Page: Power transmitted through Nozzle

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,


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

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,


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.

page fluid mechanics 2 fm2 • 63 views
modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 270 written 7 months ago by gravatar for Syedahina Mohi Syedahina Mohi0
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