Crew Parker Lab 2 Report
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School
University of California, Irvine *
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Course
107
Subject
Aerospace Engineering
Date
Dec 6, 2023
Type
Pages
8
Uploaded by DukeResolve213
MAE 107 Lab Report 2
Fall 2023
Crew Parker ID: 14186406
Instructor: Dr. Y. Wang
TA: Cody
Lab 2 Report
Summary
To perform this lab, a pot of water is brought to a boil using a heating plate, as
appropriate safety personnel watch over the area. An aluminum sphere with an embedded
thermocouple is hung from a lab stand, and submerged in the boiling water. The
thermocouple allows us to monitor and graph the sphere’s temperature as it increases. This
data is later imported into google sheets to calculate and plot specific values. The sphere is
removed once it reaches 100 degrees Celsius, and hung back on its lab stand to cool
naturally in the ambient air. The sphere is then submerged in boiling water again, removed,
and placed into a wind tunnel with the flow set to a maximum to cool. This causes a forced
convection cooling process of the sphere. The whole process is then repeated with a brass
sphere cooled naturally and through forced convection. All data is recorded and plotted.
Cooling the spheres through different methods shows that forced convection cools
faster than natural convection. Additionally, using two different materials for the spheres
shows that thermal conductivity affects the rate at which materials gain or lose heat through
heat transfer processes. Our results ended up matching quite closely with those from the
textbook, which gave us confidence in our methods and instruments. The Solidworks
simulations of the same heating and cooling of the sphere actually produced a very similar
graph to that made up of the data we collected during the forced convection cooling process
in lab.
Crew Parker
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MAE 107 Lab Report 2
Fall 2023
Questions
Q2.) Plot T
s
versus t and ln θ versus t for each run. Find a way to use the curve of ln θ
versus t to calculate h.
a.
See
Graphs
b.
In order to use the curve of ln θ versus t to calculate h, we can select any two points on
the graph, find the slope between the points, and multiply that slope by -((m*C
v
)/A
s
).
Q3.) From the development of Equation 7 and the experimental design, what do you think
may contribute to the errors (not related to the exp. equipment) in calculating h? How to
minimize the errors?
Eq. 7
After rearranging the equation given in the textbook, it can be seen that errors, without
considering faulty lab equipment, can be made if one incorrectly measures the weight of the ball,
the size of the ball, or the amount of time it took for the ball to reach a certain temperature. It can
also come from one recording the data from the graph at incorrect times. The errors can be
minimized by making sure each of the measurements are correct as well as starting the recording
of data prior to allowing the ball to heat/removing the ball from the heated medium.
Q4.) Compute the Biot number for each run. Are the Biot numbers small enough to justify
the lumped capacity system assumption?
(Biot # Equation)
Crew Parker
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MAE 107 Lab Report 2
Fall 2023
k
Brass
= 111 W/m*K
k
Aluminum
= 237 W/m*K
A
Sphere
= 0.00202683 m
2
m
Sphere
= 0.58483 kg
h
avg
(brass, forced convection) = 0.001362 W/m
2
*K
h
avg
(brass, natural convection) = 0.0009633 W/m
2
*K
h
avg
(aluminum, natural convection) = 0.0003252 W/m
2
*K
Bi
Brass, Forced Convection
= 5.1868E-8 m
Bi
Brass, Natural Convection
= 3.6740E-8 m
Bi
Aluminum, Natural Convection
= 5.8084E-9 m
These values of Bi are small enough to justify the lumped capacity system assumption as they
are all < 0.1.
Q5.) Do you expect the heat transfer coefficients h to be different for the brass and
aluminum spheres under the same conditions? If yes, why? If no, why? Compare your
results for the brass and aluminum spheres and comment.
Utilizing the following relationship:
We can see that h is proportional to C
v
. Since C
v
is different for aluminum and brass under the
same conditions, h must also be different.
h
avg
(brass, natural convection) = 0.0009633 W/m
2
*K
h
avg
(aluminum, natural convection) = 0.0003252 W/m
2
*K
Q6.) Compare your values of h to those in heat-transfer references.
The textbook states that typical values of h for air are 30-100 W/m
2
*K for forced convection.
Our values of around 75.74 W/m
2
*K for forced convection lined up with the textbook’s
estimations pretty accurately, although the margin for success is quite wide.
Q7.) From the basic measurements, if using Equation 7 and any point (Ts, t) to calculate h,
estimate δh using δT and δt. Please comment on how to minimize δh.
h = 1.9589
dh can be minimized by increasing the precision of instruments used to measure time and
temperature. This is because these values are used to calculate h, and therefore determine the
accuracy/precision of the measurement of h.
Q8.). Use Solidworks to reconstruct a sphere and run the simulation and plot the
temperature profile change with time (T(r,t)). Use one data set of cooling to compare with
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MAE 107 Lab Report 2
Fall 2023
the analytical result (Equation 8) and simulation result (using the center temperature). For
simulation, reduce the thermal conductivity by 100, please replot T(r,t) and comment on
the result.
The temperature vs time graph produced by the cooling simulation in Solidworks very closely
matched our graphed data of experimental cooling. In terms of shape and values, it most closely
matched with the forced cooling graphs.
See
Graphs
Q1.) In the sample of experimental data, estimate the uncertainties in the measurements of
time and temperature.
The approximate uncertainty is calculated by dividing the most precise decimal place of the
measurement instrument by 2. Therefore, for the stopwatch, the uncertainty of the time
measurement is 0.0005 seconds, and, for the thermocouple, the temperature uncertainty would be
5 degrees C.
Q9.) The TES tank at the central plan of UC IRVINE is a large cylinder 105 feet high and
88 feet in diameter to store the excess chilled water. Estimate the heat loss from the tank for
winds of 0 and 20 m s−1. Take the surface of the tank to be 40 °F and the air temperature
to be 120 °F. For a 12 hour period, compare energy loss to “that” stored in the tank. Would
more insulation be worth it? Use the heat transfer coefficients from the experiments.
h = 105 ft
D = 88 ft
V = 0 (Natural), 20 m/s (Forced Conv.)
T
s
= 40
℉
= 277.594 K
T
∞
= 120
℉
= 322.039 K
t = 12 hr
Crew Parker
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MAE 107 Lab Report 2
Fall 2023
More insulation would be worth adding, as this would decrease the amount of heat lost in the
process and increase the efficiency of the cycle.
Q10.) Use ~30 data points around the time constant in your measurement of forced
convection for the brass sphere cooling to calculate h (~30). Plot the frequency distribution
of h (please choose a proper bin size). Make a comment on the Histogram
This histogram represents the heat transfer coefficients for forced convection with the Brass
sphere. Our calculated heat transfer coefficients are fairly spread out but most of our values lie
between 183 and 253.
Q11.) If the cooling starts at t=10 s, instead of 0 s, please rederive the solution to Equation
1.
→
→
→
𝛳 =
𝑇
?
−𝑇
∞
𝑇
𝑖
−𝑇
∞
?θ
θ
= ?(??𝜃) =−
ℎ𝐴
?
?𝐶
𝑣
??
𝛳 = 𝐶
1
?
−
ℎ𝐴
?
?
?𝐶
𝑣
(With
𝜃
=1 and t=10)
→
→
1 = 𝐶
1
?
−
10𝐴
?
ℎ
?𝐶
𝑣
𝐶
1
= ?
−
10𝐴
?
ℎ
?𝐶
𝑣
=
𝛳 =
𝑇
?
−𝑇
∞
𝑇
𝑖
−𝑇
∞
?
−
10𝐴
?
ℎ
?𝐶
𝑣
* ?
−
ℎ𝐴
?
?
?𝐶
𝑣
Crew Parker
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MAE 107 Lab Report 2
Fall 2023
Graphs
Q2.)
a.
Brass, Natural Convection:
b.
Brass, Forced Convection:
c.
Aluminum, Natural Convection:
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MAE 107 Lab Report 2
Fall 2023
d.
Aluminum, Forced Convection:
Q8.)
Solidworks Generated Graphs
Crew Parker
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MAE 107 Lab Report 2
Fall 2023
k = 234
k = 134
Sample Calculations
Q7.)
δℎ =
−?𝐶
𝑣
?𝐴
?
δ𝑇 +
?𝐶
𝑣
??(θ)
𝐴
?
?
2
δ?
Set dT=.05 and dt=.005
= 1.958936
δℎ =
−.586(385)
(295)(.00202)
(. 005) +
(>586)(385)(−1.16)
(.002)(295
2
)
(. 05)
References
“Practical Handbook of Thermal Fluid Science”. Yun Wang.
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