An ideal fluid of density p=700 kg/m^3 flows through a horizontal tube with an i

Question

An ideal fluid of density p=700 kg/m^3 flows through a horizontal tube with an input radius of 0.24m and output radius of 0.12m (as seen below). Assume that the difference in pressure is measured to be P1-P2=1200 Pa

1. Explain why the pressure at the output radius is smaller than at the input radius.

2. Calculate the speed of the fluid and the volume flow rate as the fluid leaves the tube. Assume that the changes in gravitational potential energy are negligible.

3. How would the volume flow rate change if we replaced this fluid with water (pofwater=1000kg/m^3)? Defend your answer.

Explanation / Answer

by bernoulli's equation

P1 + 0.5 * density * velocity1^2 = P2 + 0.5 * density * velocity2^2

and by continuity equation

area1 * velocity1 = area2 * velocity2

since area of the location 2 is less the velocity at this location will be more

so,

velocity2 > velocity1

since velocity2 > velocity1 so,

P1 < P2 (by first equation)

pi * 0.24^2 * velocity1 = pi * 0.12^2 * velocity2

0.24^2 * velocity1 = 0.12^2 * velocity2

4 * velocity1 = velocity2

P1 - P2 = 0.5 * density * (velocity2^2 - velocity1^2)

1200 = 0.5 * 700 * (16 * velocity1^2 - velocity1^2)

velocity1 = 0.478 m/s

velocity2 = 4 * 0.478

velocity2 = 1.912 m/s

speed of the fluid as it leaves the tube = 1.912 m/s

volume flow rate = area * velocity

volume flow rate = pi * 0.12^2 * 1.912

volume flow rate = 0.0864 m^3 / s

if we replace fluid with water

1200 = 0.5 * 1000 * (16 * velocity1^2 - velocity1^2)

velocity1 = 0.4 m/s

velocity2 = 4 * 0.4

velocity2 = 1.6 m/s

volume flow rate = pi * 0.12^2 * 1.6

volume flow rate = 0.0723 m^3 / s

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