Chapter 11, Problem 11.95P

To determine

(a)

To prove:

The maximum power, $h_f=\frac{H}{3}$ for an impulse turbine.

Answer to Problem 11.95P

The maximum power generated by an impulse turbine is $h_f=\frac{H}{3}$.

Explanation of Solution

Given:

Referencing the below diagram:

The equation given below is also taking into consideration:

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:

Now, for determining maximum power:

$\begin{array}{l}P=\rho Qu(V_j-u)(1-\cos \beta )\\ P=\rho Q\frac{V_j}{2}(V_j-\frac{V_j}{2})(1-\cos \beta )\\ P=\rho Q\frac{V_j{}^2}{4}(1-\cos \beta )\\ \end{array}$

Where,

$\rho$

• = density of fluid
• Q = flow rate of turbine
$\beta$
• = exit angle
Jet area has an equation of,

$A_j=\frac{Q}{V_j}$

So the maximum power of turbine is reduced to,

$\begin{array}{l}P=\frac{\rho A}{4}(1-\cos \beta )V_j{}^3\mathrm{.....}(1)\\ P=C{V}_j{}^2{V}_j\text{ where C = }\frac{\rho A}{4}(1-\cos \beta )\text{ is constant}\end{array}$

For the reservoir and the outlet jet, applying steady flow energy equation:

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

Where:

• H = head of turbine
• f = friction factor
• L = length of pipe
• D = diameter of pipe
$V_{pipe}^{}$
• = velocity of pipe
Now, using continuity equation:

$\begin{array}{l}{A}_j{V}_j=A_{pipe}{V}_{pipe}\\ \frac{\pi D_j^2}{4}{V}_j=\frac{\pi D^2}{4}{V}_{pipe}\\ V_{pipe}=\frac{D_j^2}{D^2}{V}_j\mathrm{...}(3)\end{array}$

In equation (2), let us put the value of $V_{pipe}=\frac{D_j^2}{D^2}{V}_j$ from equation (3).

$V_j^2=2gH-\frac{fL{D}_j^4}{D^5}{V}_j{}^2$

Lastly, putting the value of $V_j^2=2gH-\frac{fL{D}_j^4}{D^5}{V}_j{}^2$ in equation (1),

$P=C\left[2gH{V}_j-fL\frac{D_j^4}{D^5}{V}_j{}^3\right]$

To determine maximum power, below is the condition:

$\begin{array}{l}\frac{dP}{d{V}_j}=0\\ \frac{Cd\left[2gH{V}_j-fL\frac{D_j^4}{D^5}{V}_j{}^3\right]}{d{V}_j}=0\\ 2gH-3fL\frac{D_j^4}{D^5}{V}_j{}^2=0\\ 2gH-3\frac{fL}{D}{V}_{pipe}{}^2=\mathrm{0...}(4)\end{array}$

But the friction head loss of pipe:

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

Therefore, the maximum power of an impulse turbine is $h_f=\frac{H}{3}$.

Conclusion:

The maximum power generated by an impulse turbine is $h_f=\frac{H}{3}$.

To determine

(b)

To prove:

The optimum velocity is $V_j=\sqrt{\frac{4}{3}gH}$ for given impulse turbine.

Answer to Problem 11.95P

The optimum velocity of an impulse turbine is $V_j=\sqrt{\frac{4}{3}gH}$ for given impulse turbine.

Explanation of Solution

Given:

Referencing the below diagram:

The equation given below is also taking into consideration:

Concept Used:

Jet area has an equation of,

$A_j=\frac{Q}{V_j}$

So, the maximum power of turbine is reduced to:

$\begin{array}{l}P=\frac{\rho A}{4}(1-\cos \beta )V_j{}^3\mathrm{.....}(1)\\ P=C{V}_j{}^2{V}_j\text{ where C = }\frac{\rho A}{4}(1-\cos \beta )\text{ is constant}\end{array}$

For the reservoir and the outlet jet, applying steady flow energy equation.

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

Where,

• H = head of turbine
• f = friction factor
• L = length of pipe
• D = diameter of pipe
$V_{pipe}^{}$
• = velocity of pipe
Calculation:

For getting optimum velocity, we have equation (2):

$\begin{array}{l}{V}_j^2=2gH-\frac{fL{V}_{pipe}^2}{D}\\ V_j=\sqrt{2gH-2g{h}_f}\\ V_j=\sqrt{2gH-2g\frac{H}{3}}\\ V_j=\sqrt{\frac{4}{3}gH}\end{array}$

The optimum velocity of an impulse turbine is

$V_j=\sqrt{\frac{4}{3}gH}$ proved.

Conclusion:

The optimum velocity of an impulse turbine is $V_j=\sqrt{\frac{4}{3}gH}$ for given impulse turbine.

To determine

(c)

The best nozzle diameter is $D_j={\left[\frac{D^5}{2fL}\right]}^{\frac{1}{4}}$ should be proved for given impulse turbine.

Answer to Problem 11.95P

The best nozzle diameter is $D_j={\left[\frac{D^5}{2fL}\right]}^{\frac{1}{4}}$ for given impulse turbine is proved.

Explanation of Solution

Given:

Referencing the below diagram:

The equation given below is also taking into consideration:

Concept Used:

The continuity equation:

$\begin{array}{l}{A}_j{V}_j=A_{pipe}{V}_{pipe}\\ \frac{\pi D_j^2}{4}{V}_j=\frac{\pi D^2}{4}{V}_{pipe}\\ V_{pipe}=\frac{D_j^2}{D^2}{V}_j\mathrm{...}(3)\end{array}$

The friction head loss of pipe:

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

$h_f=\frac{H}{3}$

Calculation:

For determining nozzle diameter,

$\begin{array}{l}{h}_f=\frac{fL}{D}\frac{V_{pipe}^2}{2g}\\ \frac{H}{3}=\frac{fL}{D}\frac{V_{pipe}^2}{2g}\\ V_{pipe}^2=\frac{2gDH}{3fL}\mathrm{...}(5)\end{array}$

We already have equation (3) with value of $V_{pipe}=\frac{D_j^2}{D^2}{V}_j\mathrm{...}(3)$

$\begin{array}{l}\frac{2gDH}{3fL}=\frac{D_j^2}{D^2}{V}_j{}^2\\ \frac{2gDH}{3fL}=\frac{D_j^2}{D^2}\frac{4}{3}gH\text{ where}\left(V_j=\sqrt{\frac{4}{3}gH}\right)\\ D_j={\left[\frac{D^5}{2fL}\right]}^{\frac{1}{4}}\end{array}$

Conclusion:

The best nozzle diameter is

$D_j={\left[\frac{D^5}{2fL}\right]}^{\frac{1}{4}}$ for given impulse turbine is proved.