The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of liquid through a hypodermic needle during an injection can be estimated with the following equation. Q = πRn P2 − P1 8ηL The length and radius of the needle are L and R, respectively, both of which have the dimension [L]. The pressures at opposite ends of the needle are P2 and P1, both of which have the dimensions of [M]/{[L][T]2}. The symbol η represents the viscosity of the liquid and has the dimensions of [M]/{[L][T]}. The symbol π stands for pi and, like the number 8 and the exponent n, has no dimensions. Using dimensional analysis, determine the value of n in the expression for Q.

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cryssatemp cryssatemp · Answer 1 · 2019-09-04T20:18:28+02:00

Answer: n=4

Explanation:

We have the following expression for the volume flow rate [tex]Q[/tex] of a hypodermic needle:

[tex]Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}[/tex] (1)

Where the dimensions of each one is:

Volume flow rate [tex]Q=\frac{L^{3}}{T}[/tex]

Radius of the needle [tex]R=L[/tex]

Length of the needle [tex]L=L[/tex]

Pressures at opposite ends of the needle [tex]P_{2}[/tex] and [tex]P_{1}=\frac{M}{LT^{2}}[/tex]

Viscosity of the liquid [tex]\eta=\frac{M}{LT}[/tex]

We need to find the value of [tex]n[/tex] whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

[tex]\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}[/tex] (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

[tex]\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}[/tex] (3)

[tex]\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}[/tex] (4)

As we can see [tex]n[/tex] must be 4 if we want the exponent to be 3:

[tex]\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}[/tex] (5)

Finally:

[tex]\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}[/tex] (6)