# Bernoulli's Equation Formula

The Bernoulli Equation is a different way of the conservation of energy principle, applied to flowing fluids. It relates the pressure, the kinetics energy and the gravitational potential energy of a fluid in a container or flowing in a tube.

Describes the lowering of fluid pressure in regions where the flow velocity is increased. In the high velocity flow through the constriction, kinetic energy must increase at the expense of pressure energy.

Pressure + ½ density * square of the velocity + density * gravity

acceleration* height = constant

The equation is written

P + ½ ρ v^{2} +ρ g h = constant

That says the whole formula holds along the system, each term can change but the sum is the same.

We have:

P: Pressure

v: velocity of the fluid

ρ: Density of the fluid

h: height of the container or the pipe here the fluid is flowing

Bernoulli's Equation Formula Questions:

1) We have a fluid with density 1 Kg/m^{3} that is moving through a pipe with transverse area 0.1 m^{2} and a velocity of 3.5 m/s. It is connected to another pipe of 1 m^{2} of area, both at the same height. The pressure at the beginning of the tube is 2 kPa. What is the pressure of fluid at the end of the second tube?

Answer: The first thing to note is that the tubes are both at the same height, then the Bernoulli formula only stand as,

P_{i} + ½ ρ v_{i}^{2} = P_{f} + ½ ρ v_{f}^{2}

The velocity at the end of the tube can be calculated using the mass continuity equation, and from there the velocity, is given by

A_{i }/A_{f}*v_{i} =v_{f }= 0.1 m^{2}/1 m^{2} * 3.5 m/s = 0.35 m/s

Substituting in the Bernoulli's equation,

P_{i} + ½ ρ v_{i}^{2} - ½ ρ v_{f}^{2} = P_{f}

2 KPa + ½ 1 Kg/m^{3} * (3.5 m/s)^{2} - ½ 1 Kg/m^{3} * (0.35 m/s)^{2} = P _{f}

P_{f} = 8.06 KPa.

2) Consider the same problem above. What is the pressure of fluid at the end of the second tube if the second tube is placed at 1 m over the ground?

Answer: The first thing to note is that the tubes are both at the same height, then the Bernoulli formula only stand as,

P_{i} + ½ ρ v_{i}^{2} = P_{f} + ½ ρ v_{f}^{2} + ρ g h

Using the previous result v_{f}=0.35 m/s

Substituting in the Bernoulli's equation,

P_{i} + ½ ρ (v_{i}^{2} - v_{f}^{2}) + ρ g h = P _{f}

2 KPa + ½ 1 Kg/m^{3} * ((3.5 m/s)^{2} -(0.35 m/s)^{2}) - 1 Kg/m^{3} * 9.8 m/s^{2} * 1
m = P_{f}

P_{f} = -1.74 KPa.

The minus sign implies here that the fluid cannot flow outside of the tube, but the negative pressure pushes back.

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