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ME 3322 — Thermodynamics
Name:
Fall 2009 — Evaluation Problems
Date:
Instructions
This is a closedbook, closednotes exam.
Any necessary equations you may need are listed
in the problem itself or on the formula sheet.
You will need a calculator.
1.
Write the solution for each problem separately.
2.
Use only the front side of each sheet of paper.
3.
Be organized, clear, and neat.
4.
Don’t substitute in values until you need to.
5.
Use proper units in all calculations.
6.
Be consistent with your sign convention for heat and work from the very beginning.
Honor Pledge
On my honor, I pledge that I have neither given nor received inappropriate aid in the
preparation of this assignment.
Signature
No.
Do not turn in
your work.
Yes.
Turn in your
work for grading.
For each problem
you solved, did you
get the correct
answer?
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View Full Document ME 3322 Thermodynamics – Formula Sheet
Page 1
Control Volume Balance Equations (uniform flow)
Sign convention:
,
, and
W
are
positive
going
into
the control volume
m
±
Q
±
±
c
V
and
negative
going
out
.
Mass,
m
,/
dm
mmA
V
v
dt
==
±±
Energy,
E
()
2
1
2
,
dE
QW mh V g
z EU K
E P
E
dt
=++
+
+
=+ +
±
±
±
Entropy,
S
,0
s
dS
Q
ms
dt
T
Σ
Σ
≥
±
±
Expansion Work Equations
Rate form:
d
Wp
dt
=−
±
V
Process form:
2
12
1
∫
V
d
Heat Transfer Rate Equations
Conduction
dT
Qk
A
dx
±
positive in the positive
x
direction
Convection
0
s
Qh
A
TT
±
positive from the surface at
T
s
to
the surroundings at
T
0
,
T
s
> T
0
Radiation
44
0
s
QA
T
T
εσ
±
positive from the surface at
T
s
to
the surroundings at
T
0
,
T
s
> T
0
Polytropic Process Equations
constant
n
p
=
V
or
21
12
n
p
p
⎛⎞
=
⎜⎟
⎝⎠
V
V
constant
n
pv
=
or
n
pv
=
Process work:
22
11
12
1
pp
W
n
−
−
VV
Process work:
12
1
p
vp
v
Wm
n
−
−
Constants
Units
Standard atmosphere:
1 atm = 101.3 kPa
Standard gravity:
g
= 9.807 m/s
2
Temperature relation:
T
(K) =
T
(ºC) + 273.15 K
Universal gas constant:
R
= 8.314 kJ/(kmol·K)
Gas constant, air:
R
= 0.287 kJ/(kg·K)
Molecular weight, air:
M
= 28.97 kg/kmol
StefanBoltzmann constant:
82
5.67 10 W/(m
K )
σ
−
=×
⋅
4
1 bar = 100 kPa
W = J/s
J = N·m
N = kg·m/s
2
Pa = N/m
2
Energy Transfer in a Steady Internally Reversible Flow with one inlet and one outlet
Work:
( )
/
out
out
in
in
W m
w
vdp
V
gz
V
gz
+
+
−
+
∫
±
±
Heat:
/
out
in
Qm q
T
d
s
∫
±
±
ME 3322 Thermodynamics – Formula Sheet
Page 2
Pure Simple Substance Property Models
Incompressible
Compressible
Equation of state:
y
=
tsc
(
y
1
,
y
2
)
The
y
variables represent independent,
intrinsic thermodynamic properties.
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This note was uploaded on 09/28/2010 for the course ME 3322 taught by Professor Neitzel during the Fall '07 term at Georgia Institute of Technology.
 Fall '07
 Neitzel

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