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Making connections: different-sized pipes

For incompressible fluids, the density of the fluid remains constant throughout, no matter the flow rate or the size of the opening through which the fluid flows. We say that, to ensure continuity of flow, the amount of fluid that flows past any point is constant. That amount can be measured by either volume or mass.

Flow rate has units of volume/time (m 3 /s or L/s). Mass flow rate ( Δ m Δ t ) has units of mass/time (kg/s) and can be calculated from the flow rate by using the density:

m =   ρ V

The average mass flow rate can be found from the flow rate:

Δ m Δ t = m t =   ρ V t =   ˙ ρ Q = ρ A v

Suppose that crude oil with a density of 880 kg/m 3 is flowing through a pipe with a diameter of 55 cm and a speed of 1.8 m/s. Calculate the new speed of the crude oil when the pipe narrows to a new diameter of 31 cm, and calculate the mass flow rate in both sections of the pipe, assuming the density of the oil is constant throughout the pipe.

Solution: To calculate the new speed, we simply use the continuity equation.

Since the cross section of a pipe is a circle, the area of each cross section can be found as follows:

For the larger pipe:

A 1 =   π ( d 1 2 ) 2 =   π ( 0.275 ) 2 = 0.238 m 2

For the smaller pipe:

A 2 = π ( 0.155 ) 2 =   0.0755 m 2

So the larger part of the pipe ( A 1 ) has a cross-sectional area of 0.238 m 2 , and the smaller part of the pipe ( A 2 ) has a cross-sectional area of 0.0755 m 2 . The continuity equation tells us that the oil will flow faster through the portion of the pipe with the smaller cross-sectional area. Using the continuity equation, we get

A 1 v 1 =   A 2 v 2
v 2 =   ( A 1 A 2 ) v 1 =   ( 0.238 0.0755 ) ( 1.8 ) =   5.7   m / s

So we find that the oil is flowing at a speed of 1.8 m/s through the larger section of the pipe ( A 1 ), and it is flowing much faster (5.7 m/s) through the smaller section ( A 2 ).

The mass flow rate in both sections should be the same.

For the larger portion of the pipe:

( Δ m Δ t ) 1 =   ρ A 1 v 1 = ( 880 ) ( 0.238 ) ( 1.8 ) =   380   kg / s  

For the smaller portion of the pipe:

( Δ m Δ t ) 2 =   ρ A 2 v 2 = ( 880 ) ( 0.75538 ) ( 5.7 ) =   380   kg / s

And so mass is conserved throughout the pipe. Every second, 380 kg of oil flows out of the larger portion of the pipe, and 380 kg of oil flows into the smaller portion of the pipe.

The solution to the last part of the example shows that speed is inversely proportional to the square of the radius of the tube, making for large effects when radius varies. We can blow out a candle at quite a distance, for example, by pursing our lips, whereas blowing on a candle with our mouth wide open is quite ineffective.

In many situations, including in the cardiovascular system, branching of the flow occurs. The blood is pumped from the heart into arteries that subdivide into smaller arteries (arterioles) which branch into very fine vessels called capillaries. In this situation, continuity of flow is maintained but it is the sum of the flow rates in each of the branches in any portion along the tube that is maintained. The equation of continuity in a more general form becomes

n 1 A 1 v ¯ 1 = n 2 A 2 v ¯ 2 , size 12{n rSub { size 8{1} } A rSub { size 8{1} } {overline {v rSub { size 8{1} } }} =n rSub { size 8{2} } A rSub { size 8{2} } {overline {v rSub { size 8{2} } }} } {}

where n 1 size 12{n rSub { size 8{1} } } {} and n 2 size 12{n rSub { size 8{2} } } {} are the number of branches in each of the sections along the tube.

Questions & Answers

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Practice Key Terms 2

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