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BigorU [14]
2 years ago
13

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.​
Physics
1 answer:
Oksana_A [137]2 years ago
7 0

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

<h3>What is gauge pressure?</h3>

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

#SPJ1

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