Chapter 11, Problem 11.97P

To determine

(a)

The maximum power generated by the wheel at 80% efficiency.

Answer to Problem 11.97P

The maximum power generated by the wheel at 80% efficiency is 9.36 MW.

Explanation of Solution

Given:

Referencing the below diagram:

We have:

$\begin{array}{l}H=450m\\ L=5\text{ km}\\ \text{D =1}\text{.2 m}\\ {\text{D}}_j=20\text{ cm}\\ \epsilon \text{ = 1 mm}\\ \text{impulse wheel diameter = 3}\text{.2 m}\end{array}$

Concept Used:

The maximum power for an impulse turbine can be obtained when, $u=\frac{V_j}{2}$.

Where:

$V_j$ = jet velocity and u = vane velocity.

Calculation:

For the head, we have steady flow energy equation:

$H=h_f+\frac{V_j{}^2}{2g}$

But the friction head loss of pipe:

$h_f=\frac{fL}{D}\frac{V_{pipe}^2}{2g}$

Hence, the head is:

$\begin{array}{l}H=\frac{fL}{D}\frac{V_{pipe}^2}{2g}+\frac{V_j{}^2}{2g}\\ V_j^2=2gH-\frac{fL{V}_{pipe}^2}{D}\mathrm{...}(1)\end{array}$

Where,

• f = friction factor
$V_j$ = diameter of pipe $V_{pipe}^{}$
• = velocity of pipe
Now, using continuity equation:

$\begin{array}{l}{A}_j{V}_j=A_{pipe}{V}_{pipe}\\ \frac{\pi D_j^2}{4}{V}_j=\frac{\pi D^2}{4}{V}_{pipe}\\ V_{pipe}=\frac{D_j^2}{D^2}{V}_j\end{array}$

Where,

$D_j$

• = diameter of jet
$A_{pipe}$
• = cross section area of pipe
$A_j$
• = area of jet
• D = diameter of pipe
Now putting the values of H, L, D and $D_j$ in the equation (1).

$\begin{array}{l}{V}_j^2\left[1+\frac{f\times 5000}{1.2}\frac{{\left(0.2\right)}^4}{{\left(1.2\right)}^4}\right]=2\times 9.81\times 450\\ V_j^2(1+3.22f)=\mathrm{8829...}(2)\end{array}$

The Reynolds number:

$\begin{array}{l}\mathrm{Re}=\frac{\rho V_{j}D}{\mu }\\ \mathrm{Re}=\frac{998\times 1.2\times V_j}{0.001}\\ \mathrm{Re}=1197600V_j\end{array}$

The ratio is:

$\frac{\epsilon }{D}=\frac{0.001}{1.2}=0.000833$

From Moody’s equation:

$\begin{array}{l}\frac{1}{\sqrt{f}}=-2\text{log}\left[\frac{\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$D$}\right.}{3.7}+\frac{2.51}{{\mathrm{Re}}_d\sqrt{f}}\right]\\ \frac{1}{\sqrt{f}}=-2\text{log}\left[\frac{0.000833}{3.7}+\frac{2.51}{1197600V_j\sqrt{f}}\right]\\ \frac{1}{\sqrt{f}}=-2\text{log}\left[0.000225+\frac{0.0000021}{V_j\sqrt{f}}\right]\mathrm{...}(3)\end{array}$

By solving equation (2) & (3), we get $f=0.0189,{\text{ and V}}_j=91.24m/s$

The volume flow rate is,

$\begin{array}{l}Q=A_j{V}_j\\ Q=\frac{\pi D_j{}^2{V}_j}{4}\\ Q=\frac{\pi \times {\left(0.2\right)}^2\times 91.24}{4}\\ Q=2.87{\text{ m}}^3/s.\end{array}$

The maximum power for the impulse wheel can be obtained for $u=\frac{V_j}{2}\text{ and }\eta \text{=0}\text{.8}$ is efficiency.

For determining maximum power:

$\begin{array}{l}P=\rho Qu(V_j-u)(1-\cos \beta )\eta \\ P=\rho Q\frac{V_j{}^2}{4}(1-\cos \beta )\eta \\ P=998\times 2.87\times \frac{{\left(91.24\right)}^2}{4}(1-\cos 165)\times 0.8\\ P=9.36MW\end{array}$

The maximum power generated by the wheel at 80% efficiency is 9.36 MW.

Conclusion:

The maximum power generated by the wheel at 80% efficiency is 9.36 MW.

To determine

(b)

The best speed should be determined for given impulse wheel.

Answer to Problem 11.97P

The best speed of the wheel is 272.3 r/min.

Explanation of Solution

Given:

Referencing the below diagram:

We have,

$\begin{array}{l}H=450m\\ L=5\text{ km}\\ \text{D =1}\text{.2 m}\\ {\text{D}}_j=20\text{ cm}\\ \epsilon \text{ = 1 mm}\\ \text{impulse wheel diameter = 3}\text{.2 m}\text{.}\end{array}$

Concept Used:

The equation given below is also taking into consideration:

Calculation:

For getting speed of wheel:

$\begin{array}{l}\Omega =\frac{2u}{D_{wheel}}\\ \Omega =\frac{V_j}{D_{wheel}}\\ \Omega =\frac{91.24}{3.2}\\ \Omega =28.51\text{ rad/s}\end{array}$

Converting the rad/s unit of speed to r/min as follows:

$\begin{array}{l}\Omega =28.51\text{ rad/s}\\ \Omega =28.51\text{ rad/s}\times \frac{60\text{ r/min}}{2\pi \text{ rad/s}}\\ \Omega =272.3\text{ r/min}\end{array}$

The best speed of the wheel is 272.3 r/min.

Conclusion:

The best speed of the wheel is 272.3 r/min.