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Determine the ratio of the flow rate through capillary tubes A and B (i.e., [tex]Q_A/Q_B[/tex]).

- The length of tube A is twice that of tube B.
- The radius of tube A is one-half that of tube B.
- The pressure across both tubes is the same.

Answer :

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

[tex]\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}[/tex]

Where

P = Pressure at each point

r = Radius

[tex]\eta =[/tex] Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]

From the problem two terms are given

[tex]R_A = \frac{R_B}{2}[/tex]

[tex]L_A = 2L_B[/tex]

Replacing we have to

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}[/tex]

[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}[/tex]

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

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