UNIVERSITY OF SOUTHERN CALIFORNIA

Viterbi School of Engineering

FLUID MECHANICS

TEAM LEARNING ASSIGMENT

Marcus Vinícius Malta

Matheus Afonso Sofia da Silva

Los Angeles, CA, United States of America

December 2015

1. First Experiment

Goal: The goal for our first experiment was to determine the velocity that the water leaves

the nozzle in order to reach the height it reaches. We will ultimately use the Bernoulli Equation.

Steps: This was a simple one for us. We simply had to use a measuring tape to record the

height the water reaches. We measured from the point in which the water leaves the nozzle to the

point it reaches its maximum height and starts to drop. When we apply the Bernoulli Equation we

know that both the pressure as soon as the water leaves the nozzle and the pressure when the

water reaches its maximum height is zero due to the fact that it is open to atmosphere. We used

our initial point to be the tip of the nozzle, so Z1 had to be zero as well. Finally, the velocity of the

water at its highest point is zero due to the “parabolic” path it has, and we know that in a parabolic

motion, the velocity at the maximum height is always zero.

Figure 1 - Tutor Hall Fountain

Applying the Bernoulli Equation from the tip of the nozzle (1) to the maximum height that the

water reaches (2) we get:

𝑃!𝛾 +

𝑉!!

2𝑔 + 𝑍! =𝑃!𝛾 +

𝑉!!

2𝑔 + 𝑍!

𝑃! = 𝑍! = 𝑃! = 𝑉! = 0

𝑉! = 2𝑔𝑍!

Figure 2 - Measurements

𝑍! = 40  𝑐𝑚 = 0.4  𝑚

𝑉! = 2 ∗ 9.81 ∗ 0.4 = 2.80  𝑚/𝑠

Figure 3 - Fountain at Seely Wintersmith Mudd Hall of Philosophy

𝐹! = 𝛾!"#$"% ∗  𝑉!"#$%&'%(

𝑊 = 𝑚𝑔

The force due to the displaced fluid is equal to the weight of the displaced fluid.

𝑊 = 𝐹!

𝑚𝑔 = 𝛾!"#$% ∗  𝑉!"#$%&'(!

𝑉!"#$%&'(! =𝑚𝑔𝛾!"#$%

2. Second Experiment

Goal: Our goal with this second experiment was to determine the volume of water

displaced by the chlorine dispenser that is placed inside the fountain. We will be using the

Buoyance Equation.

Steps: This was also a simple experiment for us regarding the measurements we had to

take. In order for us to be able to use the Buoyance Equation, we simply needed to know the

weight of the chlorine dispenser. To measure the weight of the chlorine dispenser, we used a

scale, since the scale we used was a bathroom scale, it was not very sensitive, so it was not able

to measure the mass of the dispenser simply by placing it on top of the scale. To actually be able

to know the mass of the dispenser, we had to measure our mass, and then measure our mass

holding the dispenser. This would give us a difference in mass and the difference would therefore

be the mass of the chlorine dispenser. After knowing the weight of the chlorine dispenser, we were

easily able to use the Buoyance Equation since we knew the other unknowns.

Figure 5 - Reference mass and object mass

𝑚 = 90.2− 89.7 = 0.5  𝑘𝑔

𝑉!"#$%&'(! =0.5 ∗ 9.819800 = 5.005×10!!  𝑚! = 0.5005  𝐿

Figure 6 - Set up

Figure 4 - Reference mass

3. Third Experiment

Goal: Our goal with the third and last experiment was to determine the force the water exerts

on the bottom pipes that feed water into the circular pipe with all the nozzles so that the water can exit

through the various nozzles and reach this given height. We used several principals during this last

experiment to reach our final answer, but we were ultimately using the Moment Equation.

Steps: This was definitely the longest and most complicated experiment out of the three. There

were several measurements we had to take as well as a lot of calculations to reach the final

conclusion. To make it easier, instead of highlighting the steps at the beginning as we did with the

previous two experiments, we went through the steps, measurements and equations as we followed

through the experiment.

Figure 7 - Fountain at Andrus Gerontology Center

Applying the Bernoulli Equation from the tip of one nozzle (1) to the maximum height that the

water reaches (2) we get:

𝑃!𝛾 +

𝑉!!

2𝑔 + 𝑍! =𝑃!𝛾 +

𝑉!!

2𝑔 + 𝑍!

𝑃! = 𝑍! = 𝑃! = 𝑉! = 0

𝑉!"##$% = 2𝑔𝑍!

Figure 8 - Measurement

𝑍! = 61  𝑐𝑚 = 0.61  𝑚

𝑉!"##$% = 2 ∗ 9.81 ∗ 0.61 = 3.4595  𝑚/𝑠

The flow rate of the nozzle is:

𝑄!"##$% = 𝑉!"##$%𝐴!

Figure 9 - Measurement

The diameter of the nozzle is 𝑑 = 1.6  𝑐𝑚.

𝑄!"##$% = 3.4595 ∗𝜋4  0.016

! = 6.9557×10!!  𝑚!/𝑠

There are 38 nozzles on the circular pipe so the total flow rate is:

𝑄!"!#$ = 38 ∗ 𝑄!"##$% = 38 ∗ 6.9557×10!! = 26.4318×10!!  𝑚!/𝑠

We can say that the mass coming through the two pipes on the base (1) that feed the circular

pipe needs to be the same as the mass getting out through all the nozzles (2), so by conservation of

mass we have:

𝑚! = 𝑚!

𝜌𝑄!"#$# = 𝜌𝑄!"!#$

𝐵𝑢𝑡:      𝑄!"#$ =𝑄!"#$#2

𝑄!"#$ =𝑄!"!#$2 =

26.4318×10!!

2 = 13.2159×10!!  𝑚!/𝑠  

Figure 10 - Measurement

Knowing that the diameter of the circular pipe and the diameter of the base pipe are:

𝐷 = 13.5  𝑐𝑚

𝑄!"#$ = 𝑉!𝐴!

𝑉! =𝑄!"#$𝐴!

=13.2159×10!!𝜋4  0.135

!= 0.9233  𝑚/𝑠

Assuming that that there is a streamline going from the base pipe (1) to the nozzle (2), we can

write:

𝑃!𝛾 +

𝑉!!

2𝑔 + 𝑍! =𝑃!𝛾 +

𝑉!!

2𝑔 + 𝑍!

𝑍! = 𝑃! = 0

𝑃! = 𝛾!"#$%𝑉!"##$%!

2𝑔 + 𝑍! −𝑉!!

2𝑔

𝑃! = 98003.4595!

2 ∗ 9.81 + 0.08−0.9233!

2 ∗ 9.81 = 6.3362  𝑘𝑃𝑎

𝐹! = 𝑃!𝐴! = 6.3362  ×10! ∗ !!  0.135! = 90.6951  𝑁

Figure 11 - CV chart

𝜕𝜕𝑡 𝜌𝑉  𝑑𝑉

!"+ 𝜌𝑉 𝑉 ⋅ 𝑛  𝑑𝐴

!"= 𝐹!"

Assume the flow is steady.

Since this is a symmetric problem, there are no net forces on the horizontal direction, so we

write the momentum equation for the vertical direction:

−𝜌 𝑉! !  𝑉!  𝐴! = 𝑅! + 𝐹!

−999 ∗ 0.9233 ∗ 0.9233 ∗𝜋4  0.135

! = 𝑅! + 90.6951  

𝑅! = −102.8852  𝑁

The reaction force found is the force that the pipe exerts on the set of pipes, this means that

the force that the water exerts on pipes is in the opposite direction, so the reaction force on the pipes

is approximately 103  𝑁 pointing upwards.