1. Brown University Physics 0050/0070
Physics Department
FLUID FLOW THROUGH A TUBE
In this experiment we will determine how well a physical relationship (also called “law”),
namely Poiseulle’s equation, applies.
In the supplementary reading material this equation was derived as
πpa 4
Q=
8ηl
where Q is the flow rate in units of volume per second; p is the pressure difference
between the ends of the tube; a, the tube radius; l , the length of the tube, and η the
viscosity of the fluid.
We will measure in the experiment Q, as a function of these parameters mentioned above.
One parameter will be varied while the others are kept constant, and Q, as a result of
variation in this parameter, plotted on log-log paper. The reason for choosing log- log
paper is that it brings out a power law dependence, such as Q ∝ a 4 and Q ∝ l −1 much
better than linear graphs. Convince yourself that on log-log paper the exponent (4 or -1
here) just determines the slope of a straight line, while on linear paper, curved graphs
with a continuously changing slope would result. If we write the dependence of Q, on the
various parameters mentioned in terms of an exponential power, we expect that Q gets
larger when, everything else remaining the same, the pressure increases or the radius
increases; Q gets smaller when the tube length increases or the liquid gets more viscous.
One would not expect a significant effect from the material the tube wall is made from,
unless it were very rough and prevented a smooth laminar flow to take place.
Temperature could enter indirectly through the dependence of the viscosity on
temperature.
Collecting all these terms results in a formula of the kind:
Kp δ aα
Q= γ β
η l
where K is a proportionality constant.
We can determine the coefficients α, β, and γ by collecting the fluid coming out of the
tube during a given amount of time as a function of that particular parameter. For
example, to find α, one would measure the amount of water flowing through various
tubes of different diameters, everything else (length, pressure difference) remaining the
same. One then writes the equation:
⎡ Kp δ ⎤
log Q = log ⎢ γ β ⎥ + α log a
⎣η l ⎦
This is a straight line of the form: logQ = B +α log a , with a as slope in a logQ versus
loga plot. In a similar way one can test for the other parameters and thus check the
equation.
It should be kept in mind that Poiseuille’s equation holds only for laminar flow. If the
fluid moves too fast, smooth shear drag between layers disappears and turbulence sets in.
The transition is determined by Reynold’s number, Re. As mentioned in the supplement,

2. Brown University Physics 0050/0070
Physics Department
2v aρ pa 2 a3 p2 ρ
Re = or if one substitutes = v , Re = .
η 8ηl 4η 2 l
This transition occurs when Re equals 2000.
PROCEDURE
We will use the apparatus shown in the figure. The
reservoir has small openings in its side closed by
corks. By removing a cork, the water level is set at
the height, h, of the resulting opening above the
horizontal pipe. The pressure difference between
the two ends of the pipe, p, is the height of the
water column, thus given by
p = ρgh . There is a supply of pipes of various radii
and lengths. The only liquid to be used here is
water so that the viscosity η, will not be varied.
Attach the desired pipe to the reservoir; set the
water level to such a height that the Reynold’s
number is less than 2000. Adjust the water flow
from the faucet, so that the water level in the
reservoir remains stable at the desired height. Put
the graduated beaker under the opening of the tube
and measure the time it takes to collect 500 ml.
This will give you the flow rate Q.
Measure the flow rate for 4 different radii tubes (keeping the length l and pressure p
constant). Repeat the procedure for 2 different lengths and 3 different pressures. One
could plot log Q, versus the log of the radius a, the length l and the height h. The slope of
these lines would give you the various exponents. Instead of calculating the log each
time, it is much easier to plot the value directly, but on log-log graph paper.
Now choose your parameters such that Re >>2000. Measure the flow rate in this
turbulent case and observe the deviation from the lines found earlier.
There is a short second experiment designed to test Bernoulli’s equation in a simple case.
The apparatus is as shown in the figure. We use the same reservoir as was used earlier.
As water flows through the horizontal tube, the water level in the vertical tubes will rise
proportionately to the water pressure. Applying Bernoulli’s equation to points l and 2
gives:
1 2 1 2
p1 + ρv1 = p 2 + ρv 2
2 2
p1 = p 0 + ρgh1 , p 2 = p 0 + ρgh2
Thus;
1 1 2
gh1 + v12 = gh2 + v 2
2 2
where h1 and h2 refer to the height of the

3. Brown University Physics 0050/0070
Physics Department
water level in the vertical tubes above the center of the main tube.
To obtain the velocities we note that the flow rate is given by Q = Av , where A is the
cross-section of the tube.
Bernoulli’s equation gives;
2 2
1⎡Q⎤ 1⎡Q ⎤
gh1 + ⎢ ⎥ = gh2 + ⎢ ⎥
2 ⎣ A1 ⎦ 2 ⎣ A2 ⎦
Set the water level in the reservoir to about 12-cm. Measure h1, h2 and the flow rate Q.
With these values calculate the left and right hand side of the above equation. How
closely are they equal?
Bernoulli’s equation is not applicable for turbulent flow. To see this qualitatively, use a
tube with an abrupt transition between narrow and wide diameter tubes. This
discontinuity introduces turbulence. Calculate now both sides of the equation and see
how close to equal they are.
Questions:
1) What is the most likely cause for differences between Poiseuille’s theoretical formula
and your experiments?
2) What is the most likely cause in your experiment for deviations from Bernoulli’s
equation?
FLUID MECHANICS SUPPLEMENT
In Chapter 14 of the book by Bueche, you find various equations of hydrodynamics,
notably the equation of continuity and Bernoulli’s equation.
We want here to discuss the flow of water through a cylindrical tube, which is not
discussed in the book.
Along the walls of the cylinder the fluid does not flow: it is stationary. There is a gradual
increase of the fluid velocity, from zero at the wall to maximum at the center. The friction
dv
force, F, that resists the velocity, is given by F = −ηA where η = viscosity, A, the
dy
dv
contact area and the velocity gradient, that is, the change of the velocity in going from
dy
the wall (v = 0) towards the center (vmax) . The center fluid drags the next layer, which
slips somewhat behind etc., until the fluid layer at the wall, which does not move at all.
This type of fluid flow is called laminar flow.
This friction is the cause of a pressure drop along a cylindrical tube. Poiseuille derived a
formula that relates the quantity of liquid, Q, (the volume ) flowing per second through a

4. Brown University Physics 0050/0070
Physics Department
cylindrical tube of radius a and length l, with the pressure difference, p, between both
ends. This formula is:
πpa 4
Q= (1)
8ηl
The constant η is the viscosity defined earlier. We will derive this equation. Let us
consider a cylinder with radius r, where r < a. At the entrance surface is a force +πr2p1,
and at the exit surface —πr2p2. The force at the cylindrical surface, the friction force, is
dv
given by − η 2πrl according to the definition for η. Since the liquid does not
dr
experience acceleration but flows with constant velocity, the sum of these three forces
dv
must equal zero. Since is negative if we define r in the usual way as the distance from
dr
the center, we obtain:
dv
πr 2 p1 − πr 2 p 2 + 2πrηl =0 (2)
dr
p − p2
This yields dv = − 1 rdr and after integration:
2ηl
− p(r 2 + c)
v= (3)
4ηl
1
where c represents the integration coefficient stemming from ∫ rdr = r 2 + c . We know
2
that v = 0 at the wall, that is, when r = a. p1—p2 has been written as p, for short.
Application of the boundary condition gives us c and results in:
p(a 2 − r 2 )
v= (4)
4ηl
This velocity profile is sketched in the figure. The total volume of liquid, Q, flowing
through a cross section per second can be obtained from:
p (a 2 − r 2 )2πrdr
a
Q=∫ (5)
0
4ηl
This is namely the sum of the fluid flowing through rings bounded by radius r and r + dr.
One obtains, using:
a
1
a
1 πpa 4
∫
0
rdr = a 2 and ∫ r 3 dr = a 4 ; Q =
2 0
4 8ηl
We will now discuss a few applications of this formula. The average velocity of the
liquid, v can be obtained from:
Q pa 2
=v =
πa 2 8ηl
Reynold made a study of the applicability of Poiseuille’s equation and concluded that
above a certain v laminar flow breaks up and turbulence sets in. This happens when:
2v r ρ
= Reynold’s number Re is > 2000.
η

5. Brown University Physics 0050/0070
Physics Department
In turbulent flow there vortex formation, the flow is irregular and noisy. Blood flow
through the narrow capillaries is laminar as one verifies easily. However, through the
aorta, with v = 0.6m/sec, r = 0.012 m, η= 0.003 kg/m sec and a density
ρ=1 gram/cm3 = 1000kg/m3, one calculates Re = 4800. So blood flow through the aorta is
turbulent. There are several other important medical implications of equation 1.
If one requires rapid blood transfusion, increasing the height of the bottle with blood by a
factor of two increases the flow by that same factor, but increasing the diameter of the
tube by two increases the flow by a factor of sixteen.
In the case of a shock victim, the person’s body temperature often drops. This results in a
significant increase of the viscosity and thus a decrease in blood flow. To keep the body
temperature up, covering of the victim with a blanket and if appropriate giving warm
drinks helps.
Finally, it should be remarked that our simple model could only approximate blood flow.
Veins are not solid tubes, but are elastic, expanding somewhat with increased pressure.
Also blood is not as simple as water in its viscosity. For example, in narrow capillaries
the blood cells line up, reducing the viscosity.