This preview shows pages 1–4. Sign up to view the full content.
Chapter 9
•
Compressible Flow
9.1
An ideal gas flows adiabatically
through a duct. At section 1,
p
1
=
140 kPa,
T
1
=
260
°
C, and
V
1
=
75 m/s. Farther
downstream,
p
2
=
30 kPa and
T
2
=
207
°
C.
Calculate
V
2
in m/s and
s
2
−
s
1
in J/(kg
⋅
K) if
the gas is (a) air,
k
=
1.4, and (b) argon,
k
=
1.67.
Fig. P9.1
Solution:
(a) For air, take
k
=
1.40,
R
=
287 J/kg
⋅
K, and c
p
=
1005 J/kg
⋅
K. The adiabatic
steadyflow energy equation (9.23) is used to compute the downstream velocity:
+=
=
+
=
+
22
2
p
2
11
1
c T
V
constant
1005(260)
(75)
1005(207)
V
or
.
2
Ans
2
m
V
335
s
≈
21
p
2
1
2
1
207
273
30
Meanwhile, s
s
c ln(T /T )
R ln(p /p )
1005ln
287ln
,
260
273
140
+
±²
±
²
−=
−
=
−
³´
³
´
µ¶
µ
¶
+
or s
2
−
s
1
=
−
105
+
442
≈
337 J/kg
⋅
K
Ans
. (a)
(b) For argon, take
k
=
1.67,
R
=
208 J/kg
⋅
K, and c
p
=
518 J/kg
⋅
K. Repeat part (a):
2
p2
1
c T
V
518(260)
(75)
518(207)
V ,
solve
.
2
Ans
+
=
+
2
m
V
246
s
=
+
±
²
−
=
−+ ≈
³
´
µ
¶
+
207
273
30
s
s
518ln
208ln
54
320
. (b)
260
273
140
Ans
⋅
266 J/kg K
9.2
Solve Prob. 9.1 if the gas is steam. Use two approaches: (a) an ideal gas from Table A.4;
and (b) real steam from the steam tables [15].
Solution:
For steam, take
k
=
1.33,
R
=
461 J/kg
⋅
K, and c
p
=
1858 J/kg
⋅
K. Then
2
1
c T
V
1858(260)
(75)
1858(207)
V ,
solve
. (a)
2
Ans
+
=
+
2
m
V
450
s
≈
+
±
²
−
=
− + ≈
³
´
µ
¶
+
207
273
30
s
s
1858ln
461ln
195
710
. (a)
260
273
140
Ans
⋅
515 J/kg K
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 636
Solutions Manual
•
Fluid Mechanics, Fifth Edition
(b) For real steam, we look up each enthalpy and entropy in the Steam Tables:
1
2
J
at 140 kPa and 260 C,
read h
2.993E6
;
kg
J
at 30 kPa and 207 C,
h
2.893E6
kg
°=
+=
+
=
+
22
2
2
11
1
Then
h
V
2.993E6
(75)
2.893E6
V ,
solve
. (b)
2
Ans
2
m
V
453
s
≈
12
JJ
at 140 kPa and 260 C,
read s
7915
,
at 30 kPa and 207 C,
s
8427
kg K
kg K
°
=
⋅⋅
−=
−
≈
21
Thus
s
s
8427
7915
. (b)
Ans
⋅
512 J/kg K
These are within
±
1% of the ideal gas estimates (a). Steam is nearly ideal in this range.
9.3
If 8 kg of oxygen in a
closed tank
at 200
°
C and 300 kPa is heated until the pressure
rises to 400 kPa, calculate (a) the new temperature; (b) the total heat transfer; and (c) the
change in entropy.
Solution:
For oxygen, take
k
=
1.40,
R
=
260 J/kg
⋅
K, and c
v
=
650 J/kg
⋅
K. Then
ρρ
±²
=∴
=
=+
=
≈
³´
µ¶
2
1
400
,
T
T (p /p )
(200
273)
631 K
. (a)
300
Ans
358 C
°
v
Q
mc
T
(8)(650)(358
200)
. (b)
Ans
=∆
=
−
≈
8.2E5 J
+
=
≈
+
v
2
1
358
273
s
s
mc ln(T /T )
(8)(650)ln
. (c)
200
273
Ans
J
1500
K
9.4
Compressibility becomes important when the Mach number
>
0.3. How fast can a
twodimensional cylinder travel in sealevel standard air before compressibility becomes
important
somewhere
in its vicinity?
Solution:
For sealevel air,
T
=
288 K,
a
=
[1.4(287)(288)]
1/2
=
340
m/s. Recall from
Chap. 8 that incompressible theory predicts
V
max
=
2
U
∞
on a cylinder. Thus
∞
∞
===
=
≈
=
max
max
2
0.3(340)
0.3
when
.
340
2
V
U
Ma
U
Ans
a
mf
t
51
167
ss
Chapter 9
•
Compressible Flow
637
9.5
Steam enters a nozzle at 377
°
C, 1.6 MPa, and a steady speed of 200 m/s and
accelerates isentropically until it exits at saturation conditions. Estimate the exit velocity
and temperature.
Solution:
At saturation conditions, steam is
not ideal
. Use the Steam Tables:
At 377
°
C and 1.6 MPa, read
h
1
=
3.205E6 J/kg
and
s
1
=
7153 J/kg
⋅
K
At
saturation
for s
1
=
s
2
=
7153,
read p
2
=
185 kPa,
T
2
=
118
°
C
, and h
2
=
2.527E6 J/kg
22
2
2
11
1
Then
h
V
3.205E6
(200)
2.527E6
V ,
solve
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 04/07/2008 for the course MAW 101a taught by Professor Sarkar during the Winter '08 term at UCSD.
 Winter '08
 Sarkar

Click to edit the document details