Chapter 6, Problem 6.73P

To determine

(a)

That the flow rate is equal to $0.03m^3/s$.

Answer to Problem 6.73P

Yes, the computed flow rate is correct and it’s equal to $0.03m^3/s$

Explanation of Solution

Given information:

The fluid is water at $20°C$

The diameter of the pipe is $10cm$

$\frac{\Delta P}{L}=1000Pa/m$

The pressure drop is defined as,

$\Delta P=\rho g{h}_f=\rho g\left(f\frac{L}{D}\left(\frac{V^2}{2g}\right)\right)$

In above equation,

$\rho$ - Density of the fluid

$h_f$ - Head loss

$L$ - Length of pipe

$D$ - Hydraulic diameter of the pipe

The volumetric flow rate Q can be defined as below,

$Q=VA$

In above equation,

$V$ - Velocity of the flow

$A$ -Area

Calculation:

According to the explanation given,

$\Delta P=\rho g{h}_f=\rho g\left(f\frac{L}{D}\left(\frac{V^2}{2g}\right)\right)$

Assume,

$\begin{array}{l}f=0.02\\ \rho =998kg/m^3\end{array}$

$\begin{array}{l}1000Pa/m=\left(998kg/m^3\right)\left(\left(0.02\right)\frac{1}{\left(0.1m\right)}\left(\frac{{\left(\frac{Q}{\pi {\left(0.05\right)}^2}\right)}^2}{2}\right)\right)\\ Q=0.03m^3/s\end{array}$

Conclusion:

Yes, the flow rate is equal to $0.03m^3/s$

(b)

The diameter that will triple the flow rate.

$d=14.7cm$

Given information:

The fluid is water at $20°C$

The flow rate is equal to $0.03m^3/s$

The power law friction factor relation is defined as,

$\Delta p\approx 0.241L{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}$

The power law friction factor relation is defined as,

$\Delta p\approx 0.241L{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}$

In above equation,

$\Delta p$ - Pressure drop

$L$ - Length of the pipe

$\rho$ -Density of the fluid

$\mu$ - Dynamic viscosity of the fluid

$d$ - Diameter of the pipe

$Q$ - Flow rate

Calculation:

To find the diameter,

According to the explanation given,

$\begin{array}{l}\Delta p\approx 0.241L{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}\\ \frac{\Delta P}{L}\approx 0.241{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}\\ 1000Pa/m=\left(0.241\right){\left(998kg/m^3\right)}^{3/4}{\left(1\times {10}^{-3}Ns/m\right)}^{1/4}{d}^{-4.75}{\left(0.09m^3/s\right)}^{1.75}\\ d=0.147m\\ d=14.7cm\end{array}$

Conclusion:

The pipe should have a diameter of $14.7cm$ to triple the flow rate.

(c)

Estimate the diameter that will triple the flow rate by given data.

$d=15cm$

Given information:

The fluid is water at $20°C$

The flow rate is equal to $0.03m^3/s$

The more exact friction factor relation can be defined as,

$\frac{1}{f^{1/2}}=2.0\log \left({\mathrm{Re}}_d{f}^{1/2}\right)-0.8$

$\frac{1}{f^{1/2}}=2.0\log \left({\mathrm{Re}}_d{f}^{1/2}\right)-0.8$

In above equation,

$f$ - Friction factor

${\mathrm{Re}}_d$ - Reynolds’s number

The Reynolds’s number can be defined as,

${\mathrm{Re}}_d=\frac{\rho Vd}{\mu }$

The friction factor can be defined as,

$f=\left(\frac{\Delta P}{L}\right)\left(\frac{2d}{\rho V^2}\right)$

Calculation:

According to the explanation,

We can rewrite the friction factor relation as,

$\begin{array}{l}\frac{1}{f^{1/2}}=2.0\log \left({\mathrm{Re}}_d{f}^{1/2}\right)-0.8\\ \frac{1}{{\left(\left(\frac{\Delta P}{L}\right)\left(\frac{2d}{\rho V^2}\right)\right)}^{1/2}}=2.0\log \left(\left(\frac{\rho Vd}{\mu }\right){\left(\left(\frac{\Delta P}{L}\right)\left(\frac{2d}{\rho V^2}\right)\right)}^{1/2}\right)-0.8\end{array}$

Substitute the values for above equation,

$\begin{array}{l}\frac{1}{{\left(15.46{\pi }^2{d}^5\right)}^{1/2}}=2.0\log \left(\left(\frac{\left(998kg/m^3\right)\left(\frac{0.09m^3/s}{\pi \left(\frac{d^2}{4}\right)}\right)d}{0.001}\right){\left(15.46{\pi }^2{d}^5\right)}^{1/2}\right)-0.8\\ d=0.15m\\ =15cm\end{array}$

Conclusion:

According to the friction factor relation, the pipe should have a diameter of $15cm$ to triple the flow rate.

To determine

(b)

The diameter that will triple the flow rate.

Answer to Problem 6.73P

$d=14.7cm$

Explanation of Solution

Given information:

The fluid is water at $20°C$

The flow rate is equal to $0.03m^3/s$

The power law friction factor relation is defined as,

$\Delta p\approx 0.241L{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}$

The power law friction factor relation is defined as,

$\Delta p\approx 0.241L{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}$

In above equation,

$\Delta p$ - Pressure drop

$L$ - Length of the pipe

$\rho$ -Density of the fluid

$\mu$ - Dynamic viscosity of the fluid

$d$ - Diameter of the pipe

$Q$ - Flow rate

Calculation:

To find the diameter,

According to the explanation given,

$\begin{array}{l}\Delta p\approx 0.241L{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}\\ \frac{\Delta P}{L}\approx 0.241{\rho }^{3/4}{\mu }^{1/4}{d}^{-4.75}{Q}^{1.75}\\ 1000Pa/m=\left(0.241\right){\left(998kg/m^3\right)}^{3/4}{\left(1\times {10}^{-3}Ns/m\right)}^{1/4}{d}^{-4.75}{\left(0.09m^3/s\right)}^{1.75}\\ d=0.147m\\ d=14.7cm\end{array}$

Conclusion:

The pipe should have a diameter of $14.7cm$ to triple the flow rate.

To determine

(c)

Estimate the diameter that will triple the flow rate by given data.

Answer to Problem 6.73P

$d=15cm$

Explanation of Solution

Given information:

The fluid is water at $20°C$

The flow rate is equal to $0.03m^3/s$

The more exact friction factor relation can be defined as,

$\frac{1}{f^{1/2}}=2.0\log \left({\mathrm{Re}}_d{f}^{1/2}\right)-0.8$

$\frac{1}{f^{1/2}}=2.0\log \left({\mathrm{Re}}_d{f}^{1/2}\right)-0.8$

In above equation,

$f$ - Friction factor

${\mathrm{Re}}_d$ - Reynolds’s number

The Reynolds’s number can be defined as,

${\mathrm{Re}}_d=\frac{\rho Vd}{\mu }$

The friction factor can be defined as,

$f=\left(\frac{\Delta P}{L}\right)\left(\frac{2d}{\rho V^2}\right)$

Calculation:

According to the explanation,

We can rewrite the friction factor relation as,

$\begin{array}{l}\frac{1}{f^{1/2}}=2.0\log \left({\mathrm{Re}}_d{f}^{1/2}\right)-0.8\\ \frac{1}{{\left(\left(\frac{\Delta P}{L}\right)\left(\frac{2d}{\rho V^2}\right)\right)}^{1/2}}=2.0\log \left(\left(\frac{\rho Vd}{\mu }\right){\left(\left(\frac{\Delta P}{L}\right)\left(\frac{2d}{\rho V^2}\right)\right)}^{1/2}\right)-0.8\end{array}$

Substitute the values for above equation,

$\begin{array}{l}\frac{1}{{\left(15.46{\pi }^2{d}^5\right)}^{1/2}}=2.0\log \left(\left(\frac{\left(998kg/m^3\right)\left(\frac{0.09m^3/s}{\pi \left(\frac{d^2}{4}\right)}\right)d}{0.001}\right){\left(15.46{\pi }^2{d}^5\right)}^{1/2}\right)-0.8\\ d=0.15m\\ =15cm\end{array}$

Conclusion:

According to the friction factor relation, the pipe should have a diameter of $15cm$ to triple the flow rate.