How can you calculate the rate of flow of a viscous liquid in a very narrow tube? The obvious answer is to solve the Navier-Stokes equation (post 17.15) for this type of flow. But that involves a lot of mathematics. We can get some idea of what the resulting equation must look like by extending the ideas described in post 17.41.
The problem is related to an object moving in a stationary liquid because whether it is the liquid that is moving or the object (or tube) that is moving depends on the motion of the observer (post 16.4) or, in other words, on the frame of reference in which the motion is observed (post 16.12). So a drag force limits the rate of flow; the drag force depends on the dimensions of the object (or tube) and the viscosity, η, of the liquid (post 17.17). The dimensions of the tube that could influence the flow are its radius, r (or its diameter or its cross-sectional area, since they all depend on each other) and its length, L.
In the previous paragraph, we were concerned with the force opposing flow. But there must also be a force which causes a stationary fluid to flow through a tube (post 16.13). When we think of forces in liquids, it is usually easier to think of pressure – the force perpendicular to a unit area (posts 17.5 and 17.15). So we will think about the pressure difference, P, between the ends of the tube – the greater this pressure difference, the faster we expect the fluid to flow.
We will think of the flow rate as the volume of liquid flowing through the tube in unit time (1 s), represented by R. From what we’ve seen above, we expect that
R = kηarbLcPd
where k represents any numbers that appear in the equation. The problem is to find the values of a, b, c and d to find how R depends on η, r, L and P.
In this equation, the combined units on the right-hand side must be the same as the units used to measure R, for the reasons given in post 17.41.
In post 17.17, we saw that viscosity is measured in Pa.s where
1 Pa.s = 1 (N.m-2).s = 1 (kg.m.s-2).m-2.s-1 = 1 kg.m-1.s-1.
Expressing the unit of force (N) in kg, m and s was explained in post 17.41. If we wanted to be completely general about the units, we would say that viscosity is measured in ML-1T-1 where M are units of mass, L are units of length and T are units of time (post 17.41).
Now let’s look at the right-hand side of the equation. k is a number and so isn’t measured in any units. r has the units of length, L. L also has the units of length, L. (Be careful not to confuse L, the length, with L, the units used to measure length). P is measured in Pa (post 17.17) and so is measured in the units of viscosity (Pa.s in the SI system) divided by the units of time; from the previous paragraph it must have the units ML-1T-2.
R is a volume divided by time and so is measured in units of L3.T-1. If our equation for R makes sense, we can write that
L3.T-1 = (ML-1T-1)a(L)b(L)c(ML-1T-2)d
using the idea developed in post 17.41. M, L and T must be raised to the same power, on both sides of this equation, for it to be true.
From M: 0 = a + d
From L: 3 = – a + d + c – d
From T: -1 = – a – 2d.
We have a problem – we have more unknown values (a, b, c and d) than we have equations that relate them. So we cannot obtain a unique set of values for the unknowns. What can we do?
To obtain a unique solution, we need to make a further assumption. If we doubled the length of the tube, we would expect the drag to double – so we would need to double P to compensate, if we wanted to keep R the same; if we made the tube three times longer we would expect to triple the drag – so we would need to triple P to maintain the same value of R. This means that we expect R to depend on P/L and so can write
R = kηarb(P/L)c.
Now the relationship between the units is
L3.T-1 = (ML-1T-1)a(L)b[(ML-1T-2)/L]c = (ML-1T-1)a(L)b(ML-2T-2)c.
To find the values of a, b and c, we equate the powers for M, L and T, on both sides of the equation, as before.
From M: 0 = a + c (1)
From L: 3 = – a + b -2c (2)
From T: -1 = –a – 2c (3)
Solving these equations (see the appendix) gives a = – 1, b = 4 and c = 1. So we deduce that
R = kη-1r4(P/L) = kr4P/(ηL).
If we had solved the Navier-Stokes equation for laminar flow (post 17.15) in the tube, we would have obtained the same result with k = 1/8. The complete result is called Poiseuille’s equation. Alternatively, we could perform experiments to find the value of k that made our equation give the correct results.
This method for developing equations that describe what happens in the physical world is called dimensional analysis. It is not a substitute for an analysis based on the underlying principles of a problem but can still be very useful.
Adding equations 1 and 3 gives – 1 = 0 – c or c = 1.
Substituting this value for c into equation 1 gives a = – 1.
Substituting these values for a and c into equation 2 gives 3 = 1 + b – 2 or b = 4.
We can check that these results are correct.
From equation 1: 0 = – 1 + 1
From equation 2: 3 = 1 + 4 – 2
From equation3: -1 = 1 – 2