Derive by the method of dimensions, an expression for the volume of a liquid flowing out per second through a narrow pipe. Asssume that the rate of flow of liwquid depends on
(i) the coeffeicient of viscosity `eta` of the liquid
(ii) the radius 'r' of the pipe and
(iii) the pressure gradient `(P)/(l)` along the pipte. Take `K=(pi)/(8)`.

Explanation

Consider a liquid flowing steadily through a horizontal capillary tube. Let `v=((V)/(t))` be the volume of the liquid flowing out per second through a capillary tube. It depends on (i) coefficient of viscosity `(eta)` of the liquid, (ii) radius of the tube (r), and (iii) the pressure gradient `((P)/(l))` Then,
`vpropeta^(a)r^(b)((P)/(l))^(c)`
`v=keta^(a)r^(b)((P)/(l))^(c)" "...(1)`
Where, k is a dimensionless constant. Therefore,
`[v]=("volume")/("time")=[L^(3)T^(-1)]`
`[(dP)/(dx)]=("pressure")/("distance")=[ML^(-2)T^(-2)]`,
`[eta]=[ML^(-1)T^(-1)]and[r]=[L]`
substituting in equation (1)
`[L^(3)T^(-1)]=[ML^(-1)T^(-1)]^(a)[L]^(b)[ML^(-2)T^(-2)]^(c)`
`M^(0)L^(3)T^(-1)=M^(a+b)L^(-a+b-2c)T^(-a-2c)`
So, equating the powers of M, L, and T on both sides, we get
`a+c=0,-a+b-2c=3,and-a-2c=-1`
On solving three equations, we get
`a=-1,b=4andc=1`
Therefore, equation (1) becomes,
`v=keta^(-1)r^(4)((P)/(l))^(1)`
Experimentally, the value of k is shown to be `(pi)/(8)`, we have
`v=(pir^(4)P)/(8etal)`
The above equation is known as Poiseuille's equation for the flow of liquid through a narrow tube or a capillary tube.