Chapter 7, Problem 7.23P

To determine

Boundary layer thickness $\delta$ at the end of the board.

Answer to Problem 7.23P

$0.3in\langle \delta \rangle 1.9in$

Explanation of Solution

Given information:

Width $b=4ft=1.22m$, length $L=8ft=2.44m$,

$U=35\frac{mi}{h}=15.6\frac{m}{s}$, take density of air $\rho =1.2\frac{kg}{m^3}$,

dynamic viscosity $\mu =1.8E-5\frac{kg}{m-s}$

Calculation:

We know that,

${\mathrm{Re}}_L=\frac{\rho UL}{\mu }=\frac{1.2(15.6)(2.44)}{1.8E-5}=2.55E6$

This shows flow is in between the Laminar and turbulent conditions.

Now we know that:

For laminar $\begin{array}{l}\frac{\delta }{L}=\frac{5}{\sqrt{2.55E6}}=0.00313\\ \delta =0.00765m=0.3in\end{array}$

For turbulent $\begin{array}{l}\frac{\delta }{L}=\frac{0.16}{{(2.55E6)}^{\frac{1}{7}}}=0.0195\\ \delta =0.047m=1.9in\end{array}$.

To determine

(b)For laminar flow Drag force $F_D$.

(b)For laminar flow $F_D$ = 0.73 N

Given information:

Width $b=4ft=1.22m$, length $L=8ft=2.44m$,

$U=35\frac{mi}{h}=15.6\frac{m}{s}$, take density of air $\rho =1.2\frac{kg}{m^3}$,

dynamic viscosity $\mu =1.8E-5\frac{kg}{m-s}$

Calculation:

We know that:

${\mathrm{Re}}_L=\frac{\rho UL}{\mu }=\frac{1.2(15.6)(2.44)}{1.8E-5}=2.55E6$

This shows flow is in between the Laminar and turbulent condition.

Area $=2bL$ for both side of flow

For laminar flow, $C_D=\frac{1.328}{\sqrt{2.55E6}}$

$\begin{array}{l}{F}_D=C_D\frac{\rho }{2}{U}^2(2bL)=\frac{1.328}{\sqrt{2.55E6}}\frac{1.2}{2}{(}^{15.6}(2\times 1.22\times 2.44)\\ F_D=0.73N\end{array}$.

(c)For turbulent flow Drag force $F_D$.

(c)For turbulent flow $F_D$ = 3.3N

Given information:

Width $b=4ft=1.22m$, length $L=8ft=2.44m$,

$U=35\frac{mi}{h}=15.6\frac{m}{s}$, take density of air $\rho =1.2\frac{kg}{m^3}$,

dynamic viscosity $\mu =1.8E-5\frac{kg}{m-s}$

Calculation:

We know that,

${\mathrm{Re}}_L=\frac{\rho UL}{\mu }=\frac{1.2(15.6)(2.44)}{1.8E-5}=2.55E6$

This shows flow is in between the Laminar and turbulent condition.

Area $=2bL$ for both sides of flow.

For turbulent flow:

$C_D=\frac{0.031}{{(2.55E6)}^{\frac{1}{7}}}$

$\begin{array}{l}{F}_D=C_D\frac{\rho }{2}{U}^2(2bL)=\frac{0.031}{{(2.55E6)}^{\frac{1}{7}}}\frac{1.2}{2}{(}^{15.6}(2\times 1.22\times 2.44)\\ F_D=3.3N\end{array}$.

Answer to Problem 7.23P

(b)For laminar flow $F_D$ = 0.73 N

Explanation of Solution

Given information:

Width $b=4ft=1.22m$, length $L=8ft=2.44m$,

$U=35\frac{mi}{h}=15.6\frac{m}{s}$, take density of air $\rho =1.2\frac{kg}{m^3}$,

dynamic viscosity $\mu =1.8E-5\frac{kg}{m-s}$

Calculation:

We know that:

${\mathrm{Re}}_L=\frac{\rho UL}{\mu }=\frac{1.2(15.6)(2.44)}{1.8E-5}=2.55E6$

This shows flow is in between the Laminar and turbulent condition.

Area $=2bL$ for both side of flow

For laminar flow, $C_D=\frac{1.328}{\sqrt{2.55E6}}$

$\begin{array}{l}{F}_D=C_D\frac{\rho }{2}{U}^2(2bL)=\frac{1.328}{\sqrt{2.55E6}}\frac{1.2}{2}{(}^{15.6}(2\times 1.22\times 2.44)\\ F_D=0.73N\end{array}$.

To determine

(c)For turbulent flow Drag force $F_D$.

(c)For turbulent flow $F_D$ = 3.3N

Given information:

Width $b=4ft=1.22m$, length $L=8ft=2.44m$,

$U=35\frac{mi}{h}=15.6\frac{m}{s}$, take density of air $\rho =1.2\frac{kg}{m^3}$,

dynamic viscosity $\mu =1.8E-5\frac{kg}{m-s}$

Calculation:

We know that,

${\mathrm{Re}}_L=\frac{\rho UL}{\mu }=\frac{1.2(15.6)(2.44)}{1.8E-5}=2.55E6$

This shows flow is in between the Laminar and turbulent condition.

Area $=2bL$ for both sides of flow.

For turbulent flow:

$C_D=\frac{0.031}{{(2.55E6)}^{\frac{1}{7}}}$

$\begin{array}{l}{F}_D=C_D\frac{\rho }{2}{U}^2(2bL)=\frac{0.031}{{(2.55E6)}^{\frac{1}{7}}}\frac{1.2}{2}{(}^{15.6}(2\times 1.22\times 2.44)\\ F_D=3.3N\end{array}$.

Answer to Problem 7.23P

(c)For turbulent flow $F_D$ = 3.3N

Explanation of Solution

Given information:

Width $b=4ft=1.22m$, length $L=8ft=2.44m$,

$U=35\frac{mi}{h}=15.6\frac{m}{s}$, take density of air $\rho =1.2\frac{kg}{m^3}$,

dynamic viscosity $\mu =1.8E-5\frac{kg}{m-s}$

Calculation:

We know that,

${\mathrm{Re}}_L=\frac{\rho UL}{\mu }=\frac{1.2(15.6)(2.44)}{1.8E-5}=2.55E6$

This shows flow is in between the Laminar and turbulent condition.

Area $=2bL$ for both sides of flow.

For turbulent flow:

$C_D=\frac{0.031}{{(2.55E6)}^{\frac{1}{7}}}$

$\begin{array}{l}{F}_D=C_D\frac{\rho }{2}{U}^2(2bL)=\frac{0.031}{{(2.55E6)}^{\frac{1}{7}}}\frac{1.2}{2}{(}^{15.6}(2\times 1.22\times 2.44)\\ F_D=3.3N\end{array}$.