0,0 +1,3 @@ +
+The elevation difference between the liquid at the center and the wall, during rotation = 0.051 m.
+
\ No newline at end of file diff --git a/629/CH4/EX4.4/example4_4.sce b/629/CH4/EX4.4/example4_4.sce new file mode 100644 index 000000000..2feeece3d --- /dev/null +++ b/629/CH4/EX4.4/example4_4.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 4.4 SURFACE PROFILE OF ROTATING LIQUID
+g=9.81; //[m/s^2]
+d=0.5; //[m]
+w=4; //(rad/s)
+//(p1/gamma)+z1-(w^2*r1^2/(2*g))=(p2/gamma)+z2-(w^2*r2^2/(2*g))
+//p1=p2
+r1=0; //[m]
+r2=d/2; //[m]
+delz=(w^2*(r2^2-r1^2)/(2*g)) //delz=(z2-z1),[m]
+printf("\nThe elevation difference between the liquid at the center and the wall, during rotation = %.3f m.\n",delz)
\ No newline at end of file diff --git a/629/CH4/EX4.5/ex4_5.txt b/629/CH4/EX4.5/ex4_5.txt new file mode 100644 index 000000000..45d806291 --- /dev/null +++ b/629/CH4/EX4.5/ex4_5.txt @@ -0,0 +1,2 @@ +
+The new levels of water in the tube are z1 = 0.021 m, z2 = 0.339 m.
\ No newline at end of file diff --git a/629/CH4/EX4.5/example4_5.sce b/629/CH4/EX4.5/example4_5.sce new file mode 100644 index 000000000..dabd3d8a8 --- /dev/null +++ b/629/CH4/EX4.5/example4_5.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 4.5 ROTATING MANOMETER TUBE
+g=9.81; //[m/s^2]
+r1=0.18; //[m]
+r2=0.36; //[m]
+//Initial water levels
+h1=0.18; //[m]
+h2=0.18; //[m]
+w=8; //angular speed[rad/s]
+//-z1+z2=w^2*(r2^2-r1^2)/(2*g)
+//z1+z2=h1+h2
+A=[-1 1;1 1];
+B=[w^2*(r2^2-r1^2)/(2*g); h1+h2]
+//Az=B, z=(A^-1)*B, z=[z1;z2]
+z=inv(A)*B
+z1=z(1) //[m]
+z2=z(2) //[m]
+printf("\nThe new levels of water in the tube are z1 = %.3f m, z2 = %.3f m.\n",z1,z2)
\ No newline at end of file diff --git a/629/CH4/EX4.6/ex4_6.txt b/629/CH4/EX4.6/ex4_6.txt new file mode 100644 index 000000000..829524b52 --- /dev/null +++ b/629/CH4/EX4.6/ex4_6.txt @@ -0,0 +1,2 @@ +
+The velocity in the throat section = 3.62 m/s.
\ No newline at end of file diff --git a/629/CH4/EX4.6/example4_6.sce b/629/CH4/EX4.6/example4_6.sce new file mode 100644 index 000000000..4c542a50c --- /dev/null +++ b/629/CH4/EX4.6/example4_6.sce @@ -0,0 +1,10 @@ +clear
+clc
+//Example 4.6 VELOCITY IN A VENTURI SECTION
+h1=1; //[m]
+h2=0.5; //[m]
+g=9.81; //[m/s^2]
+//h1-h2=(V2^2-V1^2)/(2*g), V2=2*V1
+V1=sqrt(2*g*(h1-h2)/3) //[m/s]
+V2=2*V1 //[m/s]
+printf("\nThe velocity in the throat section = %.2f m/s.\n",V2)
\ No newline at end of file diff --git a/629/CH4/EX4.7/ex4_7.txt b/629/CH4/EX4.7/ex4_7.txt new file mode 100644 index 000000000..74c3803cd --- /dev/null +++ b/629/CH4/EX4.7/ex4_7.txt @@ -0,0 +1,3 @@ +
+The velocity of the liquid in the drain port = 14 m/s.
+
\ No newline at end of file diff --git a/629/CH4/EX4.7/example4_7.sce b/629/CH4/EX4.7/example4_7.sce new file mode 100644 index 000000000..348b16f21 --- /dev/null +++ b/629/CH4/EX4.7/example4_7.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 4.7 OUTLET VELOCITY FROM DRAINING TANK
+g=9.81; //[m/s^2]
+delz=10; //delz=(z1-z2),[m]
+//(p1/gamma)+z1-(V1^2/(2*g))=(p2/gamma)+z2-(V2^2/(2*g)), p1=p2
+V1=0;//[m]
+V2=sqrt(V1^2+2*g*delz) //[m/s]
+printf("\nThe velocity of the liquid in the drain port = %.f m/s.\n",V2)
\ No newline at end of file diff --git a/629/CH4/EX4.8/ex4_8.txt b/629/CH4/EX4.8/ex4_8.txt new file mode 100644 index 000000000..77f02612c --- /dev/null +++ b/629/CH4/EX4.8/ex4_8.txt @@ -0,0 +1,3 @@ +
+ The kerosene velocity in the pipe = 24.3 ft/s.
+
\ No newline at end of file diff --git a/629/CH4/EX4.8/example4_8.sce b/629/CH4/EX4.8/example4_8.sce new file mode 100644 index 000000000..0e7fd3ca6 --- /dev/null +++ b/629/CH4/EX4.8/example4_8.sce @@ -0,0 +1,17 @@ +clear
+clc
+//Example 4.8 APPLICATION OF PITOT EQUATION WITH MANOMETER
+g=32.2; //[ft/s^2]
+y=7/12; //[ft]
+//Specific gravities
+S_kero=0.81;
+S_Hg=13.55;
+//Specific weights
+g_water=62.4; //[lbf/ft^3]
+g_Hg=S_Hg*g_water //[lbf/ft^3]
+g_kero=S_kero*g_water //[lbf/ft^3]
+rho_kero=g_kero/g //density[lbm/ft^3]
+//Manometer equation, pz1-pz2=y*(gamma_Hg-gamma_kero)
+//Pitot-static tube equation, V=[2*(pz1-pz2)/rho]^(1/2)
+V=(2*y*(g_Hg-g_kero)/rho_kero)^(1/2) //[ft/s]
+printf("\n The kerosene velocity in the pipe = %.1f ft/s.\n",V)
\ No newline at end of file diff --git a/629/CH4/EX4.9/ex4_9.txt b/629/CH4/EX4.9/ex4_9.txt new file mode 100644 index 000000000..96318a86d --- /dev/null +++ b/629/CH4/EX4.9/ex4_9.txt @@ -0,0 +1,3 @@ +
+The velocity of air in the tunnel = 35.4 m/s.
+
\ No newline at end of file diff --git a/629/CH4/EX4.9/example4_9.sce b/629/CH4/EX4.9/example4_9.sce new file mode 100644 index 000000000..cf0c9d0e5 --- /dev/null +++ b/629/CH4/EX4.9/example4_9.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 4.9 PITOT TUBE APPLICATION WITH PRESSURE GAGE
+p=98*10^3; //pressure[N/m^3]
+R=287; //gas constant[J/kg.K]
+T=20+273; //temperature[K]
+rho=p/(R*T) //density[kg/m^3]
+del_p=730; //[N/m^2]
+//Pitot-static tube equation,
+V=sqrt(2*del_p/rho) //velocity[m/s]
+printf("\nThe velocity of air in the tunnel = %.1f m/s.\n",V)
\ No newline at end of file diff --git a/629/CH5/EX5.1/ex5_1.txt b/629/CH5/EX5.1/ex5_1.txt new file mode 100644 index 000000000..5d4c47e12 --- /dev/null +++ b/629/CH5/EX5.1/ex5_1.txt @@ -0,0 +1,6 @@ +
+The disharge in the pipe in both units is 2.42 m^3/s and 85.4 cfs.
+
+
+The mean velocity in the pipe in both units is 34.2 m/s and 112 ft/s.
+
\ No newline at end of file diff --git a/629/CH5/EX5.1/example5_1.sce b/629/CH5/EX5.1/example5_1.sce new file mode 100644 index 000000000..e67160c8c --- /dev/null +++ b/629/CH5/EX5.1/example5_1.sce @@ -0,0 +1,14 @@ +clear
+clc
+//Example 5.1 VOLUME FLOW RATE AND MEAN VELOCITY
+m=3; //mass flow rate[kg/s]
+rho=1.24; //density[kg/m^3]
+Q=m/rho //discharge[m^3/s]
+//1m^3=35.31ft^3
+printf("\nThe disharge in the pipe in both units is %.2f m^3/s and %.1f cfs.\n\n",Q,Q*35.31)
+
+d=0.3; //diameter[m]
+A=(%pi*d^2)/4 //area[m^2]
+V=Q/A //mean velocity[m/s]
+//1ft=0.3048 m
+printf("\nThe mean velocity in the pipe in both units is %.1f m/s and %.f ft/s.\n",V,V/0.3048)
\ No newline at end of file diff --git a/629/CH5/EX5.2/ex5_2.txt b/629/CH5/EX5.2/ex5_2.txt new file mode 100644 index 000000000..676a44d19 --- /dev/null +++ b/629/CH5/EX5.2/ex5_2.txt @@ -0,0 +1,3 @@ +
+The discharge per meter width of the channel = 6.24 m^3/s per meter.
+
\ No newline at end of file diff --git a/629/CH5/EX5.2/example5_2.sce b/629/CH5/EX5.2/example5_2.sce new file mode 100644 index 000000000..686cc137b --- /dev/null +++ b/629/CH5/EX5.2/example5_2.sce @@ -0,0 +1,8 @@ +clear
+clc
+//Example 5.2 FLOW IN SLOPING CHANNEL
+A=0.6; //depth[m]
+theta=30; //slope(degrees)
+V=12; //velocity[m/s]
+Q=V*cosd(theta)*A //discharge per meter[m^2/s]
+printf("\nThe discharge per meter width of the channel = %.2f m^3/s per meter.\n",Q)
\ No newline at end of file diff --git a/629/CH5/EX5.3/ex5_3.txt b/629/CH5/EX5.3/ex5_3.txt new file mode 100644 index 000000000..db6fab098 --- /dev/null +++ b/629/CH5/EX5.3/ex5_3.txt @@ -0,0 +1,2 @@ +
+The discharge in the channel = 20 m^3/s.
\ No newline at end of file diff --git a/629/CH5/EX5.3/example5_3.sce b/629/CH5/EX5.3/example5_3.sce new file mode 100644 index 000000000..885166ba8 --- /dev/null +++ b/629/CH5/EX5.3/example5_3.sce @@ -0,0 +1,10 @@ +clear
+clc
+//Example 5.3 DISCHARGE IN CHANNEL WITH NON-UNIFORM VELOCITY DISTRIBUTION
+d=2; //depth[m]
+w=5; //width[m]
+umax=3; //max velocity[m/s]
+//Discharge equation Q=integrate('u','A',0,d)
+//dA=w*dy
+Q=integrate('w*umax*sqrt(y/d)','y',0,2) //[m^3/s]
+printf("\nThe discharge in the channel = %.f m^3/s.\n",Q)
\ No newline at end of file diff --git a/629/CH5/EX5.4/ex5_4.txt b/629/CH5/EX5.4/ex5_4.txt new file mode 100644 index 000000000..97260e151 --- /dev/null +++ b/629/CH5/EX5.4/ex5_4.txt @@ -0,0 +1,2 @@ +
+The rate of water accumulating in the tank = 14.5 kg/s.
\ No newline at end of file diff --git a/629/CH5/EX5.4/example5_4.sce b/629/CH5/EX5.4/example5_4.sce new file mode 100644 index 000000000..263b0d244 --- /dev/null +++ b/629/CH5/EX5.4/example5_4.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 5.4 MASS ACCUMULATION IN A TANK
+A=0.0025; //area[m^2]
+V=7; //velocity[m/s]
+rho=1000; //density[kg/m^3]
+mi=rho*V*A //inlet mass flow rate[kg/s]
+Q=0.003; //[m^3/s]
+mo=rho*Q //outlet mass flow rate[kg/s]
+//Continuity equation, mcv+mo-mi=0
+mcv=mi-mo //accumulation rate[kg/s]
+printf("\nThe rate of water accumulating in the tank = %.1f kg/s.\n",mcv)
\ No newline at end of file diff --git a/629/CH5/EX5.5/ex5_5.txt b/629/CH5/EX5.5/ex5_5.txt new file mode 100644 index 000000000..729ac1660 --- /dev/null +++ b/629/CH5/EX5.5/ex5_5.txt @@ -0,0 +1,2 @@ +
+The rate of rise of water in the reservoir is 0.484 ft/hr
\ No newline at end of file diff --git a/629/CH5/EX5.5/example5_5.sce b/629/CH5/EX5.5/example5_5.sce new file mode 100644 index 000000000..ea79e81fb --- /dev/null +++ b/629/CH5/EX5.5/example5_5.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 5.5 RATE OF WATER RISE IN RESERVOIR
+A=40; //area[mi^2]
+Q1=400000; //discharge rate into the reservoir[ft^3/s]
+Q2=250000; //outflow rate[cfs]
+//mcv=mi-mo
+//(rho*Q2)+(rho*Qrise)=rho*Q1
+Qrise=Q1-Q2 //[cfs]
+//1mi=5280ft, 1hr=3600sec
+Vrise=Qrise*3600/(A*(5280)^2) //rise rate[ft/hr]
+printf("\nThe rate of rise of water in the reservoir is %.3f ft/hr.\n",Vrise)
\ No newline at end of file diff --git a/629/CH5/EX5.6/ex5_6.txt b/629/CH5/EX5.6/ex5_6.txt new file mode 100644 index 000000000..222418eef --- /dev/null +++ b/629/CH5/EX5.6/ex5_6.txt @@ -0,0 +1,2 @@ +
+The time elapsed for that drop in water tank = 31.9 s.
\ No newline at end of file diff --git a/629/CH5/EX5.6/example5_6.sce b/629/CH5/EX5.6/example5_6.sce new file mode 100644 index 000000000..6abde25b5 --- /dev/null +++ b/629/CH5/EX5.6/example5_6.sce @@ -0,0 +1,14 @@ +clear
+clc
+//Example 5.6 WATER LEVEL DROP RATE IN DRAINING TANK
+D1=0.1; //diameter of outlet[m]
+DT=1; //diameter of tank[m]
+A1=(%pi*D1^2)/4 //[m^2]
+AT=(%pi*DT^2)/4 //[m^2]
+g=9.81; //[m/s^2]
+ho=2; //[m]
+hf=0.5; //[m]
+//mi=0,mo=rho*A1*V1=rho*sqrt(2gh)*A1, mcv=mi-mo
+//continuity equation, mi=d(rho*AT*h)/dt
+t=integrate('(-AT)*(A1*(sqrt(2*g*h)))^(-1)','h',ho,hf)
+printf("\nThe time elapsed for that drop in water tank = %.1f s.\n",t)
\ No newline at end of file diff --git a/629/CH5/EX5.7/ex5_7.txt b/629/CH5/EX5.7/ex5_7.txt new file mode 100644 index 000000000..87133a618 --- /dev/null +++ b/629/CH5/EX5.7/ex5_7.txt @@ -0,0 +1,3 @@ +
+The time elapsed for the absolute pressure drop = 85766 s.
+
\ No newline at end of file diff --git a/629/CH5/EX5.7/example5_7.sce b/629/CH5/EX5.7/example5_7.sce new file mode 100644 index 000000000..2f0afbc59 --- /dev/null +++ b/629/CH5/EX5.7/example5_7.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 5.7 DEPRESSURIZATION OF GAS IN TANK
+V=10; //volume[m^3]
+po=500; //initial pressure[kPa]
+pf=400; //final pressure[kPa]
+R=518; //gas constant[J/kg.K]
+T=300; //[K]
+A=10^-7; //area[m^2]
+//mi=0,mo=0.66*p*A/sqrt(R*T), mcv=mi-mo
+//continuity equation, mi=V*d(rho)/dt
+t=integrate('-V*(0.66*A*p*sqrt(R*T))^(-1)','p',po,pf)
+printf("\nThe time elapsed for the absolute pressure drop = %.f s.\n",t)
\ No newline at end of file diff --git a/629/CH5/EX5.8/ex5_8.txt b/629/CH5/EX5.8/ex5_8.txt new file mode 100644 index 000000000..79990b816 --- /dev/null +++ b/629/CH5/EX5.8/ex5_8.txt @@ -0,0 +1,2 @@ +
+The water speed in the 60cm pipe = 8 m/s.
\ No newline at end of file diff --git a/629/CH5/EX5.8/example5_8.sce b/629/CH5/EX5.8/example5_8.sce new file mode 100644 index 000000000..e539dcdf2 --- /dev/null +++ b/629/CH5/EX5.8/example5_8.sce @@ -0,0 +1,9 @@ +clear
+clc
+//Example 5.8 VELOCITY IN A VARIABLE-AREA PIPE
+A1=120^2; //[cm^2]
+A2=60^2; //[cm^2]
+V1=2; //speed in 120cm pipe[m/s]
+//flow rates Q1=Q2
+V2=V1*(A1/A2) //speed in 60 cm pipe[m/s]
+printf("\nThe water speed in the 60cm pipe = %.f m/s.\n",V2)
\ No newline at end of file diff --git a/629/CH5/EX5.9/ex5_9.txt b/629/CH5/EX5.9/ex5_9.txt new file mode 100644 index 000000000..05b3c9617 --- /dev/null +++ b/629/CH5/EX5.9/ex5_9.txt @@ -0,0 +1,2 @@ +
+The pressure difference recorded by the pressure gage = 150 kPa.
\ No newline at end of file diff --git a/629/CH5/EX5.9/example5_9.sce b/629/CH5/EX5.9/example5_9.sce new file mode 100644 index 000000000..af6ea6273 --- /dev/null +++ b/629/CH5/EX5.9/example5_9.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 5.9 WATER FLOW THROUGH A VENTURIMETER
+//Bernoulli equation, p1+(g*z1)+(rho*V1^2)/2=p2+(g*z2)+(rho*V2^2)/2
+rho=1000; //density[kg/m^3]
+V1=10; //velocity[m/s]
+A21=0.5; //(A21=A2/A1)
+V12=A21 //(V12=V1/V2)
+del_pz=(rho*V1^2/2)*((1/V12)^2-1)/10^3 //change in piezometric pressure[kPa]
+del_pg=del_pz //pressure change in gage[kPa]
+printf("\nThe pressure difference recorded by the pressure gage = %.f kPa.\n",del_pg)
\ No newline at end of file diff --git a/629/CH6/EX6.1/ex6_1.txt b/629/CH6/EX6.1/ex6_1.txt new file mode 100644 index 000000000..f5f532c42 --- /dev/null +++ b/629/CH6/EX6.1/ex6_1.txt @@ -0,0 +1,3 @@ +
+ The force acting on the beam that supports the rocket, Fb = 7.56 N.
+
\ No newline at end of file diff --git a/629/CH6/EX6.1/example6_1.sce b/629/CH6/EX6.1/example6_1.sce new file mode 100644 index 000000000..3cadf3587 --- /dev/null +++ b/629/CH6/EX6.1/example6_1.sce @@ -0,0 +1,14 @@ +clear
+clc
+//Example 6.1 THRUST OF ROCKET
+g=9.81; //[m/s^2]
+m=0.04; //mass[kg]
+D=0.01; //[m]
+A=%pi*D^2/4 //area[m^2]
+rho=0.5; //density[kg/m^3]
+v=450; //[m/s]
+//Sum of forces, Fz=-Fb-m.g
+Mo=-rho*A*v^2 //momentum outflow[N]
+Fz=Mo //[N]
+Fb=-Fz-m*g //Force on beam[N]
+printf("\n The force acting on the beam that supports the rocket, Fb = %.2f N.\n",Fb)
\ No newline at end of file diff --git a/629/CH6/EX6.10/ex6_10.txt b/629/CH6/EX6.10/ex6_10.txt new file mode 100644 index 000000000..84df14937 --- /dev/null +++ b/629/CH6/EX6.10/ex6_10.txt @@ -0,0 +1,3 @@ +
+ The frictional force acting on the block = 313 N.
+
\ No newline at end of file diff --git a/629/CH6/EX6.10/example6_10.sce b/629/CH6/EX6.10/example6_10.sce new file mode 100644 index 000000000..1d9a468f5 --- /dev/null +++ b/629/CH6/EX6.10/example6_10.sce @@ -0,0 +1,10 @@ +clear
+clc
+//Example 6.10 JET IMPINGING ON MOVING BLOCK
+A=5*10^-4; //area[m^2]
+rho=1000; //density[kg/m^3]
+vj=50; //[m/s]
+vb=25; //[m/s]
+//Force on cart
+F=rho*A*(vj-vb)^2 //[N]
+printf("\n The frictional force acting on the block = %.f N.\n",F)
\ No newline at end of file diff --git a/629/CH6/EX6.11/ex6_11.txt b/629/CH6/EX6.11/ex6_11.txt new file mode 100644 index 000000000..a24bc3c63 --- /dev/null +++ b/629/CH6/EX6.11/ex6_11.txt @@ -0,0 +1,2 @@ +
+ The ratio of propellant mass to initial mass to achieve orbital velocity = 0.907.
\ No newline at end of file diff --git a/629/CH6/EX6.11/example6_11.sce b/629/CH6/EX6.11/example6_11.sce new file mode 100644 index 000000000..21df5043f --- /dev/null +++ b/629/CH6/EX6.11/example6_11.sce @@ -0,0 +1,10 @@ +clear
+clc
+//Example 6.11 PROPELLANT MASS RATIO FOR ACHIEVING ORBITAL VELOCITY
+Vbo=7600; //orbital velocity[m/s]
+Isp=3200; //specific impulse[m/s]
+//Vbo=Isp*log(mi/mf)
+mif=exp(Vbo/Isp) //mif=mi/mf
+//mp=mi-mf
+mpi=1-1/mif //mpi=mp/mi
+printf("\n The ratio of propellant mass to initial mass to achieve orbital velocity = %.3f.\n",mpi)
\ No newline at end of file diff --git a/629/CH6/EX6.12/ex6_12.txt b/629/CH6/EX6.12/ex6_12.txt new file mode 100644 index 000000000..592bbab54 --- /dev/null +++ b/629/CH6/EX6.12/ex6_12.txt @@ -0,0 +1,3 @@ +
+ The maximum pressure that develops at the downstream end = 303 psig.
+
\ No newline at end of file diff --git a/629/CH6/EX6.12/example6_12.sce b/629/CH6/EX6.12/example6_12.sce new file mode 100644 index 000000000..5c8c05bc1 --- /dev/null +++ b/629/CH6/EX6.12/example6_12.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 6.12 PRESSURE RISE DUE TO WATER HAMMER EFFECT
+rho=1.94; //[slugs/ft^3]
+Ev=3.2*10^5; //[lbf/in^2]
+V=4; //[ft/s]
+//Sound speed
+c=sqrt(Ev*144/rho) //[ft/s]
+L=3000; //[ft]
+tc=2*L/c //[s]
+//Closure time of 1sec is less than tc
+//Pressure rise
+delp=rho*V*c/144 //[psi]
+pi=40; //initial pressure[psi]
+pmax=pi+delp //[psi]
+printf("\n The maximum pressure that develops at the downstream end = %.f psig.\n",pmax)
\ No newline at end of file diff --git a/629/CH6/EX6.13/ex6_13.txt b/629/CH6/EX6.13/ex6_13.txt new file mode 100644 index 000000000..95196410e --- /dev/null +++ b/629/CH6/EX6.13/ex6_13.txt @@ -0,0 +1,2 @@ +
+ The moment that the support system must resist, M = 3.61 kN.m.
\ No newline at end of file diff --git a/629/CH6/EX6.13/example6_13.sce b/629/CH6/EX6.13/example6_13.sce new file mode 100644 index 000000000..e0eee6d1c --- /dev/null +++ b/629/CH6/EX6.13/example6_13.sce @@ -0,0 +1,32 @@ +clear
+clc
+//Example 6.13 RESISTING MOMENT ON REDUCING BEND
+r1=0.15; //[m]
+r2=0.475; //[m]
+d1=0.3; //[m]
+d2=0.15; //[m]
+A1=%pi*d1^2/4 //[m^2]
+A2=%pi*d2^2/4 //[m^2]
+p1=150*10^3; //[Pa]
+p2=59.3*10^3; //[Pa]
+//Torque due to pressure
+Mp=(r1*p1*A1)+(r2*p2*A2) //[N.m]
+
+rho=998; //density[kg/m^3]
+Q=0.25; //discharge[m^3/s]
+m=rho*Q //mass flow rate[kg/s]
+v1=Q/A1 //[m/s]
+v2=Q/A2 //[m/s]
+Mi=-m*r1*v1 //moment due to inflow
+Mo=m*r2*v2 //moment due to outflow
+//Moment of momentum flow
+Mm=Mo-Mi
+
+d=0.2; //[m]
+W=1420; //weight[N]
+//Moment due to weight
+Mw=d*W
+
+//Moment exerted by support
+M=(-Mp-Mm+Mw)*10^-3 //[kN.m]
+printf("\n The moment that the support system must resist, M = %.2f kN.m.\n",-M)
\ No newline at end of file diff --git a/629/CH6/EX6.14/ex6_14.txt b/629/CH6/EX6.14/ex6_14.txt new file mode 100644 index 000000000..8977ab4ce --- /dev/null +++ b/629/CH6/EX6.14/ex6_14.txt @@ -0,0 +1,2 @@ +
+ The power produced by the turbine = 343 kW.
\ No newline at end of file diff --git a/629/CH6/EX6.14/example6_14.sce b/629/CH6/EX6.14/example6_14.sce new file mode 100644 index 000000000..8d7746258 --- /dev/null +++ b/629/CH6/EX6.14/example6_14.sce @@ -0,0 +1,17 @@ +clear
+clc
+//Example 6.14 POWER DELIVERED BY A FRANCIS TURBINE
+D=1; //[m]
+l=0.04; //[m]
+Q=0.5; //discharge[m^3/s]
+rho=1000; //[kg/m^3]
+m=rho*Q //mass flow rate[kg/s]
+//Radial velocity
+Vr=Q/(%pi*D*l) //[m/s]
+theta=70; //degrees
+//Tangential velocity
+Vt=Vr*tand(theta) //[m/s]
+T=m*(D/2)*Vt //Torque[N.m]
+w=1200*2*%pi/60 //angular speed (rad/s)
+P=T*w/10^3 //Power[kW]
+printf("\n The power produced by the turbine = %.f kW.\n",P)
\ No newline at end of file diff --git a/629/CH6/EX6.2/ex6_2.txt b/629/CH6/EX6.2/ex6_2.txt new file mode 100644 index 000000000..92e21759e --- /dev/null +++ b/629/CH6/EX6.2/ex6_2.txt @@ -0,0 +1,6 @@ +
+ The tension in the cable = 233 lbf.
+
+
+ The weight recorded by the scale = 1203 lbf.
+
\ No newline at end of file diff --git a/629/CH6/EX6.2/example6_2.sce b/629/CH6/EX6.2/example6_2.sce new file mode 100644 index 000000000..6c82f8cd5 --- /dev/null +++ b/629/CH6/EX6.2/example6_2.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 6.2 CONCRETE FLOWING INTO CART
+A=1; //[ft^2]
+rho=150; //density[lbm/ft^3]
+v=10; //speed[ft/s]
+theta=60; //degrees
+//Momentum accumulation=0, outflow=0
+//Momentum inflow
+mi_x=rho*A*v^2*cosd(theta) //x-direction
+mi_z=rho*A*v^2*sind(theta) //z-direction
+//1slug=32.2 lbm
+//Tension in cable
+T=mi_x/32.2 //[lbf]
+printf("\n The tension in the cable = %.f lbf.\n\n",T)
+W=800; //weight[lbf]
+N=W+(mi_z/32.2) //[lbf]
+printf("\n The weight recorded by the scale = %.f lbf.\n",N)
\ No newline at end of file diff --git a/629/CH6/EX6.3/ex6_3.txt b/629/CH6/EX6.3/ex6_3.txt new file mode 100644 index 000000000..520c8f094 --- /dev/null +++ b/629/CH6/EX6.3/ex6_3.txt @@ -0,0 +1,5 @@ +
+ The air speed at the exit of the nozzle = 77.9 m/s.
+
+
+ The force on the flange = 9.90 N.
\ No newline at end of file diff --git a/629/CH6/EX6.3/example6_3.sce b/629/CH6/EX6.3/example6_3.sce new file mode 100644 index 000000000..1d9d2d663 --- /dev/null +++ b/629/CH6/EX6.3/example6_3.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 6.3 FORCE ON A NOZZLE
+d1=0.06; //[m]
+A1=%pi*d1^2/4 //area[m^2]
+d2=0.01; //[m]
+rho=1.22; //density[kg/m^3]
+p1=3.7*1000; //[Pa]
+//Bernoulli equation, p1+rho*v1^2/2=rho*v2^2/2
+v2=sqrt(2*p1/(rho*(1-(d2/d1)^4))) //Exit velocity[m/s]
+printf("\n The air speed at the exit of the nozzle = %.1f m/s.\n\n",v2)
+v1=v2*(d2/d1)^2 //Inlet velocity[m/s]
+m=rho*A1*v1 //mass flow rate[kg/s]
+//Force on flange
+F=m*(v2-v1)-p1*A1 //[N]
+printf("\n The force on the flange = %.2f N.\n",-F)
\ No newline at end of file diff --git a/629/CH6/EX6.4/ex6_4.txt b/629/CH6/EX6.4/ex6_4.txt new file mode 100644 index 000000000..8604e0d25 --- /dev/null +++ b/629/CH6/EX6.4/ex6_4.txt @@ -0,0 +1,2 @@ +
+ The force exerted by the jet on the vane, F = (53.0 lbf)i+(91.8 lbf)j.
\ No newline at end of file diff --git a/629/CH6/EX6.4/example6_4.sce b/629/CH6/EX6.4/example6_4.sce new file mode 100644 index 000000000..89516873a --- /dev/null +++ b/629/CH6/EX6.4/example6_4.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 6.4 WATER DEFLECTED BY A VANE
+r=0.0417; //[ft]
+A=%pi*r^2 //area[ft^2]
+v=100; //velocty[ft/s]
+rho=1.94; //[slugs/ft^3]
+m=rho*A*v; //mass flow rate[slugs/s]
+theta=60;//degrees
+//Momentum outflow vector, mo=[mo_x mo_y]
+mo=[m*v*cosd(theta) -m*v*sind(theta)]
+//Momentum inflow vector, mi=[mi_x mi_y]
+mi=[m*v 0]
+//Force vector, F=[Fx Fy]
+F=mi-mo //[lbf]
+printf("\n The force exerted by the jet on the vane, F = (%.1f lbf)i+(%.1f lbf)j.\n",F(1),F(2))
\ No newline at end of file diff --git a/629/CH6/EX6.6/ex6_6.txt b/629/CH6/EX6.6/ex6_6.txt new file mode 100644 index 000000000..1b58651f2 --- /dev/null +++ b/629/CH6/EX6.6/ex6_6.txt @@ -0,0 +1,3 @@ +
+The net force required to hold the bend in place, F =(-8.53 kN)i+(-31.8 kN)j+(15.1 kN)k.
+
\ No newline at end of file diff --git a/629/CH6/EX6.6/example6_6.sce b/629/CH6/EX6.6/example6_6.sce new file mode 100644 index 000000000..ea9d6a092 --- /dev/null +++ b/629/CH6/EX6.6/example6_6.sce @@ -0,0 +1,24 @@ +clear
+clc
+//Example 6.6 FORCES ACTING ON A PIPE BEND
+p=75*10^3 //[Pa]
+r=0.5; //[m]
+A=%pi*r^2 //area[m^2]
+S=0.94;
+rho=S*1000 //drnsity[kg/m^3]
+Gamma=S*9.81 //specific weight of oil[kN/m^3]
+V=1.2; //volume of oil[m^3]
+Q=2; //[m^3/s]
+m=rho*Q //mass flow rate[kg/s]
+v=2.55; //[m/s]
+theta=30; //degrees
+//Reaction force
+//Rx+p*A-p*A*cos(theta)=m*v*cos(theta)-m*v
+Rx=-(p*A+m*v)*(1-cosd(theta))/10^3 //[kN]
+//Ry+p*A*sin(theta)=-m*v*sin(theta)
+Ry=-(p*A+m*v)*sind(theta)/10^3 //[kN]
+We=4; //empty weight of bend[kN]
+Rz=(Gamma*V)+We //[kN]
+//Reaction force vector
+R=[Rx Ry Rz] //[kN]
+printf("\n The net force required to hold the bend in place, F =(%.2f kN)i+(%.1f kN)j+(%.1f kN)k.\n",R(1),R(2),R(3))
\ No newline at end of file diff --git a/629/CH6/EX6.7/ex6_7.txt b/629/CH6/EX6.7/ex6_7.txt new file mode 100644 index 000000000..295731dbc --- /dev/null +++ b/629/CH6/EX6.7/ex6_7.txt @@ -0,0 +1,2 @@ +
+ The components of force, F required to hold the bend in place are Fx = -16.1 kN, Fz = 1.48 kN.
\ No newline at end of file diff --git a/629/CH6/EX6.7/example6_7.sce b/629/CH6/EX6.7/example6_7.sce new file mode 100644 index 000000000..67d05ba75 --- /dev/null +++ b/629/CH6/EX6.7/example6_7.sce @@ -0,0 +1,27 @@ +clear
+clc
+//Example 6.7 WATER FLOW THROUGH REDUCING BEND
+d1=0.3; //[m]
+d2=0.15; //[m]
+A1=%pi*d1^2/4 //[m^2]
+A2=%pi*d2^2/4 //[m^2]
+rho=1000; //[kg/m^3]
+Q=0.25; //[m^3/s]
+v1=Q/A1 //inlet speed[m/s]
+v2=Q/A2 //outlet speed[m/s]
+p1=150; //[kPa]
+g_w=9810; //specific weight of water
+del_z=0.325; //(del_z=z1-z2)[m]
+//Bernoulli equation, p1+(rho*v1^2/2)+(g_w*z1)=p2+(rho*v2^2/2)+(g_w*z2)
+p2=p1+(rho*(v1^2-v2^2)/2+g_w*del_z)*10^-3 //[kPa]
+//Pressure forces
+Fp=(p1*A1)+(p2*A2) //[kN]
+//Momentum flux
+Fm=rho*Q*(v1+v2)*10^-3 //[kN]
+Wb=500; //weight of metal in the bend[N]
+V=0.1; //bend volume[m^3]
+Wf=g_w*V //weight of water[N]
+//Reaction force components
+Rx=-Fp-Fm //[kN]
+Rz=(Wb+Wf)*10^-3 //[kN]
+printf("\n The components of force, F required to hold the bend in place are Fx = %.1f kN, Fz = %.2f kN.\n",Rx,Rz)
\ No newline at end of file diff --git a/629/CH6/EX6.8/ex6_8.txt b/629/CH6/EX6.8/ex6_8.txt new file mode 100644 index 000000000..f9985969f --- /dev/null +++ b/629/CH6/EX6.8/ex6_8.txt @@ -0,0 +1,2 @@ +
+ The drag force on the device and support panes = 304 N.
\ No newline at end of file diff --git a/629/CH6/EX6.8/example6_8.sce b/629/CH6/EX6.8/example6_8.sce new file mode 100644 index 000000000..591d0f328 --- /dev/null +++ b/629/CH6/EX6.8/example6_8.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 6.8 DRAG FORCE ON WIND-TUNNEL MODEL
+ro=0.5; //radius of tunnel[m]
+A1=%pi*ro^2 //[m^2]
+p1=1.5*10^3; //[Pa]
+p2=10^3; //[Pa]
+v1=30; //velocity at inlet[m/s]
+rho=1; //[kg/m^3]
+//velocity profile, v=v1*K(r/ro)
+//Q1=Q, A1*v1=A*v
+K=A1*v1*(integrate('v1*(r/ro)*2*%pi*r','r',0,ro))^-1
+F1=rho*A1*v1^2 //momentum at cross-section-1
+F2=integrate('rho*(v1*K*(r/ro))^2*2*%pi*r','r',0,ro) //at cross-section-2
+Fx=F2-F1
+//From momentum equation in x-direction, Fx=p1*A-p2*A-Fd
+Fd=(p1-p2)*A1-Fx //Drag force[N]
+printf("\n The drag force on the device and support panes = %.f N.\n",Fd)
\ No newline at end of file diff --git a/629/CH6/EX6.9/ex6_9.txt b/629/CH6/EX6.9/ex6_9.txt new file mode 100644 index 000000000..62e4e7387 --- /dev/null +++ b/629/CH6/EX6.9/ex6_9.txt @@ -0,0 +1,5 @@ +
+ The flow rate under the sluice gate = 2008 ft^3/s.
+
+ The force on the sluice gate = 66.6 tons.
+
\ No newline at end of file diff --git a/629/CH6/EX6.9/example6_9.sce b/629/CH6/EX6.9/example6_9.sce new file mode 100644 index 000000000..c70d35a89 --- /dev/null +++ b/629/CH6/EX6.9/example6_9.sce @@ -0,0 +1,25 @@ +clear
+clc
+//Example 6.9 FORCE ON A SLUICE GATE
+g=32.2; //[ft/s^2]
+d1=20; //[ft]
+d2=3; //[ft]
+w=20; //gate width[ft]
+v2=sqrt((2*g*(d1-d2))/(1-(d2/d1)^2)) //[ft/s]
+v1=d2*v2/d1 //[ft/s]
+rho=1.94; //[slugs/ft^3]
+Q=v2*d2*w //discharge[ft^3/s]
+printf("\n The flow rate under the sluice gate = %.f ft^3/s.\n",Q)
+m=rho*Q //mass flow rate[slugs/s]
+Gamma=62.4; //[lbf/ft^3]
+F1=Gamma*w*(d1^2)/2
+F2=Gamma*w*(d2^2)/2
+//momentum inflow
+mi=m*v1 //[lbf]
+//momentum outflow
+mo=m*v2 //[lbf]
+Fx=mo-mi //[lbf]
+//Sum of forces in x-direction, Fx=F1-F2-Fg
+//1ton=2000 lbf
+Fg=(F1-F2-Fx)/2000 //tons
+printf("\n The force on the sluice gate = %.1f tons.\n",Fg)
\ No newline at end of file diff --git a/629/CH7/EX7.1/ex7_1.txt b/629/CH7/EX7.1/ex7_1.txt new file mode 100644 index 000000000..48208fce0 --- /dev/null +++ b/629/CH7/EX7.1/ex7_1.txt @@ -0,0 +1,2 @@ +The kinetic-energy correction factor is 2.
+
\ No newline at end of file diff --git a/629/CH7/EX7.1/example7_1.sce b/629/CH7/EX7.1/example7_1.sce new file mode 100644 index 000000000..fe37a4056 --- /dev/null +++ b/629/CH7/EX7.1/example7_1.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 7.1 KINETIC ENERGY CORRECTION FACTOR FOR LAMINAR FLOW
+Vmax=1; //max velocity[m/s](say)
+ro=1; //radius of pipe[m](say)
+A=%pi*ro^2 //area[m^2]
+//Given, V=Vmax(1-(r/ro)^2)
+//dA=2*pi*r*dr
+Vbar=(1/A)*integrate('Vmax*(1-(r/ro)^2)*2*%pi*r','r',0,ro) //mean velocity[m/s]
+alpha=(1/A)*integrate('((Vmax*(1-(r/ro)^2))/Vbar)^3*2*%pi*r','r',0,ro) //kinetic-energy correction factor
+printf("\nThe kinetic-energy correction factor is %.f.\n",alpha)
\ No newline at end of file diff --git a/629/CH7/EX7.2/ex7_2.txt b/629/CH7/EX7.2/ex7_2.txt new file mode 100644 index 000000000..da3ebc2c0 --- /dev/null +++ b/629/CH7/EX7.2/ex7_2.txt @@ -0,0 +1 @@ +The pressure in the pipe at L=2000m is = 418 kPa.
\ No newline at end of file diff --git a/629/CH7/EX7.2/example7_2.sce b/629/CH7/EX7.2/example7_2.sce new file mode 100644 index 000000000..6d2594101 --- /dev/null +++ b/629/CH7/EX7.2/example7_2.sce @@ -0,0 +1,21 @@ +clear
+clc
+//Example 7.2 PRESSURE IN A PIPE
+//Energy equation, (p1/gamma)+(alpha1*V1^2/2g)+hp=(p2/gamma)+(alpha2*V2^2/2g)+ht+hL
+p1=0; //pressure at top of reservoir is p_atm=0
+ht=0;
+hp=0;
+V1=0;
+Gamma=9810; //specific weight[N/m^3]
+alpha2=1;
+z1=100; //[m]
+z2=20; //[m]
+L=2000; //[m]
+D=0.2; //diameter[m]
+A=%pi*D^2/4 //area[m^2]
+Q=0.06; //rate of flow[m^3/s]
+g=9.81; //[m/s^2]
+V2=Q/A //[m/s]
+hL=(0.02*(L/D)*V2^2)/(2*g) //head loss[m]
+p2=p1+Gamma*((z1-z2)+hp-ht-hL-(alpha2*V2^2)/(2*g))/10^3 //pressure at L[kPa]
+printf("\nThe pressure in the pipe at L=2000m is = %.f kPa.\n",p2)
\ No newline at end of file diff --git a/629/CH7/EX7.3/ex7_3.txt b/629/CH7/EX7.3/ex7_3.txt new file mode 100644 index 000000000..0ccc901a6 --- /dev/null +++ b/629/CH7/EX7.3/ex7_3.txt @@ -0,0 +1,5 @@ +
+Power that must be supplied to the flow by the pump
+ in kilowatts = 204 kW
+ in horsepower = 273 hp.
+
\ No newline at end of file diff --git a/629/CH7/EX7.3/example7_3.sce b/629/CH7/EX7.3/example7_3.sce new file mode 100644 index 000000000..f64a938a0 --- /dev/null +++ b/629/CH7/EX7.3/example7_3.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 7.3 POWER NEEDED BY A PUMP
+//Energy equation, V1=V2,(p1/gamma)+hp=(p2/gamma)+ht+hL
+ht=0;
+hL=3; //[m]
+p1=70000; //[N/m^2]
+p2=350000; //[N/m^2]
+z1=30; //[m]
+z2=40; //[m]
+Gamma=9810; //specific weight[N/m^3]
+hp=(p2-p1)/Gamma+(z2-z1)+hL //pump head[m]
+Q=0.5; //rate of flow[m^3/s]
+P=Gamma*Q*hp/10^3 //power[kW]
+//1hp=0.746 kW
+printf("\nPower that must be supplied to the flow by the pump \n in kilowatts = %.f kW \n in horsepower = %.f hp.\n",P,P/0.746)
\ No newline at end of file diff --git a/629/CH7/EX7.4/ex7_4.txt b/629/CH7/EX7.4/ex7_4.txt new file mode 100644 index 000000000..2787da343 --- /dev/null +++ b/629/CH7/EX7.4/ex7_4.txt @@ -0,0 +1,2 @@ +
+The rate of power generation is 7.16 MW.
\ No newline at end of file diff --git a/629/CH7/EX7.4/example7_4.sce b/629/CH7/EX7.4/example7_4.sce new file mode 100644 index 000000000..4cf0221d0 --- /dev/null +++ b/629/CH7/EX7.4/example7_4.sce @@ -0,0 +1,17 @@ +clear
+clc
+//Example 7.4 POWER PRODUCED BY A TURBINE
+//Energy equation, V1=V2=0,(p1/gamma)+hp=(p2/gamma)+ht+hL
+hp=0;
+hL=1.5; //head loss[m]
+p1=0;
+p2=0;
+z1=61; //[m]
+z2=0; //[m]
+Gamma=9810; //specific weight[N/m^3]
+ht=(p1-p2)/Gamma+(z1-z2)-hL //turbine head[m]
+Q=14.1; //rate of flow[m^3/s]
+Pi=Gamma*Q*ht/10^6 //power input[MW]
+eta=0.87; //efficiency
+Po=eta*Pi //power output[MW]
+printf("\nThe rate of power generation is %.2f MW.\n",Po)
\ No newline at end of file diff --git a/629/CH7/EX7.5/ex7_5.txt b/629/CH7/EX7.5/ex7_5.txt new file mode 100644 index 000000000..ecc3f45fb --- /dev/null +++ b/629/CH7/EX7.5/ex7_5.txt @@ -0,0 +1,3 @@ +
+The horizontal force required to hold the transition in place = 8.16 kN,in -ve x direction.
+
\ No newline at end of file diff --git a/629/CH7/EX7.5/example7_5.sce b/629/CH7/EX7.5/example7_5.sce new file mode 100644 index 000000000..2b7f2344f --- /dev/null +++ b/629/CH7/EX7.5/example7_5.sce @@ -0,0 +1,24 @@ +clear
+clc
+//Example 7.5 FORCE ON A CONTRACTION IN A PIPE
+D1=0.3; //[m]
+D2=0.2; //[m]
+A1=%pi*D1^2/4 //area[m^2]
+A2=%pi*D2^2/4 //area[m^2]
+Q=0.707; //rate of flow[m^3/s]
+V1=Q/A1; //velocity[m/s]
+V2=Q/A2; //velocity[m/s]
+ht=0;
+hp=0;
+hL=2.58; //[m]
+alpha1=1;
+alpha2=1;
+rho=1000; //density[kg/m^3]
+Gamma=9810; //specific weight[N/m^3]
+g=9.81; //[m/s^2]
+p1=250000; //pressure[Pa]
+p2=p1+Gamma*(hp-ht-hL-(alpha2*V2^2-alpha1*V1^2)/(2*g)) //pressure at L[Pa]
+//Momentum equation
+//p1*A1-p2*A2+Fx=m*V2-m*V1, m=rho*Q
+Fx=(rho*Q*(V2-V1)+p2*A2-p1*A1)/10^3 //force[kN]
+printf("\nThe horizontal force required to hold the transition in place = %.2f kN,in -ve x direction.\n",-Fx)
\ No newline at end of file diff --git a/629/CH7/EX7.6/ex7_6.txt b/629/CH7/EX7.6/ex7_6.txt new file mode 100644 index 000000000..1b39b43a8 --- /dev/null +++ b/629/CH7/EX7.6/ex7_6.txt @@ -0,0 +1,4 @@ +
+The head supplied by the pump = 178 ft.
+
+The power supplied to the flow = 159 hp.
\ No newline at end of file diff --git a/629/CH7/EX7.6/example7_6.sce b/629/CH7/EX7.6/example7_6.sce new file mode 100644 index 000000000..85205f957 --- /dev/null +++ b/629/CH7/EX7.6/example7_6.sce @@ -0,0 +1,22 @@ +clear
+clc
+//Example 7.6 EGL AND HGL FOR A SYSTEM
+//Energy equation, (p1/gamma)+(alpha1*V1^2/2g)+hp=(p2/gamma)+(alpha2*V2^2/2g)+ht+hL ,V1=V2=0
+p1=0;
+p2=0;
+ht=0;
+Gamma=62.4; //specific weight[lbf/ft^3]
+z1=520; //[ft]
+z2=620; //[ft]
+L=5000; //[ft]
+D=1; //diameter[ft]
+A=%pi*D^2/4 //area[ft^2]
+Q=7.85; //rate of flow[ft^3/s]
+V=Q/A //[ft/s]
+g=32.2; //[ft/s^2]
+hL=(0.01*(L/D)*V^2)/(2*g) //head loss[ft]
+hp=round((p2-p1)/Gamma+(z2-z1)+hL) //pump head[ft]
+printf("\nThe head supplied by the pump = %.f ft.\n",hp)
+//1hp.s= 550ft.lbf
+Wp=round(Gamma*Q*hp/550)///power in hp
+printf("\nThe power supplied to the flow = %.f hp.\n",Wp)
\ No newline at end of file diff --git a/629/CH8/EX8.10/ex8_10.txt b/629/CH8/EX8.10/ex8_10.txt new file mode 100644 index 000000000..7270b143c --- /dev/null +++ b/629/CH8/EX8.10/ex8_10.txt @@ -0,0 +1,5 @@ +
+ The velocity in the prototype = 8.4 m/s.
+
+ The water flow rate in the model = 0.89 m^3/s.
+
\ No newline at end of file diff --git a/629/CH8/EX8.10/example8_10.sce b/629/CH8/EX8.10/example8_10.sce new file mode 100644 index 000000000..c9179eecd --- /dev/null +++ b/629/CH8/EX8.10/example8_10.sce @@ -0,0 +1,13 @@ +clear
+clc
+//Example 8.10 MODELING FLOOD DISCHARGE OVER A SPILLWAY
+Lmp=1/49; //Lmp=(Lm/Lp)
+//Froude-number similitude
+//Vm/(g*Lm)=Vp/(g*Lp)
+Vm=1.2; //[m/s]
+Vp=Vm/(Lmp^(1/2))
+printf("\n The velocity in the prototype = %.1f m/s.\n",Vp)
+Amp=(Lmp)^2 //Ratio of areas, Amp=(Am/Ap)
+Qp=15000; //[m^3/s]
+Qm=Qp*Amp*Vm/Vp //[m^3/s]
+printf("\n The water flow rate in the model = %.2f m^3/s.\n",Qm)
\ No newline at end of file diff --git a/629/CH8/EX8.4/ex8_4.txt b/629/CH8/EX8.4/ex8_4.txt new file mode 100644 index 000000000..e7619c99e --- /dev/null +++ b/629/CH8/EX8.4/ex8_4.txt @@ -0,0 +1,2 @@ +
+ The air speed in the wind tunnel for scaled model and dynamically similar conditions, V = 100 m/s.
\ No newline at end of file diff --git a/629/CH8/EX8.4/example8_4.sce b/629/CH8/EX8.4/example8_4.sce new file mode 100644 index 000000000..f8b6306a0 --- /dev/null +++ b/629/CH8/EX8.4/example8_4.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 8.4 REYNOLDS-NUMBER SIMILITUDE
+//p-prototype, m-model
+Vp=10; //speed[m/s]
+//Reynolds-number similitude, Rem=Rep
+//Vm*Lm/vm=Vp*Lp/vp
+//vm=vp
+vmp=1; //vmp=vm/vp
+Lmp=1/10; //Lmp=Lm/Lp
+Vm=Vp*vmp/Lmp //speed[m/s]
+printf("\n The air speed in the wind tunnel for scaled model and dynamically similar conditions, V = %.f m/s.\n",Vm)
\ No newline at end of file diff --git a/629/CH8/EX8.5/ex8_5.txt b/629/CH8/EX8.5/ex8_5.txt new file mode 100644 index 000000000..53b76e949 --- /dev/null +++ b/629/CH8/EX8.5/ex8_5.txt @@ -0,0 +1,2 @@ +
+ The flow rate required for the model, Q = 117 cfs.
\ No newline at end of file diff --git a/629/CH8/EX8.5/example8_5.sce b/629/CH8/EX8.5/example8_5.sce new file mode 100644 index 000000000..8b3ce150b --- /dev/null +++ b/629/CH8/EX8.5/example8_5.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 8.5 REYNOLDS-NUMBER SIMILITUDE OF A VALVE
+//p-prototype, m-model
+Lmp=1/6; //Lmp=(Lm/Lp)
+//Vm*Lm/vm=Vp*Lp/vp, vm=vp
+Vmp=1/Lmp //Vmp=(Vm/Vp)
+Qp=700; //[cfs]
+Amp=(Lmp)^2 //Ratio of areas, Amp=(Am/Ap)
+//Discharge
+Qm=Qp*Vmp*Amp //[cfs]
+printf("\n The flow rate required for the model, Q = %.f cfs.\n",Qm)
diff --git a/629/CH8/EX8.6/ex8_6.txt b/629/CH8/EX8.6/ex8_6.txt new file mode 100644 index 000000000..02572fa11 --- /dev/null +++ b/629/CH8/EX8.6/ex8_6.txt @@ -0,0 +1,3 @@ +
+ The pressure difference between two points on the prototype = 178 Pa.
+
\ No newline at end of file diff --git a/629/CH8/EX8.6/example8_6.sce b/629/CH8/EX8.6/example8_6.sce new file mode 100644 index 000000000..32d9dfb9c --- /dev/null +++ b/629/CH8/EX8.6/example8_6.sce @@ -0,0 +1,12 @@ +clear
+clc
+//Example 8.6 APPLICATION OF PRESSURE COEFFICIENT
+//p-prototype, m-model
+Lmp=1/10; //Lmp=(Lm/Lp)
+//Vm*Lm/vm=Vp*Lp/vp, vm=vp
+Vpm=Lmp //Vpm=(Vp/Vm)
+Pm=17.8; //[kPa]
+//Pressure difference
+//Pm/(rho_m*Vm^2/2)=Pp/(rho_p*Vp^2/2)
+Pp=Pm*10^3*Vpm^2 //[Pa]
+printf("\n The pressure difference between two points on the prototype = %.f Pa.\n",Pp)
\ No newline at end of file diff --git a/629/CH8/EX8.7/ex8_7.txt b/629/CH8/EX8.7/ex8_7.txt new file mode 100644 index 000000000..0c04479df --- /dev/null +++ b/629/CH8/EX8.7/ex8_7.txt @@ -0,0 +1,3 @@ +
+ The expected drag force on the prototype = 1530 N.
+
\ No newline at end of file diff --git a/629/CH8/EX8.7/example8_7.sce b/629/CH8/EX8.7/example8_7.sce new file mode 100644 index 000000000..7755fca6f --- /dev/null +++ b/629/CH8/EX8.7/example8_7.sce @@ -0,0 +1,11 @@ +clear
+clc
+//Example 8.7 DRAG FORCE FROM WIND TUNNEL TESTING
+//p-prototype, m-model
+Lmp=1/10; //Lmp=(Lm/Lp)
+//Vm*Lm/vm=Vp*Lp/vp, vm=vp
+Vpm=Lmp //Vpm=(Vp/Vm)
+//Fm/(rho_m*Lm^2*Vm^2/2)=Fp/(rho_p*Lp^2*Vp^2/2)
+Fm=1530; //force[N]
+Fp=Fm*Vpm^2/Lmp^2 //[N]
+printf("\n The expected drag force on the prototype = %.f N.\n",Fp)
\ No newline at end of file diff --git a/629/CH8/EX8.8/ex8_8.txt b/629/CH8/EX8.8/ex8_8.txt new file mode 100644 index 000000000..4679a232c --- /dev/null +++ b/629/CH8/EX8.8/ex8_8.txt @@ -0,0 +1,2 @@ +
+ The pressure drop in the actual nozzle = 0.0625 psid.
\ No newline at end of file diff --git a/629/CH8/EX8.8/example8_8.sce b/629/CH8/EX8.8/example8_8.sce new file mode 100644 index 000000000..65566fd13 --- /dev/null +++ b/629/CH8/EX8.8/example8_8.sce @@ -0,0 +1,21 @@ +clear
+clc
+//Example 8.8 MEASURING HEAD LOSS IN NOZZLE IN REVERSE FLOW
+//p-prototype, m-model
+Vm=20; //[ft/s]
+Dm=3/12; //[ft]
+vm=1.22*10^-5; //[ft^2/s]
+Vp=5; //[ft/s]
+Dp=3; //[ft]
+vp=1.41*10^-5; //[ft^2/s]
+//Reynolds numbers
+Rem=Vm*Dm/vm
+Rep=Vp*Dp/vp
+rho=1.94; //[slugs/ft^3]
+Pm=1; //pressure [lbf/in^2]
+//Pressure coefficient
+CPm=Pm*144/(rho*Vm^2/2)
+//Both Rep, Rem >10^3, therefore
+CPp=CPm
+Pp=(CPp*rho*Vp^2/2)/144 //[lbf/in^2]
+printf("\n The pressure drop in the actual nozzle = %.4f psid.\n",Pp)//
\ No newline at end of file diff --git a/629/CH8/EX8.9/ex8_9.txt b/629/CH8/EX8.9/ex8_9.txt new file mode 100644 index 000000000..99b8de7d2 --- /dev/null +++ b/629/CH8/EX8.9/ex8_9.txt @@ -0,0 +1,3 @@ +
+ The minimum required wind tunnel speed = 3.8 m/s.
+
\ No newline at end of file diff --git a/629/CH8/EX8.9/example8_9.sce b/629/CH8/EX8.9/example8_9.sce new file mode 100644 index 000000000..d98e2c7b1 --- /dev/null +++ b/629/CH8/EX8.9/example8_9.sce @@ -0,0 +1,19 @@ +clear
+clc
+//Example 8.9 MODEL TESTS FOR DRAG FORCE ON AN AUTOMOBILE
+c=1235; //[km/hr]
+//Vm*Lm/vm=Vp*Lp/vp, vm=vp
+Lp=0.4; //[m]
+Lm=0.4; //[m]
+Vp=100; //[km/hr]
+Vm=Vp*Lp/Lm //[km/hr]
+//Mach number
+Mm=Vm/c
+//Mm is too high for test models and results in unwanted compressibility effects.
+mu=1.51*10^-5; //[m^2/s]
+Rep=Vp*0.278*Lp/mu
+//CFm=CFp, if Rem>=10^5
+Rem=10^5;
+//Wind tunnel speed
+Vm=Rem*mu/Lm //[m/s]
+printf("\n The minimum required wind tunnel speed = %.1f m/s.\n",Vm)
diff --git a/629/CH9/EX9.1/ex9_1.txt b/629/CH9/EX9.1/ex9_1.txt new file mode 100644 index 000000000..b61066503 --- /dev/null +++ b/629/CH9/EX9.1/ex9_1.txt @@ -0,0 +1,2 @@ +
+The shear stress on the plates = 333 N/m^2.
\ No newline at end of file diff --git a/629/CH9/EX9.1/example9_1.sce b/629/CH9/EX9.1/example9_1.sce new file mode 100644 index 000000000..0a5ae501e --- /dev/null +++ b/629/CH9/EX9.1/example9_1.sce @@ -0,0 +1,8 @@ +clear
+clc
+//Example 9.1 SHEAR STRESS IN COUETTE FLOW
+U=1; //speed of upper plate [m/s]
+L=3*10^-4; //distance between the plates [m]
+mu=1*10^-1; //[N.s/m^2]
+tau=mu*U/L //shear stress [N/m^2]
+printf("\nThe shear stress on the plates = %.f N/m^2.\n",tau)
\ No newline at end of file diff --git a/629/CH9/EX9.2/ex9_2.txt b/629/CH9/EX9.2/ex9_2.txt new file mode 100644 index 000000000..ae264b44f --- /dev/null +++ b/629/CH9/EX9.2/ex9_2.txt @@ -0,0 +1,3 @@ +
+The pressure gradient, (dp/ds) = 5448 N/m^2 per meter.
+
\ No newline at end of file diff --git a/629/CH9/EX9.2/example9_2.sce b/629/CH9/EX9.2/example9_2.sce new file mode 100644 index 000000000..a28332092 --- /dev/null +++ b/629/CH9/EX9.2/example9_2.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 9.2 PRESSURE GRADIENT FOR FLOW BETWEEN PARALLEL PLATES
+q=0.01; //discharge per meter [m^2/s]
+rho=800; //density [kg/m^3]
+mu=2*10^-2; //[N.s/m^2]
+Re=q*rho/mu
+//Re<1000. Hence, flow is luminar and equations apply.
+
+v=mu/rho //viscosity [m^2/s]
+B=0.01; //[m]
+g=9.81; //[m/s^2]
+Gamma=0.8*9810; //specific weight [N/m^3]
+dhds=-12*v*q/(g*B^3) //Piezometric head gradient (dh/ds)
+dzds=-1; //(dz/ds)
+//(dh/ds)=(d(p/gamma)/ds)+(dz/ds)
+dpds=Gamma*(dhds-dzds) //pressure gradient (dp/ds), [N/m^3]
+printf("\nThe pressure gradient, (dp/ds) = %.f N/m^2 per meter.\n",dpds)
\ No newline at end of file diff --git a/629/CH9/EX9.3/ex9_3.JPG b/629/CH9/EX9.3/ex9_3.JPG Binary files differnew file mode 100644 index 000000000..ba784d1d6 --- /dev/null +++ b/629/CH9/EX9.3/ex9_3.JPG diff --git a/629/CH9/EX9.3/example9_3.sce b/629/CH9/EX9.3/example9_3.sce new file mode 100644 index 000000000..c3b4bbc88 --- /dev/null +++ b/629/CH9/EX9.3/example9_3.sce @@ -0,0 +1,16 @@ +clear
+clc
+//Example 9.3 LAMINAR BOUNDARY-LAYER THICKNESS AND SHEAR STRESS
+Uo=1; //[ft/s]
+v=10^-4; //[ft^2/s]
+S=0.86; //specific gravity
+rho=1.94*S //density [slugs/ft^3]
+mu=rho*v //[lbf.s/ft^2]
+//Graph plots
+x=linspace(0.1,7,15); //distance[ft]
+delta=5*(sqrt(v*x/Uo))*12; //boundary layer thickness [inch]
+To=0.332*mu*Uo*sqrt(Uo*(v*x)^-1)*10^2; //sheer stress in 100 [psf]
+plot(x,delta,"+-")
+plot(x,To,"o-")
+xtitle("delta,To vs x","Distance,ft ","delta (inches), Tox100 (psf)");
+legend("Boundary layer thickness (delta)","Surface sheer stress (To)");
\ No newline at end of file diff --git a/629/CH9/EX9.4/ex9_4.txt b/629/CH9/EX9.4/ex9_4.txt new file mode 100644 index 000000000..bd965be6b --- /dev/null +++ b/629/CH9/EX9.4/ex9_4.txt @@ -0,0 +1,3 @@ +
+The shear resistance on one side of the plate = 0.108 lbf.
+
\ No newline at end of file diff --git a/629/CH9/EX9.4/example9_4.sce b/629/CH9/EX9.4/example9_4.sce new file mode 100644 index 000000000..f484bd771 --- /dev/null +++ b/629/CH9/EX9.4/example9_4.sce @@ -0,0 +1,18 @@ +clear
+clc
+//Example 9.4 RESISTANCE CALCULATION FOR LAMINAR BOUNDARY LAYER ON A FLAT PLATE
+//To find Approx Value
+function [A]= approx (V,n)
+ A= round(V*10^n)/10^n; //V-Value, n-to what place
+ funcprot (0)
+endfunction
+L=6; //[ft]
+v=10^-4; //viscosity [ft^2/s]
+Uo=1; //[ft/s]
+Re=Uo*L/v //Reynolds number
+Cf=approx(1.33/Re^(1/2),4)
+B=4; //[ft]
+S=0.86;
+rho=S*1.94 //[slugs/ft^3]
+Fs=approx((Cf*B*L*rho*Uo^2)/2,3) //shear force [lbf]
+printf("\nThe shear resistance on one side of the plate = %.3f lbf.\n",Fs)
\ No newline at end of file diff --git a/629/CH9/EX9.5/ex9_5.txt b/629/CH9/EX9.5/ex9_5.txt new file mode 100644 index 000000000..5f8494bd8 --- /dev/null +++ b/629/CH9/EX9.5/ex9_5.txt @@ -0,0 +1,9 @@ +
+The velocity of water as determined by
+(a)Logarithmic velocity distribution = 14.1 ft.
+(b)Velocity defect law = 14.4 ft.
+(c)Power-law formula = 14.39 ft.
+
+
+The nominal thickness of the viscous sublayer = 2.55*10^(-3) inches.
+
\ No newline at end of file diff --git a/629/CH9/EX9.5/example9_5.sce b/629/CH9/EX9.5/example9_5.sce new file mode 100644 index 000000000..752d1a219 --- /dev/null +++ b/629/CH9/EX9.5/example9_5.sce @@ -0,0 +1,26 @@ +clear
+clc
+//Example 9.5 TURBULENT BOUNDARY-LAYER PROPERTIES
+to=0.896; //[lbf/ft^3]
+rho=1.94; //[slugs/ft^3]
+uo=sqrt(to/rho) //shear velocity [ft/s]
+
+//Logarithmic velocity distribution
+y=0.0088; //[ft]
+v=1.22*10^-5; //[ft^2/s]
+uL=uo*(2.44*log(y*uo/v)+5.56) //velocity of water [ft/s]
+
+delta=0.088; //[ft]
+yn=y/delta //non-dimensional distance
+
+//velocity defect law
+//(Uo-u)/uo=8.2, from fig 9.11
+Uo=20; //[ft/s]
+ud=Uo-8.2*uo //velocity [ft/s]
+
+//Power-law formula
+up=Uo*yn^(1/7) //velocity [ft/s]
+printf("\nThe velocity of water as determined by\n(a)Logarithmic velocity distribution = %.1f ft.\n(b)Velocity defect law = %.1f ft.\n(c)Power-law formula = %.2f ft.\n\n",uL,ud,up)
+
+deltaN=(11.84*v/uo)*12*10^3 //nominal thickness in 10^-3 inches
+printf("\nThe nominal thickness of the viscous sublayer = %.2f*10^(-3) inches.\n",deltaN)
\ No newline at end of file diff --git a/629/CH9/EX9.6/ex9_6.txt b/629/CH9/EX9.6/ex9_6.txt new file mode 100644 index 000000000..69493ca2f --- /dev/null +++ b/629/CH9/EX9.6/ex9_6.txt @@ -0,0 +1,12 @@ +
+Average shear-stress coefficient, Cf, for the plate = 0.00294.
+
+
+Total shear force on one side of plate = 4.77 N.
+
+
+Shear force due to laminar part = 0.256 N.
+
+
+Shear force due to turbulent part = 4.51 N.
+
\ No newline at end of file diff --git a/629/CH9/EX9.6/example9_6.sce b/629/CH9/EX9.6/example9_6.sce new file mode 100644 index 000000000..d3d30928f --- /dev/null +++ b/629/CH9/EX9.6/example9_6.sce @@ -0,0 +1,30 @@ +clear
+clc
+//Example 9.6 LAMINAR/TURBULENT BOUNDARY LAYER ON FLAT PLATE
+Uo=30; //[m/s]
+L=3; //[m]
+v=1.51*10^-5; //viscosity [m^2/s]
+Re=Uo*L/v //Reynolds number
+
+//Average shear-stress coefficient
+Cf=(0.523/(log(0.06*Re))^2)-(1520/Re)
+printf("\nAverage shear-stress coefficient, Cf, for the plate = %.5f.\n\n",Cf)
+
+//Total shear force
+B=1; //[m]
+rho=1.2; //[kg/m^3]
+Fs=Cf*B*L*rho*(Uo^2/2) //[N]
+printf("\nTotal shear force on one side of plate = %.2f N.\n\n",Fs)
+
+Re_tr=5*10^5;
+xtr=Re_tr*v/Uo //transition point [m]
+//Laminar average shear-stress coefficient
+Cfl=1.33/Re_tr^(1/2)
+
+//Laminar shear force
+Fsl=(Cfl*B*xtr*rho*Uo^2)/2 //[N]
+printf("\nShear force due to laminar part = %.3f N.\n\n",Fsl)
+
+//Turbulent shear force
+Fst=Fs-Fsl //[N]
+printf("\nShear force due to turbulent part = %.2f N.\n",Fst)
\ No newline at end of file diff --git a/629/CH9/EX9.7/ex9_7.txt b/629/CH9/EX9.7/ex9_7.txt new file mode 100644 index 000000000..cbc17dc96 --- /dev/null +++ b/629/CH9/EX9.7/ex9_7.txt @@ -0,0 +1,3 @@ +
+Total shear resistance on both sides of plate = 28.6 N.
+
\ No newline at end of file diff --git a/629/CH9/EX9.7/example9_7.sce b/629/CH9/EX9.7/example9_7.sce new file mode 100644 index 000000000..492de2129 --- /dev/null +++ b/629/CH9/EX9.7/example9_7.sce @@ -0,0 +1,15 @@ +clear
+clc
+//Example 9.7 RESISTANCE FORCE WITH TRIPPED BOUNDARY LAYER
+L=6; //[m]
+B=3; //[m]
+mu=1.81*10^-5; //[N.s/m^2]
+rho=1.2; //[Kg/m^3]
+Uo=20; //[m/s]
+ReL=rho*Uo*L/mu //Reynold's number
+//Average shear-stress coefficient
+Cf=0.032/ReL^(1/7)
+A=L*B //area [m^2]
+//Resistance force
+Fs=2*Cf*A*rho*(Uo^2)/2 //[N]
+printf("\nTotal shear resistance on both sides of plate = %.1f N.\n",Fs)
\ No newline at end of file |
