Question

The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross-section is A, Calculate the speed and pressure at another end where the area of cross-section is A/3.​

Answer 1

The speed and pressure at another end will be 9 m/sec and -31500 pascals.

What is gauge pressure?

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

For horizontal pipe, Z₁=Z₂

Pressure at end 1, P₁ = 4500Pa

Speed at end 1, V₁ = 3 m/sec

Speed at end 2, V₂ = ? m/sec

Area of cross-section at end 1,A₁ =A

Area of cross-section at end 1,A₂ = A/3

From the continuity equation;

A₁V₁=A₂V₂

A× 3  = (A/3)×V₂

V₂ = 9 m/sec

From Bernoulli's equation;

$\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2 \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} = \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} = \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 = - \ 31,500 \ pa$

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

To learn more about the gauge pressure, refer to the link;

brainly.com/question/14012416

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