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Internal Mouthpiece

Discharge and Pressure in an Internal Mouthpiece

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Contents

Introduction
Discharge Through An Internal Mouthpiece
Pressure In An Internal Mouthpiece
Page Comments

Key Facts

Gyroscopic Couple: The rate of change of angular momentum ( $\tau$ ) = $I\omega\Omega$ (In the limit).

= Moment of Inertia.
$\omega$ = Angular velocity
$\Omega$ = Angular velocity of precession.

Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.

Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.

Introduction

An internal mouthpiece, extending into the fluid (i.e., inside the vessel) is known as Re-entrant or Borda's mouthpiece. There are mainly two types of internal mouthpieces depending upon their nature of discharge:

Mouthpiece running free
Mouthpiece running full

In this page we will discuss about the discharge and pressure in an internal mouthpiece when it runs free and full.

Discharge Through An Internal Mouthpiece

Mouthpiece Running Free

If the jet, after contraction, does not touch the sides of the mouthpiece, it is said to be running free as shown in fig-1. It has been experimentally found that if the length of the mouthpiece extending into the fluid is less than 3 times the diameter of the orifice, it will run free.

Consider a mouthpiece running free. Let,

= Height of liquid above the mouthpiece,
= Area of orifice or mouthpiece,
= Area of the contracted jet,
= Velocity of the liquid
= Specific weight of the liquid

We know that pressure of the liquid on the mouthpiece,

Force acting on the mouthpiece = Pressure $\times$ Area

(2)

Mass of the liquid flowing per second $= \frac{wQ}{g} = \frac{wa_cv}{g}$

$\therefore$ Momentum of the flowing liquid = Mass $\times$ Velocity

$=\frac{wa_cv\times v}{g}=\frac{wa_cv^2}{g}$

(3)

Since the water is initially at rest, therefore initial momentum = 0

and change of momentum

$= \frac{wa_cv^2}{g}$

(4)

According to Newton's second Law of Motion, the force is equal to the rate of change of momentum. Therefore equating equations (2) and (4),

$wHa = \frac{wa_cv^2}{g}$

$\therefore Ha = \frac{a_cv^2}{g}$

$\Rightarrow \frac{v^2}{2g}\times a = \frac{a_cv^2}{g}$

$\Rightarrow a = 2a_c \Rightarrow \frac{a_c}{a} = \frac{1}{2} = 0.5$

$\therefore$ Coefficient of contraction,

= $\frac{a_c}{a}$ = 0.5

We know that coefficient of discharge (assuming

equal to 1),

$C_d = C_v \times C_c = 1 \times 0.5 = 0.5$

$\therefore$ Discharge, $Q = C_d.a\sqrt {2gH} = 0.5a \sqrt {2gH}$

Mouthpiece Running Full

If the jet, after contraction, expands and fills up the whole mouthpiece, it is said to be running full as shown in fig-2. If the length of the mouthpiece is more than 3 times the diameter of the orifice, it will run full.

Consider a mouthpiece running full. Let,

= Height of liquid above the mouthpiece,
= Area of the flow at the vena contracta,
= Velocity of the liquid at the outlet
= Velocity of the liquid at the vena contracta

Since the liquid is flowing continuously, therefore

$v_c.a_c = v.a \Rightarrow v_c = \frac{v.a}{a_c}$

Previously we have seen that the coefficient of contraction in an internal mouthpiece is 0.5.

Therefore substituting the value of $\frac{a_c}{a}$ = 0.5 in the above equation,

(5)

We see that the jet after passing through section 1-1 suddenly enlarges at section 2-2. Therefore there will be a loss of head due to sudden enlargement. We know that the loss of head due to sudden enlargement,

$h_c = \frac{(v_c-v)^2}{2g} = \frac{(2v-v)^2}{2g} = \frac{v^2}{2g}$

$\therefore$ Applying Bernoulli's equation to points A and B,

$H = \frac{v^2}{2g} + Losses$

$\Rightarrow H = \frac{v^2}{2g} + \frac{v^2}{2g} = \frac{2v^2}{2g} = \frac{v^2}{g}$

$\Rightarrow v = \sqrt {gH}$

(6)

$\therefore$ Actual Discharge = $a\times \sqrt {gH}$

We know that theoretical discharge $= a\sqrt {2gH}$

$\therefore$ Coefficient of discharge,C_d = Actual discharge / Theoretical discharge

$\Rightarrow C_d = \frac{a\sqrt {gH}}{a\sqrt {2gH}} = 0.707$

(7)

and discharge,

$Q = C_da\sqrt {2gH} = 0.707a\sqrt {2gH}$

Example:

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Example - Discharge through an Internal Mouthpiece running free

Problem

A Borda's mouthpiece of 50mm diameter is provided on one side of a tank containing water up to a height of 3m above the center line or the orifice. Find the discharge through the mouthpiece, if the mouthpiece is running free.

Workings

Given,

$d$ = 50 mm = 0.05m
$H$ = 3m

The area of the mouthpiece,

$a = \frac{\pi}{4} \times (d)^2 = \frac{\pi}{4} \times (0.05)^2 = 1.964\times 10^{-3}$

and discharge through the mouthpiece,

$Q = 0.5a\sqrt {2gH}$

$\Rightarrow 0.5\times (1.964\times 10^{-3})\times \sqrt {2\times 9.81\times 3} m^3/s$

$\Rightarrow Q = 7.53\times 10^{-3}m^3/s = 7.53\;liters/s$

Solution

The discharge through the mouthpiece = 7.53 liters/s

Pressure In An Internal Mouthpiece

Consider a vessel, open to atmosphere at its top, having an orifice with an internal mouthpiece, as shown in fig-3. We know that the jet of the liquid, on entering the mouthpiece, will first contract up to vena contracta C and then expand and fill up the whole orifice as shown in figure.

Let,

= Atmospheric pressure head
= Height of liquid above the mouthpiece
= Absolute pressure head at vena contracta
= Coefficient of contraction
= Velocity of liquid at outlet
= Velocity of liquid at vena contracta

Applying Bernoulli's equation to point A and C,

$H_a + H = H_c + \frac{{v_c}^2}{2g}$

$H_c = H_a + H - \frac{v_c^2}{2g} = H_a + H - (\frac{v}{0.5})^2\times \frac{1}{2g}$

$\Rightarrow H_c = H_a + H - \frac{4v^2}{2g} = H_a + H - \frac{2v^2}{g}$

(8)

From equation (6) we know that,

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

substituting the value of $\frac{v^2}{g}$ in equation (8)

This shows that the absolute pressure head at vena contracta is less than the atmospheric pressure by an amount equal to height of the liquid above the vena contracta. A little consideration will show that the absolute pressure head at vena contracta will be zero, when

of water

It means that if this condition is reached (i.e., height of water above the mouthpiece reaches 10.3m), the flow of the liquid would be no longer steady. This means that the mouthpiece will not run full at its outlet and the jet will flow straight without touching the walls of the mouthpiece. But, in actual practice, this condition also reaches at 7.5 m.

Example:

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Example - Pressure in an Internal Mouthpiece

Problem

A re-entrant mouthpiece of 75mm diameter is discharging water under a constant head of 3.5m. Determine the absolute pressure head at vena contracta, if the atmosphere pressure head be 10.3 m of water.

Workings

Given,

$d$ = 75 mm = 0.075 m
$H$ = 3.5 m
$H_a$ = 10.3 m

We know that the absolute pressure head at the vena contracta of an internal mouthpiece.

H_c = H_a - H = 10.3 - 3.5 = 6.8 m

Solution

Absolute pressure head = 6.8 m