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An ideal fluid is flowing with a speed of 12 cm/s through a pipe of diameter 5 cm. The pipe splits into three smaller pipes, each with a diameter of 2 cm. What is the speed of the fluid in the smaller pipes?

User Gooseman
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Answer:

The speed of the fluid through the smaller pipes is:
v_s = 25 \, (cm)/(s)

Step-by-step explanation:

To solve this problem, we use the conservation of volumetric flow.

the volumetric flow rate through a pipe is:


Q= v\cdot A

where
Q is the flow rate,
v is the speed of the fluid and
A is the cross section of the pipe. for a cylindrical pipe we have:


A= \pi \cdot  (D^2)/(4)

Where D stands for Diameter.

This equation together with the previous one give us:


Q=v\cdot \pi \cdot  (D^2)/(4)

Now, we know that all the flow through the first pipe must also come out the other three pipes, that is, the amount of fluid is conserved, no fluid disappears or appears out of thin air.

This is expressed mathematically by:


Q_T=Q_1+Q_2+Q_3

where
Q_T is the flow through the
5 \,cm pipe, while
Q_1 ,
Q_2 and
Q_3 represent the flow through the three smaller pipes.

now, as the three smaller pipes are the same, the flow through each of them will be the same as the flow through any of the others. That is:


Q_1 = Q_2 = Q_3= Q_s where s stands for small.

FInally, we have:


Q_T=Q_1+Q_2+Q_3


Q_T=Q_s+Q_s+Q_s = 3 \, Q_s

Now, using our formula for Q in terms of v and D:


v_t*\pi*(D_t^2)/(4)=3\,v_s*\pi*(D_s^2)/(4)\\v_tD_t^2=3\,v_sD_s^2\\v_s= (v_tD_t^2)/(3D_s^2)

Now, we replace
D_s=2\, cm\\D_t=5\, cm\\v_t = 12 (cm)/(s)

And get
v_s = 25 \, (cm)/(s)

User Glarkou
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