A water pipe tapers down from an initial radius of R 1 = 0.2 m to a final radius

Question

A water pipe tapers down from an initial radius of R1 = 0.2 m to a final radius of R2 = 0.09 m. The water flows at a velocity v1 = 0.87 m/s in the larger section of pipe.

1.

What is the volume flow rate of the water?

2.

What is the velocity of the water in the smaller section?

3.

Using this water supply, how long would it take to fill up a swimming pool with a volume of V = 134 m3? (give your answer in minutes)

4.

The water pressure in the center of the larger section of the pipe is P1 = 253250 Pa. Assume the density of water is 103 kg/m3.

What is the pressure in the center of the smaller section of the pipe?

5.

If the pipe was turned vertical and the volume flow rate in the larger section is kept the same, which answers would change?

a) the speed of water in the smaller section

b) the volume flow rate in the smaller section

c) the pressure in the smaller section

Explanation / Answer

This is a problem of fluid mechanics

Part 1)

We write the equation of continuity

A v = cte

Data

R1 = 0.2 m

v1 = 0.87 m/s

R2= 0.09 m

A1 = 0.22

A1 = 0.126 m²

Flow

Q = A1 v1

Q = 0.126 0.87

Q = 0.1096 m³/s

Part 2)

A1 v1 = A2 v2

A2 = 0.092 = 0.02545 m²

v2 = A1/A2 v1

v2 = (0.1096/0.025445) 0.87

v2 = 3.7474 m/s

Part 3)

Flow is 0.1096 m³/s

0.1096 m³–--- 1 s

134 m³–---- t

t = 134/0.1096

t = 1222.63 s

t = 1222.63 s (1 min/60s)

t = 20.38 min

Pat 4)

For this part we use the Bernoulli equation

P + ½ v2+ g y = cte

assume that this horizontal pipe y1=y2

P1 + ½ v12 = P2 + ½ v22

P2 = P1 + ½ v12 - ½ v22

P2 = P1 + ½ (v12 - v22 )

P2 = 253250 + ½ 103 (0.872 – 3.74742 )

P2 = 253250 +51.5 (- 13.2861) = 253250 – 684.2342

P2 = 252565.77 Pa

Part 5)

In the case of pipe vertically and the distances they are different

case a) As used dec Bernoulli equation for speed and this may be less the flow must be reduced

case b) we use the Bernoulli equation to calculate the speed, which is less than the horizontal pipe and you have to work against the height and pressure

case c) the new pressure is subtracted from the high pressure

p2' = p2 – g h = P2 – 103 9.8 h

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