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Unformatted text preview: Problem 13.1 Given: Air extracted from a large tank Find: [2] Mass flow rate Solution:
Basic equations: mrate = ρ⋅ V⋅ A h1 + V1 2 V2 2 p = h2 +
2 2 ρ k = const The mass flow rate is given by T0 = ( 70 + 273) ⋅ K p0 = 101⋅ kPa
J
cp = 1004⋅
kg⋅ K k = 1.4 mrate = ρ⋅ A⋅ V A= We need the density and velocity at the nozzle. In the tank From the isentropic relation = const p = 25⋅ kPa D = 15⋅ cm Given or available data T⋅ p ( 1−k)
k p⎞
ρ = ρ0⋅ ⎛ ⎟
⎜p
⎝ 0⎠ ρ0 = 1
k R = 286.9⋅ 2 π⋅ D
4 J
kg⋅ K 2 A = 0.0177 m p0 ρ0 = 1.026 R ⋅ T0 kg
3 m kg ρ = 0.379 3 m We can apply the energy equation between the tank (stagnation conditions) and the point in the nozzle to find the velocity
2 h0 = h + V
2 V= ( ) 2⋅ h 0 − h = ⎛ p0 ⎞
T = T0⋅ ⎜ ⎟
⎝p⎠ Fot T we again use insentropic relations ( Then V= 2⋅ cp⋅ T0 − T The mass flow rate is mrate = ρ⋅ A⋅ V Note that the flow is supersonic at this point
Hence we must have a convergingdiverging nozzle ) V = 476 ( 2⋅ cp⋅ T0 − T ) ( 1−k)
k T = 230.167 K T = −43.0⋅ °C m
s kg
mrate = 3.18
s
c= k⋅ R⋅ T 304
c= m
s M= V
c M = 1.57 Problem 13.3 [2] Given: Steam flow through a nozzle Find: Speed and Mach number; Mass flow rate; Sketch the shape Solution:
Basic equations: mrate = ρ⋅ V⋅ A h1 + V1 2 V2 2 = h2 +
2 2 Assumptions: 1) Steady flow 2) Isentropic 3) Uniform flow 4) Superheated steam can be treated as ideal gas
T0 = ( 450 + 273) ⋅ K p0 = 6⋅ MPa p = 2⋅ MPa D = 2⋅ cm Given or available data k = 1.30 R = 461.4⋅ J
kg⋅ K (Table A.6) From the steam tables (try finding interactive ones on the Web!), at stagnation conditions
J
s0 = 6720⋅
kg⋅ K Hence at the nozzle section h0 = 3.302 × 10 ⋅ 6J J
and
s = s0 = 6720⋅
kg⋅ K p = 2 MPa
T = 289 °C From these values we find from the steam tables that ( 2⋅ h 0 − h ) Hence the first law becomes V= The mass flow rate is given by mrate = ρ⋅ A⋅ V = Hence mrate = For the Mach number we need c= A⋅ V
v k⋅ R ⋅ T V = 781
A⋅ V
v kg 6J h = 2.997 × 10 ⋅ m
s
2 A= kg 3 v = 0.1225⋅ π⋅ D
4 A = 3.14 × 10 −4 2 m kg
mrate = 2.00
s
c = 581 The flow is supersonic starting from rest, so must be convergingdiverging m
s M= V
c M = 1.35 m
kg Problem 13.4 Given: Air flow in a passage Find: [2] Mach number; Sketch shape Solution:
Basic equations: p0
p ⎛
⎝ = ⎜1 + k − 1 2⎞
⋅M ⎟
2
⎠ k
k−1 c= k⋅ R⋅ T
m
s T1 = ( 10 + 273) ⋅ K p1 = 150⋅ kPa V1 = 120⋅ p2 = 50⋅ kPa k = 1.4 R = 286.9⋅ The speed of sound at state 1 is c1 = m
c1 = 337
s Hence V1
M1 =
c1 Given or available data k⋅ R⋅ T1 M1 = 0.356 For isentropic flow stagnation pressure is constant. Hence at state 2 Hence Solving for M2 p0 = p1⋅ ⎛ 1 +
⎜ ⎝ M2 = k−1
2⎞
⋅ M1 ⎟
2
⎠ p0
p2 k−1
2⎞
= ⎛1 +
⋅ M2 ⎟
⎜
2
⎝
⎠ k
k−1 k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k − 1 p2
⎣⎝ ⎠
⎦ p0 = 164 kPa M2 = 1.42 Hence, as we go from subsonic to supersonic we must have a convergingdiverging nozzle k
k−1 J
kg⋅ K Problem 13.5 [2] Given: Data on flow in a passage
Find: Pressure at downstream location Solution:
ft·lbf/lbm·oR R=
k=
T1 = 53.33
1.4
560 p1 = The given or available data is: 30 psi
ft/s o R V1 = 1750 M2 = 2.5 c1 = 1160 M1 = 1.51 p 01 = 111 psi p 02 = 111 psi p2 = 6.52 psi Equations and Computations:
From T 1 and Eq. 12.18 Then ft/s From M 1 and p 1, and Eq. 13.7a
(using builtin function Isenp (M ,k )) For isentropic flow (p 01 = p 02) From M 2 and p 02, and Eq. 13.7a
(using builtin function Isenp (M ,k )) Problem 13.6 [3] Given: Data on flow in a nozzle
Find: Mass flow rate; Throat area; Mach numbers Solution:
J/kg·K R=
k=
T0 =
p1 = 286.9
1.4
523
200 K
kPa A= The given or available data is: 1 cm2 p2 = 50 kPa Equations and Computations:
We don't know the two Mach numbers. We do know for each that Eq. 13.7a applies: Hence we can write two equations, but have three unknowns ( 1, M 2, and p 0)!
M
We also know that states 1 and 2 have the same area. Hence we can write Eq. 13.7d twice: We now have four equations for four unknowns (A *, M 1, M 2, and p 0)!
We make guesses (using Solver) for M 1 and M 2, and make the errors in computed A * and p 0 zero.
M1 = 0.512 from Eq. 13.7a: p0 = 239 and from Eq. 13.7d: A* = 0.759 cm For: M2 = 1.68 kPa p0 = 239 kPa 0.00% 2 A* = 0.759 cm2 0.00% Note that the throat area is the critical area Sum The stagnation density is then obtained from the ideal gas equation
ρ0 = 1.59 kg/m3 The density at critical state is obtained from Eq. 13.7a (or 12.22c)
ρ* = Errors 1.01 3 kg/m The velocity at critical state can be obtained from Eq. 12.23) V* = 418 m/s m rate = 0.0321 kg/s The mass flow rate is ρ*V *A * 0.00% Problem 13.8 Given: Air flow in a passage Find: [3] Speed and area downstream; Sketch flow passage Solution:
Basic equations: T0
T k−1 2
⋅M
2 = 1+ c= ⎛ 1 + k − 1 ⋅ M2 ⎞
⎟
A
1⎜
2
=
⋅⎜
⎟
k +1
Acrit
M⎜
⎟
2
⎝
⎠ k ⋅ R⋅ T T1 = ( 32 + 460) ⋅ R p1 = 25⋅ psi M1 = 1.75 T2 = ( 225 + 460) ⋅ R Given or available data k = 1.4 Rair = 53.33⋅
2 D1 = 3⋅ ft
Hence A1 = T0 = T1⋅ ⎛ 1 +
⎜ ⎝ π⋅ D 1 T0 = 793 R k −1
2⎞
⋅ M1 ⎟
2
⎠ 4 A1 = 7.07 ft ft⋅ lbf
lbm⋅ R 2 T0 = 334 °F For isentropic flow stagnation conditions are constant. Hence
M2 = ⎞
2 ⎛ T0
⋅⎜
− 1⎟
k − 1 T2
⎝ We also have c2 = Hence Acrit = ft
c2 = 1283
s V2 = M2⋅ c2 From state 1 M2 = 0.889 ⎠ k⋅ Rair⋅ T2 V2 = 1141
A1⋅ M1 ⎛ 1 + k − 1 ⋅M 2 ⎞
⎜
1⎟
2
⎜
⎟
k+ 1
⎜
⎟
2
⎝
⎠ Hence at state 2 k+ 1
2⋅ ( k−1) k−1
2⎞
⎛
Acrit ⎜ 1 + 2 ⋅ M2 ⎟
A2 =
⋅⎜
⎟
k+ 1
M2 ⎜
⎟ ⎝ 2 ft
s
Acrit = 5.10 ft k+ 1
2⋅ ( k−1) A2 = 5.15 ft ⎠ Hence, as we go from supersonic to subsonic we must have a convergingdiverging diffuser 2 2 k +1
2⋅ ( k −1) Problem 13.10 [2] Given: Data on flow in a passage
Find: Flow rate; area and pressure at downstream location; sketch passage shape Solution:
R=
k=
A1 = 286.9
1.4
0.25 T1 = 283 K p1 = 15 kPa V1 = 590 m/s T2 = 410 M2 = The given or available data is: 0.75 J/kg.K
m2 Equations and Computations:
From T 1 and Eq. 12.18 (12.18)
c1 = Then 337 M1 = 1.75 m/s Because the flow decreases isentropically from supersonic to subsonic
the passage shape must be convergentdivergent From p 1 and T 1 and the ideal gas equation
ρ1 = 0.185 kg/m3 m rate = 27.2 kg/s The mass flow rate is m rate = ρ1A 1V 1 From M 1 and A 1, and Eq. 13.7d
(using builtin function IsenA (M ,k )) (13.7d) A* = 0.180 m2 A2 = 0.192 m2 From M 2 and A *, and Eq. 13.7d
(using builtin function IsenA (M ,k )) From M 1 and p 1, and Eq. 13.7a
(using builtin function Isenp (M ,k )) (13.7a)
p 01 = 79.9 kPa p 02 = 79.9 kPa p2 = 55.0 kPa For isentropic flow (p 01 = p 02) From M 2 and p 02, and Eq. 13.7a
(using builtin function Isenp (M ,k )) Problem 13.11 [3] Given: Flow in a converging nozzle to a pipe
Find: Plot of mass flow rate Solution:
The given or available data is R = 287 J/kg·K
k = 1.4
T 0 = 293 K
p 0 = 101 kPa
Dt = 1 cm
2
A t = 0.785 cm Equations and Computations:
The critical pressure is given by p * = 53.4 kPa
Hence for p = 100 kPa down to this pressure the flow gradually increases; then it is constant
c V = M ·c ρ = p /RT
3
(m/s) (m/s)
(kg/m )
343
41
1.19
342
58
1.18
342
71
1.18
341
82
1.17
341
92
1.16
340
101
1.15
337
138
1.11
335
168
1.06
332
195
1.02
329
219
0.971
326
242
0.925
322
264
0.877
318
285
0.828
315
306
0.778
313
313
0.762
313
313
0.762
313
313
0.762
313
313
0.762
313
313
0.762 Flow
(kg/s)
0.00383
0.00539
0.00656
0.00753
0.00838
0.0091
0.0120
0.0140
0.0156
0.0167
0.0176
0.0182
0.0186
0.0187
0.0187
0.0187
0.0187
0.0187
0.0187 Using critical conditions, and Eq. 13.9 for mass flow rate:
53.4
1.000
244
313
313
0.762
0.0185
(Note: discrepancy in mass flow rate is due to roundoff error) Flow Rate in a Converging Nozzle
0.020
0.018
0.016
Flow Rate (kg/s) p
M
T (K)
(kPa) (Eq. 13.7a) (Eq. 13.7b)
100
0.119
292
99
0.169
291
98
0.208
290
97
0.241
290
96
0.270
289
95
0.297
288
90
0.409
284
85
0.503
279
80
0.587
274
75
0.666
269
70
0.743
264
65
0.819
258
60
0.896
252
55
0.974
246
53.4
1.000
244
53
1.000
244
52
1.000
244
51
1.000
244
50
1.000
244 0.014
0.012
0.010
0.008
0.006
0.004
0.002
0.000
50 60 70 80
p (kPa) 90 100 Problem 13.12 [2] Given: Flow in a convergingdiverging nozzle to a pipe
Find: Plot of mass flow rate Solution:
J/kg·K R=
k=
T0 = 286.9
1.4
293 p0 = 101 kPa Dt = 1 At = 0.785 cm
cm2 p* = The given or available data is 53.4 kPa K
De = 2.5 Ae = 4.909 cm
cm2 Equations and Computations:
The critical pressure is given by This is the minimum throat pressure For the CD nozzle, we can compute the pressure at the exit required for this to happen
A* =
A e/A * = 0.785 6.25
M e = 0.0931
p e = 100.4 cm2 (= A t)
or 3.41 (Eq. 13.7d) or 67.2 kPa (Eq. 13.7a) Hence we conclude flow occurs in regimes iii down to v (Fig. 13.8); the flow is ALWAYS choked! p*
M
T * (K)
c*
V * = c * ρ = p /RT
(kPa) 13.7a) (Eq. 13.7b) (m/s)
(m/s)
(kg/m3)
53.4
1.000
244
313
313
0.762
(Note: discrepancy in mass flow rate is due to roundoff error) Flow
(kg/s)
0.0187
0.0185 (Using Eq. 13.9) Problem 13.13 [3] Given: Data on tank conditions; isentropic flow
Find: Plot crosssection area and pressure distributions Solution:
R=
k=
T0 = The given or available data is: 53.33
1.4
500 ft·lbf/lbm·oR
o R p0 = 45 psia pe = 14.7 psia m rate = 2.25 lbm/s Equations and Computations:
From p 0, p e and Eq. 13.7a (using builtin function IsenMfromp (M,k)) (13.7a)
Me = 1.37 Because the exit flow is supersonic, the passage must be a CD nozzle
We need a scale for the area.
From p 0, T 0, m flow, and Eq. 13.10c (13.10c)
Then At = A* = 0.0146 ft2 For each M , and A *, and Eq. 13.7d
(using builtin function IsenA (M ,k ) (13.7d) we can compute each area A .
From each M , and p 0, and Eq. 13.7a
(using builtin function Isenp (M ,k )
we can compute each pressure p . L (ft) A (ft 2) 1.00
1.25
1.50
1.75
2.00
2.50
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
14.6
16.00
17.00
18.00
19.00
20.00 p (psia) 0.1234
0.0989
0.0826
0.0710
0.0622
0.0501
0.0421
0.0322
0.0264
0.0227
0.0201
0.0183
0.0171
0.0161
0.0155
0.0150
0.0147
0.0146
0.0146
0.0147
0.0149
0.0152
0.0156
0.0161 M
0.069
0.086
0.103
0.120
0.137
0.172
0.206
0.274
0.343
0.412
0.480
0.549
0.618
0.686
0.755
0.823
0.892
0.961
1.000
1.098
1.166
1.235
1.304
1.372 44.9
44.8
44.7
44.5
44.4
44.1
43.7
42.7
41.5
40.0
38.4
36.7
34.8
32.8
30.8
28.8
26.8
24.9
23.8
21.1
19.4
17.7
16.2
14.7 Area Variation in Passage
0.14
0.12 A (ft2) 0.10
0.08
0.06
0.04
0.02
0.00
0 5 10 15 20 L (ft) p (psia) Pressure Variation in Passage
50
45
40
35
30
25
20
15
10
5
0
0 2 4 6 8 10
L (ft) 12 14 16 18 20 Problem 13.14 Given: Air flow in a converging nozzle Find: [2] Mass flow rate Solution:
Basic equations: mrate = ρ⋅ V⋅ A Given or available data pb = 35⋅ psi pb
p0 = 0.583 T p0 = 60⋅ psi k = 1.4 Since T0 p = ρ⋅ R ⋅ T Rair = 53.33⋅ Mt = and Tt = ft⋅ lbf
lbm⋅ R ct = ρt = k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k−1
⎣⎝ pt ⎠
⎦ T0
1+ k−1
2 2 ⋅ Mt k⋅ Rair⋅ Tt
pt
Rair⋅ Tt mrate = ρt⋅ At⋅ Vt k−1 2
⋅M
2 T0 = ( 200 + 460) ⋅ R is greater than 0.528, the nozzle is not choked and Hence = 1+ At = π2
⋅D
4t pt = pb Mt = 0.912 Tt = 566⋅ R Tt = 106⋅ °F Vt = ct Vt = 1166⋅ ρt = 5.19 × 10 ft
s − 3 slug
⋅
3 ft slug
mrate = 0.528⋅
s lbm
mrate = 17.0⋅
s p0
p ⎛
⎝ = ⎜1 + k − 1 2⎞
⋅M ⎟
2
⎠ Dt = 4⋅ in
At = 0.0873⋅ ft 2 k
k −1 Problem 13.15 Given: Isentropic air flow in converging nozzle Find: [2] Pressure, speed and Mach number at throat Solution:
Basic equations: T0
T p0 k −1 2
⋅M
2 =1+
p ⎛
⎝ k − 1 2⎞
⋅M ⎟
2
⎠ = ⎜1 + m
s p1 = 350⋅ kPa V1 = 150⋅ k = 1.4 Given or available data R = 286.9⋅ k
k −1 M1 = 0.5 pb = 250⋅ kPa J
kg⋅ K The flow will be choked if pb/p0 < 0.528
p0 = p1 ⋅ ⎛ 1 +
⎜ ⎝ Hence so p0
pt k −1
2⎞
⋅ M1 ⎟
2
⎠ k −1
2⎞
= ⎛1 +
⋅ Mt ⎟
⎜
2
⎝
⎠ Mt = k
k −1 where Also V1 = M1⋅ c1 = M1⋅ k⋅ R⋅ T1 Then T0 = T1⋅ ⎛ 1 +
⎜ Hence Tt = Then ct = Finally Vt = Mt⋅ ct ⎝ 2 T0
k−1
2
⋅ Mt
1+
2
k⋅ R⋅ Tt p0 = 0.602 (Not choked) k
k −1 k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k−1
⎣⎝ pt ⎠
⎦ k−1 pb p0 = 415 kPa 2⎞
⋅ M1 ⎟ ⎠ pt = pb pt = 250 kPa Mt = 0.883 or 1 ⎛ V1 ⎞
T1 =
⋅⎜
⎟
k⋅ R M1
⎝⎠ 2 T1 = 224 K T0 = 235 K T0 = −37.9 °C Tt = 204 K Tt = −69.6 °C m
ct = 286
s
Vt = 252 m
s T1 = −49.1 °C Problem 13.16 [3] Given: Data on three tanks
Find: Mass flow rate; Pressure in second tank Solution:
The given or available data is: R=
k=
At = 286.9
1.4
1 J/kg.K
cm2 We need to establish whether each nozzle is choked. There is a large total pressure drop so this is likely.
However, BOTH cannot be choked and have the same flow rate. This is because Eq. 13.9a, below
(13.9b) indicates that the choked flow rate depends on stagnation temperature (which is constant) but also
stagnation pressure, which drops because of turbulent mixing in the middle chamber. Hence BOTH nozzles
cannot be choked. We assume the second one only is choked (why?) and verify later.
T 01 =
p 01 =
p e1 =
We make a guess at the pressure at the first nozzle exit:
NOTE: The value shown is the final answer! It was obtained using Solver !
p 02 =
This will also be tank 2 stagnation pressure:
p3 =
Pressure in tank 3:
Temperature and pressure in tank 1: 308
650
527 K
kPa
kPa 527
65 kPa
kPa Equations and Computations:
From the p e1 guess and Eq. 13.17a:
Then at the first throat (Eq.13.7b):
The density at the first throat (Ideal Gas) is:
Then c at the first throat (Eq. 12.18) is:
Then V at the first throat is:
Finally the mass flow rate is: M e1 =
T e1 = 0.556
290 ρ e1 = 6.33
341
190
0.120 c e1 =
V e1 =
m rate = K
kg/m3
m/s
m/s
kg/s First Nozzle! For the presumed choked flow at the second nozzle we use Eq. 13.9a, with T 01 = T 02 and p 02:
m rate = 0.120 kg/s For the guess value for p e1 we compute the error between the two flow rates:
Δm rate =
0.000
Use Solver to vary the guess value for p e1 to make this error zero!
Note that this could also be done manually. kg/s Second Nozzle! Problem 13.17 [2] Problem 13.19 [2] Given: Data on converging nozzle; isentropic flow
Find: Pressure and Mach number; throat area; mass flow rate Solution:
R=
k=
A1 = 286.9
1.4
0.05 J/kg.K T1 = 276.3 K V1 = 200 m/s p atm = The given or available data is: 101 kPa m2 Equations and Computations:
From T 1 and Eq. 12.18 (12.18)
c1 = Then 333 M1 = 0.60 m/s To find the pressure, we first need the stagnation pressure.
If the flow is just choked
pe =
p atm = p* = 101 kPa From p e = p * and Eq. 12.22a (12.22a)
p0 = 191 kPa From M 1 and p 0, and Eq. 13.7a
(using builtin function Isenp (M ,k ) (13.7a)
Then p1 = 150 kPa The mass flow rate is m rate = ρ1A 1V 1
Hence, we need ρ1 from the ideal gas equation.
ρ1 = 1.89 kg/m3 m rate = 18.9 kg/s The mass flow rate m rate is then The throat area A t = A * because the flow is choked.
From M 1 and A 1, and Eq. 13.7d
(using builtin function IsenA (M ,k ) (13.7d) A* =
Hence 0.0421 m2 At = 0.0421 m2 Problem 13.20 [2] Problem 13.21 [2] Problem 13.23 [2] Given: Temperature in and mass flow rate from a tank
Find: Tank pressure; pressure, temperature and speed at exit Solution:
R=
k=
T0 = 286.9
1.4
273 At = The given or available data is: 0.001
2 m rate = J/kg.K
K
m2
kg/s Equations and Computations:
Because p b = 0
Hence the flow is choked! pe = p* Hence Te = T* From T 0, and Eq. 12.22b
(12.22b)
T* =
Te = 228
228
45.5 Also Me = Hence Ve = K
K
o C 1
V* = From T e and Eq. 12.18 ce
(12.18) ce =
Then 302 m/s Ve = 302 m/s To find the exit pressure we use the ideal gas equation
after first finding the exit density.
The mass flow rate is m rate = ρeA eV e
ρe = Hence 6.62 kg/m3 432 kPa From the ideal gas equation p e = ρeRT e
pe =
From p e = p * and Eq. 12.22a
(12.22a)
p0 = 817 kPa We can check our results:
From p 0, T 0, A t, and Eq. 13.9a (13.9a)
Then m choked =
m choked = 2.00
m rate kg/s
Correct! Problem 13.24 [2] Given: Isentropic air flow into a tank Find: Initial mass flow rate; Ts process; explain nonlinear mass flow rate Solution:
Basic equations: T0
T = 1+ p0 k−1 2
⋅M
2 p Then p0 = 101⋅ kPa pb = p0 − 10⋅ kPa k = 1.4 Given or available data R = 286.9⋅ A= π2
⋅D
4 Avena = 65⋅ %⋅ A
pb The flow will be choked if pb/p0 < 0.528 Hence p0 p0 k − 1 2⎞
= ⎛1 +
⋅M ⎟
⎜
pvena ⎝
2
⎠ = 0.901 Mvena = Then Tvena = Then cvena = and Vvena = Mvena⋅ cvena Also ρvena = k − 1 2⎞
⋅M ⎟
2
⎠ mrate = ρ⋅ A⋅ V pb = 91⋅ kPa T0 = ( 20 + 273) ⋅ K D = 5⋅ mm
2 Avena = 12.8⋅ mm (Not choked) k
k −1 k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜
− 1⎥
⎟
k − 1 pvena
⎣⎝
⎠
⎦ so Finally J
kg⋅ K ⎛
⎝ = ⎜1 + k
k−1 T0
k−1
2
⋅ Mvena
1+
2
k⋅ R⋅ Tvena pvena
R⋅ Tvena mrate = ρvena⋅ Avena⋅ Vvena where pvena = pb pvena = 91⋅ kPa Mvena = 0.389 Tvena = 284 K Tvena = 11.3⋅ °C m
cvena = 338
s
Vvena = 131 m ρvena = 1.12 kg s 3 m − 3 kg mrate = 1.87 × 10 s The Ts diagram will be a vertical line (T decreases and s = const). After entering the tank there will be turbulent mixing (s increases) and th
comes to rest (T increases). The mass flow rate versus time will look like the curved part of Fig. 13.6b; it is nonlinear because V AND ρ va Problem 13.25 Given: Spherical cavity with valve Find: [3] Time to reach desired pressure; Entropy change Solution:
Basic equations: T0
T = 1+ k−1 2
⋅M
2 p0
p ⎛
⎝ = ⎜1 + k − 1 2⎞
⋅M ⎟
2
⎠ p = ρ⋅ R ⋅ T Then the inlet area is p0 = 101⋅ kPa Tatm = ( 20 + 273) ⋅ K pf = 45⋅ kPa Given or available data c= Tf = Tatm At = π2
⋅d
4 k⋅ R⋅ T ρf = 2 ρf = 0.535 k ⎛2⎞
We have choked flow so mrate = At⋅ p0⋅
⋅⎜
⎟
R ⋅ T0 ⎝ k + 1 ⎠
Δt = 3 d = 1⋅ mm R = 286.9⋅ pb = pf pb so D = 50⋅ cm J
kg⋅ K and tank volume is V = p0 k+ 1
2⋅ ( k−1) J
cp = 1004⋅
kg⋅ K π3
⋅D
3 3 V = 0.131 m = 0.446 (Choked) and final mass is M = ρf ⋅ V M = 0.0701 kg m Since the mass flow rate is constant (flow is always choked) Hence kg k ⎛2⎞
mchoked = At⋅ p0⋅
⋅⎜
⎟
R ⋅ T0 ⎝ k + 1 ⎠ T0 = Tatm k = 1.4 At = 0.785 mm pf
R ⋅ Tf ⎛ T2 ⎞
⎛ p2 ⎞
Δs = cp⋅ ln ⎜ ⎟ − R⋅ ln ⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠ mrate = ρ⋅ A⋅ V The flow will be choked if pb/p0 < 0.528; the MAXIMUM back pressure is
The final density is k
k−1 M k+ 1
2⋅ ( k−1) Δt = or M
mrate − 4 kg mrate = 1.873 × 10 Δt = 374 s mrate M = mrate⋅ Δt s Δt = 6.23 min The air in the tank will be cold when the valve is closed. Because ρ =M/V is constant, p = ρRT = const x T, so as the temperature rises to
ambient, the pressure will rise too. ⎛ T2 ⎞
⎛ p2 ⎞
For the entropy change during the charging process is given by Δs = cp⋅ ln ⎜ ⎟ − R⋅ ln ⎜ ⎟ where T1 = Tatm
⎝ T1 ⎠
⎝ p1 ⎠
and p1 = p0 p2 = pf Hence ⎛ T2 ⎞
⎛ p2 ⎞
Δs = cp⋅ ln ⎜ ⎟ − R⋅ ln ⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠ Δs = 232 T2 = Tatm J
kg⋅ K Problem 13.26 [3] Problem 13.27 [3] Problem 13.28 [3] Problem 13.29 [3] Given: Airdriven rocket in space
Find: Tank pressure; pressure, temperature and speed at exit; initial acceleration Solution:
R=
k=
T0 = 286.9
1.4
398 At =
M=
m rate = 25
25
0.05 Because p b = 0
Hence the flow is choked! pe = p* Hence Te = T* The given or available data is: J/kg.K
K
mm2
kg
kg/s Equations and Computations: From T 0, and Eq. 12.22b
(12.22b)
T* = 332 Te = 332 K 58.7 o Also Me = Hence Ve = K C 1
V* = From T e and Eq. 12.18 ce
(12.18) ce =
Then 365 m/s Ve = 365 m/s To find the exit pressure we use the ideal gas equation
after first finding the exit density.
The mass flow rate is m rate = ρeA eV e
ρe = 0.0548 pe = Hence 5.21 kg/m3 From the ideal gas equation p e = ρeRT e
kPa From p e = p * and Eq. 12.22a
(12.22a)
p0 = 9.87 kPa We can check our results:
From p 0, T 0, A t, and Eq. 13.9a
(13.9a) m choked = Then m choked = 0.050
m rate kg/s
Correct! The initial acceleration is given by:
(4.33) which simplifies to: pe At − Max = mrateV
ax = or:
1.25 ax =
m/s2 m rate V + p e At
M Problem 13.30 Given: Gas cylinder with broken valve Find: [3] Mass flow rate; acceleration of cylinder Solution:
Basic equations: T0
T = 1+ k−1 2
⋅M
2 p0
p ⎛
⎝ = ⎜1 + k − 1 2⎞
⋅M ⎟
2
⎠ k
k−1 p = ρ⋅ R ⋅ T c= k⋅ R⋅ T mrate = ρ⋅ A⋅ V (4.33) Given or available data patm = 101⋅ kPa p0 = 20⋅ MPa d = 10⋅ mm pb = patm T0 Ve = ce The exit pressure is pe = Ve = 313
p0 ⎛1 + k − 1 ⎞
⎜
⎟
2⎠
⎝
Then k
k−1 mrate = ρe⋅ Ae⋅ Ve The momentum equation (Eq. 4.33) simplifies to Hence ax = so Te = 244 K ⎛1 + k − 1 ⎞
⎜
⎟
2⎠
⎝ The exit speed is Ae = so the nozzle area is The flow will be choked if pb/p0 < 0.528: The exit temperature is Te = T0 = ( 20 + 273) ⋅ K p0 = 5.05 × 10 Ae = 78.5⋅ mm
−3 J
kg⋅ K R = 286.9⋅
2 MCV = 65⋅ kg (Choked: Critical conditions) Te = −29⋅ °C ce = and exit density is ρe = k⋅ R⋅ Te m
s pe = 10.6⋅ MPa kg
mrate = 3.71
s (pe − patm)⋅ Ae − MCV⋅ ax = −Ve⋅ mrate (pe − patm)⋅ Ae + Ve⋅ mrate
MCV pb π2
⋅d
4 k = 1.4 m
ax = 30.5
2
s The process is isentropic, followed by nonisentropic expansion to atmospheric pressure pe
R ⋅ Te ρe = 151 kg
3 m Problem 13.32 Given: Spherical air tank Find: [4] Air temperature after 30s; estimate throat area Solution:
Basic equations: T0 = 1+ T k−1 2
⋅M
2 p
ρ k ⌠ → ⎯→
⎯
⎮
∂⌠
⎮
ρ dVCV + ⎮ ρ⋅ V dACS = 0
∂t ⎮
⌡
⌡ = const (4.12) Assumptions: 1) Large tank (stagnation conditions) 2) isentropic 3) uniform flow
patm = 101⋅ kPa p1 = 2.75⋅ MPa T1 = 450⋅ K D = 2⋅ m ΔM = 30⋅ kg Given or available data Δt = 30⋅ s k = 1.4 R = 286.9⋅ pb = patm The flow will be choked if pb/p1 < 0.528: so pb
p1 = 0.037 V= π3
⋅D
6 3 V = 4.19⋅ m J
kg⋅ K (Initially choked: Critical conditions) We need to see if the flow is still choked after 30s
ρ1 = The initial (State 1) density and mass are For an isentropic process ρ
The final temperature is k = const T2 = p2
ρ2⋅ R To estimate the throat area we use ρ1 = 21.3 R ⋅ T1 so kg
3 M1 = ρ1⋅ V ⎛ ρ2 ⎞
p2 = p1⋅ ⎜ ⎟
⎝ ρ1 ⎠ M2 = 59.2 kg k p2 = 1.55⋅ MPa T2 = 382 K ρ2 =
pb
p2 M2
V ρ2 = 14.1 or = 0.0652 At = The average stagnation pressure is p0ave = T1 + T2
2
p1 + p2
2 3 (Still choked) ΔM
Δt⋅ ρtave⋅ Vtave where we use average values of density and speed at the throat.
T0ave = kg
m T2 = 109⋅ °C ΔM
= mtave = ρtave⋅ At⋅ Vtave
Δt The average stagnation temperature is M1 = 89.2 kg m M2 = M1 − ΔM The final (State 2) mass and density are then p p1 T0ave = 416 K
p0ave = 2.15⋅ MPa Hence the average temperature and pressure (critical) at the throat are
T0ave Ttave = Hence Finally ⎛1 + k − 1 ⎞
⎜
⎟
2⎠
⎝ Vtave = At = k⋅ R⋅ Ttave ΔM
Δt⋅ ρtave⋅ Vtave Ttave = 347 K and p0ave ptave = ⎛1 + k − 1 ⎞
⎜
⎟
2⎠
⎝
Vtave = 373 m
s At = 2.35 × 10 ρtave = −4 2 m ptave
R⋅ Ttave At = 235⋅ mm 2 This corresponds to a diameter Dt = 4⋅ At
π Dt = 0.0173 m Dt = 17.3⋅ mm The process is isentropic, followed by nonisentropic expansion to atmospheric pressure k
k−1 ptave = 1.14⋅ MPa ρtave = 11.4 kg
3 m Problem 13.33 Given: Ideal gas flow in a converging nozzle Find: [4] Exit area and speed Solution:
T0 Basic equations: T Given or available data = 1+ p0 k−1 2
⋅M
2 p1 = 35⋅ psi p ρ1 = 0.1⋅ lbm
ft Check for choking: c1 = Hence 3 V1
M1 =
c1 ⎝ k−1
2⎞
⋅ M1 ⎟
2
⎠ ⎛ k + 1⎞
⎜
⎟
⎝2⎠ M2 = From M1 we find Finally from continuity p2 = 25⋅ psi c1 = k⋅ p1 ft
c1 = 1424
s ρ1 k = 1.25 p0 = 37.8 psi
pcrit = 21.0 psi Hence p2 > pcrit, so NOT choked k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k − 1 p2
⎣⎝ ⎠
⎦ ⎛ 1 + k − 1 ⋅M 2 ⎞
⎜
1⎟
2
⎜
⎟
k+ 1
⎜
⎟
2
⎝
⎠
For isentropic flow 2 A1 = 1⋅ ft k
k−1 k
k−1 M1⋅ A1 Acrit = ft
s ⎛ 1 + k − 1 ⋅ M2 ⎞
⎟
A
1⎜
2
=
⋅⎜
⎟
k+ 1
Acrit
M⎜
⎟
2
⎝
⎠ M1 = 0.351 p0 The critical pressure is then pcrit = Then we have V1 = 500⋅ k − 1 2⎞
⋅M ⎟
2
⎠ k⋅ R⋅ T1 or, replacing R using the ideal gas equation p0 = p1⋅ ⎛ 1 +
⎜ Then ⎛
⎝ = ⎜1 + k
k−1 k+ 1
2⋅ ( k−1) k p⋅ ρ = const ρ⋅ A⋅ V = const k+ 1
2⋅ ( k−1) so so M2 = 0.830 Acrit = 0.557 ft 2 k−1
2⎞
⎛
Acrit ⎜ 1 + 2 ⋅ M2 ⎟
A2 =
⋅⎜
⎟
k+ 1
M2 ⎜
⎟ ⎝ ⎛ p1 ⎞
ρ2 = ρ1⋅ ⎜ ⎟
⎝ p2 ⎠ 2 1
k A1⋅ ρ1
V2 = V1⋅
A2⋅ ρ2 ρ2 = 0.131 lbm
ft V2 = 667 ft
s 3 ⎠ k+ 1
2⋅ ( k−1) A2 = 0.573 ft 2 Problem 13.34 [4] Part 1/3 Problem 13.34 [4] Part 2/3 Problem 13.34 [4] Part 3/3 Problem 13.35 [4] Problem 13.36 Given: CD nozzle attached to large tank Find: [2] Flow rate Solution:
Basic equations: T0
T = 1+ k−1 2
⋅M
2 p0
p ⎛
⎝ k − 1 2⎞
⋅M ⎟
2
⎠ = ⎜1 + p0 = 150⋅ kPa T0 = ( 35 + 273) ⋅ K k = 1.4 Given or available data R = 286.9⋅ For isentropic flow Me = Then Te = Also ce = ρe = Finally J
kg⋅ K k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k − 1 pe
⎣⎝ ⎠
⎦ mrate = ρe⋅ Ve⋅ Ae pe = 101⋅ kPa
Ae = π2
⋅D
4 D = 2.75⋅ cm
2 Ae = 5.94 cm Te = 275 K m
ce = 332
s ρe = 1.28 kg
3 m kg
mrate = 0.195
s Te = 1.94 °C Ve = Me⋅ ce ⎛ 1 + k − 1 ⋅ M 2⎞
⎜
e⎟
2
⎝
⎠ pe
R ⋅ Te mrate = ρ⋅ V⋅ A Me = 0.773 T0 k⋅ R⋅ Te k
k−1 Ve = 257 m
s Problem 13.37 [2] Given: Design condition in a convergingdiverging nozzle
Find: Tank pressure; flow rate; throat area Solution:
R=
k=
T0 = 53.33
1.4
560 Ae =
pb =
Me = 1
14.7
2 pe = 14.7 ft.lbf/lbm.oR pb pe = The given or available data is: o R in2
psia Equations and Computations:
At design condition psia From M e and p e, and Eq. 13.7a
(using builtin function Isenp (M ,k )
(13.7a) p0 = 115 psia From M e and A e, and Eq. 13.7d
(using builtin function IsenA (M ,k ) (13.7d) A* =
Hence 0.593 in2 At = 0.593 in2 From p 0, T 0, A t, and Eq. 13.10a (13.10a)
m choked = 1.53 lb/s Problem 13.38 [4] Part 1/2 Problem 13.38 [4] Part 2/2 Problem 13.39 [2] Problem 13.43 [3] Part 1/2 Problem 13.43 [3] Part 2/2 Problem 13.44 Given: Rocket motor on test stand Find: [3] Mass flow rate; thrust force Solution:
Basic equations: T0
T = 1+ k−1 2
⋅M
2 p0
p (patm − pe)⋅ Ae + Rx = mrate⋅ Ve
Given or available data pe = 75⋅ kPa patm = 101⋅ kPa d = 25⋅ cm From the pressures The exit speed is Then T0 ⎛ 1 + k − 1 ⋅ M 2⎞
⎜
e⎟
2
⎝
⎠ Ve = Me⋅ ce p0 = 4⋅ MPa ( ) c= k⋅ R⋅ T mrate = ρ⋅ A⋅ V T0 = 3250⋅ K
Ae = π2
⋅d
4 k = 1.25 R = 300⋅ J
kg⋅ K 2 Ae = 491⋅ cm Me = 3.12 Te = 1467 K Ve = 2313 mrate = ρe⋅ Ae⋅ Ve p = ρ⋅ R ⋅ T Momentum for pressure pe and velocity Ve at exit; Rx is the reaction for so the nozzle exit area is The momentum equation (Eq. 4.33) simplifies to Hence k − 1 2⎞
⋅M ⎟
2
⎠ k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k − 1 pe
⎣⎝ ⎠
⎦ Me = The exit temperature is Te = ⎛
⎝ = ⎜1 + k
k−1 m
s ce = k⋅ R⋅ Te and kg
mrate = 19.3
s (pe − patm)⋅ Ae − MCV⋅ ax = −Ve⋅ mrate Rx = pe − patm ⋅ Ae + Ve⋅ mrate Rx = 43.5⋅ kN m
ce = 742
s ρe = pe
R ⋅ Te ρe = 0.170⋅ kg
3 m Problem 13.47 [3] Problem 13.48 [4] Given: Compressed CO2 in a cartridge expanding through a nozzle Find: Throat pressure; Mass flow rate; Thrust; Thrust increase with diverging section; Exit area Solution:
Basic equations: Assumptions: 1) Isentropic flow 2) Stagnation in cartridge 3) Ideal gas 4) Uniform flow
J
kg⋅ K k = 1.29 R = 188.9⋅ p0 = 35⋅ MPa Given or available data: T0 = ( 20 + 273) ⋅ K
p0 From isentropic relations pcrit = ⎛1 + k − 1 ⎞
⎜
⎟
2⎠
⎝
Since pb << pcrit, then pt = pcrit Throat is critical so k
k −1 patm = 101⋅ kPa
dt = 0.5⋅ mm
pcrit = 19.2 MPa mrate = ρt⋅ Vt⋅ At
Tt = Vt = pt = 19.2 MPa T0
k−1
1+
2
k⋅ R⋅ Tt Tt = 256 K Vt = 250 m
s 2 At =
ρt = π⋅ dt
4
pt
R ⋅ Tt mrate = ρt⋅ Vt⋅ At At = 1.963 × 10
ρt = 396 −7 2 kg
3 m kg
mrate = 0.0194
s m Rx − ptgage⋅ At = mrate⋅ Vt For 1D flow with no body force the momentum equation reduces to
Rx = mrate⋅ Vt + ptgage⋅ At ptgage = pt − patm Rx = 8.60 N When a diverging section is added the nozzle can exit to atmospheric pressuree = patm
p k−1
⎤⎤
⎡
⎡
⎢
⎢
⎥⎥
k
⎢ 2 ⎢⎛ p0 ⎞
⎥⎥
Me = ⎢
⋅ ⎢⎜ ⎟
− 1⎥⎥
⎣ k − 1 ⎣⎝ pe ⎠
⎦⎦ Hence the Mach number at exit is Te = ce = T0
1+ k−1
2
⋅ Me
2 1
2 Me = 4.334 Te = 78.7 K
m
ce = 138
s k⋅ R⋅ Te Ve = Me⋅ ce Ve = 600 m
s The mass flow rate is unchanged (choked flow)
From the momentum equation Rx = mrate⋅ Ve The percentage increase in thrust is 11.67⋅ N − 8.60⋅ N
= 35.7 %
8.60⋅ N The exit area is obtained from mrate = ρe⋅ Ve⋅ Ae Ae = T mrate
ρe⋅ Ve and ρe = pe T0
pt
Tt
Conv.
Nozzle
CD
Nozzle Te
s ρe = 6.79 R ⋅ Te Ae = 4.77 × 10 p0 pb Rx = 11.67 N kg
3 m
−6 2 m Ae = 4.77 mm 2 Problem 13.49 [3] Given: CO2 cartridge and convergent nozzle
Find: Tank pressure to develop thrust of 15 N Solution:
J/kg·K R=
k=
T0 = 188.9
1.29
293 pb = 101 kPa Dt = 0.5 mm At = The given or available data is: 0.196 mm2 K Equations and Computations: The momentum equation gives
R x = m flowV e
Hence, we need m flow and V e
pe = pb pe = For isentropic flow 101 kPa If we knew p 0 we could use it and p e, and Eq. 13.7a, to find M e.
Once M e is known, the other exit conditions can be found.
Make a guess for p 0, and eventually use Goal Seek (see below).
p0 = 44.6 MPa From p 0 and p e, and Eq. 13.7a
(using builtin function IsenMfromp (M ,k ) (13.7a)
Me = 4.5 From M e and T 0 and Eq. 13.7b
(using builtin function IsenT (M ,k ) (13.7b) Te = 74.5 From T e and Eq. 12.18 K
(12.18) ce = m/s Ve = Then 134.8
606 m/s The mass flow rate is obtained from p 0, T 0, A t, and Eq. 13.10a (13.10a)
m choked = 0.0248 kg/s Finally, the momentum equation gives
R x = m flowV e
=
15.0
We need to set R x to 15 N. To do this use Goal Seek
to vary p 0 to obtain the result! N Problem 13.50 Given: Air flow in an insulated duct Find: [2] Mass flow rate; Range of choked exit pressures Solution:
T0 Basic equations: T = 1+ k−1 2
⋅M
2 c= ⎛ 1 + k − 1 ⋅ M2 ⎞
⎟
A
1⎜
2
=
⋅⎜
⎟
k+ 1
Acrit
M⎜
⎟
2
⎝
⎠ k⋅ R⋅ T T0 = ( 80 + 460) ⋅ R p0 = 14.7⋅ psi k = 1.4 Given or available data Rair = 53.33⋅ p1 = 13⋅ psi
ft⋅ lbf
lbm⋅ R A= π⋅ D
4 k+ 1
2⋅ ( k−1) D = 1⋅ in 2 2 A = 0.785⋅ in Assuming isentropic flow, stagnation conditions are constant. Hence M1 = k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k − 1 p1
⎣⎝ ⎠
⎦ c1 =
Also ρ1 = M1 = 0.423
m
c1 = 341
s k⋅ Rair⋅ T1
p1 ρ1 = 0.0673⋅ Rair⋅ T1 mrate = ρ1⋅ V1⋅ A When flow is choked hence 1
M2 = We also have c2 = From continuity ρ1⋅ V1 = ρ2⋅ V2 k⋅ Rair⋅ T2 1+ k−1
2 2 ⋅ M1 T1 = 521⋅ R T1 = 61.7⋅ °F m
s V1 = 144 T2 = 450⋅ R T2 = −9.7⋅ °F V2 = c2 V2 = 1040⋅ lbm
3 lbm
mrate = 0.174⋅
s
T0
T2 =
k−1
1+
2
ft
c2 = 1040⋅
s
V1
ρ2 = ρ1⋅
V2 p2 = ρ2⋅ Rair⋅ T2 T0 V1 = M1⋅ c1 ft Hence Hence T1 = ρ2 = 0.0306⋅ ft
s lbm
ft 3 p2 = 5.11⋅ psi The flow will therefore choke for any back pressure (pressure at the exit) less than or equal to this pressure
(From Fanno line function p1
pcrit = 2.545 at M1 = 0.423 so pcrit = p1
2.545 pcrit = 5.11 psi Check!) Problem 13.51 [4] Given: Air flow from converging nozzle into pipe
Find: Plot Ts diagram and pressure and speed curves Solution:
ft·lbf/lbm·oR R=
k=
cp = 53.33
1.4
0.2399 T0 = 187
710 p0 = 25 psi pe = 24 psi Me = 0.242 Using builtin function IsenT (M ,k ) Te = 702 Using p e, M e, and function Fannop (M ,k ) p* = 5.34 Using T e, M e, and function FannoT (M ,k ) T* = 592 The given or available data is: Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using builtin function IsenMfromp (M ,k )) o
Btu/lbm· R
o ft·lbf/lbm· R
R o o R psi
o R We can now use Fannoline relations to compute values for a range of Mach numbers: M T /T * 0.242
0.25
0.26
0.27
0.28
0.29
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
0.43
0.44
0.45 1.186
1.185
1.184
1.183
1.181
1.180
1.179
1.177
1.176
1.174
1.173
1.171
1.170
1.168
1.166
1.165
1.163
1.161
1.159
1.157
1.155
1.153 o
T ( R) 702
701
701
700
699
698
720
697
697 700
696
695 680
694
660
T (o693
R)
692 640
691
690 620
689 600
688
687 580
686
0
685
684
682 c (ft/s)
1299
1298
1298
1297
1296
1296
1295
1294
1293
1292
1292
1291
1290
1289
1288
1287
1286
1285
1284
1283
1282
1281 Δs
o
(ft·lbf/lbm· R)
Eq. (12.11b)
315
4.50
24.0
0.00
325
4.35
23.2
1.57
337
4.19
22.3
3.50
350 Ts Curve (Fanno)
4.03
21.5
5.35
363
3.88
20.7
7.11
376
3.75
20.0
8.80
388
3.62
19.3
10.43
401
3.50
18.7
11.98
414
3.39
18.1
13.48
427
3.28
17.5
14.92
439
3.19
17.0
16.30
452
3.09
16.5
17.63
464
3.00
16.0
18.91
477
2.92
15.6
20.14
489
2.84
15.2
21.33
502
2.77
14.8
22.48
514
2.70
14.4
23.58
527
2.63
14.0
24.65
2.56
13.7 30
25.68
10539
20
40
552
2.50
13.4
26.67
.
o
s
564
2.44 (ft lbf/lbm R)
13.0
27.63
576
2.39
12.7
28.55
V (ft/s) p /p * p (psi) 50 0.46
0.47
0.48
0.49
0.5
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.6
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.7
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1 1.151
1.149
1.147
1.145
1.143
1.141
1.138
1.136
1.134
1.132
1.129
1.127
1.124
1.122
1.119
1.117
1.114
1.112
1.109
1.107
1.104
1.101
1.098
1.096
1.093
1.090
1.087
1.084
1.082
1.079
1.076
1.073
1.070
1.067
1.064
1.061
1.058
1.055
1.052
1.048
1.045
1.042
1.039
1.036
1.033
1.029
1.026
1.023
1.020
1.017
1.013
1.010
1.007
1.003
1.000 681
1280
589
2.33
12.4
29.44
680
1279
601
2.28
12.2
30.31
679
1277
613
2.23
11.9
31.14
677
1276
625
Velocity V 2.18
Versus M11.7
(Fanno) 31.94
676
1275
638
2.14
11.4
32.72
675
1274
650
2.09
11.2
33.46
1400
674
1273
662
2.05
11.0
34.19
672 1200 1271
674
2.01
10.7
34.88
671
1270
686
1.97
10.5
35.56
669 1000 1269
698
1.93
10.3
36.21
668
1267
710
1.90
10.1
36.83
800
667
1266
722
1.86
9.9
37.44
V (ft/s)
665
733
1.83
9.8
38.02
600 1265
664
1263
745
1.80
9.6
38.58
400 1262
662
757
1.76
9.4
39.12
661
1260
769
1.73
9.2
39.64
200
659
1259
781
1.70
9.1
40.14
658
792
1.67
8.9
40.62
0 1258
656
1256
804
1.65
8.8
41.09
0.2
0.3
0.4
0.5
0.6
0.7
0.8
655
1255
815
1.62
8.6
41.53
M8.5
653
1253
827
1.59
41.96
652
1252
839
1.57
8.4
42.37
650
1250
850
1.54
8.2
42.77
648
1248
861
1.52
8.1
43.15
647
1247
873
1.49
8.0
43.51
645
1245
884
Pressure p 1.47
Versus M7.8Fanno)43.85
(
643
1244
895
1.45
7.7
44.18
642
1242
907
1.43
7.6
44.50
30
640
1240
918
1.41
7.5
44.80
638
1239
929
1.38
7.4
45.09
25
636
1237
940
1.36
7.3
45.36
635 20
1235
951
1.35
7.2
45.62
633
1234
962
1.33
7.1
45.86
631 15
1232
973
1.31
7.0
46.10
p (psi)
629
1230
984
1.29
6.9
46.31
628 10
1228
995
1.27
6.8
46.52
626
1227
1006
1.25
6.7
46.71
624
1225
1017
1.24
6.6
46.90
5
622
1223
1027
1.22
6.5
47.07
620
1221
1038
1.20
6.4
47.22
0
619
1219
1049
1.19
6.3
0.2
0.3
0.4
0.5
0.6
0.7 47.37
0.8
617
1218
1059
1.17
6.3
47.50
M 6.2
615
1216
1070
1.16
47.63
613
1214
1080
1.14
6.1
47.74
611
1212
1091
1.13
6.0
47.84
609
1210
1101
1.11
6.0
47.94
607
1208
1112
1.10
5.9
48.02
605
1206
1122
1.09
5.8
48.09
603
1204
1132
1.07
5.7
48.15
601
1202
1142
1.06
5.7
48.20
600
1201
1153
1.05
5.6
48.24
598
1199
1163
1.04
5.5
48.27
596
1197
1173
1.02
5.5
48.30
594
1195
1183
1.01
5.4
48.31
592
1193
1193
1.00
5.3
48.31 0.9 1.0 0.9 1.0 Problem 13.52 [4] Given: Air flow from convergingdiverging nozzle into pipe
Find: Plot Ts diagram and pressure and speed curves Solution:
R=
k=
cp = 53.33
1.4
0.2399 T0 =
p0 =
pe = 187
710
25
2.5 Me = Using builtin function IsenT (M ,k ) Te = 368 Using p e, M e, and function Fannop (M ,k ) p* = 6.84 Using T e, M e, and function FannoT (M ,k ) T* = 592 ft·lbf/lbm·oR 2.16 The given or available data is: Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using builtin function IsenMfromp (M ,k )) Btu/lbm·oR
ft·lbf/lbm·oR
R o psi
psi o R psi
o R We can now use Fannoline relations to compute values for a range of Mach numbers: M T /T * T (oR) 2.157
2
1.99
1.98
1.97
1.96
1.95
1.94
1.93
1.92
1.91
1.9
1.89
1.88
1.87
1.86
1.85
1.84
1.83
1.82
1.81
1.8
1.79
1.78
1.77
1.76
1.75
1.74
1.73
1.72
1.71
1.7 0.622
0.667
0.670
0.673
0.676
0.679
0.682
0.685
0.688
0.691
0.694
0.697
0.700
0.703
0.706
0.709
0.712
0.716
0.719
0.722
0.725
0.728
0.731
0.735
0.738
0.741
0.744
0.747
0.751
0.754
0.757
0.760 368
394
396
398
400
402
403
405
407
409
410
o
412
T ( R)
414
416
418
420
421
423
425
427
429
431
433
435
436
438
440
442
444
446
448
450 c (ft/s) 650
600
550
500
450
400
350
300 V (ft/s) 940
974
976
978
980
982
985
987
989
991
993
996
998
1000
1002
1004
1007
1009
0 1011 5
1013
1015
1018
1020
1022
1024
1027
1029
1031
1033
1036
1038
1040 2028
1948
1942
1937
1931
1926
1920
1914
1909
1903
1897
1892
1886
1880
1874
1868
1862
1856
1850
10
1844
1838
1832
1826
1819
1813
1807
1801
1794
1788
1781
1775
1768 Δs
(ft·lbf/lbm·oR)
Eq. (12.11b)
0.37
2.5
0.00
0.41
2.8
7.18
0.41
2.8
7.63
0.41
2.8
8.07
Curve (Fanno)
0.42
2.9
8.51
0.42
2.9
8.95
0.42
2.9
9.38
0.43
2.9
9.82
0.43
2.9
10.25
0.43
3.0
10.68
0.44
3.0
11.11
0.44
3.0
11.54
0.44
3.0
11.96
0.45
3.1
12.38
0.45
3.1
12.80
0.45
3.1
13.22
0.46
3.1
13.64
0.46
3.1
14.05
0.46
15
20 3.2
25 14.46 30
0.47
3.2 o
14.87
.
s
0.47 (ft lbf/lbm R)
3.2
15.28
0.47
3.2
15.68
0.48
3.3
16.08
0.48
3.3
16.48
0.49
3.3
16.88
0.49
3.3
17.27
0.49
3.4
17.66
0.50
3.4
18.05
0.50
3.4
18.44
0.50
3.5
18.82
0.51
3.5
19.20
0.51
3.5
19.58
p /p * Ts p (psi) 35 40 1.69
1.68
1.67
1.66
1.65
1.64
1.63
1.62
1.61
1.6
1.59
1.58
1.57
1.56
1.55
1.54
1.53
1.52
1.51
1.5
1.49
1.48
1.47
1.46
1.45
1.44
1.43
1.42
1.41
1.4
1.39
1.38
1.37
1.36
1.35
1.34
1.33
1.32
1.31
1.3
1.29
1.28
1.27
1.26
1.25
1.24
1.23
1.22
1.21
1.2
1.19
1.18
1.17
1.16
1.15
1.14
1.13
1.12
1.11
1.1
1.09
1.08
1.07
1.06
1.05
1.04
1.03
1.02
1.01
1 0.764
0.767
0.770
0.774
0.777
0.780
0.784
0.787
0.790
0.794
0.797
0.800
0.804
0.807
0.811
0.814
0.817
0.821
0.824
0.828
0.831
0.834
0.838
0.841
0.845
0.848
0.852
0.855
0.859
0.862
0.866
0.869
0.872
0.876
0.879
0.883
0.886
0.890
0.893
0.897
0.900
0.904
0.907
0.911
0.914
0.918
0.921
0.925
0.928
0.932
0.935
0.939
0.942
0.946
0.949
0.952
0.956
0.959
0.963
0.966
0.970
0.973
0.976
0.980
0.983
0.987
0.990
0.993
0.997
1.000 452
1042
454
1045
456
1047
458
1049
460
1051
462
1054
464
1056
466
1058
468
1060
470
1063
472
1065
2500
474
1067
476
1069
478 2000 1072
480
1074
482 1500 1076
484
1078
V (ft/s)
486
1080
1000
488
1083
490
1085
500 1087
492
494
1089
496
0 1092
498
1094
2.0
500
1096
502
1098
504
1101
506
1103
508
1105
510
1107
512
1110
514
1112
516
1114
8
518
1116
520 7
1118
522
1121
6
524
1123
527 5
1125
529 4
1127
p (psi)
531
1129
533 3
1132
535 2
1134
537
1136
539 1
1138
541 0
1140
543
2.0 1143
545
1145
547
1147
549
1149
551
1151
553
1153
555
1155
557
1158
559
1160
561
1162
564
1164
566
1166
568
1168
570
1170
572
1172
574
1174
576
1176
578
1179
580
1181
582
1183
584
1185
586
1187
588
1189
590
1191
592
1193 1761
0.52
3.5
19.95
1755
0.52
3.6
20.32
1748
0.53
3.6
20.69
1741
0.53
3.6
21.06
1735
0.53
3.7
21.42
1728
0.54
3.7
21.78
1721
0.54
3.7
22.14
1714
0.55
3.7
22.49
1707
Velocity V 0.55
Versus M 3.8
(Fanno) 22.84
1700
0.56
3.8
23.18
1693
0.56
3.8
23.52
1686
0.57
3.9
23.86
1679
0.57
3.9
24.20
1672
0.58
3.9
24.53
1664
0.58
4.0
24.86
1657
0.59
4.0
25.18
1650
0.59
4.0
25.50
1642
0.60
4.1
25.82
1635
0.60
4.1
26.13
1627
0.61
4.1
26.44
1620
0.61
4.2
26.75
1612
0.62
4.2
27.05
1605
0.62
4.3
27.34
1597
0.63
4.3
27.63
1.8
1.6
1.4
1589
0.63
4.3
27.92
M4.4
1582
0.64
28.21
1574
0.65
4.4
28.48
1566
0.65
4.5
28.76
1558
0.66
4.5
29.03
1550
0.66
4.5
29.29
1542
Pressure p 0.67
Versus M4.6Fanno)29.55
(
1534
0.68
4.6
29.81
1526
0.68
4.7
30.06
1518
0.69
4.7
30.31
1510
0.69
4.8
30.55
1502
0.70
4.8
30.78
1493
0.71
4.8
31.01
1485
0.71
4.9
31.24
1477
0.72
4.9
31.46
1468
0.73
5.0
31.67
1460
0.74
5.0
31.88
1451
0.74
5.1
32.09
1443
0.75
5.1
32.28
1434
0.76
5.2
32.48
1426
0.76
5.2
32.66
1417
0.77
5.3 1.4
32.84
1.8
1.6
1408
0.78
5.3
33.01
M 5.4
1399
0.79
33.18
1390
0.80
5.4
33.34
1381
0.80
5.5
33.50
1372
0.81
5.6
33.65
1363
0.82
5.6
33.79
1354
0.83
5.7
33.93
1345
0.84
5.7
34.05
1336
0.85
5.8
34.18
1327
0.86
5.9
34.29
1318
0.87
5.9
34.40
1308
0.87
6.0
34.50
1299
0.88
6.0
34.59
1290
0.89
6.1
34.68
1280
0.90
6.2
34.76
1271
0.91
6.2
34.83
1261
0.92
6.3
34.89
1251
0.93
6.4
34.95
1242
0.94
6.5
34.99
1232
0.96
6.5
35.03
1222
0.97
6.6
35.06
1212
0.98
6.7
35.08
1203
0.99
6.8
35.10
1193
1.00
6.8
35.10 1.2 1.2 1.0 1.0 Problem 13.53 [2] Problem 13.54 Given: Air flow in a converging nozzle and insulated duct Find: [3] Pressure at end of duct; Entropy increase Solution:
T0 Basic equations: T = 1+ k−1 2
⋅M
2 p0 ⎛
⎝ p = ⎜1 + k − 1 2⎞
⋅M ⎟
2
⎠ k
k−1 ⎛ T2 ⎞
⎛ p2 ⎞
Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠ T0 = ( 250 + 460) ⋅ R p0 = 145⋅ psi p1 = 125⋅ psi k = 1.4 Given or available data Btu
cp = 0.2399⋅
lbm⋅ R Rair = 53.33⋅ c= k⋅ R⋅ T T2 = ( 150 + 460) ⋅ R
ft⋅ lbf
lbm⋅ R Assuming isentropic flow in the nozzle
k−1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p0 ⎞
⋅ ⎢⎜ ⎟
− 1⎥
k − 1 p1
⎣⎝ ⎠
⎦ M1 = M1 = 0.465 T1 = T0
1+ k−1
2 ⋅ M1 2 ⎡⎛ T0 ⎞ ⎤
⋅ ⎢⎜ ⎟ − 1⎥
k − 1 T2 In the duct T0 (a measure of total energy) is constant, so M2 = ⎣⎝ ⎠ 2 T1 = 681⋅ R T1 = 221⋅ °F M2 = 0.905 ⎦ k⋅ Rair⋅ T1 ft
c1 = 1279⋅
s V1 = M1⋅ c1 V1 = 595⋅ k⋅ Rair⋅ T2 ft
c2 = 1211⋅
s V2 = M2⋅ c2 V2 = 1096⋅ ft
s V1
ρ2 = ρ1⋅
V2 ρ2 = 0.269⋅ lbm ρ1 = Also p1
Rair⋅ T1 ρ1 = 0.4960⋅ p2 = ρ2⋅ Rair⋅ T2 p2 = 60.8⋅ psi (Note: Using Fanno line relations, at M1 = 0.465 so Finally
T1
Tcrit
p1
pcrit Then T2
Tcrit = 1.031 so M2 = 0.907 p2
pcrit ft
s lbm
ft mrate = ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A Hence Then c1 =
c2 = At each location 3 ft ⎛ T2 ⎞
⎛ p2 ⎞
Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠
T1 = 1.150 Tcrit = = 2.306 pcrit = = 1.119 p2 = 1.119⋅ pcrit 1.150
p1
2.3060 3 Δs = 0.0231⋅ Btu
lbm⋅ R Tcrit = 329 K pcrit = 54.2⋅ psi p2 = 60.7⋅ psi Check!) Problem 13.57 [3] Given: Air flow in a CD nozzle and insulated duct Find: Temperature at end of duct; Force on duct; Entropy increase Solution: ( T0 ) Basic equations: Fs = p1 ⋅ A − p2 ⋅ A + Rx = mrate⋅ V2 − V1 Given or available data T1 = ( 100 + 460) ⋅ R p1 = 18.5⋅ psi k = 1.4 k −1 2
⋅M
2 Btu
cp = 0.2399⋅
lbm⋅ R T =1+ M1 = 2 ⎛ T2 ⎞
⎛ p2 ⎞
Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠
2 M2 = 1
Rair = 53.33⋅ A = 1⋅ in
ft⋅ lbf
lbm⋅ R Assuming isentropic flow in the nozzle
k −1
2
⋅ M1
1+
T0 T2
2
⋅
=
T1 T0
k −1
2
1+
⋅ M2
2
Also c1 = ρ1 = 1+ k⋅ Rair⋅ T1 V1 = M1⋅ c1
p1
Rair⋅ T1 mrate = ρ1⋅ V1⋅ A ρ1 = 0.0892⋅ V1 = 2320⋅
lbm
ft 3 ) 1+ ft
s 2
k −1
2 ⋅ M1 2 ⋅ M2 2 T2 = 840⋅ R V2 = 1421⋅ ft
s V1
ρ 2 = ρ 1⋅
V2 ρ2 = 0.146⋅ lbm so p2 = ρ2⋅ Rair⋅ T2 ( ) Rx = p2 − p1 ⋅ A + mrate⋅ V2 − V1 Rx = −13.3⋅ lbf Finally ⎛ T2 ⎞
⎛ p2 ⎞
Δs = cp⋅ ln ⎜ ⎟ − Rair⋅ ln ⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠ Δs = 0.0359⋅ M1 = 2 T1
Tcrit
p1
pcrit = = T1
T2
p1
p2 = 0.6667 = 0.4083 ft p2 = 45.3⋅ psi Hence (Note: Using Fanno line relations, at T2 = 380⋅ °F k ⋅ Rair⋅ T2 V2 = M2⋅ c2 c2 = mrate = ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A2 lbm
mrate = 1.44⋅
s ( T2 = T1⋅ so k −1 (Force is to the right) Btu
lbm⋅ R
T2 = p2 = p1
0.4083 T1
0.667 p2 = 45.3⋅ psi T2 = 840⋅ R Check!) 3 Problem 13.59 [4] Problem 13.61 [4] Problem 13.62 [2] Problem 13.63 Given: Air flow in a converging nozzle and insulated duct Find: [2] Length of pipe Solution:
Basic equations: Fannoline flow equations, and friction factor Given or available data T0 = ( 250 + 460) ⋅ R p0 = 145⋅ psi p1 = 125⋅ psi T2 = ( 150 + 460) ⋅ R D = 2⋅ in k = 1.4 Btu
cp = 0.2399⋅
lbm⋅ R Rair = 53.33⋅ From isentropic relations k−1
⎤⎤
⎡
⎡
⎢
⎢
⎥⎥
k
⎢ 2 ⎢⎛ p0 ⎞
⎥⎥
M1 = ⎢
⋅ ⎢⎜ ⎟
− 1⎥⎥
⎣ k − 1 ⎣⎝ p1 ⎠
⎦⎦ T0
T1 = 1+ k−1
2
⋅ M1
2 1
2 so M1 = 0.465 T1 = T0 p1 pcrit = Also, for T2
Tcrit p1
2.3044 = 1.031 T1 = 221⋅ °F = 1.150 Tcrit = ⎡
⎤
fave⋅ Lmax1
1 − M1
k + 1 ⎢ ( k + 1) ⋅ M1
⎥
=
+
⋅ ln ⎢
= 1.3923
2
Dh
2⋅ k
⎛ k − 1 ⋅ M 2⎞ ⎥
k⋅ M1
2⋅ ⎜ 1 +
⎢
1 ⎟⎥
2
⎣⎝
⎠⎦
k+ 1 ⎛ ⎞ ⎟
2
1⎜
=
=
⋅⎜
⎟
pcrit
p2
M1
k−1
2
⎜ 1+
⋅ M1 ⎟
2
⎝
⎠
p1 T1 = 681⋅ R ⎛ 1 + k − 1 ⋅ M 2⎞
⎜
1⎟
2
⎝
⎠ 2 Then for Fannoline flow ft⋅ lbf
lbm⋅ R 2 1
2 T1 = 2.3044 Tcrit pcrit = 54.2⋅ psi T2
Tcrit 1+ =
1+ k−1
2
⋅ M1
2 Tcrit = 592⋅ R k+ 1
2 = k+ 1
2 k−1
2
⋅ M2
2 leads to M2 = 2
k−1 T1
1.150 Tcrit = 132⋅ °F ⎛ k + 1 Tcrit ⋅⎜ ⎝2 Then ρ1 = T2 ⎞ − 1⎟ ⎠ M2 = 0.906 ⎡
⎤
fave⋅ Lmax2
1 − M2
k + 1 ⎢ ( k + 1) ⋅ M2
⎥
=
+
⋅ ln
= 0.01271
⎢ ⎛ k−1
2
2⋅ k
Dh
2⎞ ⎥
k⋅ M2
⋅ M2 ⎟ ⎥
2⋅ ⎜ 1 +
⎢
2
⎣⎝
⎠⎦ Also ⋅ 2 p1
Rair⋅ T1 2 ρ1 = 0.496 lbm
ft 3 V1 = M1⋅ k⋅ Rair⋅ T1 V1 = 595 ft
s For air at T1 = 221 °F, from Table A.9 (approximately) μ = 4.48 × 10 − 7 lbf ⋅ s
⋅
2 Re1 = and Re1 = 3.41 × 10 ft For commercial steel pipe (Table 8.1) Hence at this Reynolds number and roughness (Eq. 8.37) Combining results e
−4
= 9 × 10
D e = 0.00015⋅ ft These calculations are a LOT easier using the Excel Addins! μ f = 0.01924 2
⋅ ft
fave⋅ Lmax2 fave⋅ Lmax1 ⎞
12
D⎛
L12 = ⋅ ⎜
−
⋅ ( 1.3923 − 0.01271)
⎟=
Dh
Dh
f
.01924 ⎝ ρ1⋅ V1⋅ D so ⎠ L12 = 12.0⋅ ft 6 Problem 13.65 [3] Part 1/2 Problem 13.65 [3] Part 2/2 Problem 13.67
Example 13.7 Example 13.7 [3] Problem 13.68 [3] Problem 13.69 [4] Part 1/2 Problem 13.69 [4] Part 2/2 Problem 13.70 Given: Air flow through a CD nozzle and tube. Find: [2] Average friction factor; Pressure drop in tube Solution:
Assumptions: 1) Isentropic flow in nozzle 2) Adiabatic flow in tube 3) Ideal gas 4) Uniform flow
J
kg⋅ K k = 1.40 R = 286.9⋅ p0 = 1.35⋅ MPa Given or available data: T0 = 550⋅ K k −1
⎤⎤
⎡
⎡
⎢
⎢
⎥⎥
k
⎢ 2 ⎢⎛ p0 ⎞
⎥⎥
From isentropic relations M1 = ⎢
⋅ ⎢⎜ ⎟
− 1⎥⎥
⎣ k − 1 ⎣⎝ p1 ⎠
⎦⎦ p1 = 15⋅ kPa where State 1 is the nozzle exit D = 2.5⋅ cm L = 1.5⋅ m 1
2 M1 = 3.617 Then for Fannoline flow (for choking at the exit) ⎡
⎤
fave⋅ Lmax
1 − M1
k + 1 ⎢ ( k + 1) ⋅ M1
⎥
=
+
⋅ ln ⎢
= 0.599
2
Dh
2⋅ k
k−1
2⎞ ⎥
⎛1 +
k⋅ M1
⋅ M1 ⎟ ⎥
⎢ 2⋅ ⎜
2
⎣⎝
⎠⎦
2 ⎡ Hence 2 ⎡ 2 2 ⎤⎤ k + 1 ⎢ ( k + 1) ⋅ M1
⎥⎥
D ⎢ 1 − M1
fave = ⋅
+
⋅ ln
⎢
⎢ ⎛ k−1
2
2⋅ k
L k⋅ M
2⎞ ⎥⎥
⋅ M1 ⎟ ⎥⎥
⎢
⎢ 2⋅ ⎜ 1 +
1
2
⎣
⎣⎝
⎠ ⎦⎦ ⎛ k+ 1 ⎞ ⎟
2
1⎜
=
=
⋅⎜
⎟
pcrit
p2
M1
k−1
2
⎜ 1+
⋅ M1 ⎟
2
⎝
⎠
p1 p2 = p1 p1
1⎤
⎡
⎢
2⎥
k+ 1
⎢
⎞⎥
⎛
⎟⎥
2
⎢1 ⎜
⋅⎜
⎟
⎢M
k−1
2⎥
⎢ 1 ⎜ 1 + 2 ⋅ M1 ⎟ ⎥
⎣
⎝
⎠⎦ Δp = p1 − p2
These calculations are a LOT easier using the Excel Addins! fave = 0.0100 1
2 = 0.159 p2 = 94.2 kPa Δp = −79.2 kPa Problem 13.71 Given: Air flow in a CD nozzle and insulated duct Find: [3] Duct length; Plot of M and p Solution:
Basic equations: Fannoline flow equations, and friction factor Given or available data T1 = ( 100 + 460) ⋅ R p1 = 18.5⋅ psi M1 = 2 Btu
cp = 0.2399⋅
lbm⋅ R k = 1.4 2 M2 = 1 A = 1⋅ in Rair = 53.33⋅ ft⋅ lbf
lbm⋅ R Then for Fannoline flow at M1 = 2
1
2 k+ 1
⎞
⎛
⎜
⎟
2
1
=
=
⋅⎜
⎟ = 0.4082
pcrit
p2
M1
k−1
2
⎜ 1+
⋅ M1 ⎟
2
⎝
⎠ p1 so p1 pcrit = p1 ⎡
⎤
fave⋅ Lmax1
1 − M1
k + 1 ⎢ ( k + 1) ⋅ M1
⎥
=
+
⋅ ln ⎢
= 0.3050
2
Dh
2⋅ k
k−1
2⎞ ⎥
⎛1 +
k⋅ M1
⋅ M1 ⎟ ⎥
⎢ 2⋅ ⎜
2
⎣⎝
⎠⎦
2 2 pcrit = 45.3⋅ psi 0.4082 and at M2 = 1 ⎡
⎤
fave⋅ Lmax2
1 − M2
k + 1 ⎢ ( k + 1) ⋅ M2
⎥
=
+
⋅ ln ⎢
=0
2
Dh
2⋅ k
k−1
2⎞ ⎥
⎛1 +
k⋅ M2
⋅ M2 ⎟ ⎥
⎢ 2⋅ ⎜
2
⎣⎝
⎠⎦ Also ρ1 = 2 p1 lbm
ρ1 = 0.089⋅
3
Rair⋅ T1
ft For air at T1 = 100⋅ °F, from Table A.9 2 V1 = M1⋅ k⋅ Rair⋅ T1
μ = 3.96 × 10 − 7 lbf ⋅ s
⋅
2 V1 = 2320⋅ e = 0.00015⋅ ft Hence at this Reynolds number and roughness (Eq. 8.37) e
−3
= 1.595 × 10
D 4⋅ A
D = 1.13⋅ in
π D= ρ1⋅ V1⋅ D so Re1 = and Re1 = 1.53 × 10 ft For commercial steel pipe (Table 8.1) ft
s μ
6 f = .02222
1.13 Combining results ⋅ ft
12
D ⎛ fave⋅ Lmax2 fave⋅ Lmax1 ⎞
L12 = ⋅ ⎜
−
=
⋅ ( 0.3050 − 0)
⎟
Dh
Dh
f
⎝
⎠ .02222 L12 = 1.29⋅ ft L12 = 15.5⋅ in These calculations are a LOT easier using the Excel Addins! The M and p plots are shown in the associated Excel workbook Problem 13.71 (In Excel) Given: Air flow in a CD nozzle and insulated duct Find: [3] Duct length; Plot of M and p Solution:
The given or available data is: M
2.00
1.95
1.90
1.85
1.80
1.75
1.70
1.65
1.60
1.55
1.50
1.45
1.40
1.35
1.30
1.25
1.20
1.15
1.10
1.05
1.00 f = 0.0222
p * = 45.3 kPa
D=
1.13 in fL max/D ΔfL max/D x (in) p /p * p (psi)
0.305
0.290
0.274
0.258
0.242
0.225
0.208
0.190
0.172
0.154
0.136
0.118
0.100
0.082
0.065
0.049
0.034
0.021
0.010
0.003
0.000 0.000
0.015
0.031
0.047
0.063
0.080
0.097
0.115
0.133
0.151
0.169
0.187
0.205
0.223
0.240
0.256
0.271
0.284
0.295
0.302
0.305 0
0.8
1.6
2.4
3.2
4.1
4.9
5.8
6.7
7.7
8.6
9.5
10.4
11.3
12.2
13.0
13.8
14.5
15.0
15.4
15.5 0.408
0.423
0.439
0.456
0.474
0.493
0.513
0.534
0.557
0.581
0.606
0.634
0.663
0.695
0.728
0.765
0.804
0.847
0.894
0.944
1.000 18.49
19.18
19.90
20.67
21.48
22.33
23.24
24.20
25.22
26.31
27.47
28.71
30.04
31.47
33.00
34.65
36.44
38.37
40.48
42.78
45.30 Fanno Line Flow Curves(M and p ) 45 2.0
1.9
1.8
1.7
1.6
M 1.5
1.4
1.3
1.2
1.1
1.0 40
35
30 p (psi)
25
M
Pressure 20
15 0 4 8
x (in) 12 16 Problem *13.73 [2] Given: Isothermal air flow in a duct Find: Downstream Mach number; Direction of heat transfer; Plot of Ts diagram Solution: 2 V1 2 T0 V2
δQ
+
= h2 +
2
dm h1 + Given or available data T1 = ( 20 + 273) ⋅ K p1 = 350⋅ kPa M1 = 0.1 From continuity mrate = ρ1⋅ V1⋅ A = ρ2⋅ V2⋅ A so ρ1⋅ V1 = ρ2⋅ V2 Also p = ρ⋅ R ⋅ T M= Hence continuity becomes p1
R ⋅ T1 2 T and ⋅ M1⋅ c1 = =1+ k −1 2
⋅M
2 Basic equations: V
c mrate = ρ⋅ V⋅ A
p2 = 150⋅ kPa V = M⋅ c or p2
⋅M ⋅c
R ⋅ T2 2 2 Since T1 = T2 c1 = c2 Hence p1
M2 =
⋅M
p2 1 From energy ⎛
⎞⎜
⎞
V2 ⎟ ⎛
V1 ⎟
δQ ⎜
= ⎜ h2 +
− ⎜ h1 +
= h02 − h01 = cp⋅ T02 − T01
dm ⎝
2⎟ ⎝
2⎟
⎠
⎠ M2 = 0.233
2 But at each state p1⋅ M1 = p2⋅ M2 so 2 ( T0
T = 1+ k−1
2 2 ⋅M or T0 = T⋅ ⎛ 1 +
⎜ ⎝ k−1
2 ) 2⎞
⋅M ⎟ ⎠ p02
Since T = const, but M2 > M1, then T02 > T01, and
δQ
>0
dm so energy is ADDED to the system T p01
T 01
p1 T02
p2 s Problem *13.74 Given: Isothermal air flow in a pipe Find: [5] Mach number and location at which pressure is 500 kPa Solution:
f ⋅ Lmax 2 1 − k⋅ M ) 2 p = ρ⋅ R ⋅ T T1 = ( 15 + 273) ⋅ K p1 = 1.5⋅ MPa V1 = 60⋅ D = 15⋅ cm k = 1.4 R = 286.9⋅ From continuity ρ1⋅ V1 = ρ2⋅ V2 or Since T1 = T2 and V = M⋅ c = M⋅ k⋅ R⋅ T p1
M2 = M1⋅
p2 c1 = m
c1 = 340
s V1
M1 =
c1 M1 = 0.176 Given or available data Then At M1 = 0.176 At M2 = 0.529 Hence k⋅ R⋅ T1 p1
M2 = M1⋅
p2
f ⋅ Lmax1
D D f ⋅ L12
D 1 − k⋅ M1 2 = 1 − k⋅ M2
2 k⋅ M2 = f ⋅ Lmax2 L12 = 18.2⋅ D D
f T1 − 2
+ ln ⎛ k⋅ M1 ⎞ = 18.819 ⎝ 2 k⋅ M1 f ⋅ Lmax2 p1 M2 = 0.529
2 = D ⎠ 2
+ ln ⎛ k⋅ M2 ⎞ = 0.614 ⎝ f ⋅ Lmax1
D ⎠ = 18.819 − 0.614 = 18.2 L12 = 210 m = ( mrate = ρ⋅ V⋅ A Basic equations: 2 + ln k⋅ M k⋅ M
m
s ⋅ V1 = f = 0.013 J
kg⋅ K p2
T2 ⋅ V2 p2 = 500⋅ kPa Problem *13.75 [2] Problem *13.76 [4] Part 1/2 Problem *13.76 [4] Part 2/2 Problem 13.78 [4] Given: Air flow from converging nozzle into heated pipe
Find: Plot Ts diagram and pressure and speed curves Solution:
R=
k=
cp = 53.33
1.4
0.2399 T0 =
p0 =
pe= 187
710
25
24 Me = Using builtin function IsenT (M ,k ) Te = 702 Using p e, M e, and function Rayp (M ,k ) p* = 10.82 Using T e, M e, and function RayT (M ,k ) T* = 2432 ft·lbf/lbm·oR 0.242 The given or available data is: Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using builtin function IsenMfromp (M ,k )) Btu/lbm·oR
ft·lbf/lbm·oR
R o psi
psi o R psi
o R We can now use Rayleighline relations to compute values for a range of Mach numbers: M T /T * 0.242
0.25
0.26
0.27
0.28
0.29
0.3
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.4
0.41
0.42
0.43
0.44
0.45
0.46 0.289
0.304
0.325
0.346
0.367
0.388
0.409
0.430
0.451
0.472
0.493
0.514
0.535
0.555
0.576
0.595
0.615
0.634
0.653
0.672
0.690
0.708
0.725 T (oR)
702
740
790
841
892
943
3000
994
1046
2500
1097
11492000
1200
1250
T (oR) 1500
1301
13511000
1400
1448 500
1496
0
1543
1589
0
1635
1679
1722
1764 c (ft/s)
1299
1334
1378
1422
1464
1506
1546
1586
1624
1662
1698
1734
1768
1802
1834
1866
1897
1926
1955
1982
2009
2035
2059 V (ft/s)
315
334
358
384Ts
410
437
464
492
520
548
577
607
637
667
697
728
759
790
50 821
852
884
916
947 Δs
(ft·lbf/lbm·oR)
Eq. (12.11b)
2.22
24.0
0.00
2.21
23.9
10.26
2.19
23.7
22.81
2.18
23.6
34.73
Curve (Rayleigh)
2.16
23.4
46.09
2.15
23.2
56.89
2.13
23.1
67.20
2.12
22.9
77.02
2.10
22.7
86.40
2.08
22.5
95.35
2.07
22.4
103.90
2.05
22.2
112.07
2.03
22.0
119.89
2.01
21.8
127.36
2.00
21.6
134.51
1.98
21.4
141.35
1.96
21.2
147.90
1.94
21.0
154.17
20.8
160.17
100 1.92
150
200
250
1.91 .
20.6
165.92
o
s
1.89 (ft lbf/lbm R)
20.4
171.42
1.87
20.2
176.69
1.85
20.0
181.73
p /p * p (psi) 300 0.47
0.48
0.49
0.5
0.51
0.52
0.53
0.54
0.55
0.56
0.57
0.58
0.59
0.6
0.61
0.62
0.63
0.64
0.65
0.66
0.67
0.68
0.69
0.7
0.71
0.72
0.73
0.74
0.75
0.76
0.77
0.78
0.79
0.8
0.81
0.82
0.83
0.84
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
1 0.742
0.759
0.775
0.790
0.805
0.820
0.834
0.847
0.860
0.872
0.884
0.896
0.906
0.917
0.927
0.936
0.945
0.953
0.961
0.968
0.975
0.981
0.987
0.993
0.998
1.003
1.007
1.011
1.014
1.017
1.020
1.022
1.024
1.025
1.027
1.028
1.028
1.029
1.029
1.028
1.028
1.027
1.026
1.025
1.023
1.021
1.019
1.017
1.015
1.012
1.009
1.006
1.003
1.000 1805
2083
979
1.83
19.8
186.57
1845
2106
1011
1.81
19.6
191.19
1884
2128
1043
19.4
195.62
Velocity V V1.80
ersus M (Rayleigh)
1922
2149
1075
1.78
19.2
199.86
1958
2170
1107
1.76
19.0
203.92
3000
1993
2189
1138
1.74
18.8
207.80
2027
2208
1170
1.72
18.6
211.52
2500
2060
2225
1202
1.70
18.4
215.08
2091
2242
1233
1.69
18.2
218.48
2000
2122
2258
1265
1.67
18.0
221.73
2150 1500 2274
1296
1.65
17.9
224.84
V (ft/s)
2178
2288
1327
1.63
17.7
227.81
2204 1000 2302
1358
1.61
17.5
230.65
2230
2315
1389
1.60
17.3
233.36
2253
1420
1.58
17.1
235.95
500 2328
2276
2339
1450
1.56
16.9
238.42
2298
1481
1.54
16.7
240.77
0 2350
2318
2361 0.3 1511
1.53
16.5
0.2
0.4
0.5
0.6
0.7 243.010.8
2337
2370
1541
1.51
16.3
245.15
M
2355
2379
1570
1.49
16.1
247.18
2371
2388
1600
1.47
15.9
249.12
2387
2396
1629
1.46
15.8
250.96
2401
2403
1658
1.44
15.6
252.70
2415
2409
1687
1.42
15.4
254.36
2427
2416
1715
1.41
255.93
Pressure p Versus M 15.2
(Rayleigh)
2438
2421
1743
1.39
15.0
257.42
2449
2426
1771
1.37
14.9
258.83
30
2458
2431
1799
1.36
14.7
260.16
2466
2435
1826
1.34
14.5
261.41
25
2474
2439
1853
1.33
14.4
262.59
2480
2442
1880
1.31
14.2
263.71
20
2486
2445
1907
1.30
14.0
264.75
2447
1933
1.28
13.9
265.73
p 2490 15
(psi)
2494
2449
1959
1.27
13.7
266.65
2497 10
2450
1985
1.25
13.5
267.50
2499
2451
2010
1.24
13.4
268.30
2501 5
2452
2035
1.22
13.2
269.04
2502
2452
2060
1.21
13.1
269.73
2502 0
2452
2085
1.19
12.9
270.36
2501
2452
2109
1.18
12.8
0.2
0.3
0.4
0.5
0.6
0.7 270.94
0.8
2500
2451
2133
1.17
12.6
271.47
M 12.5
2498
2450
2156
1.15
271.95
2495
2449
2180
1.14
12.3
272.39
2492
2448
2203
1.12
12.2
272.78
2488
2446
2226
1.11
12.0
273.13
2484
2444
2248
1.10
11.9
273.43
2479
2441
2270
1.09
11.7
273.70
2474
2439
2292
1.07
11.6
273.92
2468
2436
2314
1.06
11.5
274.11
2461
2433
2335
1.05
11.3
274.26
2455
2429
2356
1.04
11.2
274.38
2448
2426
2377
1.02
11.1
274.46
2440
2422
2398
1.01
10.9
274.51
2432
2418
2418
1.00
10.8
274.52 0.9 1.0 0.9 1.0 Problem 13.79 [4] Given: Air flow from convergingdiverging nozzle into heated pipe
Find: Plot Ts diagram and pressure and speed curves Solution:
R=
k=
cp = 53.33
1.4
0.2399 T0 =
p0 =
pe = 187
710
25
2.5 Me = Using builtin function IsenT (M ,k ) Te = 368 Using p e, M e, and function Rayp (M ,k ) p* = 7.83 Using T e, M e, and function RayT (M ,k ) T* = 775 ft·lbf/lbm·oR 2.16 The given or available data is: Equations and Computations:
From p 0 and p e, and Eq. 13.7a
(using builtin function IsenMfromp (M ,k )) Btu/lbm·oR
ft·lbf/lbm·oR
R o psi
psi o R psi
o R We can now use Rayleighline relations to compute values for a range of Mach numbers: M T /T * T (oR) 2.157
2
1.99
1.98
1.97
1.96
1.95
1.94
1.93
1.92
1.91
1.9
1.89
1.88
1.87
1.86
1.85
1.84
1.83
1.82
1.81
1.8
1.79
1.78
1.77
1.76
1.75
1.74
1.73
1.72
1.71
1.7 0.475
0.529
0.533
0.536
0.540
0.544
0.548
0.552
0.555
0.559
0.563
0.567
0.571
0.575
0.579
0.584
0.588
0.592
0.596
0.600
0.605
0.609
0.613
0.618
0.622
0.626
0.631
0.635
0.640
0.645
0.649
0.654 368
410
413
416
418
421
424
427
430
433
436
o
440
T ( R)
443
446
449
452
455
459
462
465
468
472
475
479
482
485
489
492
496
499
503
507 c (ft/s) 800
750
700
650
600
550
500
450
400
350
300 940
993
996
1000
1003
1007
1010
1014
1017
1021
1024
1028
1032
1035
1039
1043
1046
1050
0 1054 10
1057
1061
1065
1069
1073
1076
1080
1084
1088
1092
1096
1100
1104 V (ft/s)
2028
1985
1982
1979Ts
1976
1973
1970
1966
1963
1960
1957
1953
1950
1946
1943
1939
1936
1932
1928
20
1925
1921
1917
1913
1909
1905
1901
1897
1893
1889
1885
1880
1876 p /p *
0.32
0.36
0.37
0.37
Curve
0.37
0.38
0.38
0.38
0.39
0.39
0.39
0.40
0.40
0.40
0.41
0.41
0.41
0.42
0.42
30
0.43
s
0.43
0.43
0.44
0.44
0.45
0.45
0.45
0.46
0.46
0.47
0.47
0.48 Δs
(ft·lbf/lbm·oR)
Eq. (12.11b)
2.5
0.00
2.8
13.30
2.9
14.15
2.9
14.99
(Rayleigh)
2.9
15.84
2.9
16.69
3.0
17.54
3.0
18.39
3.0
19.24
3.0
20.09
3.1
20.93
3.1
21.78
3.1
22.63
3.2
23.48
3.2
24.32
3.2
25.17
3.2
26.01
3.3
26.86
40 3.3
50 27.70 60
3.3 o
28.54
.
(ft lbf/lbm R)
3.4
29.38
3.4
30.22
3.4
31.06
3.5
31.90
3.5
32.73
3.5
33.57
3.6
34.40
3.6
35.23
3.6
36.06
3.7
36.89
3.7
37.72
3.7
38.54
p (psi) 70 80 1.69
1.68
1.67
1.66
1.65
1.64
1.63
1.62
1.61
1.6
1.59
1.58
1.57
1.56
1.55
1.54
1.53
1.52
1.51
1.5
1.49
1.48
1.47
1.46
1.45
1.44
1.43
1.42
1.41
1.4
1.39
1.38
1.37
1.36
1.35
1.34
1.33
1.32
1.31
1.3
1.29
1.28
1.27
1.26
1.25
1.24
1.23
1.22
1.21
1.2
1.19
1.18
1.17
1.16
1.15
1.14
1.13
1.12
1.11
1.1
1.09
1.08
1.07
1.06
1.05
1.04
1.03
1.02
1.01
1 0.658
0.663
0.668
0.673
0.677
0.682
0.687
0.692
0.697
0.702
0.707
0.712
0.717
0.722
0.727
0.732
0.737
0.742
0.747
0.753
0.758
0.763
0.768
0.773
0.779
0.784
0.789
0.795
0.800
0.805
0.811
0.816
0.822
0.827
0.832
0.838
0.843
0.848
0.854
0.859
0.865
0.870
0.875
0.881
0.886
0.891
0.896
0.902
0.907
0.912
0.917
0.922
0.927
0.932
0.937
0.942
0.946
0.951
0.956
0.960
0.965
0.969
0.973
0.978
0.982
0.986
0.989
0.993
0.997
1.000 510
1107
514
1111
517
1115
521
1119
525
1123
529
1127
532
1131
536
1135
540
1139
544
1143
548
1147
2500
551
1151
555
1155
559 2000 1159
563
1164
567 1500 1168
571
1172
V (ft/s)
575
1176
1000
579
1180
583
1184
500 1188
587
591
1192
595
0 1196
599
1200
2.0
603
1204
607
1208
612
1213
616
1217
620
1221
624
1225
628
1229
632
1233
636
1237
9
641
1241
645 8
1245
649 7
1249
653
1253
6
657
1257
662 5
1261
p (psi)
666 4
1265
670 3
1269
674
1273
2
678
1277
682 1
1281
686 0
1285
690
2.0 1288
694
1292
699
1296
703
1300
706
1303
710
1307
714
1310
718
1314
722
1318
726
1321
730
1324
733
1328
737
1331
741
1334
744
1337
747
1341
751
1344
754
1347
757
1349
761
1352
764
1355
767
1358
769
1360
772
1362
775
1365 1872
1867
1863
1858
1853
1849
1844
1839
1834
Velocity V
1829
1824
1819
1814
1809
1803
1798
1793
1787
1782
1776
1770
1764
1758
1752
1.8
1746
1740
1734
1728
1721
1715
1708
Pressure p
1701
1695
1688
1681
1674
1667
1659
1652
1645
1637
1629
1622
1614
1606
1.81598
1589
1581
1573
1564
1555
1546
1537
1528
1519
1510
1500
1491
1481
1471
1461
1451
1441
1430
1420
1409
1398
1387
1376
1365 0.48
3.8
39.36
0.48
3.8
40.18
0.49
3.8
41.00
0.49
3.9
41.81
0.50
3.9
42.62
0.50
3.9
43.43
0.51
4.0
44.24
0.51
4.0
45.04
45.84
V0.52
ersus M (4.1
Rayleigh)
0.52
4.1
46.64
0.53
4.1
47.43
0.53
4.2
48.22
0.54
4.2
49.00
0.54
4.3
49.78
0.55
4.3
50.56
0.56
4.3
51.33
0.56
4.4
52.10
0.57
4.4
52.86
0.57
4.5
53.62
0.58
4.5
54.37
0.58
4.6
55.12
0.59
4.6
55.86
0.60
4.7
56.60
0.60
4.7
57.33
1.6
1.4
0.61
4.8
58.05
M4.8
0.61
58.77
0.62
4.9
59.48
0.63
4.9
60.18
0.63
5.0
60.88
0.64
5.0
61.56
0.65
5.1
62.24
Versus M (Rayleigh)
0.65
5.1
62.91
0.66
5.2
63.58
0.67
5.2
64.23
0.68
5.3
64.88
0.68
5.3
65.51
0.69
5.4
66.14
0.70
5.5
66.76
0.71
5.5
67.36
0.71
5.6
67.96
0.72
5.6
68.54
0.73
5.7
69.11
0.74
5.8
69.67
0.74
5.8
70.22
0.75
5.9
70.75
0.76
6.0 1.4
71.27
1.6
0.77
6.0
71.78
M 6.1
0.78
72.27
0.79
6.2
72.75
0.80
6.2
73.21
0.80
6.3
73.65
0.81
6.4
74.08
0.82
6.4
74.50
0.83
6.5
74.89
0.84
6.6
75.27
0.85
6.7
75.63
0.86
6.7
75.96
0.87
6.8
76.28
0.88
6.9
76.58
0.89
7.0
76.86
0.90
7.1
77.11
0.91
7.1
77.34
0.92
7.2
77.55
0.93
7.3
77.73
0.94
7.4
77.88
0.95
7.5
78.01
0.97
7.6
78.12
0.98
7.6
78.19
0.99
7.7
78.24
1.00
7.8
78.25 1.2 1.2 1.0 1.0 Problem 13.80 [2] Problem 13.81 [2] Problem 13.82 Given: Frictionless air flow in a pipe Find: [2] Heat exchange per lb (or kg) at exit, where 500 kPa Solution:
Basic equations: mrate = ρ⋅ V⋅ A p = ρ⋅ R ⋅ T Given or available data T1 = ( 15 + 273) ⋅ K ρ1 = ) ( p1 − p2 = ρ1⋅ V1⋅ V2 − V1 (Energy) ) p1 = 1⋅ MPa p1 R = 286.9⋅ c1 = m
c1 = 340
s kg
3 k⋅ R⋅ T1 m
V1 = 119 From momentum V2 = + V1 From continuity ρ1⋅ V1 = ρ2⋅ V2 m
s V1
ρ 2 = ρ 1⋅
V2 ρ2 = 3.09 T2 = 564 K p2 J
kg⋅ K m
s V2 = 466 V1 = M1⋅ c1 ρ1⋅ V1 J
cp = 1004⋅
kg⋅ K (Momentum) p2 = 500⋅ kPa ρ1 = 12.1 R ⋅ T1 p1 − p2 M1 = 0.35 k = 1.4 D = 5⋅ cm At section 1 ( δQ
= cp⋅ T02 − T01
dm T2 = 291 °C Hence T2 = and T02 = T2⋅ ⎛ 1 +
⎜ k −1 with T01 = T1⋅ ⎛ 1 +
⎜ k −1 Then kg
m 3 δQ
Btu
kJ
= cp⋅ T02 − T01 = 164⋅
= 383⋅
dm
lbm
kg ρ2⋅ R ⎝
⎝ ( 2 2⎞
⋅ M2 ⎟ T02 = 677 K T02 = 403 °C 2⎞
⋅ M1 ⎟ T01 = 295 K T01 = 21.9 °C ⎠ 2 ⎠ ) (Note: Using Rayleigh line functions, for M1 = 0.35 T0
T0crit
so = 0.4389 T0crit = T01
0.4389 T0crit = 672 K close to T2 ... Check!) M2 = 1 Problem 13.83 Given: Frictionless flow of Freon in a tube Find: [2] Heat transfer; Pressure drop NOTE: ρ2 is NOT as stated; see below Solution:
Basic equations: mrate = ρ⋅ V⋅ A ( p = ρ⋅ R ⋅ T Q = mrate⋅ h02 − h01 Btu
Given or available data h1 = 25⋅
lbm ρ1 = 100⋅ lbm
ft π2
⋅D
4 3 ) 2 ( h0 = h + V
2 p1 − p2 = ρ1⋅ V1⋅ V2 − V1 h2 = 65⋅ Btu
lbm ρ2 = 0.850⋅ mrate
V1 =
ρ1⋅ A ft
V1 = 8.03
s ft
V2 = 944
s V2 3 lbm
mrate = 1.85⋅
s h01 = h1 +
2 mrate
V2 =
ρ2⋅ A Then A= lbm
ft 2 D = 0.65⋅ in ) A = 0.332 in 2 ( ) The heat transfer is Q = mrate⋅ h02 − h01 The pressure drop is Δp = ρ1⋅ V1⋅ V2 − V1 ( V1 h02 = h2 +
2 Q = 107 ) Btu
s Δp = 162 psi h01 = 25.0 Btu
lbm h02 = 82.8 Btu
lbm 2 (74 Btu/s with the wrong ρ2!) (1 psi with the wrong ρ2!) Problem 13.84 [3] Problem 13.86 [3] Problem 13.87 [3] Given: Frictionless flow of air in a duct Find: Heat transfer without choking flow; change in stagnation pressure Solution:
Basic equations: T0
T p0 π2
⋅D
4 2 A = 78.54 cm
ρ1 = From continuity p1 − p2 = k = 1.4 ( 3 m
s then ) 2 From continuity ρ1⋅ V1 = or p1
p1
⋅ M1⋅ c1 =
⋅ M ⋅ k⋅ R⋅ T1 =
R ⋅ T1
R ⋅ T1 1 k⋅ R⋅ T1 T2 T02 = T2⋅ ⎛ 1 +
⎜ k−1 ⎝ p02 = p2⋅ ⎛ 1 +
⎜ 2
k−1 ⎝ ( 2 2⎞
⋅ M2 ⎟ 2⎞
⋅ M2 ⎟ p02 = 58.8 kPa ) δQ
MJ
= cp⋅ T02 − T01 = 1.12⋅
dm
kg (Using Rayleigh functions, at M1 = 0.215 T01
T0crit = T01 2 2 p
2
2
⋅ k⋅ R⋅ T⋅ M = k⋅ p⋅ M
R⋅ T
p2 = 31.1 kPa k p1⋅ M1
⋅
= ρ2⋅ V2 =
R
T1 k
k−1 ⎠ 2 J
kg⋅ K M1 = 0.215 ⎛ 1 + k⋅ M 2 ⎞
1⎟
p2 = p1⋅ ⎜
⎜ 1 + k⋅ M 2 ⎟
2⎠
⎝ T02 = 1394 K ⎠ V1
M1 =
c1
ρ ⋅ V = ρ⋅ c ⋅ M = but ⎛ p2 M2 ⎞
T2 = T1⋅ ⎜ ⋅
⎟
⎝ p1 M1 ⎠ p2⋅ M2 T1 Finally m
c1 = 331
s kg mrate
2
2
⋅ V2 − V1 = ρ2⋅ V2 − ρ1⋅ V1
A p1 − p2 = k⋅ p2⋅ M2 − k⋅ p1⋅ M1 Then c1 = V1 = 71.2 2 = R = 286.9⋅ m Hence Hence J
cp = 1004⋅
kg⋅ K ρ1 = 0.894 R ⋅ T1 p1⋅ M1 D = 10⋅ cm δQ
= cp⋅ T02 − T01
dm mrate
V1 =
ρ1⋅ A From momentum kg
mrate = 0.5⋅
s M2 = 1 ) p1 At state 1 ) mrate = ρ⋅ A⋅ V p1 = 70⋅ kPa ( Given or available data T1 = ( 0 + 273) ⋅ K
A= k − 1 2⎞
⋅M ⎟
2
⎠ p = ρ⋅ R ⋅ T p mrate
⋅ V2 − V1
A p1 − p2 = ⎛
⎝ = ⎜1 + ( k−1 2
⋅M
2 = 1+ k
k−1 k p2⋅ M2
⋅
R
T2 2 T2 = 1161 K
T01 = T1⋅ ⎛ 1 +
⎜ ⎝ p01 = p1⋅ ⎛ 1 +
⎜ Δp0 = p02 − p01 T01
= 0.1975 T02 =
T02
0.1975 ⎝ T2 = 888 °C
k−1 ⎠ 2
k−1
2 2⎞
⋅ M1 ⎟ 2⎞
⋅ M1 ⎟ ⎠ T01 = 276 K k
k−1 p01 = 72.3 kPa Δp0 = −13.5 kPa T02 = 1395 K and ditto for p02
...Check!) Problem 13.88 [3] Problem 13.93 [3] Given: Data on flow through gas turbine combustor
Find: Maximum heat addition; Outlet conditions; Reduction in stagnation pressure; Plot of process Solution:
R=
k=
cp =
T1 =
p1 =
M1 = The given or available data is: 286.9
1.4
1004
773
1.5
0.5 p02 J/kg·K T02
J/kg·K
K
MPa p2
T p01
T01
T1 Equations and Computations:
From p1 = ρ 1 RT1 ρ1= 6.76 V1 = M 1 kRT1 V1 = 279 p1 kg/m3 From T2 m/s s
Using builtin function IsenT (M,k):
T 01 /T 1 = 1.05 T 01 = 812 K Using builtin function Isenp (M,k):
p 01 /p 1 = 1.19 p 01 = 1.78 MPa M2 = For maximum heat transfer: 1 Using builtin function rayT0 (M,k), rayp0 (M,k), rayT (M,k), rayp (M,k), rayV (M,k):
*
T 01 /T 0* =
T0 =
0.691
1174 K
* p 01 /p 0 =
T /T * =
* p /p = ρ /ρ =
* * p0 =
T* = 1.114
0.790 * 1.60 MPa ( = p 02) 978 K ( = T 02)
( = p 2) 1.778 p= 0.844 0.444 ρ= 3.01 MPa
3
kg/m 182 kPa * Note that at state 2 we have critical conditions!
Hence: From the energy equation: p 012 – p 01 = δQ
dm 0.182 MPa = c p (T02 − T01 ) δ Q /dm = 364 kJ/kg ( = T 02) ( = ρ 2) Problem 13.94 [3] Problem 13.95 [3] Problem 13.96 [3] Problem 13.97 [4] Part 1/2 Problem 13.97 [4] Part 2/2 Problem 13.98 [4] Part 1/2 Problem 13.98 [4] Part 2/2 Problem 13.99 Given: Normal shock due to explosion Find: Shock speed; temperature and speed after shock [3] V
Shock speed Vs S hift coordinates: (Vs – V) ( Vs) Solution:
Shock at rest 2
M1 +
k−1
2 Basic equations: 2 M2 = p2
p1 = V = M⋅ c = M⋅ k⋅ R⋅ T ⎛ 2⋅ k ⎞ ⋅ M 2 − 1
⎜
⎟1
⎝ k − 1⎠ ⎛ 1 + k − 1 ⋅ M 2⎞ ⋅ ⎛ k⋅ M 2 − k − 1 ⎞
⎜
⎟
1⎟⎜
1
2
2⎠
⎠⎝
=⎝
2
T1
⎛ k + 1 ⎞ ⋅M 2
⎜
⎟1
⎝2⎠
T2 2⋅ k
2 k−1
⋅ M1 −
k+1
k+1 k = 1.4 From the pressure ratio M1 = Then we have ⎛ 1 + k − 1 ⋅ M 2⎞ ⋅ ⎛ k⋅ M 2 − k − 1 ⎞
⎜
⎟
1⎟⎜
1
2⎠
2
⎠⎝
T2 = T1⋅ ⎝
2
⎛ k + 1 ⎞ ⋅M 2
⎜
⎟1
⎝2⎠ M2 = R = 286.9⋅ J
kg⋅ K Given or available data ⎛ k + 1 ⎞ ⋅ ⎛ p2 + k − 1 ⎞
⎜
⎟⎜
⎟
⎝ 2⋅ k ⎠ ⎝ p1 k + 1 ⎠ p2 = 30⋅ MPa p1 = 101⋅ kPa T1 = ( 20 + 273) ⋅ K T2 = 14790 K T2 = 14517⋅ °C M1 = 16.0 2
2
M1 +
k−1 M2 = 0.382 ⎛ 2⋅ k ⎞ ⋅ M 2 − 1
⎜
⎟1
⎝ k − 1⎠
m Then the speed of the shock (Vs = V1) is V1 = M1⋅ k⋅ R⋅ T1 V1 = 5475 After the shock (V2) the speed is V2 = M2⋅ k⋅ R⋅ T2 V2 = 930 m V = Vs − V2 V = 4545 m But we have V2 = Vs − V s Vs = V1 Vs = 5475 s s These results are unrealistic because at the very high postshock temperatures experienced, the specific heat ratio will NOT
be constant! The extremely high initial air velocity and temperature will rapidly decrease as the shock wave expands in a
spherical manner and thus weakens. m
s Problem 13.100 [3] Given: CD nozzle with normal shock Find: Mach numbers at the shock and at exit; Stagnation and static pressures before and after the shock Solution:
⎛ 1 + k − 1 ⋅ M2 ⎞
⎟
A
1⎜
2
=
⋅⎜
⎟
k+ 1
Acrit
M⎜
⎟
2
⎝
⎠ Basic equations: Isentropic flow Given or available data 2
2
M1 +
k−1 2 M2 = Normal shock k+ 1
2⋅ ( k−1) Rair = 53.33⋅
2 At = 1.5⋅ in p1 = p02 2⋅ k
2 k−1
⋅ M1 −
k+1
k+1 ⎛
⎝ = ⎜1 + p p2 ⎛ 2⋅ k ⎞ ⋅ M 2 − 1
⎜
⎟1
⎝ k − 1⎠ k = 1.4 p0 p01 = k − 1 2⎞
⋅M ⎟
2
⎠ k
k−1 ⎛ k + 1 ⋅M 2 ⎞
1⎟
⎜
2
⎜
⎟
⎜ 1 + k − 1 ⋅ M12 ⎟
2
⎝
⎠ k
k−1 ⎛ 2⋅ k ⋅ M 2 − k − 1 ⎞
⎜
⎟
1
k + 1⎠
⎝k + 1 ft⋅ lbf
lbm⋅ R T0 = ( 175 + 460) ⋅ R (Shock area) 2 As = 2.5⋅ in p01 = 125⋅ psi 1
k−1 Ae = 3.5⋅ in 2 Because we have a normal shock the CD must be accelerating the flow to supersonic so the throat is at critical state.
Acrit = At At the shock we have As = 1.667 ⎛ 1 + k − 1 ⋅M 2 ⎞
1⎟
1⎜
2
At this area ratio we can find the Mach number before the shock from the isentropic relation
=
⋅⎜
⎟
k+ 1
Acrit
M1 ⎜
⎟
2
⎝
⎠
Acrit k+ 1
2⋅ ( k−1) As Solving iteratively (or using Excel's Solver, or even better the function isenMsupfromA from the Web site!) M1 = 1.985 The stagnation pressure before the shock was given: p01 = 125 psi The static pressure is then p1 = p01 ⎛ 1 + k − 1 ⋅ M 2⎞
⎜
1⎟
2
⎝
⎠ k
k−1 p1 = 16.4 psi After the shock we have Also M2 = 2
2
M1 +
k−1 M2 = 0.580 ⎛ 2⋅ k ⎞ ⋅ M 2 − 1
⎜
⎟1
⎝ k − 1⎠ ⎛ k + 1 ⋅M 2 ⎞
1⎟
⎜
2
⎜
⎟
⎜ 1 + k − 1 ⋅ M12 ⎟
2
⎝
⎠
p02 = p01⋅ k
k−1 ⎛ 2⋅ k ⋅ M 2 − k − 1 ⎞
⎜
⎟
1
k + 1⎠
⎝k + 1 and p02 = 91.0 psi 1
k−1 2⋅ k
2 k − 1⎞
p2 = p1⋅ ⎛
⋅ M1 −
⎜
⎟
k + 1⎠
⎝k + 1 p2 = 72.4 psi Finally, for the Mach number at the exit, we could find the critical area change across the shock; instead we find the
new critical area from isentropic conditions at state 2.
− ⎛ 1 + k − 1 ⋅M 2 ⎞
⎜
2⎟
2
Acrit2 = As⋅ M2⋅ ⎜
⎟
k+ 1
⎜
⎟
2
⎝
⎠ At the exit we have Ae
Acrit2 k+ 1
2⋅ ( k−1)
2 Acrit2 = 2.06 in = 1.698 ⎛ 1 + k − 1 ⋅M 2 ⎞
e⎟
1⎜
2
At this area ratio we can find the Mach number before the shock from the isentropic relation
=
⋅⎜
⎟
k+ 1
Acrit2
Me ⎜
⎟
2
⎝
⎠ k+ 1
2⋅ ( k−1) Ae Solving iteratively (or using Excel's Solver, or even better the function isenMsubfromA from the Web site!) These calculations are obviously a LOT easier using the Excel functions available on the Web site! Me = 0.369 Problem 13.101 Given: Normal shock near pitot tube Find: [2] Air speed Solution:
Basic equations: ( p1 − p2 = ρ1⋅ V1⋅ V2 − V1 ) Given or available data T1 = 285⋅ R Rair = 53.33⋅
k −1
⎤
⎡
⎢
⎥
k
⎥
2 ⎢⎛ p02 ⎞
⋅ ⎢⎜
− 1⎥
⎟
k −1
⎣⎝ p2 ⎠
⎦ At state 2 M2 = From momentum p1 − p2 = ρ2⋅ V2 − ρ1⋅ V1 2 2 2 2 ⎤
1 ⎡ p2 ⎛
2
⋅ ⎢ ⋅ 1 + k⋅ M2 ⎞ − 1⎥
⎝
⎠
k p1
⎣ Also c1 = Then ⎦ k⋅ Rair⋅ T1 V1 = M1⋅ c1 k − 1 2⎞
⋅M ⎟
2
⎠ p02 = 10⋅ psi M2 = 0.574 2 2 p1 2 p
2
2
⋅ k ⋅ R⋅ T⋅ M = k ⋅ p ⋅ M
R⋅ T ρ ⋅ V = ρ⋅ c ⋅ M = or 2
2
p1⋅ ⎛ 1 + k⋅ M1 ⎞ = p2⋅ ⎛ 1 + k⋅ M2 ⎞
⎝
⎠
⎝
⎠ M1 = 2.01
ft
c1 = 827
s
ft
s V1 = 1822⋅
p2 p2 = 8⋅ psi ft⋅ lbf
lbm⋅ R V1 = 1666 Note: With p1 = 1.5 psi we obtain (Using normal shock functions, for ⎛
⎝ = ⎜1 + but p1 − p2 = k⋅ p2⋅ M2 − k⋅ p1⋅ M1 M1 = p p1 = 1.75⋅ psi k = 1.4 Hence p0 (Momentum) k
k−1 = 4.571 we find M1 = 2.02 ft
s
M2 = 0.573 Check!) Problem 13.103 [2] Problem 13.104 [2] Given: Normal shock
Find: Speed and temperature after shock; Entropy change Solution:
R=
k=
cp = The given or available data is: 53.33
1.4
0.2399 T 01 = 1250 p1 = 20 M1 = 300.02 V1 = 764 Using builtin function IsenT (M,k):
T 01 /T 1 = 0.0685 Btu/lbm·R Btu/lbm·R
R o 2.5 ρ1 = ft·lbf/lbm·R 2.25 psi Equations and Computations:
From p1 = ρ1 RT1 kg/m3
m/s o R o F o R 728 T1 = o F 143 psi 556
96 Using builtin function NormM2fromM (M,k):
M2 = 0.513 Using builtin function NormTfromM (M,k):
T 2 /T 1 = 2.14 Using builtin function NormpfromM (M,k):
p 2 /p 1 =
From V 2 = M 2 kRT 2 From ⎛T
Δ s = c p ln ⎜ 2
⎜T
⎝1 T2 = p2 = 7.13 V2 = 867 Δs = 0.0476
37.1 ft/s ⎛p ⎞
⎞
⎟ − R ln ⎜ 2 ⎟
⎜p ⎟
⎟
⎝ 1⎠
⎠ Btu/lbm·R
ft·lbf/lbm·R 1188 Problem 13.106 [2] Given: Normal shock
Find: Pressure after shock; Compare to isentropic deceleration Solution:
R=
k=
T 01 = 286.9
1.4
550 p 01 = 650 M1 = The given or available data is: 2.5 J/kg·K
K
kPa Equations and Computations: Using builtin function Isenp (M,k):
p 01 /p 1 = 17.09 Using builtin function NormM2fromM (M,k):
M2 = 0.513 Using builtin function NormpfromM (M,k):
p 2 /p 1 = 7.13 p1 = 38 kPa p2 = 271 kPa p2 = 543 kPa Using builtin function Isenp (M,k) at M 2:
p 02 /p 2 =
But for the isentropic case:
Hence for isentropic deceleration: 1.20 p 02 = p 01 Problem 13.107 [2] Given: Normal shock
Find: Speed and Mach number after shock; Change in stagnation pressure Solution:
R=
k=
T1 = 53.33
1.4
445 p1 = 5 V1 = 2000 mph c1 = 1034 ft/s M1 = The given or available data is: 2.84 ft·lbf/lbm·R
o 0.0685 Btu/lbm·R R psi
2933 ft/s 793 ft/s Equations and Computations:
From c1 = kRT1 Then Using builtin function NormM2fromM (M,k):
M2 = 0.486 Using builtin function NormdfromM (M,k):
ρ 2 /ρ 1 = 3.70 Using builtin function Normp0fromM (M,k):
p 02 /p 01 = 0.378 Then V2 = ρ1
V1
ρ2 V2 = 541 mph Using builtin function Isenp (M,k) at M 1:
p 01 /p 1 = 28.7 From the above ratios and given p 1:
p 01 = 143 psi p 02 = 54.2 psi p 01 – p 02 = 89.2 psi Problem 13.108 [2] Given: Normal shock
Find: Speed; Change in pressure; Compare to shockless deceleration Solution:
R=
k=
T1 = 53.33
1.4
452.5 p1 = 14.7 psi V1 = 1750 mph c1 = 1043 ft/s M1 = The given or available data is: 2.46 ft·lbf/lbm·R
o 0.0685 Btu/lbm·R R
2567 ft/s p2 = 101 psi p2 – p1 = 86.7 psi 781 ft/s p2 = 197 psi p2 – p1 = 182 psi Equations and Computations:
From c1 = kRT1 Then Using builtin function NormM2fromM (M,k):
M2 = 0.517 Using builtin function NormdfromM (M,k):
ρ 2 /ρ 1 = 3.29 Using builtin function NormpfromM (M,k):
p 2 /p 1 = 6.90 Then V2 = ρ1
V1
ρ2 V2 = 532 mph Using builtin function Isenp (M,k) at M 1:
p 01 /p 1 = 16.1 Using builtin function Isenp (M,k) at M 2:
p 02 /p 2 = 1.20 From above ratios and p 1, for isentropic flow (p 0 = const): Problem 13.109 [2] Problem 13.111 [2] Problem 13.112 Given: Normal shock Find: [4] RankineHugoniot relation Solution:
2 Momentum: p1 + ρ1⋅ V1 = p2 + ρ2⋅ V2 Energy: Basic equations: h1 + 2 Mass:
Ideal Gas: 1
1
2
2
⋅ V1 = h2 + ⋅ V2
2
2 ( ) ( ) ρ1 V1 = ρ2⋅ V2
p = ρ⋅ R ⋅ T 2 2 ( )( From the energy equation 2⋅ h2 − h1 = 2⋅ cp⋅ T2 − T1 = V1 − V2 = V1 − V1 ⋅ V1 + V2 From the momentum equation p2 − p1 = ρ1⋅ V1 − ρ2⋅ V2 = ρ1⋅ V1⋅ V1 − V2 Hence V1 − V2 = Using this in Eq 1 2 ( 2 ) ) (1) where we have used the mass equation p2 − p1
ρ1⋅ V1 p2 − p1
p2 − p1 ⎛
V2 ⎞ p2 − p1 ⎛
2⋅ cp⋅ T2 − T1 =
⋅ V1 + V2 =
⋅ ⎜1 +
⋅ ⎜1 +
⎟=
ρ1⋅ V1
ρ1
ρ1
⎝ V1 ⎠
⎝ ( ) ( ) ρ1 ⎞ 1⎞
⎛1
⎟ = (p2 − p1)⋅ ⎜ + ⎟
ρ2
⎠
⎝ ρ1 ρ2 ⎠ where we again used the mass equation
Using the idea gas equation ⎛ p2 2⋅ cp⋅ ⎜ ⎝ ρ2⋅ R − p1 ⎞ 1⎞
⎛1
⎟ = (p2 − p1)⋅ ⎜ + ⎟
ρ1⋅ R
⎠
⎝ ρ1 ρ2 ⎠ Dividing by p1 and multiplying by ρ2, and using R = cp  cv, k = cp/cv
2⋅ Collecting terms cp ⎛ p2 ρ2 ⎞
⎞ ⎛ ρ2 ⎞
k ⎛ p2 ρ2 ⎞ ⎛ p2
⋅⎜
−
⋅⎜
−
⎟ = 2⋅
⎟ = ⎜ − 1⎟ ⋅ ⎜ + 1⎟
R p1 ρ1
k − 1 p1 ρ1
⎝
⎠
⎝
⎠ ⎝ p1 ⎠ ⎝ ρ1 ⎠ p2 ⎛ 2⋅ k
⋅⎜
−1−
p1 k − 1
⎝ p2
p1 For an infinite pressure ratio = ρ2 ⎞ 2⋅ k ρ2 ρ2
⋅
−
−1
k − 1 ρ1 ρ1 ⎛ 2⋅ k
−1−
⎜
⎝k − 1 ( k + 1) − ( k − 1) ⋅ 2⋅ k ρ2 ρ2 ⋅
−
−1
⎟=
ρ1
⎠ k − 1 ρ1 ρ1 ρ2
ρ1 ρ2 ⎞ ⎟
ρ1
⎠
=0 = ( k + 1) ρ2
⋅
−1
( k − 1) ρ1
( k + 1)
( k − 1) or − ρ2 or p2
p1 ( k + 1) ⋅
= ρ1 ρ1 − ( k − 1) ( k + 1) − ( k − 1) ⋅ ρ1
ρ2 ρ2 = k+1
k−1 ρ2
ρ1 (= 6 for air) Problem 13.113 [3] Problem 13.114 [3] Problem 13.115 [3] Problem 13.117 [3] Problem 13.119 [4] Problem 13.120 [2] Problem 13.121 [2] Problem 13.124 [3] Given: Normal shock in CD nozzle
Find: Exit pressure; Throat area; Mass flow rate Solution:
R=
k=
T 01 =
p 01 =
M1 = 286.9
1.4
550
700
2.75 A1 = 25 cm2 Ae = The given or available data is: 40 cm2 J/kg·K
K
kPa Equations and Computations (assuming State 1 and 2 before and after the shock): Using builtin function Isenp (M,k):
p 01 /p 1 = 25.14 p1 = 28 kPa Using builtin function IsenT (M,k):
T 01 /T 1 = 2.51 T1 = 219 K Using builtin function IsenA (M,k):
A 1 /A 1* = 3.34 A 1* = A t = 7.49 cm2 p 02 = 284 kPa Me= 0.279 pe= 269 Then from the Ideal Gas equation: ρ1 =
c1 =
V1 = Also:
So:
Then the mass flow rate is: 0.4433
297
815 m rate =
m rate = kg/m3
m/s
m/s ρ 1 V 1A 1
0.904 kg/s For the normal shock:
Using builtin function NormM2fromM (M,k):
M2 = 0.492 Using builtin function Normp0fromM (M,k) at M 1:
p 02 /p 01 =
0.41
For isentropic flow after the shock:
Using builtin function IsenA (M,k):
A 2 /A 2* =
A2 =
But: 1.356
A1 A 2* = 18.44 Hence: Using builtin function IsenAMsubfromA (Aratio,k):
A e /A 2* =
For:
2.17
Using builtin function Isenp (M,k):
p 02 /p e = 1.06 cm2 kPa Problem 13.125 [2] Problem 13.128 [3] Problem 13.129 [3] Problem 13.131 [3] Problem 13.132 [4] Problem 13.133 [3] Problem *13.135 [5] Problem *13.136 [2] Problem *13.137 [4] Problem 13.138 [3] Given: Normal shock Find: Approximation for downstream Mach number as upstream one approaches infinity Solution:
Basic equations: 2 M2n = 2
2
M1n +
k −1 (13.48a) ⎛ 2⋅ k ⎞ ⋅ M 2 − 1
⎜
⎟ 1n
⎝ k − 1⎠ M2n = M2⋅ sin ( β − θ) (13.47b) 2
2
M1n +
k −1 Combining the two equations M2n
M2 =
=
sin( β − θ) 1+
M2 = ⎛ 2⋅ k ⎞ ⋅ M 2 − 1
⎜
⎟ 1n
⎝ k − 1⎠
=
sin( β − θ) 2
2
M1n +
k −1 ⎡⎛ 2⋅ k ⎞ ⋅ M 2 − 1⎤ ⋅ sin( β − θ) 2
⎢⎜
⎟ 1n
⎥
⎣⎝ k − 1 ⎠
⎦ 2
( k − 1) ⋅ M1n 2 ⎡⎛ 2⋅ k ⎞ − 1 ⎥ ⋅ sin( β − θ) 2
⎤
⎢⎜
⎟
⎝ k − 1 ⎠ M 2⎥
⎢
1n ⎦
⎣ As M1 goes to infinity, so does M1n, so
M2 = 1 ⎛ 2⋅ k ⎞ ⋅ sin ( β − θ) 2
⎜
⎟
⎝ k − 1⎠ M2 = k−1
2⋅ k⋅ sin ( β − θ) 2 Problem 13.139 [3] Given: Data on an oblique shock
Find: Mach number and pressure downstream; compare to normal shock Solution:
R=
k=
p1 = 286.9
1.4
80 M1 = 2.5 β= The given or available data is: 35 J/kg.K
kPa
o Equations and Computations:
From M 1 and β M 1n = 1.43 M 1t = 2.05 From M1n and p1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) p2 = 178.6 V t1 = V t2 The tangential velocity is unchanged Hence c t1 M t1 =
(T 1 ) 1/2 c t2 M t2 M t1 = (T 2)1/2 M t2
M 2t = (T 1/T 2)1/2 M t1 From M1n, and Eq. 13.48c
(using builtin function NormTfromM (M ,k ))
T 2/T 1 =
Hence 1.28 M 2t = 1.81 kPa Also, from M1n, and Eq. 13.48a
(using builtin function NormM2fromM (M ,k )) (13.48a) M 2n = 0.726 The downstream Mach number is then
M 2 = (M 2t2 + M 2n2)1/2
M2 = 1.95 Finally, from geometry
V 2n = V 2sin(β  θ)
Hence θ = β  sin1(V 2n/V 2) or θ = β  sin1(M 2n/M 2)
θ= 13.2 o 570 kPa For the normal shock:
From M1 and p1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
p2 =
Also, from M1, and Eq. 13.48a
(using builtin function NormM2fromM (M ,k ))
M2 = 0.513 For the minimum β :
The smallest value of β is when the shock is a Mach wave (no deflection)
β = sin1(1/M 1)
β= 23.6 o Problem 13.140 [3] Given: Oblique shock in flow at M = 3
Find: Minimum and maximum β, plot of pressure rise across shock Solution:
The given or available data is: R=
k=
M1 = 286.9
1.4
3 J/kg.K Equations and Computations:
The smallest value of β is when the shock is a Mach wave (no deflection)
β = sin1(1/M 1)
β=
The largest value is 19.5 o β= 90.0 o The normal component of Mach number is
M 1n = M 1sin(β) (13.47a) For each β, p2/p1 is obtained from M1n, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) Computed results:
β (o) M 1n p 2/p 1 19.5
20
30
40
50
60
70
75
80
85
90 1.00
1.03
1.50
1.93
2.30
2.60
2.82
2.90
2.95
2.99
3.00 1.00
1.06
2.46
4.17
5.99
7.71
9.11
9.63
10.0
10.3
10.3 Pressure Change across an Oblique Shock
12.5
10.0 p 2/p 1 7.5
5.0
2.5
0.0
0 30 60
β()
o 90 Problem 13.141 [3] Given: Velocities and deflection angle of an oblique shock
Find: Shock angle β; pressure ratio across shock Solution:
R=
k=
V1 = 286.9
1.4
1250 V2 = The given or available data is: 650 θ= 35 J/kg.K
m/s
m/s
o Equations and Computations:
From geometry we can write two equations for tangential velocity:
For V 1t V 1t = V 1cos(β) (1) For V 2t V 2t = V 2cos(β  θ) (2) For an oblique shock V 2t = V 1t, so Eqs. 1 and 2 give
V 1cos(β) = V 2cos(β  θ)
Solving for β (3) β = tan1((V 1  V 2cos(θ))/(V 2sin(θ)))
β= (Alternatively, solve Eq. 3 using Goal Seek !) 62.5 o For p 2/p 1, we need M 1n for use in Eq. 13.48d (13.48d) We can compute M 1 from θ and β, and Eq. 13.49
(using builtin function Theta (M ,β, k )) (13.49) θ= 35.0 o β= 62.5 o M1 = For 3.19 This value of M 1 was obtained by using Goal Seek :
Vary M 1 so that θ becomes the required value.
(Alternatively, find M 1 from Eq. 13.49 by explicitly solving for it!)
We can now find M 1n from M 1. From M 1 and Eq. 13.47a
M 1n = M 1sin(β)
Hence M 1n = 2.83 Finally, for p 2/p 1, we use M 1n in Eq. 13.48d
(using builtin function NormpfromM (M ,k )
p 2 /p 1 = 9.15 (13.47a) Problem 13.142 Given: Data on an oblique shock
Find: Deflection angle θ; shock angle β ; Mach number after shock Solution:
The given or available data is: R=
k=
M1 =
T1 =
p 2 /p 1 = 286.9
1.4
3.25
283
5 J/kg.K K Equations and Computations:
From p 2/p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k )
and Goal Seek or Solver )
(13.48d) p 2 /p 1 = 5.00 M 1n = For 2.10 From M 1 and M 1n, and Eq 13.47a
M 1n = M 1sin(β )
β= 40.4 (13.47a)
o From M 1 and β , and Eq. 13.49
(using builtin function Theta (M ,β , k ) (13.49) θ= 23.6 o To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using builtin function NormM2fromM (M ,k )) (13.48a) M 2n = 0.561 The downstream Mach number is then obtained from
from M 2n, θ and β , and Eq. 13.47b
M 2n = M 2sin(β  θ)
Hence M2 = 1.94 (13.47b) Problem 13.143 [4] Given: Airfoil with included angle of 20o
Find:
Mach number and speed at which oblique shock forms Solution:
R=
k=
T1 = 286.9
1.4
288 θ= The given or available data is: 10 J/kg.K
K
o Equations and Computations: From Fig. 13.29 the smallest Mach number for which an oblique shock exists
at a deflection θ = 10o is approximately M 1 = 1.4.
By trial and error, a more precise answer is
(using builtin function Theta (M ,β, k )
M1 = 1.42 β=
θ= 67.4
10.00 c1 =
V1 = 340
483 A suggested procedure is:
1) Type in a guess value for M 1
2) Type in a guess value for β o
o m/s
m/s 3) Compute θ from Eq. 13.49
(using builtin function Theta (M ,β, k )) (13.49)
4) Use Solver to maximize θ by varying β
o
5) If θ is not 10 , make a new guess for M 1
6) Repeat steps 1  5 until θ = 10 o Computed results:
β() θ() 67.4
56.7
45.5
39.3
35.0
31.9
27.4
22.2
19.4
17.6
16.4 10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0 o M1
1.42
1.50
1.75
2.00
2.25
2.50
3.00
4.00
5.00
6.00
7.00 o Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0% Sum: 0.0% To compute this table:
1) Type the range of M 1
2) Type in guess values for β
3) Compute θ from Eq. 13.49
(using builtin function Theta (M ,β, k )
o
Compute the absolute error between each θ and θ = 10
4)
5) Compute the sum of the errors
6) Use Solver to minimize the sum by varying the β values
(Note: You may need to interactively type in new β values
if Solver generates β values that lead to no θ, or to
β values that correspond to a strong rather than weak shock) Oblique Shock Angle as a Function of
Aircraft Mach Number
90
75
60
β (o) 45
30
15
0
1 2 3 4
M 5 6 7 Problem 13.144 [4] Given: Airfoil with included angle of 60o
Find: Plot of temperature and pressure as functions of angle of attack Solution:
R=
k=
T1 =
p1 =
V1 = The given or available data is: 286.9
1.4
276.5
75
1200 δ= 60 c1 = 333 M1 = J/kg.K
K
kPa
m/s 3.60 o Equations and Computations:
From T 1
Then m/s Computed results:
α( ) β() θ ( ) Needed θ() 0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00
22.00
24.00
26.00
28.00
30.00 47.1
44.2
41.5
38.9
36.4
34.1
31.9
29.7
27.7
25.7
23.9
22.1
20.5
18.9
17.5
16.1 30.0
28.0
26.0
24.0
22.0
20.0
18.0
16.0
14.0
12.0
10.0
8.0
6.0
4.0
2.0
0.0 30.0
28.0
26.0
24.0
22.0
20.0
18.0
16.0
14.0
12.0
10.0
8.0
6.0
4.0
2.0
0.0 o o o o Sum: Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0% M 1n p 2 (kPa) T 2 (oC) 2.64
2.51
2.38
2.26
2.14
2.02
1.90
1.79
1.67
1.56
1.46
1.36
1.26
1.17
1.08
1.00 597
539
485
435
388
344
304
267
233
202
174
149
126
107
90
75 357
321
287
255
226
198
172
148
125
104
84
66
49
33
18
3 597 357 Max: To compute this table:
1)
2)
3)
4)
5)
6)
7) 8)
9)
10) Type the range of α
Type in guess values for β
Compute θNeeded from θ = δ/2  α
Compute θ from Eq. 13.49
(using builtin function Theta (M ,β, k )
Compute the absolute error between each θ and θNeeded
Compute the sum of the errors
Use Solver to minimize the sum by varying the β values
(Note: You may need to interactively type in new β values
if Solver generates β values that lead to no θ)
For each α, M 1n is obtained from M 1, and Eq. 13.47a
For each α, p 2 is obtained from p 1, M 1n, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
For each α, T 2 is obtained from T 1, M 1n, and Eq. 13.48c
(using builtin function NormTfromM (M ,k )) Pressure on an Airfoil Surface
as a Function of Angle of Attack
700 p 2 (kPa) 600
500
400
300
200
100
0
0 5 10 15 20 25 30 25 30 α( )
o Temperature on an Airfoil Surface
as a Function of Angle of Attack
400
350
o
T 2 ( C) 300
250
200
150
100
50
0
0 5 10 15
α( )
o 20 Problem 13.145 [4] Given: Airfoil with included angle of 60o
Find: Angle of attack at which oblique shock becomes detached Solution:
R=
k=
T1 =
p1 =
V1 = The given or available data is: 286.9
1.4
276.5
75
1200 δ= 60 c1 = 333 M1 = J/kg.K
K
kPa
m/s 3.60 o Equations and Computations:
From T 1
Then m/s From Fig. 13.29, at this Mach number the smallest deflection angle for which
an oblique shock exists is approximately θ = 35o. By using Solver , a more precise answer is
(using builtin function Theta (M ,β, k )
M1 = 3.60 β=
θ= 65.8
37.3 o
o A suggested procedure is:
1) Type in a guess value for β
2) Compute θ from Eq. 13.49
(using builtin function Theta (M ,β, k )) (13.49)
3) Use Solver to maximize θ by varying β
For a deflection angle θ the angle of attack α is
α = θ  δ/2
α=
7.31 o Computed results:
α (o) β (o) θ (o) Needed θ (o) 0.00
1.00
2.00
3.00
4.00
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.31 47.1
48.7
50.4
52.1
54.1
57.4
58.1
58.8
59.5
60.4
61.3
62.5
64.4
65.8 30.0
31.0
32.0
33.0
34.0
35.5
35.8
36.0
36.3
36.5
36.8
37.0
37.3
37.3 30.0
31.0
32.0
33.0
34.0
35.5
35.7
36.0
36.2
36.5
36.7
37.0
37.2
37.3
Sum: Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0% M 1n
2.64
2.71
2.77
2.84
2.92
3.03
3.06
3.08
3.10
3.13
3.16
3.19
3.25
3.28 0.0% Max: p 2 (kPa)
597
628
660
695
731
793
805
817
831
845
861
881
910
931 T 2 (oC)
357
377
397
418
441
479
486
494
502
511
521
533
551
564 931 564 To compute this table:
Type the range of α
Type in guess values for β
Compute θNeeded from θ = α + δ/2
Compute θ from Eq. 13.49
(using builtin function Theta (M ,β, k )
Compute the absolute error between each θ and θNeeded
Compute the sum of the errors
Use Solver to minimize the sum by varying the β values
(Note: You may need to interactively type in new β values
if Solver generates β values that lead to no θ)
For each α, M 1n is obtained from M 1, and Eq. 13.47a
For each α, p 2 is obtained from p 1, M 1n, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
For each α, T 2 is obtained from T 1, M 1n, and Eq. 13.48c
(using builtin function NormTfromM (M ,k )) 1)
2)
3)
4)
5)
6)
7) 8)
9)
10) Pressure on an Airfoil Surface
as a Function of Angle of Attack
1000 p 2 (kPa) 900
800
700
600
500
0 2 4 6 8 6 8 o
α( ) Temperature on an Airfoil Surface
as a Function of Angle of Attack
600 500 o T 2 ( C) 550 450
400
350
300
0 2 4
α (o) Problem 13.146 Given: Data on airfoil flight
Find: Lift per unit span Solution:
R=
k=
p1 = 286.9
1.4
70 M1 = 2.75 δ=
c= The given or available data is: 7
1.5 J/kg.K
kPa
o m Equations and Computations:
The lift per unit span is
L = (p L  p U)c (1) (Note that p L acts on area c /cos(δ), but its
normal component is multiplied by cos(δ))
For the upper surface:
pU = p1 pU = 70.0 kPa For the lower surface:
We need to find M 1n
θ= δ The deflection angle is θ= 7 o From M 1 and θ, and Eq. 13.49
(using builtin function Theta (M , β,k ))
(13.49) θ= 7.0 o β= 26.7 o M 1n = 1.24 For (Use Goal Seek to vary β so that θ = δ)
From M 1 and β From M 1n and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k )) (13.48d)
p2 =
pL = p2 pL =
From Eq 1 113 113 kPa L= 64.7 kN/m kPa Problem 13.147 [3] Given: Data on airfoil flight
Find: Lift per unit span Solution:
R=
k=
p1 = 286.9
1.4
75 M1 = 2.75 δU = 5 o δL =
c= 15
2 o The given or available data is: J/kg.K
kPa m Equations and Computations:
The lift per unit span is
L = (p L  p U)c (1) (Note that each p acts on area c /cos(δ), but its
normal component is multiplied by cos(δ))
For the upper surface:
We need to find M 1n(U)
θU = δU θU = The deflection angle is 5 o From M 1 and θU, and Eq. 13.49
(using builtin function Theta (M , β,k ))
(13.49) For θU = 5.00 o βU = 25.1 o (Use Goal Seek to vary βU so that θU = δU)
From M 1 and βU M 1n(U) = 1.16 From M 1n(U) and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) p2 = 106 pU = p2 pU = 106 θL = δL θL = 15 o θL = 15.00 o βL = 34.3 o kPa kPa For the lower surface:
We need to find M 1n(L)
The deflection angle is From M 1 and θL, and Eq. 13.49
(using builtin function Theta (M , β,k ))
For (Use Goal Seek to vary βL so that θL = δL)
From M 1 and βL M 1n(L) = 1.55 From M 1n(L) and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
p2 =
pL = p2 pL =
From Eq 1 198 198 kPa L= 183 kN/m kPa Problem 13.148 [3] Given: Oblique shock Mach numbers
Find: Deflection angle; Pressure after shock Solution:
k=
p1 =
M1 = 1.4
75
4 M2 = The given or available data is: 2.5 β= 33.6 kPa Equations and Computations:
We make a guess for β: o From M 1 and β, and Eq. 13.49 (using builtin function Theta (M , β,k ))
(13.49)
θ=
From M 1 and β
From M 2, θ, and β 21.0 M 1n =
M 2n = 2.211
0.546 o (1) We can also obtain M 2n from Eq. 13.48a (using builtin function normM2fromM (M ,k )) (13.48a) M 2n = 0.546 (2) We need to manually change β so that Eqs. 1 and 2 give the same answer.
Alternatively, we can compute the difference between 1 and 2, and use
Solver to vary β to make the difference zero
Error in M 2n = 0.00% Then p 2 is obtained from Eq. 13.48d (using builtin function normpfromm (M ,k ))
(13.48d) p2 = 415 kPa Problem 13.149 [4] Given: Air flow into engine
Find: Pressure of air in engine; Compare to normal shock Solution:
k=
p1 = 1.4
50 M1 = The given or available data is: 3 θ= 7.5 kPa
o Equations and Computations:
Assuming isentropic flow deflection
p 0 = constant
p 02 = p 01 For p 01 we use Eq. 13.7a (using builtin function Isenp (M , k ))
(13.7a) p 01 = kPa p 02 =
For the deflection 1837
1837 kPa θ= 7.5 o From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
(13.55)
ω1 =
Deflection =
Applying Eq. 1 49.8 ω2  ω1 = ω(M 2)  ω(M 1) o (1) ω2 = ω1  θ ω2 = 42.3 (Compression!)
o From ω2, and Eq. 13.55 (using builtin function Omega (M , k ))
ω2 = 42.3 M2 = For 2.64 o (Use Goal Seek to vary M 2 so that ω2 is correct)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
For the normal shock (2 to 3) 86.8 M2 = 2.64 kPa From M 2 and p 2, and Eq. 13.41d (using builtin function NormpfromM (M ,k ))
(13.41d) p3 = 690 kPa For slowing the flow down from M 1 with only a normal shock, using Eq. 13.41d
p= 517 kPa Problem 13.150 [3] Given: Air flow in a duct
Find: Mach number and pressure at contraction and downstream; Solution:
k=
M1 = 1.4
2.5 θ=
p1 = The given or available data is: 7.5
50 o kPa Equations and Computations:
For the first oblique shock (1 to 2) we need to find β from Eq. 13.49
(13.49) We choose β by iterating or by using Goal Seek to target θ (below) to equal the given θ
Using builtin function theta (M, β,k )
θ= 7.50 o β= 29.6 o Then M 1n can be found from geometry (Eq. 13.47a)
M 1n = 1.233 Then M 2n can be found from Eq. 13.48a)
Using builtin function NormM2fromM (M,k )
(13.48a)
M 2n = 0.822 Then, from M 2n and geometry (Eq. 13.47b)
M2 = 2.19 From M 1n and Eq. 13.48d (using builtin function NormpfromM (M ,k ))
(13.48d) p 2/p 1 = 1.61 p2 = 80.40 Pressure ratio We repeat the analysis of states 1 to 2 for states 2 to 3, to analyze the second oblique shock
We choose β for M 2 by iterating or by using Goal Seek to target θ (below) to equal the given θ
Using builtin function theta (M, β,k )
θ= 7.50 o β= 33.5 o Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a)
M 2n = 1.209 Then M 3n can be found from Eq. 13.48a)
Using builtin function NormM2fromM (M,k )
M 3n = 0.837 Then, from M 3n and geometry (Eq. 13.47b)
M3 = 1.91 From M 2n and Eq. 13.48d (using builtin function NormpfromM (M ,k ))
p 3/p 2 = 1.54 p3 = 124 Pressure ratio Problem 13.151 [3] NOTE: Angle is 30o not 50o! Given: Air flow in a duct
Find: Mach number and pressure at contraction and downstream; Solution:
k=
M1 = 1.4
2.5 β=
p1 = The given or available data is: 30
50 o kPa Equations and Computations:
For the first oblique shock (1 to 2) we find θ from Eq. 13.49
(13.49)
Using builtin function theta (M, β,k )
θ= 7.99 o Also, M 1n can be found from geometry (Eq. 13.47a)
M 1n = 1.250 Then M 2n can be found from Eq. 13.48a)
Using builtin function NormM2fromM (M,k )
(13.48a)
M 2n = 0.813 Then, from M 2n and geometry (Eq. 13.47b)
M2 = 2.17 From M 1n and Eq. 13.48d (using builtin function NormpfromM (M ,k ))
(13.48d) p 2/p 1 = 1.66 p2 = 82.8 Pressure ratio We repeat the analysis for states 1 to 2 for 2 to 3, for the second oblique shock
We choose β for M 2 by iterating or by using Goal Seek to target θ (below) to equal the previous θ
Using builtin function theta (M, β,k )
θ= 7.99 o β= 34.3 o Then M 2n (normal to second shock!) can be found from geometry (Eq. 13.47a)
M 2n = 1.22 Then M 3n can be found from Eq. 13.48a)
Using builtin function NormM2fromM (M,k )
M 3n = 0.829 Then, from M 3n and geometry (Eq. 13.47b)
M3 = 1.87 From M 2n and Eq. 13.48d (using builtin function NormpfromM (M ,k ))
p 3/p 2 = 1.58 p3 = 130 Pressure ratio Problem 13.152 [3] Given: Deflection of air flow
Find: Pressure changes Solution:
R=
k=
p=
M=
θ1 = 286.9
1.4
95
1.5
15 θ2 = The given or available data is: 15 J/kg.K
kPa
o
o Equations and Computations:
We use Eq. 13.55
(13.55) and
Deflection = ωa  ωb = ω(M a)  ω(M b) From M and Eq. 13.55 (using builtin function Omega (M , k ))
ω= 11.9 θ1 = ω1  ω ω1 = θ1 + ω ω1 = 26.9 o For the first deflection:
Applying Eq. 1 From ω1, and Eq. 13.55
(using builtin function Omega (M , k )) o (1) ω1 = 26.9 M1 = 2.02 For o (Use Goal Seek to vary M 1 so that ω1 is correct)
Hence for p 1 we use Eq. 13.7a
(13.7a) The approach is to apply Eq. 13.7a twice, so that
(using builtin function Isenp (M , k ))
p 1 = p (p 0/p )/(p 0/p 1)
p1 = 43.3 kPa For the second deflection:
We repeat the analysis of the first deflection
Applying Eq. 1
θ2 + θ1 = ω2  ω ω2 = θ2 + θ1 + ω ω2 = 41.9 o (Note that instead of working from the initial state to state 2 we could have
worked from state 1 to state 2 because the entire flow is isentropic)
From ω2, and Eq. 13.55
(using builtin function Omega (M , k ))
ω2 = 41.9 M2 = For 2.62 o (Use Goal Seek to vary M 2 so that ω2 is correct)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p (p 0/p )/(p 0/p 2)
p2 = 16.9 kPa Problem 13.153 [3] Given: Deflection of air flow
Find: Mach numbers and pressures Solution
R=
k=
p2 =
M2 = 286.9
1.4
10
4 θ1 = 15 o θ2 = The given or available data is: 15 o J/kg.K
kPa Equations and Computations:
We use Eq. 13.55
(13.55) and
Deflection = ωa  ωb = ω(M a)  ω(M b) From M and Eq. 13.55 (using builtin function Omega (M , k ))
ω2 = 65.8 o For the second deflection:
Applying Eq. 1
ω1 = ω2  θ2 ω1 = 50.8 o From ω1, and Eq. 13.55
(using builtin function Omega (M , k ))
ω1 = 50.8 M1 = For 3.05 o (Use Goal Seek to vary M 1 so that ω1 is correct) (1) Hence for p 1 we use Eq. 13.7a
(13.7a) The approach is to apply Eq. 13.7a twice, so that
(using builtin function Isenp (M , k ))
p 1 = p 2(p 0/p 2)/(p 0/p 1)
p1 = 38.1 kPa For the first deflection:
We repeat the analysis of the second deflection
Applying Eq. 1
θ2 + θ1 = ω2  ω ω = ω2  (θ2 + θ1)
ω= 35.8 o (Note that instead of working from state 2 to the initial state we could have
worked from state 1 to the initial state because the entire flow is isentropic)
From ω, and Eq. 13.55
(using builtin function Omega (M , k ))
For ω=
M= 35.8
2.36 o (Use Goal Seek to vary M so that ω is correct)
Hence for p we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p = p 2(p 0/p 2)/(p 0/p )
p= 110 kPa Problem 13.154 [4] Given: Mach number and deflection angle
Find: Static and stagnation pressures due to: oblique shock; compression wave Solution:
R=
k=
p1 = 286.9
1.4
50 M1 = 3.5 θ= 35 o θ= 35 o The given or available data is: J/kg.K
kPa Equations and Computations:
For the oblique shock:
We need to find M 1n
The deflection angle is
From M 1 and θ, and Eq. 13.49
(using builtin function Theta (M , β,k ))
(13.49) θ= 35.0 o β= For 57.2 o (Use Goal Seek to vary β so that θ = 35o)
From M 1 and β M 1n = 2.94 From M 1n and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) p2 = 496 kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using builtin function NormM2fromM (M ,k )) (13.48a) M 2n = 0.479 The downstream Mach number is then obtained from
from M 2n, θ and β, and Eq. 13.47b
M 2n = M 2sin(β  θ)
M2 = Hence (13.47b) 1.27 For p 02 we use Eq. 12.7a
(using builtin function Isenp (M , k ))
(13.7a) p 02 = p 2/(p 02/p 2)
p 02 = 1316 kPa For the isentropic compression wave:
p 0 = constant For isentropic flow p 02 = p 01 p 01 = 3814 kPa p 02 = 3814 kPa For p 01 we use Eq. 13.7a
(using builtin function Isenp (M , k )) (Note that for the oblique shock, as required by Eq. 13.48b (13.48b) p 02/p 01 =
0.345
(using builtin function Normp0fromM (M ,k ) p 02/p 01 = 0.345
(using p 02 from the shock and p 01) θ= −θ θ= For the deflection 35.0 (Compression )
o We use Eq. 13.55 (13.55)
and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 = 58.5 ω2 = ω1 + θ ω2 = 23.5 o ω2 = 23.5 o M2 = 1.90 Applying Eq. 1 o From ω2, and Eq. 13.55
(using builtin function Omega (M , k ))
For (Use Goal Seek to vary M 2 so that ω2 = 23.5o)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 = 572 kPa (1) Problem 13.155 [3] Given: Wedgeshaped airfoil
Find: Lift per unit span assuming isentropic flow Solution:
The given or available data is: R
k
p
M =
=
=
= δ=
c= 286.9
1.4
70
2.75
7
1.5 J/kg.K
kPa
o m Equations and Computations:
The lift per unit span is
L = (p L  p U)c (1) (Note that p L acts on area c /cos(δ), but its
normal component is multiplied by cos(δ))
For the upper surface:
pU = p pU = 70 θ= −δ θ= 7.0 kPa For the lower surface: o We use Eq. 13.55
(13.55) and
Deflection = ωL  ω = ω(M L)  ω(M ) (2) From M and Eq. 13.55 (using builtin function Omega (M , k ))
ω= 44.7 θ= ωL  ω ωL = θ+ω ωL = 37.7 o ωL = 37.7 o ML = 2.44 Applying Eq. 2 o From ωL, and Eq. 13.55
(using builtin function Omega (M , k ))
For (Use Goal Seek to vary M L so that ωL is correct)
Hence for p L we use Eq. 13.7a
(13.7a) The approach is to apply Eq. 13.7a twice, so that
(using builtin function Isenp (M , k ))
p L = p (p 0/p )/(p 0/p L)
pL =
From Eq 1 113 kPa L= 64.7 kN/m Problem 13.156 [4] Given: Mach number and airfoil geometry
Find: Lift and drag per unit span Solution:
R=
k=
p1 = 286.9
1.4
50 M1 = The given or available data is: 1.75 α=
c= 18
1 J/kg.K
kPa
o m Equations and Computations:
F = (p L  p U)c The net force per unit span is
Hence, the lift force per unit span is L = (p L  p U)c cos(α) (1) D = (p L  p U)c sin(α) (2) The drag force per unit span is For the lower surface (oblique shock):
We need to find M 1n
θ= α θ= The deflection angle is 18 o From M 1 and θ, and Eq. 13.49
(using builtin function Theta (M , β,k ))
(13.49) θ= 18.0 o β= For 62.9 o (Use Goal Seek to vary β so that θ is correct)
From M 1 and β M 1n = 1.56 From M 1n and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) p2 = 133.2 pL = p2 pL = 133.2 kPa kPa For the upper surface (isentropic expansion wave):
p 0 = constant For isentropic flow p 02 = p 01 For p 01 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
(13.7a) p 01 = 266 kPa p 02 = 266 kPa θ= α θ= For the deflection 18.0 (Compression )
o We use Eq. 13.55 (13.55)
and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 =
ω2 = ω1 + θ ω2 = Applying Eq. 3 19.3 37.3 o o From ω2, and Eq. 13.55 (using builtin function Omega (M , k ))
ω2 = 37.3 M2 = For 2.42 o (Use Goal Seek to vary M 2 so that ω2 is correct)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 = 17.6 pU = p2 pU = 17.6 kPa From Eq. 1 L= 110.0 kN/m From Eq. 2 D= 35.7 kN/m kPa (3) Problem 13.157 Given: Mach number and airfoil geometry Find: Plot of lift and drag and lift/drag versus angle of attack Solution:
The given or available data is:
k=
p1 = 1.4
50 M1 = 1.75 α=
c= kPa
o 12
1 m Equations and Computations:
The net force per unit span is
F = (p L  p U)c
Hence, the lift force per unit span is
L = (p L  p U)c cos(α) (1) The drag force per unit span is
D = (p L  p U)c sin(α) (2) For each angle of attack the following needs to be computed:
For the lower surface (oblique shock):
We need to find M 1n
Deflection θ= α From M 1 and θ, and Eq. 13.49
(using builtin function Theta (M , β,k ))
(13.49) β find (Use Goal Seek to vary β so that θ is the correct value)
From M 1 and β find M 1n
From M 1n and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) find p2 and pL = p2 [4] For the upper surface (isentropic expansion wave):
p 0 = constant For isentropic flow p 02 = p 01 For p 01 we use Eq. 13.7a
(using builtin function Isenp (M , k )) (13.7a) find p 02 =
θ= Deflection 266 kPa α we use Eq. 13.55
(13.55) and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) (3) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 = Applying Eq. 3 19.3 ω2 = find ω1 + θ o (4) From ω2, and Eq. 12.55 (using builtin function Omega (M , k ))
From ω2 find M2 (Use Goal Seek to vary M 2 so that ω2 is the correct value)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
pU = p2 Finally, from Eqs. 1 and 2, compute L and D
Computed results:
o
α( ) β (o) θ (o) 0.50
1.00
1.50
2.00
4.00
5.00
10.00
15.00
16.00
16.50
17.00
17.50
18.00 35.3
35.8
36.2
36.7
38.7
39.7
45.5
53.4
55.6
56.8
58.3
60.1
62.9 0.50
1.00
1.50
2.00
4.00
5.00
10.0
15.0
16.0
16.5
17.0
17.5
18.0 Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0% Sum: 0.0% M 1n p L (kPa) ω2 ( ) ω2 from M 2 ( ) 1.01
1.02
1.03
1.05
1.09
1.12
1.25
1.41
1.44
1.47
1.49
1.52
1.56 51.3
52.7
54.0
55.4
61.4
64.5
82.6
106.9
113.3
116.9
121.0
125.9
133.4 19.8
20.3
20.8
21.3
23.3
24.3
29.3
34.3
35.3
35.8
36.3
36.8
37.3 19.8
20.3
20.8
21.3
23.3
24.3
29.3
34.3
35.3
35.8
36.3
36.8
37.3 o o Error
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0%
0.0% Sum: 0.0% M2 p U (kPa) L (kN/m) D (kN/m) 1.77
1.78
1.80
1.82
1.89
1.92
2.11
2.30
2.34
2.36
2.38
2.40
2.42 48.7
47.4
46.2
45.0
40.4
38.3
28.8
21.3
20.0
19.4
18.8
18.2
17.6 2.61
5.21
7.82
10.4
20.9
26.1
53.0
82.7
89.6
93.5
97.7
102.7
110 0.0227
0.091
0.205
0.364
1.46
2.29
9.35
22.1
25.7
27.7
29.9
32.4
35.8 L/D
115
57.3
38.2
28.6
14.3
11.4
5.67
3.73
3.49
3.38
3.27
3.17
3.08 To compute this table:
1) Type the range of α
2) Type in guess values for β
3) Compute θ from Eq. 13.49
(using builtin function Theta (M ,β, k )
4) Compute the absolute error between each θ and α
5) Compute the sum of the errors
6) Use Solver to minimize the sum by varying the β values
(Note: You may need to interactively type in new β values
if Solver generates β values that lead to no θ)
7) For each α, M 1n is obtained from M 1, and Eq. 13.47a
8) For each α, p L is obtained from p 1, M 1n, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
9) For each α, compute ω2 from Eq. 4
10) For each α, compute ω2 from M 2, and Eq. 13.55
(using builtin function Omega (M ,k ))
11) Compute the absolute error between the two values of ω2
12) Compute the sum of the errors
13) Use Solver to minimize the sum by varying the M 2 values
(Note: You may need to interactively type in new M 2 values)
if Solver generates β values that lead to no θ)
14) For each α, p U is obtained from p 02, M 2, and Eq. 13.47a
(using builtin function Isenp (M , k ))
15) Compute L and D from Eqs. 1 and 2 Lift and Drag of an Airfoil
as a Function of Angle of Attack L and D (kN/m) 120
100
80
Lift
Drag 60
40
20
0
0 2 4 6 8 10 12 14 16 18 20 α (o) Lift/Drag of an Airfoil
as a Function of Angle of Attack
140
120 L/D 100
80
60
40
20
0
0 2 4 6 8 10
α (o) 12 14 16 18 20 Problem 13.158 [4] Given: Mach number and airfoil geometry
Find: Drag coefficient Solution:
R=
k=
p1 =
M1 = The given or available data is: 286.9
1.4
95
2 α=
δ= 0
10 J/kg.K
kPa
o
o Equations and Computations:
The drag force is
D = (p F  p R)cs tan(δ/2) (1) (s and c are the span and chord)
This is obtained from the following analysis
Airfoil thickness (frontal area) = 2s (c /2tan(δ/2))
Pressure difference acting on frontal area = (p F  p R)
(p F and p R are the pressures on the front and rear surfaces)
2
C D = D /(1/2ρV A ) The drag coefficient is (2) But it can easily be shown that
ρV 2 = pkM 2
Hence, from Eqs. 1 and 2
C D = (p F  p R)tan(δ/2)/(1/2pkM 2) (3) For the frontal surfaces (oblique shocks):
We need to find M 1n From M 1 and θ, and Eq. 13.49
(using builtin function Theta (M , β,k )) θ= δ/2 θ= The deflection angle is 5 o (13.49) θ=
β= 5.0
34.3 M 1n = 1.13 For o
o (Use Goal Seek to vary β so that θ = 5o)
From M 1 and β From M 1n and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) p2 = 125.0 pF = p2 pF = 125.0 kPa kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using builtin function NormM2fromM (M ,k )) (13.48a) M 2n = 0.891 The downstream Mach number is then obtained from
from M 2n, θ and β, and Eq. 13.47b
M 2n = M 2sin(β  θ)
M2 = Hence (13.47b) 1.82 For p 02 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
(13.7a) p 02 = 742 kPa For the rear surfaces (isentropic expansion waves):
Treating as a new problem
Here: M 1 is the Mach number after the shock
and M 2 is the Mach number after the expansion wave
p 01 is the stagnation pressure after the shock
and p 02 is the stagnation pressure after the expansion wave M 1 = M 2 (shock)
M1 = 1.82 p 01 = p 02 (shock)
p 01 = 742 kPa p 0 = constant For isentropic flow p 02 = p 01 p 02 = 742 θ= δ θ= For the deflection 10.0 kPa o We use Eq. 13.55 (13.55)
and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 =
ω2 = ω1 + θ ω2 = Applying Eq. 3 21.3 31.3 o o From ω2, and Eq. 13.55 (using builtin function Omega(M, k))
For ω2 =
M2 = 31.3
2.18 o (Use Goal Seek to vary M 2 so that ω2 = 31.3o)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 =
pR = p2 pR =
Finally, from Eq. 1 71.2 71.2 CD = 0.0177 kPa kPa (3) Problem 13.159 [4] FU 1
FL RU
RL Given: Mach number and airfoil geometry
Find: Lift and Drag coefficients Solution:
R=
k=
p1 =
M1 = The given or available data is: 286.9
1.4
95
2 α=
δ= 12
10 J/kg.K
kPa
o
o Equations and Computations:
Following the analysis of Example 13.14
the force component perpendicular to the major axis, per area, is
F V/sc = 1/2{(p FL + p RL)  (p FU + p RU)} (1) and the force component parallel to the major axis, per area, is
F H/sc = 1/2tan(δ/2){(p FU + p FL)  (p RU + p RL)} (2) using the notation of the figure above.
(s and c are the span and chord)
The lift force per area is
F L/sc = (F Vcos(α)  F Hsin(α))/sc (3) The drag force per area is
F D/sc = (F Vsin(α) + F Hcos(α))/sc C L = F L/(1/2ρV 2A ) The lift coefficient is (4) (5) But it can be shown that
ρV 2 = pkM 2 (6) Hence, combining Eqs. 3, 4, 5 and 6
2
C L = (F V/sc cos(α)  F H/sc sin(α))/(1/2pkM ) (7) Similarly, for the drag coefficient
C D = (F V/sc sin(α) + F H/sc cos(α))/(1/2pkM 2) (8) For surface FL (oblique shock):
We need to find M 1n
θ= α + δ/2 θ= The deflection angle is 17 o From M 1 and θ, and Eq. 13.49
(using builtin function Theta (M , β,k ))
(13.49) For θ=
β= 17.0
48.2 o
o (Use Goal Seek to vary β so that θ = 17o)
From M 1 and β M 1n = 1.49 From M 1n and p 1, and Eq. 13.48d
(using builtin function NormpfromM (M ,k ))
(13.48d) p2 = 230.6 p FL = p2 p FL = 230.6 kPa kPa To find M 2 we need M 2n. From M 1n, and Eq. 13.48a
(using builtin function NormM2fromM (M ,k )) (13.48a) M 2n = 0.704 The downstream Mach number is then obtained from
from M 2n, θ and β, and Eq. 13.47b
M 2n = M 2sin(β  θ)
Hence M2 = (13.47b) 1.36 For p 02 we use Eq. 13.7a
(using builtin function Isenp (M , k )) (13.7a) p 02 = 693 kPa For surface RL (isentropic expansion wave):
Treating as a new problem
Here: M 1 is the Mach number after the shock
and M 2 is the Mach number after the expansion wave
p 01 is the stagnation pressure after the shock
and p 02 is the stagnation pressure after the expansion wave
M 1 = M 2 (shock)
M1 = 1.36 p 01 = p 02 (shock)
p 01 = 693 kPa p 0 = constant For isentropic flow p 02 = p 01 p 02 = 693 θ= δ θ= For the deflection 10.0 kPa o We use Eq. 13.55 (13.55)
and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 =
ω2 = ω1 + θ ω2 = Applying Eq. 3 7.8 17.8 o o From ω2, and Eq. 13.55 (using builtin function Omega (M , k ))
For ω2 =
M2 = 17.8
1.70 o (Use Goal Seek to vary M 2 so that ω2 = 17.8o)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 = 141 p RL = p2 p RL = 141 kPa kPa (3) For surface FU (isentropic expansion wave):
M1 = 2.0 p 0 = constant For isentropic flow p 02 = p 01 p 01 =
p 02 = 743
743 For p 01 we use Eq. 13.7a
(using builtin function Isenp (M , k )) θ= α  δ/2 θ= For the deflection 7.0 kPa o We use Eq. 13.55
and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 =
ω2 = ω1 + θ ω2 = Applying Eq. 3 26.4 33.4 o o From ω2, and Eq. 13.55 (using builtin function Omega(M, k))
For ω2 =
M2 = 33.4
2.27 o (Use Goal Seek to vary M 2 so that ω2 = 33.4o)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 = 62.8 p FU = p2 p FU = 62.8 kPa kPa For surface RU (isentropic expansion wave):
Treat as a new problem.
Flow is isentropic so we could analyse from region FU to RU
but instead analyse from region 1 to region RU.
M1 =
For isentropic flow 2.0 p 0 = constant
p 02 = p 01 (3) p 01 =
p 02 = 743
743 θ= α + δ/2 θ= TOTAL deflection 17.0 kPa
kPa o We use Eq. 13.55
and
Deflection = ω2  ω1 = ω(M 2)  ω(M 1) From M 1 and Eq. 13.55 (using builtin function Omega (M , k ))
ω1 =
ω2 = ω1 + θ ω2 = Applying Eq. 3 26.4 43.4 o o From ω2, and Eq. 13.55 (using builtin function Omega(M, k))
ω2 =
M2 = For 43.4
2.69 o (Use Goal Seek to vary M 2 so that ω2 = 43.4o)
Hence for p 2 we use Eq. 13.7a
(using builtin function Isenp (M , k ))
p 2 = p 02/(p 02/p 2)
p2 = 32.4 p RU = p2 p RU = 32.4 kPa p FL =
p RL =
p FU =
p RU = 230.6
140.5
62.8
32.4 kPa
kPa
kPa
kPa kPa The four pressures are: From Eq 1 F V/sc = 138 kPa From Eq 2 F H/sc = 5.3 kPa From Eq 7 CL = 0.503 From Eq 8 CD = 0.127 (3) ...
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This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .
 Spring '11
 TUZLA

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