CodeCogs

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

Introduction

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 $C_V$ , The Coefficient of Contraction $C_C$ , and the Coefficient of Discharge $C_D$ , 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 $W$ . Then if the effective cross sectional area of the jet is $a$ and the velocity of discharge is $V$ , then :