1.

show that the efficiency of a free jet striking normally on a series of flat plates mounted on the periphery of a wheel can never exceed 50%?​

Answer»

Given:

A free jet striking normally on a series of flat plates mounted on the periphery of a wheel.

To Find:

Show that the EFFICIENCY of jet can never exceed 50%.

Solution:

Let we take:

\mathbf{\textrm{Density of fluid }\ = \rho \ \dfrac{kg}{m^{3}}}\\

\mathbf{\textrm{Velocity of fluid }\ = v \ \dfrac{m}{s}}

\mathbf{\textrm{Speed of wheel }\ = u \ \dfrac{m}{s}}

\mathbf{\textrm{Exit area of nozzle }\ = a \ (m^{2})}

Mass of fluid strike with blade is :

\mathbf{\textrm{Mass of fluid }\ = m =\rho \times a\times V \ \ \ \ \ \dfrac{kg}{s}}

\mathbf{\textrm{kinetic energy of fluid }\ = K.E =\dfrac{1}{2}\rho \times a\times V \times V^{2}\ \ \ \ \ (J)}      ...1)

Relative velocity of jet strike with blade with RESPECT to blade = V- u

Work DONE by jet on blade:

\mathbf{\textrm{Work done }\ = W =\rho \times a\times V(V-u)\times u\ \ \ \ \ \ (J)}        ...2)

\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{Work\ done\ by\ fluid}{Kinetic \ energy\ of\ fluid}}

\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{\rho \times a\times V(V-u)\times u}{\dfrac{1}{2}\rho \times a\times V \times V^{2}}}

On simplify:

\mathbf{\textrm{Efficiency }\ = \eta =\dfrac{2\times (V-u)\times u}{V^{2}}}         ...3)

From here, we see that efficiency of jet depend upon speed of wheel for a particular value of speed of jet:

On differentiating efficiency with respect to blade speed , we get:

\mathbf{u = \dfrac{V}{2}}                       ...4)

This is required condition of blade speed for maximum efficiency of jet:

Putting value of u from equation 4  to equation 3):

\mathbf{\textrm{Maximum Efficiency }\ = \eta _{max}=\dfrac{2\times (V-\frac{V}{2})\times \frac{V}{2}}{V^{2}}}

On solving , we get:

\mathbf{\textrm{Maximum Efficiency }\ = \eta _{max}=0.5 = 50}%

HENCE PROVE that the efficiency of a free jet striking normally on a series of flat plates mounted on the periphery of a wheel can never exceed 50%.


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