Question

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.

Answer 1

Answer: n=4

Step-by-step explanation:

We have the following expression for the volume flow rate
`Q` of a hypodermic needle:

`Q=(\pi R^(n)(P_(2)-P_(1)))/(8\eta L)` (1)

Where the dimensions of each one is:

Volume flow rate
`Q=(L^(3))/(T)`

Radius of the needle
`R=L`

Length of the needle
`L=L`

Pressures at opposite ends of the needle
`P_(2)` and
`P_(1)=(M)/(LT^(2))`

Viscosity of the liquid
`\eta=(M)/(LT)`

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

`(L^(3))/(T)=(\pi L^(n)((M)/(LT^(2))))/(8((M)/(LT)) L)` (2)

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

`(L^(3))/(T)=(\pi)/(8) ( L^(n))/(LT)` (3)

`(L^(3))/(T)=(\pi)/(8) ( L^(n-1))/(T)` (4)

As we can see
`n` must be 4 if we want the exponent to be 3:

`(L^(3))/(T)=(\pi)/(8) ( L^(4-1))/(T)` (5)

Finally:

`(L^(3))/(T)=(\pi)/(8) ( L^(3))/(T)` (6)