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-rw-r--r--629/CH9/EX9.1/ex9_1.txt2
-rw-r--r--629/CH9/EX9.1/example9_1.sce8
-rw-r--r--629/CH9/EX9.2/ex9_2.txt3
-rw-r--r--629/CH9/EX9.2/example9_2.sce18
-rw-r--r--629/CH9/EX9.3/ex9_3.JPGbin0 -> 32210 bytes
-rw-r--r--629/CH9/EX9.3/example9_3.sce16
-rw-r--r--629/CH9/EX9.4/ex9_4.txt3
-rw-r--r--629/CH9/EX9.4/example9_4.sce18
-rw-r--r--629/CH9/EX9.5/ex9_5.txt9
-rw-r--r--629/CH9/EX9.5/example9_5.sce26
-rw-r--r--629/CH9/EX9.6/ex9_6.txt12
-rw-r--r--629/CH9/EX9.6/example9_6.sce30
-rw-r--r--629/CH9/EX9.7/ex9_7.txt3
-rw-r--r--629/CH9/EX9.7/example9_7.sce15
14 files changed, 163 insertions, 0 deletions
diff --git a/629/CH9/EX9.1/ex9_1.txt b/629/CH9/EX9.1/ex9_1.txt
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+
+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
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index 000000000..0a5ae501e
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+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
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index 000000000..ae264b44f
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+
+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
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+++ b/629/CH9/EX9.2/example9_2.sce
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+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
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+++ b/629/CH9/EX9.3/ex9_3.JPG
Binary files differ
diff --git a/629/CH9/EX9.3/example9_3.sce b/629/CH9/EX9.3/example9_3.sce
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index 000000000..c3b4bbc88
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+++ b/629/CH9/EX9.3/example9_3.sce
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+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
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+++ b/629/CH9/EX9.4/ex9_4.txt
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+
+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
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+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
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+
+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
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+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
generated by cgit (git 2.34.1) at 2024-12-20 03:52:11 +0000