A flow rate of 20 cm^3/sec is obtained through a tube of length 100 cm and radiu

Question

A flow rate of 20 cm^3/sec is obtained through a tube of length 100 cm and radius 0.50 cm when there is a pressure difference of 25 cm of water along the tube. Calculate the net flow rate when the following changes are made, assuming all other factors remain the same as the original conditions.

A) A 350-cm-long tube replaces the original.

B) The tube radius is decreased to 0.40 cm.

C) The pressure difference (P1-P2) is increased to 40 cm of water,

D) The pressure difference is increased to 50 cm of water and the radius is decreased to 0.20 cm.

*Please list all steps and identify variables/equations used.

Explanation / Answer

(a) L= 350cm ; r =0.50; pressure diff = 25

Volume Flow rate =?

Volume flowrate = (Pressure diff) x (radius)4 / (8/pi) Viscosityx length = 25*(0.5)4 / 2.54 * 0.3006*350

Volume flowrate = 5.714 cm^3/sec

(b) r =0.4cm, L= 100, P=25

Again using the above formula we get,

Volume flowrate = 8.192 cm^3/sec

(c)Press Diff = 40cm, r =0.5, L= 100

Again using the above formula we get,

Volume flowrate = 32.0 cm^3/sec

(d)Press diff = 50, r =0.2 and L =100

Again using the above formula we get,

Volume flowrate = 1.02 cm^3/sec

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