Instructor: Hacker Engineering 232 Sample Exam 2 Solutions

Transcription

Instructor: Hacker
Engineering 232
Sample Exam 2 Solutions
Print your name neatly. If you forget to write your name, or if I can’t read your writing,
you can lose up to 100 points. Answer all the questions that you can. Circle your answers.
You must show your work. You will not receive credit for lucky guesses. Show your
work as clearly as you can: if I can’t understand how you got an answer, I will not give
you credit for it. Remember, I know how to solve the problem; and to make matters
worse, I have a lot of training in following logical arguments!
Warning: The definition of “little or no work” will be determined by the instructor, not
the student.
Grading rules. Except for problem 1, each problem will be graded on a scale of 0 to
10 points. A grading algorithm is given with the problem. The algorithm is based on
a series of steps that you should follow in solving the problem. You should try to learn
these steps: in the future, you may find yourself facing difficult problems with no steps
laid out for you to follow.
On any problems, clarity is as important as the correct procedure and the correct answer.
If you do not clearly label your steps, and your work within those steps, then I will grade
your work as wrong. I will not waste time struggling to read an incoherent mess that
is purported to be the solution; and after the exam is handed in and graded, I will
not improve your grade based on your explanation of what you were trying to say. One
purpose of these problems is to teach you how to lay out a logical argument that someone
else with a technical background can follow. If in doubt, write it down! I refuse to look
at it before you turn it in and tell you if there is enough detail! I also refuse to give you
any hints, or tell you if you are on the right track. If you are unable to do this, then you
should not and will not get credit for the problems.
If you solve the wrong problem, you will receive 0 points, even if your solution to the
wrong problem is correct. You are responsible for reading the problem correctly. If you
think a problem is ambiguous, you should ask the instructor for clarification.
For maximum partial credit be sure to write down:
Step 1 (4 out of 10 points): (i) identify the system, (ii) what you are given, (iii)
what you want, (iv) draw a picture, and (v) list any assumptions that you need to make
to solve the problem: for example, “the pressure is constant”, “the system is a closed
system”, etc. If there are multiple parts, be sure and label the answer for each part.
The “Give up” option: If you have no idea how to solve the problem, you can write
“Give up” and circle it instead of writing nonsense in hopes of getting partial credit. If
I see “Give up”, I will ignore everything else that you’ve written and award you 3 out of
10 points for the problem. This is your reward for knowing what you don’t know, and
for being honest about it.
Thermodynamics 232, Sample Exam 2
Copyright ©Wayne Hacker 2009. All rights reserved.2
Problem 1. (25%)
Part 1 (Using Steam Tables to Determine Properties): (20%)
This problem consists of 11 short-answer problems. No partial credit will be given for
these problems. However, you can miss one of the 11 problems without penalty. Your
grade will be based on your best 10 problems. You will not receive extra credit for getting
all 11 right.
Instructions: Determine the missing properties and the phase descriptions in the following table for water.
(1)
(2)
(3)
(4)
(5)
T (◦ C)
Ti
Ti
Ti
Tsat
Ti
P (bar)
Pi
Psat
Psat
Pi
Pi
v (m3 /kg)
v(Ti , Pi )
vf (Ti )
(1 − xi )vf + xi vg
vg (Ti )
v(Ti , Pi )
u (kJ/kg)
u(Ti , Pi )
uf (Ti )
(1 − xi )uf + xi ug
ug (Ti )
u(Ti , Pi )
x (quality)
N/A
0
xi
1
N/A
Phase description
Sub-cooled liquid
Saturated Liquid
Liquid-Vapor
Saturated Vapor
Super-heated Vapor
Note: N/A means not applicable.
You may find the following formulas useful.
v = (1 − x)vf + xvg
u = (1 − x)uf + xug
Solution:
(1) Sub-cooled liquid + Pressure and Temperature is given ⇒ use table A-5
(2) Saturated Liquid + Temperature is given ⇒ use table A-2
(3) Liquid-Vapor + Temperature + quality is given ⇒ use table A-2
(4) Saturated Vapor + specific volume is given ⇒ use table A-3
(5) Super-heated Vapor + Pressure and Temperature ⇒ use table A-4
Part 2 (Reading the compressibility chart) (5%)
No partial credit will be given for this problem
A closed rigid container full of carbon monoxide has a temperature of T = 77◦ C and a
pressure of 0.35 bar. Determine if this gas can be treated as an ideal gas by computing
PR and TR , then go to the compressibility chart and verify that Z ≈ 1. In order to get
credit you need to compute the reduced temperature and pressure: TR , PR , respectively,
and on your copy of the compressibility chart locate the point (PR , Z) on the chart and
clearly mark it.
You should find the following information useful:
Tc = 133 K (critical temperature for carbon monoxide in kelvin)
Pc = 35 bar (critical pressure for carbon monoxide in bars (1 bar = 100 kPa))
P
PR ≡ , where PR is the reduced pressure
Pc
TR ≡
T
, where TR is the reduced temperature
Tc
For exam problems 2, 3 and 4, review the example problems that we did in class and any
relevant chapter 3 homework problems.
In particular, review any problems involving
• A closed frictionless piston-cyclinder problem involving saturated liquid/and or
vapor states.
• A closed rigid container problem involving the quality of water.
• Air as a gas undergoing a cycle
An example of the kinds of facts that I’ll be giving with the problems:
kg
(molecular weight of air)
Mair = 28.97
kmol
kJ
(universal gas constant)
R = 8.314
kmol · K
Tc = 133 K (critical temperature for air in kelvin)
Pc = 37.7 bar (critical pressure for air in bars (1 bar = 100 kPa))
P
PR ≡ , where PR is the reduced pressure
Pc
T
TR ≡ , where TR is the reduced temperature
Tc
Problem 2. (25 %) Consider a closed frictionless piston-cyclinder apparatus that contains a mass m = .2 kg of saturated liquid water (the initial state). The system under
goes an isobaric process where the pressure is held at P = 100 kPa. Heat is added to the
system until the water is completely vaporized (i.e., the final state is a saturated vapor
state).
(a) Determine the initial and final temperatures.
(b) Determine the change in specific volume (in m3 /kg).
(c) Determine the amount of heat Q that was added to the water during this process.
Note: The initial state is on the saturated liquid line (the left-hand boundary) of the
vapor dome, and the final state is on the saturated vapor line (the right-hand boundary)
of the vapor dome.
2
2
2
Problem 3. (25 %) A closed, well-insulated, rigid tank having a volume of 1 m3 contains
saturated water vapor at 100◦ C (the initial state). The water is rapidly stirred until the
pressure is 1.5 bar, or equivalently 150 kPa (the final state).
(a) Determine the temperature at the final state, in ◦ C, and
(b) the work done during the process, in kJ.
Solution: See class notes for the worked out solution.
Problem 4. (An application of the ideal gas law using specific heat)
The apparatus: Two closed rigid insulated containers with volumes −
V 1 and −
V 2 are
connected by an insulated pipe with an insulated valve. The containers are filled with a
gas with a specific gas Rgas . In general, −
V 1 6= −
V 2 , the amount of gas, temperature, and
pressure in each container is different.
The given initial conditions:
Container 1: mass m1,i , temperature T1,i , and pressure P1,i .
Container 2: mass m2,i , temperature T2,i , and pressure P2,i .
The process: The valve is then opened and the gases are allowed to mix.
The final state of the system: We wait a long time until the gas reaches equilibrium.
The assumptions: We assume that the conditions are such that we can treat the gas
as an ideal gas with constant specific heat cv .
Questions:
(a) Derive a formula for the final mass distribution for each in each container for the
general case stated above.
(b) To simplify the mathematics in this second part we assume −
V1 = −
V 2 and m1,i = m2,i .
Derive a formula for the final specific internal energy u in terms of the gas constant Rgas
and the specific heat cv and use this result to determine a formula for temperature and
pressure.
Note: The simplifying assumption in part (b) is not necessary, after you work out the
special case, you should try to work out the general case.
m1,i
T1,i
P1,i
−
V1
valve
insulated pipe
container 1
m2,i
T2,i
P2,i
−
V2
container 2
Figure 1: The initial state of the system
Solution: Start by writing down a diagram of the final state.
m1,f
Tf
Pf
−
V2
container 1
valve
insulated pipe
m2,f
Tf
Pf
−
V2
container 2
Figure 2: The final state of the system


m1,i
given: initial state of container 1: T1,i


P1,i
given properties:
want:


Rgas
−
V1


−
V2


m2,i
and initial state of container 2: T2,i


P2,i
(see assumptions and observations below)


Tf = T1,f = T2,f

P = P = P
f
1,f
2,f

m1,f 6= m1,i (in general)



m2,f 6= m2,i (in general)
(see assumptions and observations below)
system: The gas contained in both containers.
Assumptions:
1. The system is closed ⇒ mtotal = m1 + m2 = constant.
2. The containers are rigid ⇒ −
V1 = −
V 1,i = −
V 1,f and −
V2 = −
V 2,i = −
V 2,f
⇒−
V total = −
V1 +−
V 2 = constant.
3. Since the containers, connecting pipe, and valve are insulated, Qsys = 0.
4. We assume that the gas in equilibrium in both the initial and final states.
5. Since there are no moving boundaries, except the tiny valve, there is negligible
work done on the system: Wsys = 0.
6. Since the containers are at rest, and the fluid is at rest in the initial and final
states of the system, we can ignore any change in K.E. and P.E. of the system:
∆KE = ∆PE = 0.
7. In the final state the gas between the two containers is well-mixed and “connected”
to one another. It follows that all of their intensive properties must be the same.
In particular, Tf = T1,f = T2,f , Pf = P1,f = P2,f , vf = v1,f = v2,f , and uf = u1,f =
u2,f . However, the extensive properties of the two containers are in general not
equal to one another: −
V 1 6= −
V 2 , m1,i 6= m2,i , m1,f 6= m2,f . This is why we always
distinguish between extensive and intensive properties!
Observations:
1. In practice, we would need to verify that we have an ideal gas by showing that the
reduced pressure and reduced temperature satisfy the standard conditions for an
ideal gas:
PR 1 (“low pressure”) and TR 1 (“high temperature”) .
A quick check of the compressibility shows that Z ≈ 1 ⇒ under these conditions.
In fact, these conditions can be relaxed to allow TR ≈ 1 provided that PR 1.
2. As already stated, since pressure and temperature are intrinsic properties of the
system, it follows from the assumption that the final state is equilibrium that the
pressure and temperature must be the same for both containers. Were this not the
case, then the pressure difference would lead to gas flowing from the high-pressure
container to the low-pressure container, violating the equilibrium assumption. Similarly, were there a temperature gradient between the two containers, then there
would be a flow of heat, which would violate the equilibrium assumption. Therefore,
we have
Pf = P1,f = P2,f and Tf = T1,f = T2,f .
3. The 1st Law of Thermodynamics reduces to the simplified equation:
∆E = ∆KE + ∆PE + ∆U = Qsys − Wsys
⇒
∆U = 0 .
4. Because our system is the sum of the gas in two separate containers, we need to
be careful how we express the internal energy U . If the system can be modeled as
an ideal gas, then U = U (T ). Since u = u(T ) is an intrinsic property, we must
have u1 (Tf ) = u2 (Tf ); however, U1 (Tf ) 6= U2 (Tf ), since U is an extrinsic property
and we have made no assumption regarding the size of the containers. Thus, our
governing equation, which is just the first law of thermo, becomes:
∆U (T ) = Uf (T ) − Ui (T ) = 0
+U
−−−−−i−→ Uf (T ) = Ui (T )
expand by container
−−−−−−−−−−−−−−−→ U1 (Tf ) + U2 (Tf ) = U1 (T1,i ) + U2 (T2,i )
express as specific quantities
−−−−−−−−−−−−−−−−−−−→ m1,f u(Tf ) + m2,f u(Tf ) = m1,i u1 (T1,i ) + m2,i u2 (T2,i )
factor
−−−−−−→ (m1,f + m2,f )u(Tf ) = m1,i u1 (T1,i ) + m2,i u2 (T2,i )
m1,i +m2,i =mtotal=m1,f +m2,f
−−−−−−−−−−−−−−−−−−−−−→ (m1,i + m2,i )u(Tf ) = m1,i u1 (T1,i ) + m2,i u2 (T2,i )
÷ (m1,i +m2,i )
m1,i u1 (T1,i ) + m2,i u2 (T2,i )
−−−−−−−−−−−→ u(Tf ) =
(m1,i + m2,i )
From the above calculation we find an expression for the intensive variable of specific
internal energy. Since it’s an intensive variable it can be applied to each container,
or to the entire system.
m1,i
m2,i
u(Tf ) =
u1 (T1,i ) +
u2 (T2,i ) .
(0.1)
m1,i + m2,i
m1,i + m2,i
Solution to Part (a)
We now work out the expression for the final mass distribution in each container. To
do this we take advantage of the fact that in the final state the gas in each container
is connected and therefore must be in equilibrium. This means that the final intensive
variables must the same in both containers. In particular, the specific volume (and
density) must be the same:
vf = v1,f = v2,f .
We can now use the definition of specific volume for the two-container system and the
above equations to get an expression for the final mass distribution in each container.
v1,f = v2,f = vf =
−
V total
mtotal
(0.2)
Lastly, applying the definition of specific volume to each container in the final state
together with the results in equation (0.2) gives the desired result:


−
V1
−
V
−
V

1
total


mtotal
= v1,f ≡


m1,f = −
m
V total
m1,f
total
⇒
(0.3)


−
V total
−
V2
−
V


2


= v2,f ≡
m2,f =
mtotal
mtotal
m2,f
−
V total
Notice that the initial mass distribution need not satisfy this relationship, in fact, it can
be anything because it is the designer that chooses the initial distribution, but once the
value is turned, nature takes over and there is only one possible outcome. This completes
part (a).
Solution to Part (b)
We now examine the special case: −
V1 = −
V 2 and m1,i = m2,i . From equation (0.3) we get
m1 = m1,i = m1,f
and m2 = m2,i = m2,f .
(∗)
Substituting equation (∗) into equation (0.1) yields
m2
m1
u(Tf ) =
u1 (T1,i ) +
u2 (T2,i )
m1 + m2
m1 + m2
·(m1 +m2 )
−−−−−−−−−→ (m1 + m2 )u(Tf ) = m1 u1 (T1,i ) + m2 u2 (T2,i )
(rearrange terms)
−−−−−−−−−−−−−→ m1 u(Tf ) − u1 (T1,i ) + m2 u(Tf ) − u2 (T2,i ) = 0
(intensive variable property)
−−−−−−−−−−−−−−−−→ m1 u1 (Tf ) − u1 (T1,i ) + m2 u2 (Tf ) − u2 (T2,i ) = 0
(definition of ∆u)
−−−−−−−−−−−−−→ m1 ∆u1 + m2 ∆u2 = 0
(0.4)
Here we used the fact that the internal energy is an intensive variable, so that it is the
same in each container in the final equilibrium state: u(Tf ) = u1 (Tf ) = u2 (Tf ).
For definiteness, we assume that T1,i < T2,i . We expect the final temperature to lie
between the minimum temperature of the two containers and the maximum temperature
of the two containers (i.e., T1,i < Tf < T2,i . This temperature range is assumed to be
moderate in order to justify our assumption that the specific heat is nearly constant over
this range, so in practice, we could approximate cv as the average over the temperature
range:
cv (T1,i ) + cv (T2,i )
.
cv ≈
2
Recall that if the specific heat is constant, then
∆u = cv ∆T, .
Substituting this result into equation (0.4) yields
m1 (cv ∆T1 ) + m2 (cv ∆T2 ) = 0
÷ cv 6=0
−−−−−−−→ m1 ∆T1 + m2 ∆T2 = 0
expand
−−−−−−−→ m1 (Tf − T1,i ) + m2 (Tf − T2,i ) = 0
solve
−−−−−−→
Tf =
m1 T1,i + m2 T2,i
.
m1 + m2
We can find the final pressure by using the ideal gas equation:
Rgas
(m1 + m2 )Tf
−
V total
sub for Tf
Rgas
(m1 T1,i + m2 T2,i )
−−−−−−−−−→ Pf =
−
V total
m1 T1,i + m2 T2,i
−
V total =V
−1 +V
−2
−−−−−−−−−−−−→ Pf = Rgas
.
−
V1 +−
V2
Pf =

Instructor: Hacker Engineering 232 Sample Exam 2 Solutions

Transcription

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