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Engineering › Fluid Mechanics › Pipes › Head Loss ›

Nozzles and mouthpieces

An analysis of the head lost in a nozzle and the resulting velocity of the jet.

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Contents

Introduction
The Head Lost In The Nozzle
Efficiency Of The Nozzle
The Power Of A Jet
Page Comments

Introduction

In this submission it is assumed that the Head H is the head behind the nozzle, and that all pipeline and valve losses have been accounted for elsewhere. There are, of course, losses in the nozzle itself, and the actual velocity of discharge will be less than the theoretical value by one to five percent. This is catered for by the use of The Coefficient of Velocity

, The Coefficient of Contraction

, and the Coefficient of Discharge

, given by:

$C_V = \frac{\text{Actual Velocity}}{\text{Theoretical Velocity}}$
(1)
$C_C = \frac{\text{Actual cross sectional area}}{\text{Geometric cross sectional Area}}$
(2)
$C_D = C_C\times C_V$
(3)

The Head Lost In The Nozzle

Let H = total head at nozzle (i.e. the sum of the pressure and velocity heads)

$H=H_1- \text{Pipe line losses}$

(4)

$(\text{Velocity})^2 = {C_{V}}^{2}\;2\,g\,H$

(5)

$H = \frac{V^2}{{C_{V}}^{2}\;2\,g}$

(6)

Head behind the Nozzle = Head after the nozzle + Head lost in Nozzle

$H = \frac{v^2}{2g} + \text{head lost }(H_l)$

(7)

substituting from (5) above

$\therefore\:H_l = H-C_{v}^{2}\:H$

(8)

$\therefore\;\;\;\text{head lost in nozzle }(H_l)= H (1 - {C_{v}}^{2})$

(9)

$H_l = H(1 - {C_{V}}^{2}) = \frac{V^2}{{C_{V}}^{2}\times2\,g}\left (1 - {C_{V}}^{2} \right )$

(10)

$\therefore \;\;\;\;\;\;\;H_l = \frac{V^2}{2\,g}\left (\frac{1}{{C_{V}}^{2}} - 1 \right )$

(11)

Efficiency Of The Nozzle

$\text{Efficiency },\eta = \frac{\dfrac{v^2}{2g}}{\text{head behind nozzle}}$

(12)

i.e.,

$\eta= \frac{v^2}{2g\:H}$

(13)

$=\frac{C_v\; \sqrt{2g\:H}}{2g\,H} = C_v^2$

(14)

Thus,

$C_V=\sqrt{Efficiency}$

(15)

The Power Of A Jet

Let the weight of fluid discharged be

. Then if the effective cross sectional area of the jet is

and the velocity of discharge is

, then :

$\text{The Kinetic Energy of the Jet} = \frac{1}{2} \frac{W}{g} V^2$

(16)

$=\frac{1}{2}\times \frac{w\,a\,V}{2\,g}\timesV^2 = \frac{w\,a\,V^3}{2\,g}$

(17)

For the Jet, the Work Done equals the change of Kinetic Energy, and hence the horse power available in the jet is:-

$\text{The horse power available} = \frac{w\,a\,V^3}{2\,g\times 550}$

(18)

Example:

[imperial]

Example - Discharge through nozzle

Problem

A Nozzle discharges 175 galls per min. under a head of 200 ft. The diameter of the nozzle is 1 in. and the diameter of the jet is 0.9 in. Find the:

a) The coefficient of velocity for the jet.
b) The head lost in the nozzle.
c) The horse power available in the jet.

Workings

(a) The Coefficient of Contraction, $C_D$ is given by:

$C_D = \left (\frac{0.9}{1.0} \right )^2 = 0.81$

(1)

The theoretical discharge Q = The nozzle area X. From equation (#8) velocity (neglecting losses) is

$\text{Velocity } = \sqrt{2\;g\;H} = \sqrt{2\times 32.2 \times 200}$

(2)

therefore

$Q = \left (\frac{\pi }{4}\times \frac{1}{12^2} \right )\times \sqrt{2\times 32.2\times 200} = 0.619\;ft.^3\;sec.^{-1}$

(3)

Since 1 gallon of water weighs 10 lb. and 1 cubic foot of water weighs 62.4 lb, then

$Q = \frac{175\times 10}{60\times 62.4} = 0.467\;ft^3\;sec^{-1}$

(4)

and the coefficient of discharge

$= \frac{0.467}{0.619} = 0.755$

(5)

From equation(#4), the Coefficient of velocity

$=\frac{C_D}{C_C} = \frac{0.755}{0.81} = 0.932$

(6)

(b) Using Equation (#8)

$H = \text{Head lost in nozzle} + \frac{V^2}{2\;g}$

(7)

i.e.

$H = H_L + {C_{V}}^{2} H$

(8)

therefore

$H_L = H\left (1 - {C_{V}}^{2} \right ) = 200(1 - 0.932^2)$

(9)

Thus

$H_L=26.28\;ft.$

(10)

(c) The horse power of the jet is dependent upon the weight of water per second and the head of water in the jet Thus

$\text{Horse power} = \frac{W\times H_J}{550}$

(11)

$=\frac{175\times 10}{60}\times (200 - 26.28)\times \frac{1}{550} = 9.21\;h.p.$

(12)

Solution

a) The coefficient of velocity for the jet = 0.932

b) The head lost in the nozzle = 26.28 ft

c) The horse power available in the jet = 9.21 h.p

Page Comments

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