NOVEMBER 19, 2020
Momentum Deficit Behind a
Cylinder Laboratory
pressure difference is related to flow velocity
profile.
Miguel Luna , Alexander Corrin, Colby Crago, James
Wu, Jonathan Snyder, Peter Yannella
Department of Mechanical and Aerospace Engineering
Rutgers University, Piscataway, New Jersey 08854
A pitot-static probe and pressure transducer inside a
wind tunnel is used to measure a voltage and to
determine the velocity profile in a free stream with
and without a 0.5-inch diameter cylinder. The reason
of this is to determine the total drag force of the same
cylinder, but against a 2-mph water flow at 40 F. The
velocity profile is used in the integration of the total
drag force in the wind tunnel, which resulted in
0.05712 N, which lead to a drag coefficient of 0.58852.
This drag coefficient was used to determine the force
per unit length in the water scenario, which is 2.986
N/m. Using this calculated ratio, the total drag force
in the water scenario for the 0.5-inch diameter and 6inch long cylindrical rod is 17.916 N.
INTRODUCTION
This laboratory presents a hypothetical
scenario to present the problem statement. It
states that an underwater installation possesses
a long 0.5 inches that is subjected to a drag force
generated by a uniform water flow moving at 2
mi/h and 40 °F normal to its axis. The objective
of the experiment is to determine the drag
force/unit length using an 0.5-inch OD model
cylinder in a wind tunnel.
How can a wind tunnel simulate the water
flow around the cylinder? The flow can be
modeled using dimensional analysis. If the
Reynolds number for the rod and the cylinder
are the same, the wind flow will be like the
water flow.
A 6 by 6-inch low speed wind tunnel is
equipped with a static pressure tap and
manometer. The wind tunnel is set to a uniform
incoming flow and a pitot-static tube lets the
LABVIEW software to measure the dynamic
pressure. In other words, it uses a pressure
transducer to convert the difference between
two pressures into a voltage. This voltage is
related to pressure difference. Consequently, the
M IGUEL LUNA
Figure 1: Wind tunnel
Once the data is collected, the properties of
the flow can be used to calculate the drag force.
The momentum deficit calculated from the
velocity profile of the undisturbed free stream
profile and velocity profile of the cylinder. With
the calculated drag force, the wind tunnel results
will be related to the water flow scenario suing
the Reynolds number as shown in equation 1,
where 𝜌 is density, μ, dynamic viscosity; v,
kinetic viscosity; V, free stream velocity; D,
diameter of the cylinder.
𝜌𝑉𝐷 𝑉𝐷
[1]
𝑅𝑒 =
=
𝜇
𝑣
The data is acquired using the pressure
transducer. The recorded voltage corresponds to
a specific pressure at every location. LABVIEW
records 200 samples at every location. In those
locations, the average voltage is calculated
using the linear curve fitting plot using the
calibration data in Table 1.
Table 1: Pressure and Voltage Calibration Data
Pressure ‘’WC
0.00001
0.24999
0.50011
0.74999
1.00001
0.75001
0.50001
0.25002
-0.00001
Output VDC
0.999
2.000
2.999
3.998
4.998
3.998
2.999
2.000
0.999
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NOVEMBER 19, 2020
Once the pressure of every location is
calculated, the corresponding velocity will be
calculated, and the velocity profile determined
using the chart Figure 2. Using the velocity
profile, the drag force can be calculated using
equation 2, where ρ is density; w, width of the
cylinder; L, distance from the center of the
cylinder to the highest point of the
measurement; U1 is the free stream velocity and
u 3(y), velocity profile of the wake of the
cylinder.
𝐿
𝐹𝐷 = ρw ∫ 𝑢3 (y)[𝑈1 − 𝑢3 (y )]d𝑦 [2]
−𝐿
Next is the calculation of the drag
coefficient as shown in equation 3, where A is
the projected area of the cylinder.
𝐹𝐷
[3]
𝐶𝐷 =
1
2
𝜌𝐴𝑉
2
The coefficient drag can be compared with
the coefficient drag found in the literature get
the percent error.
Then, the force per unit length is then
calculated for the water flow scenario.
RESULTS AND DISCUSSION
The first part of the experiment involves the
derivation of the drag force on the model
cylinder. The basics laws that apply in the
system are conservation of mass, Newton’s
second law, angular momentum principle, and
the first and second laws of thermodynamics.
For this system, mass, M, is constant
throughout the system. Therefore,
𝑑𝑀
= 0 [4]
𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
Where
𝑀𝑠𝑦𝑠𝑡𝑒𝑚 = ∫
𝑑𝑚 = ∫
𝑀(𝑠𝑦𝑠𝑡𝑒𝑚)
Where the linear momentum of the system is
𝑃⃗⃗𝑠𝑦𝑠𝑡𝑒𝑚 = ∫
𝑉⃗⃗𝑑𝑚 = ∫
𝑀(𝑠𝑦𝑠𝑡𝑒𝑚)
𝑉⃗⃗ × 𝜌𝑑 V [7]
𝑉(𝑠𝑦𝑠𝑡𝑒𝑚)
The next step is to convert the system rate
equation into an equivalent. The derivation is
long and is beyond the scope of the report.
Therefore, equation 8 will let us convert an
extensive property N to a formulation to use
with a control volume where η is an intensive
property (i.e. per unit mass).
𝑑𝑁
𝜕
⃗⃗ ∙ 𝑑𝐴
⃗⃗ [8]
= ∫ 𝜂𝜌𝑑V + ∫ 𝜂𝜌𝑉
𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚 𝜕𝑡 𝐶𝑉
𝐶𝑆
A useful case is to assume that the density is
constant (incompressible fluid), and to consider
that the control volume is nondeformable (i.e. V
is constant). Relating equation 8, with the
previous assumptions, equation 9 is obtained
which shows the conservation of mass through
a control volume.
∫ 𝜌⃗𝑉⃗ ∙ 𝑑 ⃗𝐴⃗ = 0 [9]
𝐶𝑆
In the same way, equation 8 is used to obtain
the nonaccelerating control volume form of
Newton’s second law as shown in equation 10
in vector form, where 𝐹⃗𝑆 is surface force and 𝐹⃗𝐵
is body forces, usually gravity.
𝐹⃗ = 𝐹⃗𝑆 + 𝐹⃗𝐵 =
𝜕
∫ 𝑉⃗⃗𝜌𝑑V + ∫ 𝑉⃗⃗𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ [10]
𝜕𝑡 𝐶𝑉
𝐶𝑆
Writing the expression for the x-component
(which is the direction of the drag force),
equation 11 is obtained.
𝜕
𝐹⃗𝑥 = ∫ 𝑢 𝜌𝑑V + ∫ 𝑢 𝜌 ⃗𝑉⃗ ∙ 𝑑⃗𝐴⃗ [11]
𝜕𝑡 𝐶𝑉
𝐶𝑆
Next is to decide for a control volume.
Figure 2 shows the control volume that will be
used.
𝜌𝑑V [5]
𝑉(𝑠𝑦𝑠𝑡𝑒𝑚)
Newton’s second law states that the sum of
all external forces acting on a system is equal to
the time rate of change of linear momentum.
𝑑𝑃⃗⃗
[6]
𝐹⃗𝑠𝑦𝑠𝑡𝑒𝑚 =
𝑑𝑡 𝑠𝑦𝑠𝑡𝑒𝑚
Figure 2: Drag Measurement of an Immersed Body
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NOVEMBER 19, 2020
The following assumptions are steady flow
(rate of change with respect to time is zero),
incompressible flow and static pressure is equal
around the 2D control volume. Applying
continuity equation as in equation 9.
∫ 𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ + ∫ 𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ + ∫ 𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ + ∫ 𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ = 0
𝐴𝐵
𝐵𝐶
𝐶𝐷
𝐴𝐷
If we assume that boundaries AD and BC
are far enough that velocity is the same as in the
free stream flow, then the mass flow rate of BC
and AD is equal to the sum of the mass flow rate
of AB and CD.
∫ 𝜌𝑉⃗⃗ ∙ 𝑑𝐴⃗⃗ + ∫ 𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ = 𝑚̇ 𝐵𝐶 + 𝑚̇ 𝐴𝐷
𝐵𝐶
𝐴𝐷
𝐿
∫ 𝜌𝑉⃗⃗ ∙ 𝑑𝐴⃗⃗ + ∫ 𝜌𝑉⃗⃗ ∙ 𝑑 𝐴⃗⃗ = 𝜌 ∫ (𝑈 − 𝑢)𝑤𝑑𝑦
𝐵𝐶
𝐴𝐷
−𝐿
Using Newton’s second law and assume the
body force are zero, where the negative sign is
to consider that this is the force that the cylinder
will have to resist:
−𝐹 = 𝑈𝑚̇ 𝐴𝐵 + 𝑈 𝑚̇ 𝐵𝐶 + 𝑈𝑚̇ 𝐴𝐷 + 𝑢𝑚̇ 𝐶𝐷
𝐿
𝐿
𝐿
−𝐹 = 𝑈 (∫ −𝜌𝑈𝑤𝑑𝑦) + 𝑈 (𝜌 ∫ (𝑈 − 𝑢)𝑤𝑑𝑦) + 𝑢 (∫ − 𝜌𝑢𝑤𝑑𝑦)
−𝐿
−𝐿
𝐿
−𝐿
−𝐹 = 𝜌𝑤 ∫ [−𝑈 2 + 𝑈 (𝑈 − 𝑢) + 𝑢2 ]𝑑𝑦
−𝐿
Taking into consideration that the width is
constant, it also remains outside the integral.
𝐿
𝐹 = 𝜌𝑤 ∫ 𝑢(𝑈 − 𝑢) 𝑑𝑦 [12]
−𝐿
As mentioned in the introduction, L
represents the distance from the center of the
cylinder.
Using the calibration data in Table 1, the
calibration curve is generated and shown in
figure 3 in the appendix. The linear function of
the calibration data is shown in equation 13.
𝑃 = 0.2501 × 𝑉 − 0.25 [13]
The calculation of the drag force requires
the velocity profile data. Using the acquired
data, the average voltage is used with the fit
curve to determine pressure. Next, the graphs
shown in figure 4 lets us determine the value of
the velocity in feet per minute. Finally, the
velocity is then converted to meters per second.
Tables 2 and 3 shows the velocity profile for
each location at the appendix.
M IGUEL LUNA
Assuming that the temperature of the air was
70 F, the density of air is 1.199 kg/m 3. Using the
velocity profile for the points 1.60 in to 3.20 in,
the resultant drag force is 0.057129 N.
Next, using the coefficient equation and the
calculated drag force, the drag coefficient for
the cylinder in the wind tunnel is 0.58852.
The reasoning behind only the location
between 1.60 and 3.20 in is because from that
point the average velocity is started to decrease
caused by the cylinder and the average velocity
is lower than the average free stream velocity.
According to the experimental data from
incompressible flow, Wiley Inter-science, the
drag coefficient for a Reynold’s number of 8000
(which is calculated using the given conditions)
should be approximately 1. The difference
between calculated drag coefficient and the
theoretical drag coefficient could be caused by
the non-consistent free stream flow. The data
recorded from the stream flow with cylinder on
average recorded higher values than free stream
velocity.
The force per unit length for the rod by the
2-mph water stream is calculated using the
previous drag
coefficient with
the
corresponding initial conditions of the water
flow. The density of water is 999.95 kg/m3 at 40
F. The force per unit length calculated using the
calculated drag coefficient is 2.986 N/m.
Table 4 shows the % error of the calculation
of the calculation of the drag force in the wind
tunnel and estimation of the drag force in the
water. It would be assumed that both rods have
the same length of six inches.
Table 4: Percent Error of Calculated Drag force for the 6inch rod
Air
Water
Experimental
0.05712
17.916
Theoretical
0.09707
30.4551
% Error
41.15
41.17
The error of the calculated drag coefficient
propagated to the water model, which leads to
approximately the same percent error in the
water model. It shows that the non-dimensional
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NOVEMBER 19, 2020
analysis carries over to a different scenario with
the same Reynolds number.
CONCLUSIONS
To find the total drag force of the water flow, it
was needed simulate a scenario where the
Reynolds number is the same. The calibration
curve shows is represented by the linear
function.
𝑃 = 0.2501 × 𝑉 − 0.25
Using the fitting curve, the velocity profile can
be determined. Adding up all the values for the
different sections, the total drag force can be
calculated, which 0.05712 N. This leads to the
drag coefficient 0.5882. If this value is used to
determine the drag force of the same-length rod
in the water flow, the total drag force in the
water is 17.916 N. However, if instead of the
experimental drag coefficient is used the
theoretical coefficient drag, there is a difference
of 41%. This difference could be caused of the
lack of consistent stream velocity and that the
pitot tube itself may affected the recorded data,
although it is unlikely.
REFERENCES
[1] Momentum Deficit Laboratory Manual
[2] Aerodynamics for Students, Aerospace,
Mechanical & Mechatronic Engg. 2005
University
of
Sydney,
http://wwwmdp.eng.cam.ac.uk/
[3] Experimental data from Incompressible
Flow, Wiley-Interscience, New York (1984)
[4] Pritchard, Philip J., and John W.
Mitchell. Fox and McDonald’s Introduction
to Fluid Mechanics. 9th ed., John Wiley &
Sons, 2015
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NOVEMBER 19, 2020
Appendix
Table 2: Air Velocity with No Cylinder
Table 3: Air Velocity with Cylinder
Height
(in)
Avg.
Voltage
Avg.
Pressure
Velocity
(m/s)
Height
(in)
Avg.
Voltage
Avg.
Pressure
Velocity
(m/s)
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
1.7943
1.8061
1.7998
1.8096
1.8052
1.8053
1.8050
1.8111
1.8067
1.8071
1.8064
1.8129
1.8038
1.8046
1.8056
1.8128
1.8100
1.8103
1.8122
1.8137
1.8176
0.1988
0.2017
0.2002
0.2026
0.2015
0.2015
0.2015
0.2030
0.2019
0.2020
0.2018
0.2034
0.2012
0.2014
0.2016
0.2034
0.2027
0.2028
0.2033
0.2036
0.2046
9.0706
9.1378
9.1024
9.1576
9.1330
9.1334
9.1319
9.1659
9.1411
9.1433
9.1396
9.1766
9.1252
9.1297
9.1352
9.1760
9.1600
9.1617
9.1724
9.1811
9.2027
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
1.8450
1.8638
1.8611
1.8630
1.8571
1.8570
1.8571
1.8257
1.8035
1.7782
1.6584
1.5850
1.5328
1.4937
1.6353
1.7071
1.7812
1.8236
1.8493
1.8503
1.8573
0.2114
0.2162
0.2155
0.2160
0.2145
0.2145
0.2145
0.2066
0.2011
0.1948
0.1648
0.1464
0.1334
0.1236
0.1590
0.1770
0.1955
0.2061
0.2125
0.2128
0.2145
9.3555
9.4592
9.4445
9.4548
9.4227
9.4219
9.4224
9.2482
9.1231
8.9787
8.2588
7.7855
7.4300
7.1527
8.1126
8.5585
8.9956
9.2365
9.3794
9.3848
9.4232
Sample Calculation
Drag Force Calculation
For height=1.60 in
𝑤 = 6 𝑖𝑛 = 0.1524 𝑚
𝑑𝑦 = 0.2 𝑖𝑛 = 0.00508 𝑚
𝜌 = 1.199 𝑘𝑔/𝑚3
𝐹𝑦=1.60 𝑖𝑛 = 𝜌𝑤 𝑢𝑦=1.60 𝑖𝑛 (𝑈𝑦=1.60 𝑖𝑛 − 𝑢𝑦=1.60 𝑖𝑛 ) 𝑑𝑦
𝐹𝑦=1.60 𝑖𝑛 = 1.199 × 0.1524 × 9.1411 (9.1411 − 9.1231)(0.00508)
𝐹𝑦=1.60 𝑖𝑛 = 0.0001524 𝑁
Drag Coefficient Calculation
𝐴 = 𝑤𝐷 = 0.001935 𝑚2
𝐹𝐷 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑟𝑎𝑔 𝑓𝑜𝑟𝑐𝑒
𝑉 = 𝑓𝑟𝑒𝑒 𝑠𝑡𝑟𝑒𝑎𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐹𝐷
0.05713
𝐶𝐷 =
=
= 0.5885
1
1
𝜌𝐴𝑉 2
(1.199)(0.001935)9.14742
2
2
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Figure 3: MATLAB Curve Fitting Tool
Figure 4: Voltage vs Pressure Fitting Curve
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