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The rate of flow of liquid through a capillary tube of radius r is v when pressure difference across the two ends of the capillary is p. If pressure is increased by 3p and radius is decreased to r/2, find the new rate of flow.`

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closed as off-topic by AccidentalFourierTransform, sammy gerbil, Jon Custer, Kyle Kanos, John Rennie Feb 26 '17 at 14:16

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The relationship between pressure drop and flow rate for viscous flow through a tube is $$\Delta p=\eta \left(\frac{4L}{D}\right)\left(\frac{32Q}{\pi D^3}\right)$$ where $\eta$ is the viscosity and Q is the volumetric throughput rate. If D is halved, then $D^4$ is 1/16 as high. So, 4 x (1/16) = 1/4. Q will be 1/4 as large.

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  • $\begingroup$ can you pleasetell where did I went wrong in my answer ?I would really appreciate it. $\endgroup$ – ATHARVA Feb 26 '17 at 1:13
  • $\begingroup$ You went wrong when you tried to solve it using Bernoulli's equation. For an inviscid fluid (which Bernoulli's equation describes) the pressure drop in a straight pipe or capillary is zero. $\endgroup$ – Chet Miller Feb 26 '17 at 12:09
  • $\begingroup$ I did not get it why is the pressure drop zero?. Thanks for the response. $\endgroup$ – ATHARVA Feb 26 '17 at 12:11
  • $\begingroup$ is it because,. As viscosity is zero no viscous force will be there. And viscous force is the reason for pressure drop in tube. $\endgroup$ – ATHARVA Feb 26 '17 at 12:23
  • $\begingroup$ Sure. That is the reason. $\endgroup$ – Chet Miller Feb 26 '17 at 12:25

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