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).

Derive by the method of dimensions, an expression for the volume of a

1.	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).
Answer» Solution :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.