Chapter 6, Problem 6.99P

To determine

(a)

The flow rate without diffuser.

Answer to Problem 6.99P

$Q=2.72\times {10}^{-3}{m}^3/s$

Explanation of Solution

Given information:

The pipe entrance is sharp edged.

The energy equation can be written as below for the system without the diffuser,

$z_1-z_2=\frac{V_2^2}{2g}+\frac{V_1^2}{2g}\left(f\frac{L}{D}+K_{entrance}\right)$

But, we can say,

$V_1=V_2$

Therefore, the above equation can be rewritten as,

$\Delta z=\frac{V_1^2}{2g}\left(1+f\frac{L}{D}+K_{entrance}\right)$

Calculation:

According to the explanation given,

Assume,

$f=0.02$

$K_{entrance}=0.5$

Therefore,

$\begin{array}{l}\Delta z=\frac{V_1^2}{2g}\left(1+f\frac{L}{D}+K_{entrance}\right)\\ 2m=\frac{V_1^2}{2\left(9.81m/s\right)}\left(1+\left(0.02\right)\frac{2m}{0.03m}+0.5\right)\\ V_1=3.721m/s\\ \end{array}$

Calculate the Reynolds’s number,

$R_e=\frac{\rho Vd}{\mu }=\frac{\left(998kg/m^3\right)\left(3.721m/s\right)\left(0.03m\right)}{0.001kg/ms}=111406.74$

Calculate the friction factor,

$\begin{array}{l}\zeta =fcn\left(R_e\right)\\ \zeta \approx 0.155R_e{}^{1.75}=\frac{f{\left(R_e\right)}^2}{2}\end{array}$

$\begin{array}{l}0.155{\left(111406.74\right)}^{1.75}=\frac{f{\left(111406.74\right)}^2}{2}\\ f=0.017\end{array}$

Calculate the exact value of the velocity,

$\begin{array}{l}\Delta z=\frac{V_1^2}{2g}\left(1+f\frac{L}{D}+K_{entrance}\right)\\ 2m=\frac{V_1^2}{2\left(9.81m/s\right)}\left(1+\left(0.017\right)\frac{2m}{0.03m}+0.5\right)\\ V_1=3.86m/s\end{array}$

Calculate the flow rate,

$\begin{array}{l}Q=VA\\ =\left(3.86m/s\right)\left(\pi {\left(0.015m\right)}^2\right)\\ Q=2.72\times {10}^{-3}{m}^3/s\end{array}$

Conclusion:

The flow rate is equal to $2.72\times {10}^{-3}{m}^3/s$ without diffuser.

To determine

(b)

The flow rate with diffuser.

Answer to Problem 6.99P

$Q=3.19\times {10}^{-3}{m}^3/s$

Explanation of Solution

Given information:

The pipe entrance is sharp edged.

The energy equation can be written as below for the system with the diffuser,

$z_1-z_2=\frac{V_2^2}{2g}+\frac{V_1^2}{2g}\left(f\frac{L}{D}+K_{entrance}+K_{Diffuser}\right)$

Above equation can be rewritten as,

$\Delta z=\frac{V_2^2}{2g}+\frac{V_1^2}{2g}\left(f\frac{L}{D}+K_{entrance}+K_{Diffuser}\right)$

Calculation:

Calculate $V_2$ in terms of $V_1$

By equating the flow rates, we can say,

$V_2={\left(\frac{3}{5}\right)}^2{V}_1=0.36V_1$

Assume,

$f=0.02$

$K_{entrance}=0.5$

$K_{Diffuser}=0.2$

Therefore,

$\Delta z=\frac{V_2^2}{2g}+\frac{V_1^2}{2g}\left(f\frac{L}{D}+K_{entrance}+K_{Diffuser}\right)$

$\begin{array}{l}2m=\frac{{\left(0.36V_1\right)}^2}{2\left(9.81m/s^2\right)}+\frac{V_1^2}{2\left(9.81m/s^2\right)}\left(\left(0.02\right)\frac{2m}{0.03m}+0.5+0.2\right)\\ V_1=4.25m/s\end{array}$

Calculate the Reynolds’s number

$R_e=\frac{\rho Vd}{\mu }=\frac{\left(998kg/m^3\right)\left(4.25m/s\right)\left(0.03m\right)}{0.001kg/ms}=127245$

Calculate the friction factor

$\begin{array}{l}\zeta =fcn\left(R_e\right)\\ \zeta \approx 0.155R_e{}^{1.75}=\frac{f{\left(R_e\right)}^2}{2}\end{array}$

$\begin{array}{l}0.155{\left(127245\right)}^{1.75}=\frac{f{\left(127245\right)}^2}{2}\\ f=0.0164\end{array}$

Calculate the exact value of the velocity

$\Delta z=\frac{V_2^2}{2g}+\frac{V_1^2}{2g}\left(1+f\frac{L}{D}+K_{entrance}+K_{Diffuser}\right)$

$\begin{array}{l}2m=\frac{{\left(0.36V_1\right)}^2}{2\left(9.81m/s^2\right)}+\frac{V_1^2}{2\left(9.81m/s^2\right)}\left(\left(0.0164\right)\frac{2m}{0.03m}+0.5+0.2\right)\\ V_1=4.52m/s\end{array}$

Therefore,

Calculate the flow rate

$\begin{array}{l}Q=VA\\ =\left(4.52m/s\right)\left(\pi {\left(0.015m\right)}^2\right)\\ Q=3.19\times {10}^{-3}{m}^3/s\end{array}$

Conclusion:

The flow rate is equal to $3.19\times {10}^{-3}{m}^3/s$ with diffuser.