University of Alberta
Department of Mechanical Engineering
MecE 230 Seminar Pack
Compiled by Morris R. Flynn, Larry W. Kostiuk and Andrew Martin
1
MEC E 230
Seminar A, Designing the HVAC system on a 30-passenger bus
Objective: To apply the concepts of control volumes and mass/energy conservation in the conceptual
design of a real world mechanical system.
Background: In designing the HVAC system on a bus, it is worth noting that the bus can go anywhere
in the world. Therefore the local environmental conditions that play a role in human comfort, such as
the ambient temperature, T∞ , are highly variable. By contrast, we wish to keep the cabin of the bus at
a constant temperature of, say, 20◦ C.
Questions:
1. [2 points] By considering a balance of chemical species, draw and label a control volume (CV)
that is appropriate for designing only the bus air exchange system. Thus you should account for
inflows, outflows, sources and sinks of O2 , CO2 , etc. for the open system that is the bus cabin.
2. [1 point] Assuming the air inside bus is fully mixed, is there any difference between the interior
conditions vs. the conditions measured at the outlet? (Yes/No)
3. [2 points] By considering energy conservation, draw and label a CV that is appropriate for
designing only the cabin cooling system. (Cooling is necessary whenever Tinterior > 20◦ C.) Include
the thermal contribution of the people on the bus, the bus motor, the A/C system and the
environment (including the Sun).
4. [1 point] The design process typically considers, within reason, worst-case scenario(s). In this
spirit, is it preferable when designing the cabin cooling system to assume cloudy or sunny conditions? An air exchange fan that is on or off?
5. [2 points] By considering energy conservation, draw and label a CV that is appropriate for
designing only the cabin heating system. (Heating is necessary whenever Tinterior < 20◦ C.) Include
the thermal contribution of the people on the bus, the bus motor, the heating system and the
environment (including the Sun).
6. [1 point] Is it preferable when designing the cabin heating system to assume cloudy or sunny
conditions? An air exchange fan that is on or off?
2
MEC E 230
Seminar B, Mystery control volumes and balance equations
Objective: To apply the concepts of control volumes in identifying particular devices or pieces of
mechanical equipment.
Background: Different devices or pieces of mechanical equipment have different associated flows of
mass and energy. In this seminar, we will gain valuable practice in matching such devices/pieces of
equipment to their corresponding control volume (CV) diagrams.
Directions: For each of the CVs shown below, select from the following list that particular device
or piece of mechanical equipment that is most likely operating inside that CV. Then write out expressions for conservation of mass and energy. In the diagrams below, arrows indicate the direction of mass
or energy flow into or out of the CVs.
Grading: 1 point per diagram.
Possible devices or mechanical equipment:
Electric air compressor
Heat exchanger
Balloon
Electric kitchen mixer
Boiler
Piston-cylinder
Rocket
Electric generator
Steam turbine
Diagram A: Device or equipment:
Diagram B: Device or equipment:
3
Diagram C: Device or equipment:
Diagram D: Device or equipment:
Diagram E: Device or equipment:
Diagram F: Device or equipment:
4
MEC E 230
Seminar C, Revisiting J.P. Joule’s work
Objective: To apply the concepts of energy conservation in examining the conversion from work to
heat.
Background: James Prescott Joule (1818-1889) was a pioneer in the field of thermodynamics and
is credited for advancing the idea that work can be converted into heat. A schematic of one of Joule’s
more famous experiments is given below. In this experiment, translational work associated with a falling
weight is converted into rotational shaft work, which is, in turn, turned into heat serving to increase
the temperature of the fluid inside the cavity. Joule sought to understand the relationship between the
work applied to the fluid and the corresponding rise of fluid temperature.
Figure 1: A cartoon schematic of Joule’s experiment. Image source http://javiciencias.blogspot.ca/2014/04/energy-heat-and-work.html
Equipment/materials:
Insulated cavity
Thermometer
Drill
Corn syrup
Rotor
Paper towels
Energy meter
Funnel
Data:
• Corn syrup: cv = 3000 J/(kg K), ρ = 1380 kg/m3
• Water: cv = 4200 J/(kg K), ρ = 998 kg/m3
5
Stopwatch
Time (s)
0
25
50
75
100
125
150
Temperature (◦ C)
• Corn syrup volume (L) =
• Voltage (V) =
• Average electrical current (A) =
Questions:
1. [2 points] Assume a 5 kg weight falling from a height of 12 m. The weight is dropped, causing
the paddlewheel to spin in 2.5 L of water. When the weight reaches the ground, the paddlewheel
is disengaged from the spool and the weight is wound back up to 12 m. The paddlewheel is then
reattached and the weight is dropped again. Suppose that the weight is dropped a total of 16
times. What will be the increase of the water temperature? Neglect sidewall heat loss and assume
a perfect conversion of work to heat.
2. [1 point] Repeat the above calculation but now suppose that the working fluid is corn syrup, not
water.
3. [3 points] Using the data collected during the demonstration, plot ∆T (vertical axis) vs. time, t
(horizontal axis). From the slope of the associated best fit line, estimate the rate of working on
the fluid, measured in Watts. How does this value compare to the electrical work supplied to the
drill? List one loss in the conversion of electrical work to rotational shaft work and one loss in the
conversion of rotational shaft work to heat.
6
MEC E 230
Seminar D, Heating and cooling by convection and conduction
Objective: To study cases where the lumped system approximation does and does not apply.
Background: In class, we learned that the lumped system approximation can be assumed only when
the thermal resistance due to conduction is much less than that for convection. Because the thermal resistance due to conduction depends on the thermal conductivity, k, however, the validity of the lumped
system approximation strongly depends on the material of construction.
Equipment/materials1 :
Large container
Thermometer
Ice
Thermocouples
Copper sphere
Thermocouple display
Plastic sphere
Paper towels
Data:
• Copper: k = 400 W/(m·K), cv = 385 J/(kg·K), ρ = 8960 kg/m3
• Plastic: k = 0.4 W/(m·K), cv = 1500 J/(kg·K), ρ = 900 kg/m3
• Each sphere has a volume of 6.5 × 10−5 m3
Results (copper sphere – cooling/heating):
Time (s)
1
Text (◦ C)
Tint (◦ C)
Time (s)
Text (◦ C)
Thermocouple display: Agilent, U1242A, True RMS Multimeter.
7
Tint (◦ C)
Stopwatch
Results (plastic sphere – cooling/heating):
Time (s)
Text (◦ C)
Tint (◦ C)
Time (s)
Text (◦ C)
Tint (◦ C)
Questions:
1. [1 point] How much heat must be removed from the copper sphere to reduce its temperature
from 20◦ C to 0◦ C? Repeat your calculation but now consider the plastic sphere. (Report your
answers as positive numbers).
2. [2 points] For each sphere and for both the cooling and the heating phases, draw curves showing
Text and Tint vs. time. Thus you should produce four figures total, each having two curves. (Be
careful not to spend too much time on this part).
3. [1 point] Why does the copper sphere cool/heat faster than its plastic counterpart?
4. [1 point] Why, for both spheres, does cooling occur more quickly than heating?
5. [1 point] For which sphere does the lumped system approximation more reasonably apply?
8
MEC E 230
Seminar E, Viscosity of fluids
Objective: To use Stokes’ settling law to measure the dynamic viscosity, µ, of a series of glycerol-water
solutions.
Background: In 1851, George Gabriel Stokes2 showed that the drag force exerted on a sphere of
radius a falling at a terminal speed, U , through an infinite, viscous fluid is D = 6πµaU . By balancing
this drag force with gravity (which pulls the sphere downwards) and buoyancy (which pushes the sphere
upwards), one can derive either (i) an expression for U provided the dynamic viscosity, µ, is known, or,
(ii) an expression for µ provided the terminal velocity, U , is known. Today, we will concern ourselves
with the latter alternative and thereby estimate µ for a series of glycerol-water solutions by measuring U .
Equipment/materials:
Graduated cylinder(s) or test tube(s)
Three glycerol-water solutions
Funnel
Paper towels
Tweezers
Glass spheres
Stopwatch
Ruler
Data:
• Sphere diameter: 1 mm; sphere density: 2.5 g/cm3
• Glycerol-water solution #1 Glycerol mass fraction = 87.0%; density, ρ = 1.227 g/cm3 ; kinematic
viscosity, ν = 1.071 cm2 /s
• Glycerol-water solution #2 Glycerol mass fraction = 74.4%; density, ρ = 1.193 g/cm3 ; kinematic
viscosity, ν = 2.968 × 10−1 cm2 /s
• Glycerol-water solution #3 Glycerol mass fraction = 57.1%; density, ρ = 1.145 g/cm3 ; kinematic
viscosity, ν = 7.861 × 10−2 cm2 /s
Solution
1
1
2
2
3
3
2
Time of descent (s)
Distance travelled (cm)
This is the same Stokes of Stokes’ Theorem fame.
9
Settling speed (cm/s)
Questions:
1. [1 point] Draw a free body diagram showing the three forces (drag, gravity and buoyancy) acting
on a sphere falling at its terminal velocity.
2. [2 points] By balancing the forces from Question 1, derive the following expression for µ:
µ=
2a2 g
(ρs − ρ) ,
9U
(1)
where ρs and ρ are the respective densities of the sphere and ambient fluid. Hint: As we will learn
later in this course and as Archimedes demonstrated over two thousand years ago, the buoyancy
force is given by Fb = V gρ where V is the volume of the sphere.
3. [2 points] Based on your measured data, estimate µ for each glycerol-water solution. How do your
measurements compare with the data from the previous page? (Recall that the kinematic and
dynamic viscosities are related via ν = µ/ρ.) Use the following formula to compute a percentage
error for each of your three solutions:
Percentage error =
|µeqn. 1 − µdata table |
× 100% .
µdata table
(2)
4. [1 point] Stokes’ equation applies for a sphere falling through an infinite ambient. Is this condition
satisfied here? What influence might be exerted by the side-walls of the graduated cylinder or
test tube?
10
MEC E 230
Seminar F, Surface tension of fluids
Seminar under development
11
MEC E 230
Seminar G, The coiled manometer
Objective: To use our knowledge of hydrostatics to explain counterintuitive flow behavior in a coiled
manometer.
Equipment/materials:
Circular plastic bucket
Electrical tape
Cardboard tube
1 L beaker or Erlenmeyer flask
Clear plastic tubing
Food coloring
Funnel
The plastic tubing should be wrapped around the outside of the bucket/tube at least five times and
should be affixed to the bucket/tube using the electrical tape.
Observations: We expect the fluid poured into the (elevated) funnel to drain out the other end of the
tubing. This expectation is realized when using the cardboard tube, but not the circular plastic bucket
where, instead, the fluid overflows the funnel. The figure, discussion and questions below will help to
explain this unexpected outcome.
Figure 2: A manometer with tubing in the shape of a sinusoid.
Discussion: Wrapping the tubbing around the bucket/tube forms, in essence, a sinusoidal shape as
shown in figure 2. When the second loop begins to fill, the air path from the first loop to the outlet of
the tubing becomes blocked off. Moreover the pressure of the air bubble that appears between points b
and c becomes larger than atmospheric pressure, patm . (For simplicity, we assume that patm = 0 below;
we further assume that ρair = 0). As a consequence Hd > Hc where, for instance, Hc indicates the
height of point c, measured relative to the floor, say. If any water flows into the third loop, the process
is repeated. Eventually, this buildup of pressure causes the water to back up in the first loop and flow
out of the funnel.
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Questions:
1. [1 point] If pa = patm = 0, what is the pressure at point b, pb ? Assume that the elevations of
points a and b are Ha and Hb , respectively.
2. [2 points] Derive two expressions for pc , one by relating pc to pb and the other by relating pc to
pd = patm = 0.
3. [1 point] Using your results from Questions 1 and 2, what is the relationship between Ha − Hb
and Hd − Hc ?
4. [1 point] On the basis of your calculations, which condition is associated with overflow?
(a) Ha − Hb > Hd − Hc
(b) Ha − Hb < Hd − Hc
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MEC E 230
Seminar H, Buoyancy and floating sports balls
Objective: To estimate the density of common sports balls using principles of buoyancy.
Background: By measuring the depth of submersion of an object floating at a free surface, it is
straightforward to estimate the density of said object. Here we apply this idea in computing the density
of three familiar sports balls.
Equipment/materials:
Golf ball
Small aquarium
Ping pong ball
Ruler
Tennis ball
Paper towels
Data:
Sports ball
Golf ball
Ping pong ball
Tennis ball
Mass (g)
44.28
2.32
58.59
Diameter (inches)
1.685
1.480
2.534
Depth of submersion (cm)
Fluid 1
V1
V2
Fluid 2
Figure 3: A solid object of density ρ0 and total volume V0 = V1 + V2 floating at the interface between
two fluids of densities ρ1 and ρ2 .
Questions:
1. [1 point] The buoyancy force acting on the floating solid object of figure 3 is FB = (ρ1 V1 + ρ2 V2 )g
where the volumes V1 and V2 are defined above. By balancing FB against the corresponding
gravitational force, FG , show that
ρ1 V1 + ρ2 V2
ρ0 =
.
(3)
V0
2. [3 points] In the event that the lower and upper fluids of figure 3 are water and air, respectively,
we regard the upper layer density to be vanishingly small, i.e. ρ1 ' 0. In this limit,
ρ0 ' ρ2
14
V2
.
V0
(4)
Estimate V2 /V0 for each of the three sports balls3 . Then, assuming ρ2 = 0.9982 g/cm3 , estimate
ρ0 for each of the three sports balls also.
3. [1.5 points] Compute the density of each of the three sports balls using the data provided in
tabular form above. How do these values compare to your results from Question 2? In each
case, calculate a percentage error using the following formula:
Percentage error =
|ρeqn. 1 − ρdata table |
× 100% .
ρdata table
(5)
4. [0.5 points] From Question 3, we note that the percentage error can be large in the case of
objects that are small and light. Which of the following fluid properties is most likely responsible
for this discrepancy: surface tension, dynamic viscosity or kinematic viscosity?
3
Hint: See http://keisan.casio.com/exec/system/1223382199, http://www.onlineconversion.com/object volume
partial sphere.htm and/or http://mathworld.wolfram.com/SphericalCap.html.
15
MEC E 230
Seminar I, Density of fluids
Objective: To examine the influence of density differences in the context of density-driven exchange
flow.
Background: In Seminar E, we derived an expression for the Stokes settling speed by balancing
forces. Today, we will consider an example of density-driven flow and derive an expression for speed
based on a balance of energy, rather than a balance of forces. Density-driven flow occurs in very many
natural and industrial settings and describes, for instance, the draft of cold air that rushes into your
home when you open your front door in winter months. Here we will study the manner in which density
differences set the front speed of a density-driven flow. Our analysis will consider data collected from a
numerical simulation rather than from a bench-scale experiment. (You will learn more about numerical
modeling later on e.g. in MecE 390).
(a)
(b)
U
ρ2
ρ1
b
ρ1
H
ρ2
b
U
Figure 4: Density-driven exchange flow. (a) The initial instant, t = 0. (b) t = ∆t > 0.
Questions:
1. [2 points] The above figures show the evolution of a density-driven exchange flow where the
(constant) front speed of the upper and lower gravity currents in figure 4 b is U . In contrasting
panels (a) and (b), it can be shown that the fluid of density ρ1 gains potential energy in the amount
1 2
2 b ρ1 gU ∆t where g is gravitational acceleration and ∆t is the time increment. The corresponding
gain of kinetic energy is 12 bρ1 U 3 ∆t. Conversely, the fluid of density ρ2 gains kinetic energy in
the amount 21 bρ2 U 3 ∆t but loses potential energy in the amount 12 b2 ρ2 gU ∆t. By balancing these
terms, show that the front speed is given by
p
U = 21 g 0 H .
(6)
Here g 0 = 2g(ρ2 − ρ1 )/(ρ2 + ρ1 ) ' g(ρ2 − ρ1 )/ρ1 , where, in writing the latter expression, we have
assumed that ρ2 − ρ1 is much less than either ρ1 or ρ2 .
2. [2 points] Using (6) and the data shown in tabular form below, estimate ρ2 if ρ1 = 1.00 g/cm3
and H = 20.0 cm. (As regards the tabulated data, you should plot these and determine the slope
of the corresponding best fit line). Note that in the numerical simulation, ρ2 is set to 1.02 g/cm3 ,
which gives you an indication of the accuracy of (6). Note also that the data corresponds to the
numerical simulation shown at http://websrv.mece.ualberta.ca/mrflynn/case306.avi.
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3. [1 point] Comment on the difference between the image presented in figure 4 b vs. the flow realized
by numerical simulation. In the latter case, what evidence do you have that the flow is turbulent?
Data:
Time (s)
0.0
2.5
5.0
7.5
10.0
12.5
15.0
17.5
20.0
22.5
Front position (cm)
0.0
20.3
43.0
64.8
86.9
108.2
130.3
151.8
173.2
194.7
17
MEC E 230
Seminar J, Tank draining 1
Objective: To evaluate the time required for a tank to drain under its own hydrostatic force.
Background: The draining and filling of storage tanks is a frequent operation in many industrial
processes. Here, we study the former scenario. Attention is first given to Case 1, indicated schematically below, in which water drains through a rounded-edged orifice located in the bottom surface.
Frictional effects are in this case minimal. We will then investigate the siphon system of Case 2 and
estimate whether frictional effects are likewise negligible for that flow also.
Case 1
Case 2
Air
Water
h
ve
ve
Figure 5: Tank draining without (Case 1) and with (Case 2) a siphon.
Equipment/materials:
Bucket
Stopper
Beaker
Paper towels
Acrylic tank with a rounded-edged orifice
Siphon tube (58 cm long)
Ruler
Stopwatch
Data:
Item
Acrylic tank
Rounded-edged orifice
Siphon tube
Symbol
D
d
dt
Diameter
12.4 cm
0.77 cm
0.25 inch
• Time required for the free surface to drop from 15 cm to 5 cm (Case 1):
• Time required for the free surface to drop from 15 cm to 5 cm (Case 2):
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Questions:
1. [1 point] Bernoulli’s equation describes the flow of a fluid that is isothermal, incompressible and
inviscid. Under steady conditions, Bernoulli’s equation can be written as
p1 1 2
p2 1 2
+ 2 v1 + gz1 =
+ 2 v2 + gz2 ,
ρ
ρ
(7)
where p is the fluid pressure, ρ is the fluid density, v is the fluid velocity, g is gravitational
acceleration and z is the elevation. With reference to the left-hand side panel of figure 5, suppose
that Point 1 is located immediately below the free surface. Then p1 ' patm , z1 = h and v1 ' 0
because the rate of descent of the free surface is small. Suppose also that√Point 2 is located
immediately below the opening so that p = patm and z2 ' 0. Show that v2 = 2gh.
2. [1 point] Again with reference to the left-hand side panel of figure 5, show that the water level
in the tank is given by the solution of the following ode:
p
dh
d2
= − 2gh 2
dt
D
(8)
(d and D are defined in the latter table from page 1).
3. [3 points] By integrating (8), estimate the amount of time required for the free surface to fall
from an elevation of 15 cm to an elevation of 5 cm. Using the following formula, compute the
percentage error with the corresponding measured value:
Percent error =
|tpredicted − tmeasured |
× 100% .
tpredicted
(9)
Provide one possible reason for the deviation between theory and measurement.
4. [3 points] Now consider the right-hand side panel of figure 5. If we neglect frictional losses in
the siphon tube, the analogue of (8) reads
p
dh
d2
= − 2gh t2 .
dt
D
(10)
On this basis, estimate the time for the interface to fall from 15 cm to 5 cm and also compute a
percentage error using (9). Is your error larger or smaller than before?
5. [2 points] Friction losses can be incorporated in Bernoulli’s equation by adding a right-hand side
dissipation term, i.e.
p1 1 2
p2 1 2
fL 2
+ 2 v1 + gz1 =
+ 2 v2 + gz2 + 12
v ,
(11)
ρ
ρ
dt 2
where L = 58 cm is the siphon tube length and f (' 0.1) is called the friction factor. In this case,
(10) is replaced by
s
dh
2gh d2t
=−
.
(12)
dt
1 + f L D2
dt
Estimate the new time for the interface to fall from 15 cm to 5 cm for the Case 2 experiment and
compute a new percentage error using (9).
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MEC E 230
Seminar K, Tank draining 2
Objective: To evaluate the time required for a tank to drain under its own hydrostatic force and to
estimate the impact of minor losses in piping systems.
Background: The draining and filling of storage tanks is a very frequent operation in industrial
processes. Here we will study the former scenario whereby the liquid exiting the tank flows through
a long pipe of radius a and length L. A laminar flow will be assumed. We will furthermore neglect
entrance and exit losses corresponding, respectively, to the points i and e, in figure 6. For example, an
entrance loss is associated with the “traffic jam” of fluid particles trying to squeeze into the relatively
small cross-sectional area of the pipe. Under the above assumptions, it can be shown that the volume
flux in the pipe is given by
πa4
dp
Q=
−
,
(13)
8µ
dx
where µ is the dynamic viscosity and
pi − patm
dp
'
dx
L
is the pressure gradient in which pi is the pressure at the pipe inlet.
−
(14)
air
water
h
e
i
Q
Figure 6: Tank draining from a long pipe.
Equipment:
Acrylic tank (rectangular cross-section)
Measuring tape
Acrylic pipe
Stopwatch
20
Bucket
Stopper
Data:
• Water density: ρ = 998.2 kg/m3
• Water dynamic viscosity: µ = 1.003 × 10−3 kg/(m·s)
• Pipe inner diameter: 2a = 0.635 cm
• Tank interior dimensions: 13.3 cm×13.3 cm
• Pipe length, L:
• Time, ∆t, required for free surface to fall from h1 = 20 cm to h2 = 10 cm:
Questions:
1. [2 points] Using your knowledge of hydrostatics and mass conservation, show that the height, h,
of the free surface relative to the mid-plane of the pipe satisfies the following ODE:
dh
πa4 ρgh
=−
.
dt
8µLAc
(15)
2. [2 points] By integrating (15) show that
∆t =
8µLAc ln(h1 /h2 )
.
πa4 ρg
(16)
3. [1 point] Explain why ∆t increases as L increases.
4. [1 point] Compare (∆t)theoretical and (∆t)experimental . On the basis of this comparison, was it
appropriate to neglect minor losses at the inlet and exit of the pipe?
21
MEC E 230
Seminar L, Deflecting jet
Objective: To determine the force exerted by a high-pressure jet as a function of the profile of the
downstream shape.
Background: Designers of vehicles and planes minimize drag by selecting streamlined shapes. Today
we will study such effects by examining the force that is exerted on a non-streamlined and streamlined
shape by a high-pressure jet of compressed air. According to theory, this force is given by
1
√
FNS = 2ṁV
FS = 1 −
ṁV ,
(17)
2
for the non-streamlined and streamlined shape, respectively. Here ṁ is the jet mass flux at the nozzle
exit and V is the corresponding jet velocity. The latter quantity is given by
p
V = γRT ,
(18)
where γ is the ratio of specific heats, i.e. γ = cp /cv , R is the gas constant and T is the temperature
at the nozzle exit. For air, γ = 1.4, R = 287 m2 /(s2 K) and T = 0.833T0 in which T0 = 20◦ C is the
temperature of the air in the compressed air cylinder. Conversely,
ṁ = 0.634
p0 π 2
· d ·V ,
RT0 4
(19)
in which p0 = 100 psi (gage) is the pressure of the air in the compressed air cylinder and d = 3.13 mm
is the nozzle diameter.
Equipment:
Compressed air cylinder, tubing and fittings
Non-streamlined shape
Pressure gage
Streamlined shape
Safety goggles
Spring scale
Pulley
Nozzle
Data:
Shape
Spring scale
reading (kg)
Force (N)
Non-streamlined
Streamlined
Questions:
1. [2 points] Show, using the equations and parameters summarized above that FNS = 9.00 N and
FS = 1.32 N. Recall that in imperial units, atmospheric pressure is 14.7 psi (absolute), i.e. 1 psi=
6.894 kPa.
2. [2 points] Compare the predictions of force made in Question 1 with your measured data.
Provide one possible reason for any discrepancies observed.
22