Question

water flows through a pipe of diameter 3m. the stagnation pressure and static pressure measure by a pitot static tube is 0.3 m and 0.24 m of water. the velocity of flows
a) 3.24 b) 2.17
c)0.1085 d)1.085
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Answer 1

Answer:

b) 2.17

Q =vA

Where;

Q = flow rate

v = velocity

A = cross sectional area.

Therefore v = Q/A

In your case Q is given and

A = pi(D^2)/4 = 3.14 × 0.5 × 0.5 / 4 = 0.2 cm^2

v= 3 / 0.2 = 2.17

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Answer 2

Answer:

The velocity of the water flows, V, calculated is $1.085m/s$.

Therefore, option d) $1.085m/s$ is correct.

Explanation:

Given data,

The diameter of the pipe = $3m$

The stagnation pressure of the water flow, P₁ = $0.3m$

The static pressure of the water flow, P₂ = $0.24m$

The velocity of the water flow, V =?

From the relation given below, we can find out the velocity of the water flows:

• $\frac{V^{2} }{2} = (P_{1}-P_{2}) g$

Here g = gravity = $9.81m/s^{2}$

And $P_{1} -P_{2}$ is known as the pressure-head difference.

Therefore, the equation becomes:

• $V = \sqrt{2 (P_{1}-P_{2}) g}$
• $V = \sqrt{2 (3-0.24) 9.81}$
• V = $\sqrt{1.1772}$
• V = $1.0849m/s$ ≈ $1.085m/s$

Hence, the velocity of the water flows, V = $1.085m/s$.