Chapter 7, Problem 7.21P

To determine

The stream velocity.

Answer to Problem 7.21P

Stream velocity $U=20\frac{m}{s}$

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mention in question.

For air at ${20}^{°}C,1atm$ take

Density of air ${\rho }_{air}=1.2\frac{kg}{m^3}$

Density of oil ${\rho }_{oil}=S.G\times {\rho }_{water}=0.827\times 998=825\frac{kg}{m^3}$

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

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after $y=4.5mm$, this shows the boundary layer ends here and free stream velocity starts from this location $y=4.5mm$.

So, we will consider manometer head $h_{mano}=0.0297m$ from table.

We know that:

The stream velocity $U=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}}{{\rho }_{air}}}$

Put the required value in this equation:

We get $\begin{array}{l}U=\sqrt{\frac{2(825-1.2)9.81\times 0.0297}{1.2}}\\ U=20\frac{m}{s}\end{array}$.

To determine

The boundary layer thickness.

Answer to Problem 7.21P

The boundary layer thickness $\delta =4.5mm$

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mention in question.

For air at ${20}^{°}C,1atm$ take

Density of air ${\rho }_{air}=1.2\frac{kg}{m^3}$

Density of oil ${\rho }_{oil}=S.G\times {\rho }_{water}=0.827\times 998=825\frac{kg}{m^3}$

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

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after $y=4.5mm$, this shows the boundary layer ends here and free stream velocity starts from this location $y=4.5mm$.

So, we will consider manometer head $h_{mano}=0.0297m$ from table.

We know that,

The stream velocity $U=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}}{{\rho }_{air}}}$

Put the required value in this equation,

We get $\begin{array}{l}U=\sqrt{\frac{2(825-1.2)9.81\times 0.0297}{1.2}}\\ U=20\frac{m}{s}\end{array}$ ANSWER (a)

As in above discussion boundary layer ends at $y=4.5mm$

Then this value is equal to boundary layer thickness.

$y=\delta =4.5mm$.

To determine

The wall shear stress.

The wall shear stress ${\tau }_w=0.14pa$

Given information:

The readings of y in mm and h in mm as mention in question.

For air at ${20}^{°}C,1atm$ take

Density of air ${\rho }_{air}=1.2\frac{kg}{m^3}$

Density of oil ${\rho }_{oil}=S.G\times {\rho }_{water}=0.827\times 998=825\frac{kg}{m^3}$

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

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after $y=4.5mm$, this shows the boundary layer ends here and free stream velocity starts from this location $y=4.5mm$.

So, we will consider manometer head $h_{mano}=0.0297m$ from table.

We know that,

The stream velocity $U=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}}{{\rho }_{air}}}$

Put the required value in this equation,

We get $\begin{array}{l}U=\sqrt{\frac{2(825-1.2)9.81\times 0.0297}{1.2}}\\ U=20\frac{m}{s}\end{array}$ ANSWER (a)

As in above discussion boundary layer ends at $y=4.5mm$

Then this value is equal to boundary layer thickness.

$y=\delta =4.5mm$

We know that:

The wall shear stress ${\tau }_w=\mu \frac{\partial u}{\partial y}{|}_{y=0}$

We can write ${\tau }_w\approx \mu \frac{\Delta u}{\Delta y}\approx \mu \frac{u_{y1}-0}{y1-0}$ ---------equation 1

where $y1=0.5\times {10}^{-3}m$ is the first value of y from table.

$u_{y1}=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}{}_{(y1)}}{{\rho }_{air}}}$

$=4.02\frac{m}{s}$

Put required values in equation 1

$\begin{array}{l}{\tau }_w\approx 1.8E-5\left(\frac{4.02-0}{0.0005-0}\right)\\ {\tau }_w\approx 0.14pa\\ \end{array}$.

The total friction drag.

The total friction drag $F_D=0.25N$

Given information:

The readings of y in mm and h in mm as mentioned in the question.

For air at ${20}^{°}C,1atm$ take

Density of air ${\rho }_{air}=1.2\frac{kg}{m^3}$

Density of oil ${\rho }_{oil}=S.G\times {\rho }_{water}=0.827\times 998=825\frac{kg}{m^3}$

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

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after $y=4.5mm$, this shows the boundary layer ends here and free stream velocity starts from this location $y=4.5mm$.

So, we will consider manometer head $h_{mano}=0.0297m$ from table.

We know that,

The stream velocity $U=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}}{{\rho }_{air}}}$

Put the required value in this equation:

We get $\begin{array}{l}U=\sqrt{\frac{2(825-1.2)9.81\times 0.0297}{1.2}}\\ U=20\frac{m}{s}\end{array}$ ANSWER (a)

As in above discussion boundary layer ends at $y=4.5mm$

Then this value is equal to boundary layer thickness.

$y=\delta =4.5mm$

We know that:

The wall shear stress ${\tau }_w=\mu \frac{\partial u}{\partial y}{|}_{y=0}$

We can write ${\tau }_w\approx \mu \frac{\Delta u}{\Delta y}\approx \mu \frac{u_{y1}-0}{y1-0}$ ---------equation 1

where $y1=0.5\times {10}^{-3}m$ is the first value of y from table.

$u_{y1}=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}{}_{(y1)}}{{\rho }_{air}}}$

$=4.02\frac{m}{s}$

Put required values in equation 1:

$\begin{array}{l}{\tau }_w\approx 1.8E-5\left(\frac{4.02-0}{0.0005-0}\right)\\ {\tau }_w\approx 0.14pa\end{array}$

We know that the friction drag at wall is twice of shear stress $\left({\tau }_w\right)$ times area $\left(xb\right)$.

We will use the value of x estimated in question 7.20 $\left(x=0.908m\right)$ .and take width $b=1.0m$

Now $F_D=2{\tau }_w(xb)$

Put all required values:

We get $\begin{array}{l}{F}_D=2\times 0.14\times 0.908\times 1.0\\ F_D=0.25N.\end{array}$.

Answer to Problem 7.21P

The wall shear stress ${\tau }_w=0.14pa$

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mention in question.

For air at ${20}^{°}C,1atm$ take

Density of air ${\rho }_{air}=1.2\frac{kg}{m^3}$

Density of oil ${\rho }_{oil}=S.G\times {\rho }_{water}=0.827\times 998=825\frac{kg}{m^3}$

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

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after $y=4.5mm$, this shows the boundary layer ends here and free stream velocity starts from this location $y=4.5mm$.

So, we will consider manometer head $h_{mano}=0.0297m$ from table.

We know that,

The stream velocity $U=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}}{{\rho }_{air}}}$

Put the required value in this equation,

We get $\begin{array}{l}U=\sqrt{\frac{2(825-1.2)9.81\times 0.0297}{1.2}}\\ U=20\frac{m}{s}\end{array}$ ANSWER (a)

As in above discussion boundary layer ends at $y=4.5mm$

Then this value is equal to boundary layer thickness.

$y=\delta =4.5mm$

We know that:

The wall shear stress ${\tau }_w=\mu \frac{\partial u}{\partial y}{|}_{y=0}$

We can write ${\tau }_w\approx \mu \frac{\Delta u}{\Delta y}\approx \mu \frac{u_{y1}-0}{y1-0}$ ---------equation 1

where $y1=0.5\times {10}^{-3}m$ is the first value of y from table.

$u_{y1}=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}{}_{(y1)}}{{\rho }_{air}}}$

$=4.02\frac{m}{s}$

Put required values in equation 1

$\begin{array}{l}{\tau }_w\approx 1.8E-5\left(\frac{4.02-0}{0.0005-0}\right)\\ {\tau }_w\approx 0.14pa\\ \end{array}$.

To determine

The total friction drag.

Answer to Problem 7.21P

The total friction drag $F_D=0.25N$

Explanation of Solution

Given information:

The readings of y in mm and h in mm as mentioned in the question.

For air at ${20}^{°}C,1atm$ take

Density of air ${\rho }_{air}=1.2\frac{kg}{m^3}$

Density of oil ${\rho }_{oil}=S.G\times {\rho }_{water}=0.827\times 998=825\frac{kg}{m^3}$

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

Assumption flow is a laminar flow.

Calculation:

We can see in the table that manometer head is not changing after $y=4.5mm$, this shows the boundary layer ends here and free stream velocity starts from this location $y=4.5mm$.

So, we will consider manometer head $h_{mano}=0.0297m$ from table.

We know that,

The stream velocity $U=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}}{{\rho }_{air}}}$

Put the required value in this equation:

We get $\begin{array}{l}U=\sqrt{\frac{2(825-1.2)9.81\times 0.0297}{1.2}}\\ U=20\frac{m}{s}\end{array}$ ANSWER (a)

As in above discussion boundary layer ends at $y=4.5mm$

Then this value is equal to boundary layer thickness.

$y=\delta =4.5mm$

We know that:

The wall shear stress ${\tau }_w=\mu \frac{\partial u}{\partial y}{|}_{y=0}$

We can write ${\tau }_w\approx \mu \frac{\Delta u}{\Delta y}\approx \mu \frac{u_{y1}-0}{y1-0}$ ---------equation 1

where $y1=0.5\times {10}^{-3}m$ is the first value of y from table.

$u_{y1}=\sqrt{\frac{2\Delta p}{{\rho }_{air}}}=\sqrt{\frac{2({\rho }_{oil}-{\rho }_{air})g{h}_{mano}{}_{(y1)}}{{\rho }_{air}}}$

$=4.02\frac{m}{s}$

Put required values in equation 1:

$\begin{array}{l}{\tau }_w\approx 1.8E-5\left(\frac{4.02-0}{0.0005-0}\right)\\ {\tau }_w\approx 0.14pa\end{array}$

We know that the friction drag at wall is twice of shear stress $\left({\tau }_w\right)$ times area $\left(xb\right)$.

We will use the value of x estimated in question 7.20 $\left(x=0.908m\right)$ .and take width $b=1.0m$

Now $F_D=2{\tau }_w(xb)$

Put all required values:

We get $\begin{array}{l}{F}_D=2\times 0.14\times 0.908\times 1.0\\ F_D=0.25N.\end{array}$.