A viscous fluid is flowing through a small tube at a rate of 0.200 m3/s. If the pressure difference across the tube is doubled, and the length of the tube is halved, and the radius is also doubled, then what is the new rate of flow of the fluid?
Question
A viscous fluid is flowing through a small tube at a rate of 0.200 m3/s. If the pressure difference across the tube is doubled, and the length of the tube is halved, and the radius is also doubled, then what is the new rate of flow of the fluid?
Solution
The flow rate of a viscous fluid through a tube is given by Poiseuille's law, which states:
Q = πr^4(P1 - P2) / 8ηL
where: Q is the flow rate, r is the radius of the tube, P1 and P2 are the pressures at the two ends of the tube, η is the viscosity of the fluid, and L is the length of the tube.
Given that the pressure difference is doubled, the length of the tube is halved, and the radius is doubled, the new flow rate Q' can be found by substituting these changes into the equation:
Q' = π(2r)^4(2(P1 - P2)) / 8η(0.5L) = 16πr^4(P1 - P2) / 4ηL = 4πr^4(P1 - P2) / ηL
Therefore, the new flow rate is 4 times the original flow rate:
Q' = 4Q = 4 * 0.200 m3/s = 0.800 m3/s
So, the new rate of flow of the fluid is 0.800 m3/s.
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