Answer: 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
