1. No category

Instructor: Hacker
Engineering 232
Sample Exam 1 Solutions
Answer Key
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Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.2
Chapter 1: Introduction to Thermodynamics
Introduction to Thermodynamics
Introductory Concepts and Definitions
Problem 1. Consider a refrigerator in a kitchen. Take the refrigerator and everything
in it to be our system. Which best describes the system’s surroundings?
(a) All of the air in the kitchen.
(b) Any one standing in the kitchen.
(c) The air inside the refrigerator.
*(d) Everything in the universe external to the system.
Problem 2. Circle the true statement:
*(a) a closed system is also called a control mass
(b) a closed system is also called control volume
(c) a closed system is an isolated system
(d) an open system is also called control mass
Problem 3. Which of the following is not an extensive property?
*(a) Density
(b) Volume
(c)
Mass
(d) Energy
(e)
None of these
Problem 4. Identify the SI of the expression P V , where P is pressure and V is volume.
(a) Pa
(b) Pa · m2
*(c) J
(d) N
(e) None of these
Problem 5. Determine whether equations (i) and (ii) below are dimensionally consistent.
1
2y
(ii) x = t v + at ,
(i) g = 2
t
2
where x = distance, y = distance, v = velocity, t = time, and a = acceleration.
(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent
(b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent
*(c) (i) and (ii) are both dimensionally consistent
(d) (i) and (ii) are both dimensionally inconsistent
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.3
Static Fluids
For all problems below assume the system is in static equilibrium.
Problem 6. The mass of an unknown gas mixture in a room that is 5 m × 10 m × 10
m is known to be 500 kg. What is the specific volume of the gas?
(a) 1 m3
*(b) 1 m3 /kg
(c) 500 m3
(d) 500 m3 /kg
(e) None of these
Solution:
v≡
length × width × height
5 · 10 · 10 m3
m3
volume
=
=
=1
mass
mass
500 kg
kg
Problem 7. A large storage tank is 10 m tall. The top of the storage tank is open to
the atmosphere. What is the gage pressure at the bottom of the tank? Take the density
of water to be ρ = 1000 mkg3 and g = 10 sm2 .
(a) 1 kPa
(b) 10 kPa
2
*(c) 10 kPa
(d) 103 kPa
(e) None of these
Solution: Use the equation
Pgage = ∆P = Pabs − Patm = ρgh = (103
m
N
kg
)(10 2 )(10m) = 105 2 = 105 Pa = 102 kPa
3
m
s
m
Problem 8. A submarine has a circular top hatch with a 1 m diameter. What is the
force exerted on the hatch from the pressure due to the water above the hatch when the
sub dives down to a depth of 200 m? Take the approximate density of sea water to be
ρ = 1000 mkg3 and g = 10 sm2 .
Note: we are ignoring atmospheric pressure Patm in this problem and we are assuming
the submarine is a rigid body (i.e., we can ignore the pressure from inside the sub).
(a) 2π × 105 N
(b) π × 106 N
(c) 2π × 106 N
*(d) π2 × 106 N
(e) None of these
Solution: First we’ll find the pressure on the hatch, then we’ll use the definition of
pressure to find the force Fhatch = Phatch Ahatch , where Fhatch is the force on the hatch
from the pressure of the sea water above the hatch, Phatch is the pressure on the hatch,
and Ahatch is the area of the hatch.
Phatch = ρgh = (103
kg
N
m
)(10 2 )(2 × 102 m) = 2 × 106 Pa = 2 × 106 2 .
3
m
s
m
Using the definition of pressure we find:
Fhatch = Phatch Ahatch =
2
(πrhatch
) Phatch
2
N
1
π
=π
m2 2 × 106
= × 106 N
2
2
m
2
Problem 9. The basic barometer can used to measure the height of a building. To do
this a measurement is made at the bottom of the building at ground level and a second
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.4
measurement is taken at the top of the building on the roof. One can then relate these
readings to pressure using the equation for a mercury-filled barometer: Patm (hHg ) =
ρHg ghHg , where ρHg = 13, 600 kg/m3 is the density of mercury, g is the acceleration due
to gravity, and hHg is the height of the mercury column in the barometer measured in
millimeters. The approximate height of the building can then be found by making a few
simplifying assumptions. For moderate sized buildings a reasonable assumption for the
average density of air under fair weather conditions is approximately ρair ≈ 1.18 Kg/m3 .
If the barometric readings at the top and bottom of a certain building are 730 mmHg
and 755 mmHg respectively, what is the height of the building in meters?
(a) 50 m
(b) 120 m
*(c) 280 m
(d) 320 m
(e) None of these
Solution: See handwritten solutions
Temperature
Use the equations TC = TK − 273 and TF = 59 TC + 32 to answer the following two
questions. Round your answers to the nearest integer.
Problem 10. Use the equation TF = 95 TC + 32 to convert 98◦ F to ◦ C.
(a) 32◦ C
(b) 208◦ C
(c) 20◦ C
*(d) 37◦ C
(e) None of these
5
5
5
5
5
Solution: TC = (TF −32) = (98−32)◦ C = ·66◦ C = ·22◦ C = (35+ )◦ C ≈ 37◦ C
9
9
9
3
3
Chapter 2: The first law of thermodynamics
Work, Power, Heat, and Energy
Problem 11. A skyrocket is fired upward. As it rises, the work done by gravity on the
rocket is:
(a) positive
*(b) negative
(c) zero
Solution: The rocket is moving upward; gravity is pulling downward. Since the force
is in the opposite direction of the motion, the work done by the force is negative.
Problem 12. How much power, in kilowatts, must be developed by the electric motor
of a 1600-kg car moving at 26 m/s on a level road if the forces of resistance total 720 N?
Round your answer to the nearest tenth of a kilowatt.
(a) 95.2 kW
(c) 133.2 kW
(e) None of these
(b) 36.3 kW
*(d) 18.7 kW
Solution: Use the equation for power as a product of force times velocity: P = F v.
Then P = Fmotor · vcar = (720N)(26m/s) = 18.7 kW. To find Fmotor , draw a free-body
diagram and apply Newton’s second law together with the fact that the car is moving
at a constant speed (so it’s not accelerating) to arrive at Fmotor = fk . See my solutions
on the web for the details.
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.5
Problem 13. Ricky, the 5 kg snowboarding raccoon, is
on his snowboard atop a frictionless ice-covered quarterpipe with radius 3 meters (see figure at right). If he
starts from rest at the top of the quarter-pipe, what is
his speed at the bottom of the quarter-pipe? For ease
of calculation, assume that g = 10 m/s2 . Round your
answer to the nearest 0.1 m/s.
(a) 7.0 m/s
(c) 8.5 m/s
(e) None of these
*(b)
(d)
x
Raccoon: 5 kg
3m
3m
7.7 m/s
9.3 m/s
Solution:
We are given: m = 5 Kg, R = 3 m, vi = 0 m/s, and g = 10 m/s2 .
We want: vf =?
We assume: no friction, hence the mechanical energy of the system is conserved.
Since the system, Ricky and the earth, has no external forces on it, such as friction, we
can apply the conservation of mechanical energy: MEi = MEf . Now, the initial potential
due to gravity is Ugrav = mgh = mgR and the initial kinetic energy is Ki = 0 since Ricky
is at rest. The final P.E. is zero, since Ricky is now at ground level h = 0. The final K.E.
is Kf = 12 mvf2 . Using the definition of mechanical energy: ME = PE + KE gives
1
mgR = PEi + KEi = MEi = MEf = PEf + KEf = mvf2
2
÷1m
2
−−−−
−−→ vf2 = 2Rg
√
p
1/2
−−−−−−→ vf = 2Rg = (2)(3 m)(10 m/s2 )
= 7.7 m/s
Notice that since R = h, this is the same speed that Ricky would have if he had jumped
straight down (go look back at our 1-D kinematic problems). The only difference is the
direction of his speed would have been different. If he had jumped straight down, then
his velocity would also have been straight down, and into the ground would go Ricky; but
the quarter pipe redirected his motion to being tangent to the ground, while preserving
his speed. So instead of busting his head, and winding up dead; Ricky sped off to work,
where he was a clerk. Were it not for conservation of mechanical energy he would have
been late for his job, and replaced by his competitor Rob.
Problem 14. If heat is added to a system and the temperature of a system increases,
without knowing anything else, which form of energy will be definitely increase?
(a) The kinetic energy of the system
(b) The potential energy of the system
(c) The work done by the system
*(d) The internal energy of the system
Solution: If the temperature increased, then the internal energy must have increased.
That’s all that we can say for sure. The only way the system’s kinetic energy could be
increased is if the bulk motion of system were increased, and we have no such information.
A similar statement goes for the potential of the system. We also don’t know if the system
does any work. Unless the system were a gas in a frictionless piston-cylinder mechanism,
we’d have no way of knowing if the system did work owing to it’s increase in internal
energy.
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.6
Problem 15. (The sign convention for heat transfer and work)
Consider a closed system. Suppose that heat Q is added to the system and work W is
done on the system. Use the standard sign conventions for heat and work transfer given
in class (and in the book) to determine which of the following combinations is correct.
Circle the correct answer.
(a) W > 0 and Q > 0
(b) W > 0 and Q < 0
*(c) W < 0 and Q > 0
(d) W < 0 and Q < 0
Solution: By convention: W > 0 if the system does work on the surroundings (environment) and Q > 0 if heat is added to the system. The sign convention for thermodynamics
is different from the sign convention for physics.
In physics, if energy flows into the system, then the energy is added to the system and the
system’s energy increases. This is defined to be the positive direction for energy transfer
for both heat and work.
The positive direction in thermodynamics differs from physics. The positive direction
for heat flow in thermodynamics is identical to the positive direction in physics, but
the positive direction in thermodynamics for work is the opposite direction as that for
physics.
The positive direction for energy transfer in thermodynamics comes from thermo’s history
in the study of the steam engine. Thus the positive direction is the natural direction for
the flow of energy in a steam engine. Heat is added to the cylinder full of steam (the
system), hence Q > 0 if steam is added to the system. However, the hot steam is used to
push a piston, so the work is done by the system on the piston (not part of the system).
Therefore, W > 0 if the work is done by the system, not on the system.
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.7
Work done on and by a thermodynamic system
For the following problems use the general formula for work done by a thermodynamic
system:
Z Vf
P (V ) dV
(0.1)
W =
Vi
where W = work, Vi = initial volume, Vf = final volume, and P (V ) = pressure. Notice
that, in general, pressure will be a function of volume. All of these problems will involve
questions for a gas contained within a closed frictionless piston-cyclinder assembly.
Recall: a polytropic process is one that satisfies P V n = constant for some value of n
and for all time during the process.
Problem 16. A gas is contained within a closed frictionless piston-cyclinder assembly.
The system undergoes a constant-pressure process at 5 bar beginning at an initial volume
Vi = 1 m3 and ending in a final volume Vf = 3 m3 . Using the standard sign convention
for work, determine how much work was done by the system.
(a) 10 kJ
(b) 100 kJ
*(c) 1000 kJ
(d) 200 kJ
(e) None of these
Solution: Since the process is an isobaric process (constant pressure) we have P (V ) =
P0 for all time t ∈ [ti , tf ]. The integral for work becomes:
Z
Vf
W =
P (V ) dV
Vi
Z Vf
=
P0 dV
(since P (V ) = P0 )
Vi
Z
Vf
= P0
dV
Vi
= P0 (Vf − Vi )
= P0 ∆V
Now ∆V = 3 m3 − 1 m3 = 2 m3 , and P0 = 5 bar = 5 × 105 Pa = 5 × 105
N
. Thus
m2
5 N
W = P0 ∆V = 5 × 10 2 (2 m3 ) = 106 N · m = 106 J = 1000 kJ
m
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.8
Problem 17. A vertical, frictionless piston-cylinder device contains a gas that is in
equilibrium with the weight of the piston balanced by the internal pressure for the gas.
The mass of the piston is 4 kg. The atmospheric pressure outside the cylinder is Patm =
105 Pa. The area of the piston is A = 20 cm2 . Determine the pressure of the gas contained
in the cylinder Pgas . Take the magnitude of gravity g = 10 m/s2 .
(a) 102 kPa
*(b) 120 kPa
(c) 10,020 Pa
(d) 100,200 Pa
(e) None of these
Solution: Since the piston is in mechanical equilibrium, the net force on the piston
must be zero. The upward force is supplied by the gas inside the piston-cylinder device,
and the downward for on the piston comes from the pull of gravity and the atmospheric
pressure. Let m be the mass of the piston, A be the surface area of the face of the piston,
Fgas be the force of the gas, and Fatm be the force due to atmospheric pressure. Then,
the force balance becomes:
÷A
−−−−−→
Fgas = Fatm + mg
mg
Pgas = Patm +
A
(4kg)(10 m/s2 )
5
= 10 Pa +
2
m2
103
= 105 Pa + 2 × 104 Pa
= (100 + 20) kPa = 120 kPa
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.9
Heat transfer Modes: Conduction, Radiation, and Convection
You may find at least one of the following formulas useful (in all formulas below A is
area and T is temperature):
∆T
• For heat transfer by conduction use Fourier’s Law: Q˙ x = ±κ A
, where κ is
L
the thermal conductivity coefficient, L is the thickness, and ∆T is the temperature
difference.
• For heat transfer by radiation use Stefan-Boltzmann Law:
4
4
], where 0 ≤ ε ≤ 1 is the emissivity and σ =
− Tsurroundings
Q˙ e = ±εσA[Tbody
−8 W
5.67 × 10 m2 ·K4 is the Stefan-Boltmann constant.
• For heat transfer by heat transfer by free and forced convection use Newton’s Law
of cooling: Q˙ c = ±hA[Tbody − Tsurroundings ], where h is the heat transfer coefficient.
Each formula contains a ± and it is your job to determine the correct sign for the problem
at hand. In general, the sign depends on the direction of the energy transfer with respect
to the chosen system.
Problem 18. A large 1-cm thick plate glass window in a restaurant has a thermal
W
. This coefficient is valid over a temperature range of 280 - 320
conductivity of 23 m·K
kelvin. The outside temperature is a sweltering 313 K (104◦ F), while the restaurant is
kept at a comfortable temperature of 293 K (68◦ F). Assuming steady-state conditions, it
can be shown that the temperature profile through the glass is linear. Assuming a linear
profile, and taking the inside of the restaurant as our system, what is the rate of heat
transfer via conduction through an area 1-m2 area of the plate glass?
(a) 1 kW
(b) 3/2 kW
*(c) 3 kW
(d) 7 kW
(e) None of these
Solution: Since energy flows from hot to cold, energy is flowing into the restaurant.
That is, energy is flowing into the system. Since heat is defined to be positive if energy
flows into the system, and since ∆t > 0, it follows that Q˙ x > 0. If we take ∆T = Tout −Tin ,
then ∆T > 0. Thus the correct form of the equation is
T
−
T
out
in
> 0.
Q˙ x = κA
L
We are given the following information:

W
κ = 23 m·K



A = 1 m 2
given:

L = 1 cm = .01 m



∆T = Tout − Tin = (313 − 293) K = 20 K
Q˙ x =
3 W
2 m·K
2
1m
2 · 10 K
1
m
102
= 3 × 103 W = 3 kW
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.10
Power and Refrigeration Cycles
Problem 19. Smokey the choo-choo train is an old steam engine with a heat-transfer
input of Qin = 100 kJ and a heat-transfer output of Qout = 70 kJ. What is the thermal
W
of the power cycle?
efficiency η = Qcycle
in
(a) .7
*(b) .3
(c) .7 kJ
(d) .3 kJ
(e) None of these
Solution: Using an energy balance we see that
Qin = Wcycle + Qout
−Qout
−−−−−−−→ Wcycle = Qin − Qout
Qout
7
3
Wcycle
=1−
=1−
=
⇒ η=
Qin
Qin
10
10
Problem 20. Only one of the following statements is true. For the figure below draw
arrows that indicate the positive direction for Qin , Qout , and Wcycle for a power cycle,
then use your graph to determine the true statement.
Hot Body
System
Cold Body
Figure 1: Power Cycle Template
(a) Heat flows from the cold body into the system and the “exaust” exits into to the hot
body; the system uses the input heat to do work.
(b) Heat flows from the cold body into the system and the “exaust” exits into to the hot
body; the system uses the input work to extract heat.
*(c) Heat flows from the hot body into the system and the “unused portion” exits into
to the cold body; the system uses the input heat to do work.
(d) Heat flows from the hot body into the system and the “unused portion” exits into to
the cold body; the system uses the input work to extract heat.
Thermodynamics 232, Sample Exam1-f11 Copyright ©Wayne Hacker 2011. All rights reserved.11
Problem 21. A gas is contained within a closed frictionless piston-cyclinder assembly.
Take the gas to be the system under investigation. In a careful experiment heat is
added to the system in such a way that the system remains in quasi-equilibrium and the
pressure satisfies a polytropic process of the form P V n = constant over the time that
the experiment is conducted. Take n > 0 to be known. The gas expands from an initial
volume Vi to a final volume Vf with Vi < Vf . The initial pressure was measured to be Pi
and the final pressure was measured to be Pf . Derive a formula and use it to determine
the work done by the system if Vi = 1 m3 , Pi = 100 kPa, Pf = 0.5 Pi and n = 0.5. Circle
your answer.
(a) 50 J
(b) -50 J
*(c) 200 kJ
(d) -200 kJ
(e) None of these
Solution: See handwritten notes on the website for the derivation of the formulas:
n

Pf
Vi
n
n


=
(∗)
⇒

 Pi Vi = Pf Vf
Vf
Pi
"
(n−1) #

Vi
Pi Vi


1−
, for n 6= 1 .
(∗∗)
 Wsys =
n−1
Vf
Rewriting equation (∗) as
Vi
Vf
(n−1)
=
Pf
Pi
(n−1)
n
and substituting it into (∗∗) yields
Wsys
"
#
(n−1)
Pi Vi
Pf n
=
1−
n−1
Pi
(∗ ∗ ∗)
Using the initial conditions: Vi = 1 m3 and Pi = 100 kPa (Pi Vi = 100kJ), and the
parameter values: Pf = 0.5 Pi , n = 0.5 yields
"
(−1) #
1
= 200 kJ .
Wsys = −2 · 100 kJ 1 −
2
Notice that the work is positive. This agrees with our sign convention, since the closed
system must do work on the environment in order to expand the volume.