Question

Show that maximum efficiency of a jet striking normallly on a series of flat plates arranged over periphery of runner is 50%

Answer 1

Answer:

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

To Find:

let us assummmeeeeeee

that the efficiency of jet doesnt exceed 50%.

Solution:

Let we take:

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

Velocity of fluid  = $v \ \dfrac{m}{s}}$

{Speed of wheel = $u \ \dfrac{m}{s}}$

Exit area of nozzle  = $a \ (m^{2})}$

Mass of fluid strike with blade is :

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

$\dfrac{1}{2}\rho \times a\times V \times V^{2}\ \ \ \ \ (J)}$

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

Work done by jet on blade:

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

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

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

On simplifying you will get:)

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

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:

$\mathbf{u}$ = $\dfrac{V}{2}}$                

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

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

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

On solving , you willv get:

${\textrm{Maximum Efficiency }$= $\eta _{max}=0.5$ = 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%.

Explanation:

look bro i had to all this atleast you can mark as brainliest wont you