Saved Bookmarks
| 1. |
Derive Poiseuille's formula for the volume of a liquid flowing per second through a pipe under streamlined flow. |
|
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 (l) coefficient of viscosity `(eta)` of the liquid, (2) radius of the tube (r), and (3) the pressure gradient `(P/l)` Then, 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 We have three unknowns a, b and c. We have three equations, on solving, we get a = -1, b = 4 and c = 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. This relation holds good for the fluids whose velocities are lesser than the critical velocity `(v_(c))`. |
|