Fluid Mechanics
Problem Set 7
Problems from Munson et al., 6th edition
As usual, the point here is not to make you do a lot of problems. You should simply do as many
of these problems as you need to do to become comfortable with the material and hence do well
on the exams. Also as usual, the best way to learn is to try the problems without looking at the
solutions, then to examine the solutions to see where you went wrong, and then to try the
problem again.
On the second page you will find a list of questions that have been asked in previous years
that will help you significantly with this problem set.
Simple plug-and-chug problems involving entrance length, flowrate, and Re:
8.2 – 8.8, 8.12
Laminar flow
8.13 – 8.16, 8.18 - 8.22, 8.24 – 8.25,
8.26 is boring to do, but interesting to look at the solution
8.27 – 8.29. Hint for 8.29. When they say the manometer reading
will be "reversed," they mean the figure to the right:
Turbulent flow: 8.32 – 8.33, 8.38
Major losses: 8.41 – 8.46, 8.49, 8.52, 8.54-8.55
Minor losses: 8.60-62
Non-circular conduits: 8.72 – 8.75. These problems may be particularly important if you are a
mechanical, environmental, or civil engineer. You may use non-circular conduits as you design
heating or cooling systems (e.g., ducts for airflow)
Pressure drops along pipes: 8.76 – 8.77, 8.80 – 8.81, 8.83 – 8.84, 8.86-8.87
Flowrate in a pipe: 8.93-8.95, 8.98-8.99 (see note below)
Calculating required pipe diameters: 8.100 – 8.103 (see note below).
Important Note: Problems 8.99 – 8.103 require you to use trial and error with the Moody
diagram. Have a look at one of the five problems and see how it's done. Then try to do the
same thing for the remaining four problems.
Questions from students in previous years about this problem set
1. Why is the head loss equation ( f * (l/D) * (V^2/2g) ) valid for any fully developed, steady,
incompressible pipe flow (horizontal OR on a hill) when it was derived assuming that the
pipe is horizontal?
 We derived the relationship between head loss and pressure drop (which does depend on
whether the pipe is tilted)
 We derived the relationship between head loss and shear stress (which does depend on
whether the pipe is tilted)
 We did not derive the relationship between head loss and the average velocity, which is
the relationship you have written above. Instead, I just wrote it down and stated it was
true from dimensional analysis. Not very satisfying, I agree.
 However, the equation you wrote is true regardless of whether the pipe is tilted or not
because the velocity V in your equation (which should really be the average velocity) will
be different if the pipe is tilted.
2. Also, why is the change in pressure equation ( = f * (l/D) * (pV^2/2) ) only valid for
horizontal pipes?

Hmmm.... I don't have this equation in my notes. I have these two equations:
head_loss = (delta P)/(gamma) - (length of pipe)(sin theta)
head_loss = ( f * (length of pipe/D) * (V^2/2g) )

Equation 1 is just the mechanical energy equation for a pipe tilted at angle theta with no
shaft work, with average velocity 1 = average velocity 2. Equation 2 is the same as you
wrote in your first question, obtained from dimensional analysis.

If we combine equation 1 and equation 2 we obtain:
(delta P) = (rho)(f)(length of pipe/D)(V^2)/(2) - (rho)(g)(length of pipe)(sin theta)

Notice that the second term is just the change in pressure that would be expected due to
hydrostatics! Notice that the first term looks like kinetic energy -- (rho)(V^2)/2

So the equation you wrote seems to just be the first term, without accounting for the tilt
of the pipe in the second term. I'm not sure where you got it. But you can now see why it
only applies to horizontal pipes.
3. For choosing which V to use in the minor loss equation (for example in 8.61 and 8.62), I
know you should use the larger V, but I'm still confused on why. Also, is that always the
case?
Actually it's not that you should use the larger V. It has to do with whether it's a contraction or an
expansion. Look at figures 8.26 and 8.27
4. Why do 8.61 and 8.62 ignore major loss?
Because there is no information provided in the problem that would allow you to calculate major
loss. You would need to know the surface roughness of the pipe in order to calculate major loss,
and the problems don't provide that information.
5. Why does l = 14 ft in 8.86?
I think the solutions made a bad substitution. I would have substituted the value of 8 instead of
14.
6. In 8.93, why is V defined before the nozzle, instead of at the nozzle? (This doesn't fit the
rule of using the larger V/smaller D for the V for head loss).
I think they are actually using the larger velocity but they made some algebra errors. I don't see
how they subtracted off V^2/2g from the right side. You should use the higher velocity. The
second full paragraph on page 421 of the textbook indicates that one should use the downstream
(high speed) velocity for nozzle calculations.
7. Also in 8.93, where does the 4^2 term come from in the second-to-last step?
They solved to find that V2 = 4V1. They then square V2 to get the 4^2. However, I think there
are other algebra errors in the textbook's solution. See my response to question 6 above.
8. For 8.99, how do they know which /D curve to look at (for the trial and error method)?
It's labeled in the figure as 0.01.
9. I have a question regarding the use of the trial and error method via the Moody Chart.
In the solutions manual, the numbers obtained for f, the friction factor, seem really exact
(e.g. .04 and .039). I can't seem to ever get the same numbers as the solutions manual
It may help to use a ruler, but I agree, sometimes it is difficult to get the exact same answer that
the textbook does. Just be sure to choose the correct /D curve.
10. and the Colebrook equation is very difficult to solve, and impossible for
me to solve without the use of a calculator.
You can't solve the Colebrook equation analytically. You use the Moody chart to solve the
Colebrook equation
11. Also, the initial guess seems rather arbitrary. At least, I can't figure out why they chose
what they did. The solutions manual usually get it after trial #2, but I don't think that's
possible for me. How do you suggest we estimate out initial guess? Is there a better way
than just picking a random number?
The textbook is choosing a first guess for f so that it's in the middle of the range of values for f.
Remember that's one of the best binary search algorithms -- split a pool into two, then split the
remaining half in half again, etc....
12. In problem 8.27, how do I calculate the viscosity of the asphalt? (I understand it is
80,000 times that of water, but I don't think the value for viscosity of water in the solutions
matches that from the book)
They used the viscosity of water at 120 degrees, because the problem states that the asphalt is at
120 degrees. The viscosity of water at 120 degrees is 1.164 e -5.
13. In problem 8.43 where did the Power loss due to friction equation come from?
This problem is about a horizontal pipe whose diameter does not change. Since there is a
pressure drop across its length, this pressure drop can only be attributed to the loss of volume
flow rate due to friction. Now volume [m^3] flowing per unit time [s] multiplied by the pressure
drop [N/m^2] gives a power [N.m/s= Joules/s].
Also -- If you think a bit, you can also obtain this identical equation from the equations for
Q and the head loss on the first page of your "equation sheet"
14. Equivalent length equation
This is a pipe flow related problem. Please consult Ch 8, section 8.4 (The Moody Chart) and
Minor losses to see where these eqns. of equivalent length come from.