Question

show that the efficiency of a free jet striking normally a series of flat plate mounted on the periphery of the wheel never exceeds 50%​

Answer 1

Given:

Here it is given that free jet striking normally on a series of flat plates mounted on the periphery of a wheel.

To Find:

We have to prove that the maximum efficiency of jet is 50%.

Solution:

Let us:

$\mathbf{\textrm{Fluid Density }\ = \rho \ \\ \ \ \ \dfrac{kg}{m^{3}}}$

$\mathbf{\textrm{FluidVelocity }\ = v \ \ \ \dfrac{m}{s}}$

$\mathbf{\textrm{Wheel's blade Speed }\ = u \ \ \ \ \dfrac{m}{s}}$

$\mathbf{\textrm{Cross-sectional area of jet }\ = a \ \ \ \ (m^{2})}$

Mass of fluid strike with blade per second is :

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

$\mathbf{\textrm{kinetic energy of fluid per second }\ = K.E =\dfrac{1}{2} \times mass\times speed^{2}\ \ \ \ \ (J)}$

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

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

Force on blade = Rate of mass flow× (Initial speed of jet - Final speed of jet)

$\mathbf{\textrm{Force on blade }\ = F =\rho \times a\times V\times (V-u)\ \ \ \ \ \ (N)}$

Work done by jet on blade:

Work done per second = Force× Speed of blade

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

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

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

On simplify, we get:

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

From equation 3) , we see that efficiency of free jet depend only upon speed of blade of  wheel for a particular value of speed of fluid jet:

On differentiating equation 3) means efficiency with respect to blade speed of wheel , we will get:

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

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

On putting value of u from equation 4)  to equation 3):

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

On simplify and solving , we will get:

$\mathbf{\textrm{Maximum Efficiency of free jet }\ = \eta _{max}=0.5 = 50}$%

From above it is clear 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%.