Page: Power transmitted through Nozzle
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Velocity through the nozzle

The kinetic energy of the jet at an outlet of nozzle =$\dfrac 12mv^2$

Now the mass of liquid at the outlet of nozzle per second = $\rho av$

$\therefore$ Kinetic energy of the jet at the outlet per sec

$=\dfrac 12\rho av\times v^2$

$=\dfrac 12\rho av^3$

Therefore power in KW at the outlet of nozzle

$=(KE/sec)\times \dfrac 1{1000}$

$=\dfrac {\dfrac 12\rho av^3}{1000}$

Therefore efficiency of power transmission through nozzle

$\eta=\dfrac {\text{Power at outlet of nozzle}}{\text{Power at inlet of pipe}}$

$\eta=\dfrac {\dfrac {\dfrac 12\rho av^3}{1000}}{\dfrac {\rho g.Q.H}{1000}}$

$\eta=\dfrac {\dfrac 12\rho av^3}{\rho g.Q.H}$

$\eta=\dfrac {\dfrac 12\rho av^3}{\rho g.av.H}$ [$\because Q=av$]

$\eta=\dfrac {v^2}{2gh}\left[\dfrac 1{1+\dfrac {4fL}{D}\times \dfrac {a^2}{A^2}}\right]$

Condition for maximum power transmitted:-

We know that, the total head at inlet of pipe = Total head at the outlet of the nozzle + losses

$H=\dfrac {v^2}{2g}+hf$

$H=\dfrac {v^2}{2g}+\dfrac {4fLv^2}{D\times 2g}$

$\therefore \dfrac {v^2}{2g}=H-\dfrac {4.f.L.v^2}{D\times 2g}$

but power transmitted through nozzle

$=\dfrac {\dfrac 12\rho av^3}{1000}=\dfrac {\dfrac 12\rho av}{1000}\times v^2$

$=\dfrac {\dfrac 12\rho av}{1000}\left[2g\left(H-\dfrac {4.f.L.V^2}{D\times 2g}\right)\right]$

$=\dfrac {\rho gav}{1000}\left(H-\dfrac {4.f.L.V^2}{D\times 2g}\right)........................(1)$

Now from the continuity equation,

$AV=av$

$\therefore V=\dfrac {av}A$

Substituting the value of V in equation (1), we get

Power transmitted through nozzle

$=\dfrac {\rho gav}{1000}\left(H-\dfrac {4.f.L.a^2}{D\times 2g}.\dfrac {v^2}{A^2}\right)$

The power (P) will be maximum, when

$\dfrac {d(P)}{dV}=0$

$\dfrac d{dV}\left[\dfrac {\rho gav}{1000}\left(H-\dfrac {4.f.L.a^2}{D\times 2g}.\dfrac {v^2}{A^2}\right)\right]=0$

$\dfrac d{dV}\left[\dfrac {\rho ga}{1000}\left(HV-\dfrac {4.f.L.a^2}{D\times 2g}.\dfrac {V^3}{A^2}\right)\right]=0$

$\left[\dfrac {\rho ga}{1000}\left(H-3\dfrac {4.f.L.a^2}{D\times 2g}.\dfrac {V^2}{A^2}\right)\right]=0$

$\left(H-3\dfrac {4.f.L}{D\times 2g}.v^2\right)=0$ [$\because v=\dfrac {av}{A}$]

$H-3hf=0$ $\left(\because \dfrac {4.f.Lv^2}{D\times 2g}\right)=hf$

$hf=\dfrac H3$

This equation gives the condition for maximum power transmitted through nozzle

• Diameter of nozzle for maximum transmission of power through nozzle: -

For maximum transmission of power, the condition is given by,

$hf=\dfrac H3$

But, $hf=\dfrac{4fLV^2}{D\times 2g}$

$\therefore \dfrac{4fLV^2}{D\times 2g}=\dfrac H3$

$H=3\times \dfrac {4fLv^2}{D\times 2g}$

But H is also = total head at outlet of nozzle + losses

$H=\dfrac {v^2}{2g}+hf=\dfrac {V^2}{2g}+\dfrac {4fLv^2}{D\times 2g}$

Equating the two values of H, we get

$3\times \dfrac {4fLv^2}{D\times 2g}=\dfrac {V^2}{2g}+\dfrac {4fLv^2}{D\times 2g}$

$\dfrac {12fLv^2}{D\times 2g}-\dfrac {4fLv^2}{D\times 2g}=\dfrac {V^2}{2g}$

$\dfrac {8fLv^2}{D\times 2g}=\dfrac {V^2}{2g}........................(1)$

But from continuity,

$av=AV$

or $v=\dfrac {AV}a$

Substituting this value of v in equation (1), we get

$\dfrac {8fL}{D\times 2g}\times \dfrac {A^2V^2}{a^2}=\dfrac {V^2}{2g}$

$\dfrac {8fL}{D}\times \dfrac {A^2}{a^2}=1...................(2)$

$\dfrac {8fL}{D}\times \dfrac {\left(\dfrac \pi 4D^2\right)^2}{\left(\dfrac \pi 4d^2\right)^2}=1$

$\dfrac {8fL}{D}\times \dfrac {\left(D^4\right)}{\left(d^4\right)}=1$

$d^4=\dfrac {D^5}{8fL}$

$\therefore d=\left(\dfrac {D^5}{8fL}\right)^{\dfrac 14}$

From equation (2),

$\dfrac {8fL}D=\dfrac {A^2}{a^2}$

$\dfrac Aa=\sqrt{\dfrac {8fL}{D}}...........................(3)$

This equation given the ratio of the area of the supply pipe to area of the nozzle, hence from this equation, the diameter of the nozzle can be calculated.

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