Water flows at 5 Liters/s through a horizontal pipe that narrows smoothly from 10 cm diameter to 5 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section

pressure = 53 Kpa

Explanation: Given that Water volume rate of flow is 5 Liters/s through a horizontal pipe. Since the volume of water flowing through the two pipes are equal, velocity in the small pipe will be greater than the velocity in the large pipe.

Also, since the velocity in the small pipe is greater than the velocity in the large pipe, pressure P1 in the large pipe will be greater than the pressure P2 in the small pipe.

Using Bernoulli’s equation which can be expressed as

P1 + (½ × ρ × v1^2) + (½ × ρ × g × h1) = P2 + (½ × ρ × v2^2) + (½ × ρ × g × h2) 

h1 = h2

Since the pipe is horizontal.

So subtract (½ × ρ × g × h) from both sides. This will lead to:

P1 + (½ × ρ × v1^2) = P2 + (½ × ρ × v2^2)

But  

Density ρ = 1000kg/m^3

P2 = 50 kPa = 50,000 Pa

Convert the velocity ( rate of flow ) to meters per second. And liters to cubic centimeters

Recall:

1 liter = 1000 cm^3 

5 liters = 5000 cm^3

Volume = Cross-sectional area × length

Length = Volume /Cross-sectional area

Where cross-sectional area of the pipe will be.

Area of circle = πr^2.

For 10 cm pipe,

area = π × 5^2

Length = 5000 / (π × 25) = 200/π cm

The water moves 200/π centimeters each second. 

Velocity = 200/π cm/s = 2/π m/s

For 5 cm pipe,

area A = π × 2.5^2

Length = 5000 / (π × 6.25) = 800/π cm

The water moves 800/π centimeters each second. 

Velocity = 800/π cm/s = 8/π m/s

Substitute the two velocities into Bernoulli’s equation.

P1 + [½ × 1000 × (2/π)^2] = 50,000 + [½ × 1000 × (8/π)^2]

P1 = 50,000 + [½ × 1000 × (8/π)^2] – [½ × 1000 × (2/π)^2]

P1 = 53039.6

Therefore, the reading of the pressure gauge in the wide section is 53 Kpa approximately

answer:

put more,