# Internal Mouthpiece

Discharge and Pressure in an Internal Mouthpiece

**Contents**

### Key Facts

**Gyroscopic Couple**: The rate of change of angular momentum () = (In the limit).

- = Moment of Inertia.
- = Angular velocity
- = 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**

### 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

### 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

*A*and*B*, Actual Discharge = We know that theoretical discharge Coefficient of discharge,C_{d}= Actual discharge / Theoretical discharge and discharge,

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,

- = 50 mm = 0.05m
- = 3m

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

*A*and

*C*, From equation (6) we know that, substituting the value of 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,

- = 75 mm = 0.075 m
- = 3.5 m
- = 10.3 m

Solution

Absolute pressure head = 6.8 m