# Bernoulli Equation

## Conservation of energy - non-viscous, incompressible fluid in steady flow

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The statement of conservation of energy is useful when solving problems involving fluids. For a non-viscous, incompressible fluid in a steady flow, the sum of pressure, potential and kinetic energies per unit volume is constant at any point.

A special form of the Euler’s equation derived along a fluid flow streamline is often called the **Bernoulli Equation:**

For steady state incompressible flow the Euler equation becomes (1). If we integrate (1) along the streamline it becomes (2). (2) can further be modified to (3) by dividing by gravity.

### Head of Flow

Equation (3) is often referred to the "**head"** because all elements has the unit of length.

### Dynamic Pressure

(2) and (3) are two forms of the Bernoulli Equation for steady state incompressible flow. If we assume that the gravitational body force is negligible, (3) can be written as (4). Both elements in the equation have the unit of pressure and it's common to refer the flow velocity component as the **dynamic pressure** of the fluid flow (5).

Since energy is conserved along the streamline, (4) can be expressed as (6). Using the equation we see that increasing the velocity of the flow will reduce the pressure, decreasing the velocity will increase the pressure.

This phenomena can be observed in a **venturi** **meter** where the pressure is reduced in the constriction area and regained after. It can also be observed in a **pitot** **tube** where the **stagnation** pressure is measured. The stagnation pressure is where the velocity component is zero.

### Example - Bernoulli Equation and Flow from a Tank through a small Orifice

Liquid flows from a tank through a orifice close to the bottom. The Bernoulli equation can be adapted to a streamline from the surface (1) to the orifice (2) as (e1):

Since (1) and (2)'s heights from a common reference is related as (e2), and the equation of continuity can be expressed as (e3), it's possible to transform (e1) to (e4).

### Vented tank

A special case of interest for equation (e4) is when the orifice area is much lesser than the surface area and when the pressure inside and outside the tank is the same - when the tank has an open surface or "vented" to the atmosphere. At this situation the (e4) can be transformed to (e5).

"The velocity out from the tank is equal to speed of a freely body falling the distance *h*." - also known as **Torricelli's Theorem.**

#### Example - outlet velocity from a vented tank

The outlet velocity of a tank with height *10 m* can be calculated as

V_{2}= (2 (9.81 m/s^{2}) (10 m))^{1/2}

= 14 (m/s)

### Pressurized Tank

If the tanks is pressurized so that product of gravity and height *(g h)* is much lesser than the pressure difference divided by the density, (e4) can be transformed to (e6). The velocity out of the tank depends mostly on the pressure difference.

#### Example - Outlet Velocity from a Pressurized Tank

The outlet velocity of a pressurized tank where

p_{1}= 0.2 (MN/m^{2})

p_{2}= 0.1 (MN/m^{2})

A_{2}/ A_{1}= 0.01

h = 10 (m)

can be calculated as

V_{2}= ( (2 / (1 - (0.01)^{2}) ((0.2 10^{6 }N/m^{2}) - (0.1 10^{6}N/m^{2})) / (1000 kg/m^{3}) + (9.81 m/s^{2}) (10 m)))^{1/2}

= 19.9 (m/s)

### Coefficient of Discharge - Friction Coefficient

Due to friction the real velocity will be somewhat lower than this theoretic examples. If we introduce a **friction coefficient** *c* (coefficient of discharge), (e5) can be expressed as (e5b).

The coefficient of discharge can be determined experimentally. For a sharp edged opening it may bee as low as *0.6*. For smooth orifices it may bee between *0.95* and *1*.

### Energy Loss through a Reduction Valve

When fluid flows through a reduction valve and pressure is reduced - there is a energy loss. By neglecting the change in elevation *(h _{1} = h_{2})* and the change in pipe velocity

*(v*the pressure energy before the valve and the pressure energy after the valve including the energy loss through the valve - is constant, and the Bernouilli equation can be modified to

_{1}= v_{2})*p _{1 }/ ρ = p_{2 }/ ρ + dE (7)*

*where *

*dE = energy loss through valve (J)*

(7) can be transformed to:

*dE = ( p_{1} - p_{2}) / ρ*

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