 # Instructor: Hacker Engineering 232 Sample Exam 3 Solutions

```Instructor: Hacker
Engineering 232
Sample Exam 3 Solutions
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. Each problem will be graded on a scale of 0 to 10 points. 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
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:
The basic problem set up (4 out of 10): (i) identify the control volume, (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. If it is helpful draw a Tv and/or a Pv-diagram.
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
Thermodynamics 232, Exam 3
2
The three fundamental equations for a steady-state control volume with one inlet/one
exit is
m
˙ =m
˙i=m
˙e
(Conservation of mass)
(Equation 1)
1
˙ cv − m∆
Q˙ cv = W
˙
h + V 2 + gz
(Conservation of energy)
2
vm
˙ = AV
(volumetric flow at each inlet and exit)
(Equation 2)
(Equation 3)
The idea gas model for air is
P v = Rair T,
u = u(T ),
h = h(T ),
where P is pressure, v is specific volume, T is temperature (measured in kelvin), u is
specific internal energy, h is specific enthalpy, and
Rair
kJ
8.314
R
kmol · K ≈ 0.287 kJ ,
=
=
kg
Mair
kg · K
28.97
kmol
where R is the universal gas constant, and Mair is the molecular mass.
For moderate temperatures of T ≈ 200 K, to T ≈ 440 K, the specific heats cv and cp are
approximately constant. Thus,
(
∆u ≈ cv ∆T ,
∆h ≈ cp ∆T .
Thermodynamics 232, Exam 3
3
Problem 1. (Diffuser) Air is flowing through a diffuser in a jet engine that is operating
under steady-state conditions. At the inlet: Ti = 10◦ C, Pi = 80 kPa, Vi = 200 m/s, and
the cross-sectional area of the inlet is Ai = 0.4 m2 . The air leaves the diffuser with a
velocity that is very small compared with the inlet velocity (i.e., Ve Vi ). Assuming
that we can model the air as an ideal gas (i.e., P v = Rair T , where Rair = R/Mair ), and
that the velocity is normal to cross-sectional area at the inlet and exit, determine
(a) the mass flow rate m
˙ through the diffuser and
(b) the temperature of the air leaving the diffuser Te .
Solution: We’ll start by writing down our control volume, what we’re given, what we
want, and our assumptions.
Control Volume: We’ll take our control volume around the diffuser. Recall a diffuser
acts like and looks like a nozzle in reverse.
Given:
Want:

Ti = 10◦ C
(inlet temperature)



P = 80 kPa (inlet pressure)
i
At the inlet:
Vi = 200 m/s (inlet velocity)



Ai = 4/10 m2 (inlet cross-sectional area)
(
m
˙ e =? (mass-flow rate at exit)
At the inlet:
Te =?
(exit temperature)
Assumptions:
⇒
d
d
mcv = Ecv = 0
dt
dt
2. One inlet/One exit
⇒
m
˙i=m
˙e=m
˙
3. Vi Ve
⇒
∆KE ≈ 12 Vi2
4. Can ignore any effects in the change in gravitational P.E.
5. No moving mechanical parts in CV
⇒
⇒
∆PE = 0
˙ cv = 0
W
6. Since the air is moving quickly, there is little time for significant heat transfer out
of the device ⇒ Q˙ cv = 0
7. We’ll model the air as an ideal gas
Notice that we are only given information at the inlet. Using fundamental equation 1
and 3 for a control volume with a single inlet/ single exit yields and equation for the
mass flow rate m
˙ at the exit.
m
˙e=m
˙i
(fundamental equations 1)
Ai Vi
=
(applying fundamental equation 3 at the inlet)
vi
Pi Ai Vi
=
(substituting for vi using the ideal gas law)
Rair Ti
(80 kPa)(4/10 m2 )(200 m/s)
kg
=
≈ 78.8
(0.287 kJ/(kg · K)((273 + 10) K)
s
Thermodynamics 232, Exam 3
4
˙ e = 79kg/s .
Thus, the answer to part (a) is m
For part (b) we need to find Te . In general, this would require finding two independent
properties of the system at the exit. However, since we have an ideal gas, part of the assumption of the ideal gas model is that the specific internal energy and specific enthalpy
are independent of pressure. That is, they can be completely determined by temperature.
In particular, he = h(Te ). Thus, we can use table A-22 to find Te , given he . This may
require interpolation, since we’re essentially using the table of data to invert the function
(i.e., Te = h−1 (he )).
To find he we’ll need to apply fundamental equation 2, the conservation of energy for
˙ cv = ∆gz = 0. Under
control volumes. From the assumptions, we have that Q˙ cv = W
these conditions fundamental equation 2 becomes
1 2
0 = 0 − m∆
˙
h+ V +0
2
1
÷(−m)
˙
−−−−−−−→ (he − hi ) + (Ve2 − Vi2 ) = 0
2
1 2
Solve for he
−−−−−−−−−−→ he = hi + (Vi − Ve2 )
2
1
≈ hi + Vi2
(By assumption: Ve Vi )
2
1
= h(Ti ) + Vi2
(For an ideal gas: h = h(T )) .
2
Thus,
1
he ≡ h(Te ) = h(Ti ) + Vi2 .
2
We are given the value of Vi , and we can use table A-22 to find h(Ti ) = h(283K). From
table A-22 we see that for a moderate temperature range of 200-440 K, the temperatures
and the enthalpies are approximately the same value. That there is a linear relationship
between h and T should be no surprise, since for an ideal gas over moderate temperature
ranges we have ∆h = cp ∆T , where cp ≈ 1 kJ/(kg · K).
1
he = hi + Vi2
2


kJ
m2  kg 
kJ 1

= 283 + (200)2 2 
kg 2
s  3 m2 
10 2
s
kJ
kJ
= (283 + 20) = 303 .
kg
kg
Since cp ≈ 1 over the range of temperatures considered, which is also verified by the
tables, we find that he = cp Te ≈ Te (with units of enthalphy). Replacing he by Te (with
units of he ) we find
kJ
303
he
kg
Te =
≈
= 303 K = 30◦ C .
kJ
cp
kg · K
Thermodynamics 232, Exam 3
5
Problem 2. (Steam Turbine)
˙ out = 5000 kilowatts.
The power output of an adiabatic steam turbine is known to be W


Pi = 2 MPa (inlet pressure)



◦


Ti = 400 C (inlet temperature)
The inlet conditions are: Vi = 50 m/s (inlet velocity)



zi = 10 m
(inlet height)



m
˙ i = 5.73 kg/s (inlet mass flow)

Pe = 15 kPa



x = .9
e
The exit conditions are:

V
e = 180 m/s



ze = 6 m
(exit pressure)
(exit quality)
(exit velocity)
(exit height)
Assumptions: The operating conditions for the turbine mechanism are
2. one inlet/one exit turbine
3. adiabatic turbine mechanism ⇒ No heat flow in or out of turbine
Using the above information answer the following questions:
(a) Sketch the location of state at the inlet and exit on a Pv-diagram.
(b) Determine the work done per unit of mass of the steam flowing through the turbine.
(c) In order to better understand the energy distribution throughout the turbine, compute the change in enthalpy ∆h, kinetic energy ∆KE, and potential energy ∆PE, for
the turbine. Compare the magnitudes of these quantities and rank them from most important to least important effects.
Thermodynamics 232, Exam 3
6
Problem 3. (Compressor)
Air enters a rotating one inlet/one exit compressor operating at a steady-state. The
compressor loses heat at a rate of Qcv = −1 KJ/s. The technicians have given the
following information at the inlet and exit of the compressor:

Pi = 1 bar
(inlet pressure)



T = 300 K (inlet temperature)
i
The inlet conditions are:

Vi = 10 m/s (inlet velocity)



Ai = 0.1 m2 (inlet cross-sectional area)


Pe = 10 bar
The exit conditions are: Te = 400 K


Ve = 1 m/s
(exit pressure)
(exit temperature)
(exit velocity)
˙ cv that must be supplied to
Using the ideal-gas model, determine the amount of power W
the compressor.
Thermodynamics 232, Exam 3
7
Problem 4. (The Shell-and Tube Heat Exchanger)
An isolated hotel in the wilds of Wyoming uses a geothermal heat exchanger to provide
hot water for the hotels needs. In addition to using the geothermal water to heat the tap
water, the hotel would like to use the leftover geothermal water to heat other devices.
The hotel has hired you to figure out the temperature of the geothermal water at the
exit of the inner tube T1,e .
The inner cylinder contains the hot geothermal water (the heating source) that is moving
at a steady mass flow rate of m
˙ 1 = 0.3kg/s through the inner cylinder. The initial
temperature at the inlet of the geothermal water is known to be T1,i = 140◦ C (assuming
the it is initially at the underground temperature), and the specific heat of the geothermal
water is known to be c1p = 4.31kJ/(kg·◦ C).
The outer cylinder contains the cold tap water that is to be heated. The cold water
in the outer cylinder is also moving at a steady mass flow rate of m
˙ 2 = 0.2kg/s. The
temperature at the inlet and exit of outer cylindrical pipe is measure at T2,i = 25◦ C and
T2,e = 60◦ C, respectively, and the specific heat of water is c2p = 4.18kJ/(kg·◦ C).
The outer boundary of the outer cylinder is well-insulated so that only a negligible amount
of energy is lost to the surroundings. There are no moving mechanical parts in the device
and it is assumed to operate under steady-state conditions. Each pipe can be treated
as a separate system having one inlet and one exit. We shall also ignore any change in
kinetic, or potential energies in the system as the fluid passes through each pipe.
Under these conditions and assumptions derive a formula the exit temperature of the
geothermal water.
Solution:
given:


˙ in = 0.3kg/s
m
Inner cylinder: Tin,i = 140◦ C


cp,in = 4.31kJ/(kg·◦ C)

m
˙ out = 0.2kg/s



T
◦
out,i = 25 C
Outer cylinder:
Tout,e = 60◦ C



cp,out = 4.18kJ/(kg·◦ C)
want: T1,e =? Note: As we will soon see, in order to find the temperature at the exit
we’ll need to solve for the rate of heat transfer in the heat exchanger, which will require
taking two separate control volumes: one around the inner tube and one around the
outer tube.
assumptions:
• We will assume steady-state operating conditions ⇒
dEcv
= 0.
dt
˙ cv = 0.
• No moving parts ⇒ W
• Insulated unit ⇒ Q˙ cv,ext = 0.
• Ignore any changes in K.E. and P.E. ⇒ ∆PE = ∆KE = 0.
• The fluid is incompressible ⇒ u = u(T ), ρ = constant, and cp ≈ constant.
Thermodynamics 232, Exam 3
8
m
˙ 1,i = m
˙ 1,e = m
˙1
m
˙ 2,i = m
˙ 2,e = m
˙2
(inner tube condition) ,
(outer tube condition) .
(0.1a)
(0.1b)
For each tube, the first law of thermodynamics (in the energy rate form) reduces to
Q˙ cv = m
˙ ∆h = m
˙ cp ∆T
Q˙ cv = m
˙ cp (Te − Ti ) .
(0.2)
Equation (0.2) will be our governing equation for modeling the heat flow in each pipe
separately. That is, if we take our control volumes to be the inner and outer tubes,
respectively, then we have
Q˙ 1,cv = m
˙ 1 c1p (T1,e − T1,i ) (inner tube governing equation) ,
Q˙ 2,cv = m
˙ 2 c2p (T2,e − T2,i ) (outer tube governing equation) .
(0.3a)
(0.3b)
Since the device is insulated, no heat escapes through the outer walls of the outer tube
(i.e., Q˙ cv,ext = 0). Thus, the energy transfer occurs between the inner wall of the outer
tube and the outer wall of the inner tube. That is, Q˙ 1,cv = −Q˙ 2,cv . Substituting this
result into equation (0.3a), and equating this resulting equation with (0.3b) yields
m
˙ 2 c2p (T2,e − T2,i ) = Q˙ 2,cv = −m
˙ 1 c1p (T1,e − T1,i )
m
˙ 1 c1p
(T1,e − T1,i )
⇒ T2,e = T2,i −
m
˙ 2 c2p
Substituting in the above given values into equation (0.4) gives T2,e ≈ 117.4◦ C.
(0.4)
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