Answer:
The speed of the fluid through the smaller pipes is: [tex]v_s = 25 \, \frac{cm}{s}[/tex]
Explanation:
To solve this problem, we use the conservation of volumetric flow.
the volumetric flow rate through a pipe is:
[tex]Q= v\cdot A[/tex]
where [tex]Q[/tex] is the flow rate, [tex]v[/tex] is the speed of the fluid and [tex]A[/tex] is the cross section of the pipe. for a cylindrical pipe we have:
[tex]A= \pi \cdot \frac{D^2}{4}[/tex]
Where D stands for Diameter.
This equation together with the previous one give us:
[tex]Q=v\cdot \pi \cdot \frac{D^2}{4}[/tex]
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:
[tex]Q_T=Q_1+Q_2+Q_3[/tex]
where [tex]Q_T[/tex] is the flow through the [tex]5 \,cm[/tex] pipe, while [tex]Q_1[/tex] , [tex]Q_2[/tex] and [tex]Q_3[/tex] 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:
[tex]Q_1 = Q_2 = Q_3= Q_s[/tex] where s stands for small.
FInally, we have:
[tex]Q_T=Q_1+Q_2+Q_3[/tex]
[tex]Q_T=Q_s+Q_s+Q_s = 3 \, Q_s[/tex]
Now, using our formula for Q in terms of v and D:
[tex]v_t*\pi*\frac{D_t^2}{4}=3\,v_s*\pi*\frac{D_s^2}{4}\\v_tD_t^2=3\,v_sD_s^2\\v_s= \frac{v_tD_t^2}{3D_s^2}[/tex]
Now, we replace [tex]D_s=2\, cm\\D_t=5\, cm\\v_t = 12 \frac{cm}{s}[/tex]
And get [tex]v_s = 25 \, \frac{cm}{s}[/tex]
