Page: Numericals on Flow through Nozzle
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Problem on diameter of nozzle for maximum transmission of power through Nozzle.

Q1) A nozzle is fitted at the end of a pipe of length 300m and of diameter 100mm. For the maximum transmission of power through the nozzle, find the diameter of nozzle. Take f=0.009.

Solution:- Given:

Length of pipe, L=300m

Diameter of pipe, D=100mm=0.1m

Coefficient of friction, f=0.009

Let the diameter of nozzle = d

$\therefore, d=\left(\dfrac{D^5}{8fl}\right)^{1/4}$

$\therefore, d=\left(\dfrac{0.1^5}{8\times 0.009\times 300}\right)^{1/4}$

d=26.08mm


Problem on power transmitted through nozzle

Q1) The head of water at the inlet of a pipe 2000m long and 500mm diameter is 60m. A nozzle of diameter 100mm at its outlet is fitted to the pipe. FInd the velocity of water at the outlet of the nozzle if f=0.01 for the pipe.

Solution: Given:-

Head of water at inlet of pipe, H=60m

Length of pipe, L=2000m

Diameter of pipe, D=500m=0.50m

Diameter of nozzle at outlet, d=100mm=0.1m

Coefficient of friction, f=0.01

Therefore to find the velocity,

$\dfrac {v^2}{2gH}=\left[\dfrac1{1+\dfrac{4fL}D\times \dfrac{a^2}{A^2}}\right]$

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

$v^2=\left[\dfrac{2\times9.81\times60}{1+\dfrac{4\times 0.01\times 2000}{0.5}\times \left(\dfrac{\dfrac \pi4d^2}{\dfrac \pi4D^2}\right)^2}\right]$

$v=\sqrt{\dfrac{2\times9.81\times60}{1+\dfrac{4\times 0.01\times 2000}{0.5}\times \left(\dfrac{0.1^2}{0.5^2}\right)^2}}$

v=30.61m/s


Problem on condition for maximum powertransmitted through nozzle

Q1) The rate of flow of water through a pipe of length 2000m and diameter 1m is $2m^3/s$. At the end of pipe a nozzle of outside diameter 300mm is fitted. Find the power transmitted through the nozzle if the head of water at inlet of pipe is 200m and f=0.01.

Solution: Given:

L=2000m

D=1m

$Q=2m^3/s$

d=300mm=0.3m

H=200m

f=0.01

Area of pipe,

$A=\dfrac \pi4D^2$

$A=\dfrac\pi4\times (1)^2=0.7854mm^2$

Velocity of water through pipe,

$V=\dfrac QA=\dfrac 2{0.7854}=2.546m/s$

$P=\dfrac{\rho g.a.v}{1000}\left[H-\dfrac{4fLv^2}{D\times 2g}\right]$

$P=\dfrac{1000\times 9.81\times 2.0}{1000}\left[200-\dfrac{4\times 0.01\times 2000\times(2.546)^2}{1\times 2\times 9.81}\right]$

$P=3405.43 \ KW$

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modified 2 days ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 250 written 5 days ago by gravatar for Syedahina Mohi Syedahina Mohi0
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