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**ENGIN1002 Engineering Physics, Federation University, Australia**

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**Task: **At all data points you can do 2 things:

With dynamic head use Bernoulli to determine the flow velocity at each point. The height of water in the pitot tube relates toi velocity. Since static head is also known at these points, and since the B of the channel is constant (75mm) you know the area of the flow. Use Continuity to determine the mean velocity. u = Q/A. Compare velocities from each method.

The sum of Pressure head + dynamic head + static head should be conserved at all points (within reason). Is it?

**Answer: Aim: **To determine the flow velocity at each point on a Weir using Bernoulli continuity equation.

**Equipment Required: **The following equipment is required to carry out the experiment

• Pitot tube

• Scale

• A Pump

• Vernier caliper (for thickness measurement)

• Valve regulators

**Formula used: **

Bernoulli's continuity equation:

(p/ρg) + (v^{2}/2g) + z = constant

Mean velocity:

u = Q/A

v = √2gh

**Description: **The main objective of this experiment is to determine the flow velocity at each point on the Weir. The Weir is 150mm long and 75 mm wide. The thickness of the weir is around 25mm. The flow is measured using the pitot tube. The pitot tube is a device used to measure the flow of liquid. It is based on the principle that the pressure at a point increases if the velocity of the fluid at that point decreases. This is a result of the conservation of kinetic energy into potential energy.

The pitot tube comprises a glass tube which is bent at right angles. It is in accordance with Bernoulli's equation.

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**Derivation:**

Bernoulli's equation states that,

(p/ρ) + gz + (v^{2}/2g) = constant

Multiplying ρ on both sides we get

p + q + ρgz = constant

p + q + ρgz = constant

where q = v^{2}/2g

p_{0 }+ ρgz = constant

p_{0 }= p + q

p_{0 }= sum of static pressure and dynamic pressure

Dividing by ρg we get

(p/ρg) + (v^{2}/2g) + z = constant

Total head H is given by

H = (p/ρg) + (v^{2}/2g) + z

H = (v^{2}/2g) + h

where h = p/ρg + z

**Procedure: **The dimensions of the weir are measured with the help of a scale. The weir ranges about 150 mm long and 75mm wide. The thickness of the weir is measured using the vernier caliper. The readings indicate that the weir is 25mm thick. The weir is placed in a setup which comprises a pump and a regulator.

**The image illustrates measuring the weir with the help of a scale**

The pump is capable of lifting the liquid through mechanical action. The set up also comprises of a pitot tube. The pitot tube is a device that is used to measure the flow of the fluid. The pitot is generally a glass tube that is bent at right angles. It is based on the principle of Bernoulli's equation.

**The image illustrates the measurement of dynamic and static heat using a pitot tube**

The liquid is allowed to flow over the weir. This flow rate of the liquid is regulated using a valve. Assumptions are made that the left side of the weir i.e the front edge of the weir is the initial point. The upstream to the weir is negative and downstream is considered to be positive. Using the pitot tube the dynamic head at each point is calculated. Similarly, the static head at the same point is also determined. The points are taken from the start of the weir to its end (0<x<150) with respect to its length. The static head is measured from the water surface to the channel floor on either side of the object. The values of the static head and dynamic head at each point are observed and the flow velocity at each point is calculated. Comparisons are made on the basis of the flow velocity with respect to static and dynamic heads.

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**Calculation:**

**Mean velocity**

u = Q/A

Area of the weir is given by= (Length.Breadth)

= 150(75) = 11250mm

Discharge Q = (C_{D} √2gh)/A

Q = (0.8 √2gh)/A

Q = (0.8√(19.62(90)))/11250

Q = 2.988 × 10^{3}

U = 2.65m/s

Static head

h = (p/ρg) + z

Static head = v^{2}/2g

Pressure head = p/ρg

Total head H is given by

H = (p/ρg) + (v^{2}/2g) + z

H = (v^{2}/2g) + h

(p/ρg) + (v^{2}/2g) + z = constant

Before water enters the tube

P = 1 atm v = ? h = 0

At this point

(p/ρg) = (v^{2}/2g)

(1/1000) = (v^{2}/2)

v = 4.54m/s

When the water rises to the top

P = 1 atm v = 0 h = 90

H = (v^{2}/2g) + h

Total head = 90mm

Static head

h = (p/ρg) = 10332.279/(1000(9.81)) = 1.053m

v = √(2p/ρ)

V = 4.54m/s

Total head at point 1 (initial)

H = (p/ρg) + (v^{2}/2g)

H = 1.05m

Total head at point 2 (top)

H = (p/ρg) + (v^{2}/2g)

H = 1.05m

Therefore the head at both the points are equal.

It can be clearly seen that as the velocity decreases pressure increases.

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