# 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 in-compressible 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 in-compressible 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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