Chapter 11, Problem 11.102P

To determine

The rate of water flow should be determined in gal/min for given multiblade HAWT.

Answer to Problem 11.102P

The water flow rate in the multiblade HAWT is 29.44 gal.min.

Explanation of Solution

Given Information:

Wind speed, 12 min/h

Diameter of cast iron pipe = 3 in.

Height, 10 ft

Diameter of turbine, 6 ft

First let us find out the speed of wind for the given system,

$\begin{array}{l}\text{V}=12\text{ min/h}\\ \text{V= }\frac{12\min }{h}\times \frac{5280\text{ ft}}{1\min }\times \frac{1h}{3600s}\\ \text{V=17}\text{.6 ft/s}\end{array}$

For the given turbine, the area would be,

$\begin{array}{l}A=\frac{\pi D^2}{4}\\ A=\frac{\pi \times {\left(6\right)}^2}{4}\\ A=28.27{\text{ ft}}^2\end{array}$

From this, we can determine the maximum power delivered by turbine.

$P_{\max }=C_p\frac{\rho }{2}A{V}^3$

Where,

$C_p$

• = maximum value of power coefficient
$\rho$
• = air density
• V = wind speed of given system

In above equation, putting the values for V, A, $C_p$ and p.

$\begin{array}{l}{P}_{\max }=C_p\frac{\rho }{2}A{V}^3\\ P_{\max }=0.29\times \frac{0.0023}{2}\times 28.27\times {\left(17.6\right)}^3\\ P_{\max }=51.41\text{ ft}\cdot \text{lbf/s}\end{array}$

Let us consider, that the pump efficiency is $\eta =0.8$, which give the pump power as:

$\begin{array}{l}{P}_{pump}=\eta \times 51.41\text{ ft}\cdot \text{lbf/s}\\ P_{pump}=0.8\times 51.41\text{ ft}\cdot \text{lbf/s}\\ P_{pump}\text{=41}\text{.13 ft}\cdot \text{lbf/s}\end{array}$

The relation of the pump power with system is as below:

$P_{pump}={\rho }_{water}gQ{H}_{system}\mathrm{...}(1)$

For above equation, we have density of water, ${\rho }_{water}=1.94{\text{ slug/ft}}^3.$

The head of system has a formula of;

$H_{system}=\Delta z+f\frac{L}{d}\frac{V_{pipe}^2}{2g}\mathrm{...}(2)$

Where:

$\Delta z$

• = height
• f = friction factor
$V_{pipe}^{}$ = pipe velocity
• d = diameter of the pipe

For getting the head of system, we need to find out velocity of pipe, $V_{pipe}^{}$.

$\begin{array}{l}{V}_{pipe}^{}=\frac{4Q}{\pi D^2}\\ V_{pipe}^{}=\frac{4Q}{\pi \times {\left(0.25\right)}^2}\\ V_{pipe}^{}=20.37Q\end{array}$

In equation (2), putting the values,

$\begin{array}{l}{H}_{system}=\Delta z+f\frac{L}{d}\frac{V_{pipe}^2}{2g}\\ H_{system}=10+f\frac{10}{0.25}\times \frac{{\left(20.37\times Q\right)}^2}{2\times 32.2}\\ H_{system}=10+258f{Q}^2\end{array}$

Putting the values from above in equation (1),

$\begin{array}{l}{P}_{pump}={\rho }_{water}gQ{H}_{system}\\ 41.13=1.94\times 32.2Q\left(10+258f{Q}^2\right)\\ Q\left(10+258f{Q}^2\right)=\mathrm{0.658...}(3)\end{array}$

The dynamic viscosity of water and for cast iron pipe, we have $\mu =2.09\times {10}^{-5}slug/ft\cdot s\text{ and }\epsilon \text{=0}\text{.00085 m}\text{.}$

Now determining the Reynolds number and the ratio for the system as:

$\begin{array}{l}\mathrm{Re}=\frac{\rho V_{pipe}d}{\mu }\\ \mathrm{Re}=\frac{1.94\times 20.37\times Q\times 0.25\times {10}^5}{2.09}\\ \mathrm{Re}=472701Q;\\ and\\ \frac{\epsilon }{d}=\frac{0.00085}{0.25}=0.0034\end{array}$

For the given system, applying the Moody’s equation,:

$\begin{array}{l}\frac{1}{\sqrt{f}}=-2\log \left[\frac{\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$D$}\right.}{3.7}+\frac{2.51}{{\mathrm{Re}}_d\sqrt{f}}\right]\\ \frac{1}{\sqrt{f}}=-2\log \left[\frac{0.0034}{3.7}+\frac{2.51}{472701Q\sqrt{f}}\right]\\ \frac{1}{\sqrt{f}}=-2\log \left[0.00092+\frac{0.0000053}{Q\sqrt{f}}\right]\mathrm{...}(4)\end{array}$

By solving the equations (3) & (4).

We get $f=0.0306\text{ and Q =0}{\text{.0656 ft}}^3/s.$

Converting the flow rate into gal/min.

$\begin{array}{l}Q=0.0656{\text{ ft}}^3/s\\ Q=\frac{0.0656{\text{ ft}}^3}{s}\times \frac{7.48052\text{ gal}}{1{\text{ ft}}^3}\times \frac{60\text{ s}}{\min }\\ Q=29.44\text{ gal/min}\end{array}$

So, the water flow rate in the multiblade HAWT is 29.44 gal.min.

Conclusion:

The water flow rate in the multiblade HAWT is 29.44 gal.min.