chapte r6 mekflu
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Chapter 4 / The Integral Forms of the Fundamental Laws
37© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
CHAPTER 4
The Integral Forms of the
Fundamental LawsFE-type Exam Review Problems: Problems 4-1 to 4-15
4.1 (B)
4.2 (D)2200 0.04 70 0.837 kg/s
0.287 293
pm AV AV
RT ρ π = = = × × =
× .
4.3 (A)
Refer to the circle of Problem 4.27:
2 375.7 2( 0.4 0.10 0.40 sin 75.5 ) 3 0.516 m /s.360
Q AV π ×
= = × × − × × × =
4.4 (D)
2 22 1
2
PW V V
Q gγ
−=
2 1 1200 200. .
0.040
P p p W
γ γ γ
− −+ =
×
4040 kW and energy req'd = 47.1 kW.
0.85PW ∴ = =
4.5 (A)2
20 V =
2
1 2 1
2V p p
g− −+
22
2120. 0 . 7 200 000 Pa.
2 9.8 9810 p p
γ −= + ∴ =
×
4.6 (C)
Manometer:2
21 2
2
V H p g p
gγ ρ + = + or
22
19810 0.02 .2
V p g
g ρ × + =
Energy:2 100 0007.96
. 3.15.2 9.81 9810
K K = ∴ =×
Combine the equations:2
119810 0.02 1.2 . 18.1 m/s.
2
V V × = × ∴ =
4.7 (B)
2
2
0.040. 7.96 m/s.
2 0.04 L
V p Qh K V
g Aγ π
Δ= = = = =
×
2 100 0007.96. 3.15.
2 9.81 9810K K = ∴ =
×
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Chapter 4 / The Integral Forms of the Fundamental Laws
38© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4.8 (C)
2 22 1
2
PW V V
Q gγ
−=
. p
γ
Δ+
160.040 400 16 kW. 18.0 kW.
0.89
PP
W W Q p
η = Δ = × = = =
4.9 (D)
2 24.58 7.1636.0 15 3.2 . 416 000 Pa
2 9.81 9810 2 9.81
B B
p p+ = + + ∴ =
× ×
In the above energy equation we used
2
2
0.2 with 4.42 m/s.
2 0.2 L
V Qh K V
g A π = = = =
×
4.10 (A)
V Q
A= =
×
=0 1
0419 89
2
.
..
π m / s.
Energy —surface to entrance: H V
g
p z K
V
gP = + + +
2
2
2
22
2
2 2γ .
∴ =×
+ + +×
= H P19 89
2 9 81
18050 5 6
19 89
2 9 812014
2 2.
..
.
..
000
9810 m.
∴ = = × × = / . . / . .W QH P P Pγ η 9810 0 1 201 4 0 75 263 000 W
4.11 (A)After the pressure is found, that pressure is multiplied by the area of the
window. The pressure is relatively constant over the area.
4.12 (C)
2 21 1 2 2
2 2
V p V p
g gγ γ + = +
2 2
1
(6.25 1) 12.73. 9810 3 085 000 Pa.
2 9.81 p
− ×= × =
×
21 1 2 1( ). 3 085 000 0.05 1000 0.1 12.73(6.25 1) p A F Q V V F ρ π − = − × × − = × × −
17 500 N.F ∴ =
4.13 (D) 2 1( ) 1000 0.01 0.2 50(50cos60 50) 2500 N. x x xF m V V − = − = × × × − = −
4.14 (A)
22 1( ) 1000 0.02 60 (40cos 45 40) 884 N. x r r x xF m V V π − = − = × × × × − =
Power 884 20 17 700 W. x BF V = × = × =
4.15 (A)Let the vehicle move to the right. The scoop then diverts the water to the right.
Then, 2 1( ) 1000 0.05 2 60 [60 ( 60)] 720 000 N. x xF m V V = − = × × × × − − =
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Chapter 4 / The Integral Forms of the Fundamental Laws
39© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Basic Laws
4.16 b) The energy transferred to or from the system must be zero: Q − W = 0.
4.20 b) The conservation of mass. d) The energy equation.
System-to-Control-Volume Transformation
4.24
1
1 1ˆ ˆ ˆ ˆˆ 0.707( )2 2
= − − − +n i j = i j . 2ˆ ˆˆ 0.866 0.5= −n i j . 3
ˆˆ = −n j .
1 1 1ˆ ˆ ˆˆ 10 0.707( ) 7.07 fpsnV ⎡ ⎤= ⋅ = ⋅ − + = −⎣ ⎦V n i i j
2 2 2ˆ ˆ ˆˆ 10 (0.866 0.5 ) 8.66 fpsnV = ⋅ = ⋅ − =V n i i j .
3 3 2ˆ ˆ
ˆ 10 ( ) 0nV = ⋅ = ⋅ − =V n i j
4.26
ˆ ˆ ˆˆ( ) 15(0.5 0.866 ) (10 12) A⋅ = + ⋅ ×B n i j j = × × =15 0 866 120 1559. cm 3
Volume =315 sin 60 10 12 1559 cm× × =
Conservation of Mass
4.32
Use Eq. 4.4.2 with mV representing the mass in the volume:
. .
ˆ0 V
c s
dm dAdt
ρ = + ⋅∫ n V 2 2 1 1V dm V A V
dt ρ ρ = + −
.V dm
Q mdt
ρ = + −
Finally, .V dm
m Qdt
ρ = −
4.34
A1V 1 = A2V 2. π × 1 25
144
2.× 60 = π ×
2 5
144
2. V 2. ∴V 2 = 15 ft/sec.
. .m AV = = × ρ π 1 94 1 25144
602
= 3.968 slug/sec.
Q = AV = π 1 25
144
2.× 60 = 32.045 ft /sec.
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Chapter 4 / The Integral Forms of the Fundamental Laws
40© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4.38
11 1 1 2 2 2 1 23 3
500 kg 1246 kg. 4.433 . 8.317
0.287 393 0.287 522m m
p A V A V
RT ρ ρ ρ ρ = = = = = =
× ×
4.433 π × 0.052 × 600 = 8.317 π × 0.052 V 2. ∴V 2 = 319.8 m/s.
m A V = ρ 1 1 1 = 20.89 kg/s. Q A V 1 1 1
= = 34.712 m /s. Q2 =32.512 m /s.
4.40
b) (2 × 1.5 + 1.5 × 1.5) 3 = π d 2
2
4
2
2× . ∴d 2 = 4.478 m
4.42
a) Since the area is rectangular, V = 5 m/s.
1000 0.08 0.8 5m AV ρ = = × × × =320 kg/s. Q =m
ρ = 30.32 m /s.
c) V × 0.08 = 10 × 0.04 + 5 × 0.02 + 5 × 0.02. ∴V = 7.5 m/s.
1000 0.08 0.8 7.5m AV ρ = = × × × = 480 kg/s.
Q m=
ρ =
30.48 m /s.
4.44
If dm dt / ,= 0 then ρ ρ ρ 1 1 1 2 2 2 3 3 3 A V A V A V = + . In terms of m Q2 3and this
becomes, letting ρ ρ ρ 1 2 3= = ,
1000 0 02 12 1000 0 012 2× × × = + ×π . . .m 2 5.08 kg/s.m∴ =
4.46
0.1
2
0
. 0.2 2 10 10(20 100 )2 0.1 2 10 .in outm m m y y dy m ρ ρ ρ ⎡ ⎤
= + × × × = − + × × × +⎢ ⎥⎢ ⎥⎣ ⎦∫
Note: We see that at y = 0.1 m the velocity u(0.1) = 10 m/s. Thus, we integrate to y =0.1, and between y = 0.1 and 0.2 the velocity u = 10:
44
32 ρ ρ ρ = +
⎡
⎣⎢⎤
⎦⎥ + .m ∴ = .m 0 6667 ρ = 0.82 kg/s.
4.48
( ) . . ( )/
m VdA y y y dy= = − − ×∫∫ ρ 2 2 1 3545 6 9 2 50
1 3
2
( )= − − +∫22 6 2 127 9 3 192 2 3
0
1 3
y y y y dy. ./
= 4.528 slug/sec.
V u= = × =2
3
2
32
4
3max .fps (See Prob. 4.43b).
θ R
cosθ = 1/2 θ = 60o
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Chapter 4 / The Integral Forms of the Fundamental Laws
41© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
ρ = +2 2 1 94
2
. . = 2.07 slug/ft3. ∴ = × × ×
⎛ ⎝ ⎜
⎞ ⎠⎟ ρ V A 2 07
4
35
1
3. = 4.6 slug/sec.
Thus, ρ V A m≠ since ρ = ρ ( y) and V = V ( y) so that V V ρ ρ ≠ .
4.50
3 33 23
m of H O4 m of air 2000 0.0015 9000 53 sm of air
π × × × × = 1.5 × (1.5h).
∴h = 0.565 m.
4.52
.m m min = +2 3 V 1 = 20 m/s (see Prob. 4.43c).
2 2320 1000 0.02 10 1000 0.02 .V π π × × = + × × ∴V 3 = 12.04 m/s.
4.54
The control surface is close to theinterface at the instant shown.
∴V i = interface velocity.
ρ ρ e e e i i i A V A V = .
2 280001.5 0.15 300 1 .0.287 673
iV π π × × × = × ××
∴V i = 0.244 m/s.
4.56
For an incompressible flow (low speed air flow)
udA A V A
=
∫2 2
1
. 20 0 8 0 151 5 2 20
0 2
y dy V /.
. . .× = ×
∫ π
20 0 85
60 2 0 156 5 2 2× = ×. . . .
/ π V 2 27.3 m/sV ∴ =
4.58
Draw a control volume around the entire set-up:
0 2 2 1 1= + −dm
dt V A V Atissue ρ ρ
= + −⎛
⎝ ⎜
⎞
⎠⎟ − ( tan ) m
d d h h htissue ρπ ρπ φ
22 2
2 1
2
14
or
tan .m d d
h h htissue = −
+⎡
⎣⎢
⎤
⎦⎥ ρπ φ
222
2 12
12
4
V e n̂
n̂V
i
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Chapter 4 / The Integral Forms of the Fundamental Laws
42© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4.60
0 2 2 1 1= + −dm
dt A V A V ρ ρ
= + × × − × − ( . . / ).m 1000 0 003 0 02 10 10 602 6π
4
3.99 10 kg/sm −
= ×
4.62
0 3 3 1 1 1 2= + − −dm
dt Q A V m ρ ρ where m Ah= ρ .
a) 0 1000 0 6 1000 0 6 60 1000 0 02 10 102 2= × + × − × × −π π . . / . .h
0.0111 m/s or 11.1 mm/sh∴ =
4.64
Choose the control volume to consist of the air volume inside the tank. Theconservation of mass equation is
0CV CS
d dV dAdt
ρ ρ = + ⋅∫ ∫ V n
Since the volume of the tank is constant, and for no flow into the tank, theequation is
0e e e
d V V A
dt
ρ ρ = +
Assuming air behaves as an ideal gas, p
RT ρ = . At the instant of interest,
1d dp
dt RT dt
ρ
= . Substituting in the conservation of mass equation, we get
( )( ) ( )23
3
1.8 kg/m 200 m/s 0.015 m kJ( ) 0.287 298 K
1.5 m kg K
e e eV Adp RT dt V
π ρ ⎡ ⎤= − = − ×⎢ ⎥⋅⎣ ⎦
14.5 kPa/sdp
dt∴ = −
Energy Equation
4.66
beltω μ = + + duW T pAV Ady
520 500 2 /60 400 0.4 0.5 10 1.81 10 100 0.5 0.8π −= × × + × × × + × × × ×
= + + =1047 800 0 000724 1847. W
4.68 80% of the power is used to increase the pressure while 20% increases the
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Chapter 4 / The Integral Forms of the Fundamental Laws
43© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
internal energy ( Q = 0 because of the insulation). Hence,
0.2Δ = m u W
1000 0 02 4 18 0 2 500× × = ×. . . .ΔT 0.836T C∴Δ = °
4.70 − = − ×
. .
W
mgT 40 0 89 b) 40 0.89 (90 000/60) 9.81 523 900 W= × × × = T W
4.72
V
g
p z
V
g
p z1
2
1
12
2
2
22 2
+ + = + +γ γ
.
12
2 32 26
36
64 4
2 2
2
2 2× + = +. .
.h
h
8 23620 1
22 2
..
.= +h
h
Continuity: 3 × 12 = h2 V 2. This can be solved by trial-and-error:
2 8' : 8.24 ? 8.31h = 2 7.9 ' : 8.24 ? 8.22h = ∵ h2 = 7 93. .'
2 1.8' : 8.24 ? 8.00h = 2 1.75' : 8.24 ? 8.31h = ∵ h2 = 1 76. '.
4.74
Manometer: Position the datum at the top of the right mercury level.
9810 4 98102
1000 9810 13 6 4 9810 22 22
2
1× + + + × = × × + × +. ( . ) . z p V
p
Divide by γ = 9810: . . . .42
13 6 4 222 2
2
1+ + + = × + + z p V
g
p
γ γ (1)
Energy:V
g
p z
V
g
p z1
2
1
12
2
2
22 2
+ + = + +γ γ
. (2)
Subtract (1) from (2): With z1 = 2 m,V
g
1
2
212 6 4= ×. . . ∴V 1 = 9.94 m/s
4.76
Q = 120 × 0.002228 = π × 1
12
2
1
⎛
⎝ ⎜
⎞
⎠⎟ V . ∴V 1 = 12.25 fps.
Continuity: π π ×⎛ ⎝ ⎜
⎞ ⎠⎟ = ×
⎛ ⎝ ⎜
⎞ ⎠⎟
1
12
1
12
2
1
2
2V V .5
. ∴V 2 = 5.44 fps.
Energy:V
g
p V
g
p V
g
1
2
1 2
2
2 1
2
2 20 37
2+ = + +
γ γ . .
3 ft
V 1
h2
V 2
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Chapter 4 / The Integral Forms of the Fundamental Laws
44© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
∴ = × + −⎡
⎣⎢
⎤
⎦⎥ p2
2 2
60 144 62 4 0 6312 25
64 4
5 44
64 4. .
.
.
.
. = 8702.9 psf or 60.44 psi
4.78
V Q A1 1 20 08
03= =
×/
.
.π = 28.29 m/s. ∴V 2 = 9V 1 = 254.6 m/s.
Energy:V
g
p V
g
p V
g
1
2
1 2
2
2 1
2
2 22
2+ = + +
γ γ . .
∴ =×
−×
⎡
⎣⎢
⎤
⎦⎥ p1
2 2
9810254 6
2 9 810 8
28 29
2 9 81
.
..
.
. = 32 1 106. .× Pa
4.80
a) Energy:V
g
p z
V
g
p z0
2
0
02
2
2
22 2
+ + = + +γ γ
. ∴ = = × ×V gz2 02 2 9 81 2 4. . =
6.862 m/s.
Q = AV = 0.8 × 1 × 6.862 = 35.49 m /s.
For the second geometry the pressure on the surface is zero but it increases withdepth. The elevation of the surface is 0.8 m:
∴ = + z V
gh0
2
2
2. ∴ = − = × ×V g z h2 02 2 9 81 2( ) . = 6.264 m/s.
∴Q = 0.8 × 6.264 = 35.01 m /s.
Note: z0 is measured from the channel bottom in the 2nd geometry.
∴ z0 = H + h.
4.82
V
g
p z
V
g
p z0
2
0
02
2
2
22 2
+ + = + +γ γ
.80 000
9810+ =
×4
2 9 81
2
2V
.. ∴V 2 = 19.04 m/s.
b) 22 2 0.09 19.04Q A V π = = × × =30.485 m /s.
4.84
Manometer: 1 213.6 .H z p H z pγ γ γ γ + + = + + ∴ = + p
H p1 212 6
γ γ . .
Energy:
p V
g
p V
g
1 1
2
2 2
2
2 2γ γ + = +
.
Combine energy and manometer: 12 62
2
2
1
2
. . H V V
g=
−
Continuity: V d
d V 2
1
2
2
2 1= .
42 1
1 42
12.6 2 1 .d
V H gd
⎛ ⎞∴ = × −⎜ ⎟⎜ ⎟
⎝ ⎠
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Chapter 4 / The Integral Forms of the Fundamental Laws
45© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
1/2 1/22
2 2 211 1 1 24 4 4 4
1 2 1 2
12.6 212.35
4 4 / 1
H gd H Q V d d d
d d d d
π π
⎛ ⎞ ⎛ ⎞×∴ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
4.86
a) Energy from surface to outlet:
V
g H 2
2
2 = . ∴ =V gH 22
2 .
Energy from constriction to outlet: p V
g
p V
g
1 1
2
2 2
2
2 2γ γ + = + .
Continuity: V V 1 24= . With p1 = pv = 2450 Pa and p2 = 100 000 Pa,
2450
9810
16
2 9 812
100 000
9810
1
2 9 812+
× × = +
× ×
. ..gH gH ∴ H = 0.663 m.
4.88
Energy: surface to surface: z z h L0 2= +
. ∴ = +
30 20 2 2
2
2V
g .
Continuity: V 1 = 4V 2. ∴V 12 = 160 g. ∴V 2
2 = 10 g.
Energy: surface to constriction: 30160
2
94 000
98101= +
−+
g
g z
( )
∴ z1 = −40.4 m. ∴ H = 40.4 + 20 = 60.4 m.
4.90
Velocity at exit = V e . Velocity in constriction = V 1. Velocity in pipe = V 2 .
Energy: surface to exit:
V
g H e
2
2 = . ∴ =V gH e
2
2 .
Continuity across nozzle: V D
d V e2
2
2= . Also, V V 1 24= .
Energy: surface to constriction: H V
g
pv= +12
2 γ .
a) 51
216
22 5
974
4= × × ×
⎛
⎝ ⎜
⎞
⎠⎟ +
−
g
Dg
..
550
9810 ∴ = D 0 131. m
4.92
. .m AV = = × ×⎛ ⎝ ⎜
⎞ ⎠⎟ × = ρ π 1 94
1
12120 5 079
2
slug / sec.
2 230 120 120 144 ft-lb5.079 32.2 / 0.85 12,950 or 23.5 hp.
2 32.2 62.4 secPW
⎡ ⎤− ×= × + =⎢ ⎥×⎣ ⎦
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Chapter 4 / The Integral Forms of the Fundamental Laws
46© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4.94
− = × × −
× +
−⎡
⎣⎢
⎤
⎦⎥ ×
..
.. .W T 2 1000 9 81
0 10 2
2 9 81
6000 87
2 000
9810
∴ = × . .W T 1 304 106 W
We used 2 22
2 10.2 m/s0.25
QV A π
= = =×
∴ =η T 0924.
4.96
a)2 2
2 12 12 1 2 1
2 1
( )2 γ γ
⎡ ⎤−− = + − + − + −⎢ ⎥
⎢ ⎥⎣ ⎦
v
S p p cV V
Q W mg z z T T g g
The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13.
311 2
1 2 2
85 9.81 600 9.81 20 5009.92 N/m . .
0.287 293 0.287
p g
RT T T γ γ
× ×= = = = =
×
2
2 2
600 000200 85 000 716.5( 1 500 000)=5 9.81 ( 293)2 9.81 20 500 9.92 9.81
T T
⎡ ⎤∴− − × + − + −⎢ ⎥×⎣ ⎦
∴ =T 2 572 K or 299 C .
Be careful of units. 2 600 000 Pa, 716.5 J/kg K v p c= = ⋅
4.98
Energy: surface to exit: − = − +⎡
⎣⎢
⎤
⎦⎥ . .W mg
V
g
V
gT T η
22
22
220 4 5
2
V mg Q2 215
6
13 26 15 9810 147=
×
= = = × =π
γ
.
. m / s. 150 N / s.
− × =×
− +×
⎡
⎣⎢
⎤
⎦⎥ ∴ = .
..
.
.. .W W T T 08 147 2 9 81
20 4 51326
2 9 81
2
15013.26
5390 kW
2
4.100
Choose a control volume that consists of the entire system and applyconservation of energy:
2 2
1 1 2 21 2
2 2P T L
p V p V H z H z h
g gγ γ + + + = + + + +
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Chapter 4 / The Integral Forms of the Fundamental Laws
47© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
We recognize that H T = 0, V 1 is negligible and h L = 210 V 2
/2g where V = Q/ A
and 2 2/ .V Q A= Rearranging we get:
2
2 1 22 1
2P L
p p V H z z h
gγ
−= + + − +
( )
( )
6 32
23
6.3 10 m /s (0.321)0.321 m/s 210 1.1 m
2 9.812.5 10 m
LQ
V h A
π
−
−
×= = = ⇒ = =
××
( )
6 3
2 242
6.3 10 m /s12.53 m/s
4 10 m
QV
Aπ
−
−
×= = =
×
Substituting the given values we get:
( ) ( )( )
22
3 2
95 100 kN/m 12.53 m/s0.5m 1.1 m 8.85 m6.660 kN/m 2 9.81 m/s
P H −
= + + + =
The power input to the pump is:
( ) ( )( )3 6 3/ 6660N/m 6.3 10 m /s 8.85m / 0.75 0.5 WP P PW QH γ η −= = × =
4.102
Energy: across the nozzle:2 2 2
1 1 2 22 1 12
5. 6.25 .
2 2 2
p V p V V V V
g gγ γ + = + = =
2 2 21 1
1
6.25400 000
. 4.58 m/s9810 2 9.81 2 9.81
V V
V ∴ + = ∴ =× × , 7.16 m/s AV = ,
2 28.6 m/s.V =
Energy: surface to exit:
2 2 228.6 4.58 7.1615 1.5 3.2 .
2 9.81 2 9.81 2 9.81P H + = + +
× × × 36.8 m.PH ∴ =
Section (1)
Pump
Carburetor
0.5 m
Section (2)
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Chapter 4 / The Integral Forms of the Fundamental Laws
48© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
29810 ( 0.01 ) 28.6 36.83820 W.
0.85P
PP
QH W
γ π
η
× × × ×∴ = = =
Energy: surface to “ A”:
2 27.16 7.1615 3.2 . 39 400 Pa2 9.81 9810 2 9.81
A A
p p= + + ∴ =× ×
Energy: surface to “ B”:
2 24.58 7.1636.0 15 3.2 . 416 000 Pa
2 9.81 9810 2 9.81
B B
p p+ = + + ∴ =
× ×
4.104
Depth on raised section = y 2 . Continuity: 3 3 2 2× = V y .
Energy (see Eq. 4.5.21):3
23
20 4
22
2
2g
V
g y+ = + +( . ).
∴ = + − + =3 0599
23 059 4 128 0
2
2
2 2 2
3
2
2. , . .g y
y y y
or
Trial-and-error:2
22
2.0: 0.11 ? 0 1.85 m.
1.8 : 0.05 ? 0
y y
y
= − ⎫∴ =⎬
= + ⎭
22
2
2.1: 0.1 ? 0 2.22 m.
2.3: 0.1 ? 0
y y
y
= − ⎫∴ =⎬
= + ⎭
The depth that actually occurs depends on the downstream conditions. We
cannot select a “correct” answer between the two.
4.106
The average velocity at section 2 is also 8 m/s. The kinetic-energy-correctionfactor for a parabola is 2 (see Example 4.9). The energy equation is:
V
g
p V
g
ph L
1
2
1
22
2
2
2 2+ = + +
γ α
γ .
8
2 9 81
1502
8
2 9 81
1102 2
× + =
× + +
. ..
000
9810
000
9810h L
∴ =h L 0 815. .m
4.108
a)0.01 2 2 4
2 2 2 20
1 1 20 0.01 0.0110 1 2 5 m/s
20.01 0.01 0.01 4 0.01
rV VdA rdr
Aπ
π
⎛ ⎞ ⎡ ⎤= = − = − =⎜ ⎟ ⎢ ⎥⎜ ⎟× ×⎢ ⎥⎝ ⎠ ⎣ ⎦
∫ ∫
30.01 23 3
3 2 3 20
1 110 1 2
0.01 5 0.01
rV dA rdr
AV α π
π
⎛ ⎞= = −⎜ ⎟⎜ ⎟× × ⎝ ⎠
∫ ∫
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Chapter 4 / The Integral Forms of the Fundamental Laws
49© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
=×
− ×
× +
×
× −
×
⎛
⎝ ⎜
⎞
⎠⎟ =
2000
0 01 5
0 01
2
3 0 01
4 0 01
3 0 01
6 0 01
0 01
8 0 012 00
2 3
2 4
2
6
4
8
6.
. .
.
.
.
.
..
4.110
Engine power =2 22 1
2 1
2D
V V F V m u u∞
⎛ ⎞−× + + −⎜ ⎟⎜ ⎟
⎝ ⎠
2 22 1
2 1( )2
f f D vV V
m q F V m c T T ∞⎡ ⎤−
= + + −⎢ ⎥⎢ ⎥⎣ ⎦
4.112
2 22 12 1
2 2 1 20 32
2 2
p pV V LV z z
g g gD
ν α
γ γ = + + − − − +
2 6
2
10 1800 2 0.35 32 .
2 9.81 9.81 0.02
V V − ×= − + ×
× ×
2 5 314.4 3.434 0. 0.235 m/s and 7.37 10 m /sV V V Q −+ − = ∴ = = ×
4.114
Energy from surface to surface:
H V
g
pz
V
g
pz K
V
gP = + + − − − +
22
22
12
11
2
2 2 2γ γ .
a) H Q
QP = +× × ×
= +40 50 04 2 9 81
40 50 72
2
2
π . ..
Try Q H H P P= = =0 25 43 2 58. : . (energy). (curve)
Try Q H H P P= = =0 30 44 6 48. : . (energy). (curve)
Solution: 30.32 m /sQ =
Momentum Equation
4.116
V
g
p V
g
p12
1 2
2
2
2 2+ = +
γ γ .
2
2 1 124 .
( / 2)
d V V V
d = =
b)2 2
1 116400 000 .2 9.81 9810 2 9.81
V V + =
× × 1 7.303 m/s.V ∴ =
400 000 03 1000 03 7 303 4 7 303 7 3032 2 π π × − = × × × −. . . ( . . ).F ∴ =F 679 N .
d)2 2
1 11
1630 144. 17.24 fps.
2 32.2 62.4 2 32.2
V V V
×+ = ∴ =
× ×
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Chapter 4 / The Integral Forms of the Fundamental Laws
50© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2 2 230 1.5 1.94 (1.5 /12) 17.24 (4 1). 127 lb.F F π π × × − = × × × − ∴ =
4.118
2 21 1 2 2 .
2 2
V p V p
g gγ γ + = +
0
20.01 0.006 0.15. 11.1 m/s.e eV V V π × = × × ∴ =
2 1( )x x xF m V V Σ = −
a) V V V 2
2
2 1 1
10
81= = .562 .
V V 12
1
2
2 9 81
400 2 441
2 9 81× + =
×.
.
..
000
9810
1 23.56 m/s.V ∴ =
∴ − = − p A F m V V 1 1 2 1 ( ).
2 2400 000 0.05 1000 0.05 23.56(0.562 23.56)Fπ π × − = × × × . ∴ =F 692 N .
c) V V V 2
2
2 1 1
10
46 25= = . .
V
g
V
g
1
2
1
2
2
400 39 06
2+ =
000
9810
.. 1 4.585 m/s.V ∴ =
2 2400 000 0.05 1000 0.05 4.585(5.25 4.585).Fπ π × − = × × × ∴ =F 2275 N.
4.120
V V V
g
p V
g
p V
g
p2 1
1
2
1 2
2
2 1
2
142 2
15
2= + = + ∴ = . . .
γ γ γ
b) V 12 2 9 81
15 9810400 000 53 33=
×
× × =
.. . 1 27.30 m/s, 29.2 m/s.V V ∴ = =
p A F m V V x x x1 1 2 1− = − ( ).
∴ = × + × × =F x 400 000 04 1000 04 7 3 22802 2 2π π . . . . N
F m V V y y y= − ( )2 1 =1000 04 7 3 29 2 10712π × × × =. . ( . ) . N
4.122
A V A V 1 1 2 2= . 2 2 2
20.025 4 (0.025 0.02 ) .V π π × × = −
2 11.11 m/s.V ∴ = p V
g
p V
g
1 1
2
2 2
2
2 2γ γ + = + .
2 2
1
11.11 49810 53 700 Pa.
2 9.81 p
⎛ ⎞−= =⎜ ⎟⎜ ⎟×⎝ ⎠
1 1 2 1( ). p A F m V V − = −
2 2 53 700 0.025 1000 0.025 4(11.11 4)
49.6 N.
F π π ∴ = × − × × −
=
4.124Continuity: 1 2 2 16 0.2 . 30 .V V V V = ∴ =
Energy (along bottom streamline):
F p
1 A
1
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Chapter 4 / The Integral Forms of the Fundamental Laws
51© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
V
g
p z
V
g
p z1
2
1
12
2
2
22 2
+ + = + +γ γ
2 22 2/900 6 0.2.
2 9.81 2 9.81
V V + = +
× ×
2 1 10.67, 0.36 m/s.V V ∴ = =
Momentum: F F F m V V 1 2 2 1− − = − ( )
9810 3(6 4) 9810 0.1(0.2 4) 1000 ( .2 4) 10.67(10.67 0.36)× × − × × − = × × × −F o
∴ =F 618 000 N. (F acts to the right on the gate.)
4.126
Continuity: 2 2 1 1 2 1 2 14 . 4 .V y V y V y y y= = ∴ =
Use the result of Example 4.12:
1/2
2 2
2 1 1 1 1
1 8
2 y y y y V g
⎡ ⎤⎛ ⎞⎢ ⎥
= − + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
a) 2 4 0.8 3.2 m. y = × =
1/22 2
1
1 83.2 0.8 0.8 0.8 .
2 9.81V
⎡ ⎤⎛ ⎞= − + + × ×⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦ 1 8.86 m/s.V ∴ =
4.128
Refer to Example 4.12:
1
1 1 11
603 6 6 10 10 . ( 6 10).
2
y y w w w V y
yγ γ ρ
⎛ ⎞− × × = × × − = ×⎜ ⎟
⎝ ⎠
2 11
1
6( 36) 600
2
y y
y
γ ρ
⎛ ⎞−∴ − = ⎜ ⎟
⎝ ⎠
1 1 1 1
1200or ( 6) 37.27. 3.8 ft, 15.8 fps.
32.2+ = = ∴ = = y y y V
4.130
2
1 1 2 2 2 2
0.052 . 15 30 m/s.
2 0.025V A V A V
π
π
×= = =
×
2 21 21 2
2 2
p pV V
gγ γ + = +
2 2
1 30 159810 337 500 Pa2 9.81
p −
∴ = =×.
( )2 1 1 1 1. ( ).x x xF m V V p A F m V Σ = − − = −
1 1 1
2 2 2 337 500 0.05 1000 0.05 15 4420 N.π π
∴ = +
= × + × × =
F p A mV
F 2
F 1
F
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Chapter 4 / The Integral Forms of the Fundamental Laws
52© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4.132
a) 2 1 1( ), .= − − = −∑ x x xF m V V F mV
1 21
30038.2 m/s
1000 0.05 ρ π = = =
× ×
mV
A
∴ = × =F 300 38 2 11. .460 N b) − = − −F m V V r B ( )(cos ).1 1α
∴ = × − =F 30028 2
38 238 2 10 6250
.
.( . ) . N
4.134
b) − = − −F m V V r B ( )(cos ).1 1α − = × − −700 1000 04 8 866 12
1
2π . ( ) (. ).V
1 40.24 m/s=V
∴ = = × × = . . .m A V ρ π 1 121000 04 40 24 202 kg / s
4.136
V R B = = × =0 30 15.5 m / s.
− = − − = × × × − R m V V x B ( )(cos ) . (.5 ).121 1000 025 40 25 1α π ∴ = R x 982 N.
∴ = = × × = .W R V x B10 10 982 15 147 300 W
4.138
− = − − = × × − − −F m V V x B ( )(cos ) . ( ) ( .5 ).12 2120 1 4 02 400 180 1 π
∴ = R x 365 N.
V B = × =1 2 150 180. m / s. .W = × × =15 365 180 986 000 W
The y-component force does no work.
4.140
b)1 1
1 1 2
1 1
100sin 30 sin 47 , 68.35 m/s
100cos30 40 cos
rr r
r
V V V
V
α α
α
⎫= ⎪∴ = = =⎬
− = ⎪⎭
2 22 2
2 2
sin 60 68.35 sin 38.9 m/s, 29.5
cos 60 68.35cos 40
V V
V
α α
α
⎫= ⎪= =⎬
= − ⎪⎭
22 1( ) 1000 0.015 100( 38.9cos60 100cos30 ).
7500 N
π − = − = × × − −
∴ =
x x x
x
R m V V
R
612 12 40 7500 3.60 10 WB xW V R∴ = = × × = ×
4.142 To find F , sum forces normal to the plate: ( )n out nF m V Σ = 1 .nV ⎡ ⎤−⎣ ⎦
F
V 1
V 2
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Chapter 4 / The Integral Forms of the Fundamental Laws
53© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
a) [ ]∴ = × × × − − =F 1000 02 4 40 40 60 11. . ( sin ) . 080 N (We have neglected friction)
2 2 3 3 10 ( ) 40sin 30 .t F m V m V mΣ = = + − − ×
Bernoulli: V V V 1 2 3= = .
2 12 3 1
1 2 3 3
.75 0.75 320 240 kg/s. 0 0.5 Continuity: 80 kg/s.
∴ = = × =∴ = − − ⎫⎬= + =⎭
m mm m mm m m m
4.144
2 2 21( ) 1000 0.02 0.4(40 ) sin 60 . 8(40 )sin 60 .= = × × − = −
r r n B x BF m V V F V
2 2 38 (40 ) 0.75 6(1600 80 ).= = − × = − + B x B B B B BW V F V V V V V
26(1600 160 3 ) 0. 13.33 m/s.= − + = ∴ =
B B BB
dW V V V
dV
4.146
1( )(cos 1) 90 .8 2.5 13.89 ( 13.89)( 1) 34 700 N.r BF m V V α = − − = × × × × − − =
50 100013.89 m/s 34 700 13.89 482 000 W or 647 hp
3600
×⎛ ⎞= = ∴ = × =⎜ ⎟
⎝ ⎠
BV W
4.148
To solve this problem, choose a control volume attached to the reverse thrustervanes, as shown below. The momentum equation is applied to a free bodydiagram:
Momentum:
[ ] ( )2 3 10.5 ( ) ( )− = + − × x r x r x r xR m V V m V
Assume the pressure in the gasesequals the atmospheric pressure
and that V r1 = V r2 = V r3. Hence,
1 1( ) 800 m/s= =r x rV V , and
2 3 1( ) ( ) sinα = = − ×r x r x rV V V
where α = 20° . Then, momentumis
[ ]
( )
1 1
1
0.5 2 sin
sin 1
α
α
− = − − ×
= − +
x r r
r
R m V m V
mV
Momentum:
[ ] ( )1 1 10.5 2 sin sin 1α α − = − − × = − + x r r rR m V m V mV
The mass flow rate of the exhaust gases is
( )1 1 40 100 1.025 102.5 kg/sair fuel air m m m m= + = + = × =
R x
V r 2
V r 3
V r 1
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Chapter 4 / The Integral Forms of the Fundamental Laws
54© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Substituting the given values we calculate the reverse thrust:
( ) ( )( )102.5 kg/s 800 m/s 1 sin 20 110 kNxR = + = Note that the thrust acting on the engine is in the opposite direction to Rx, and
hence it is referred to as a reverse thrust; its purpose is to decelerate the airplane.
4.150
For this steady-state flow, we fix the boat and move the upstream air. This provides us with the steady-state flow of Fig. 4.17. This is the same as observingthe flow while standing on the boat.
1 1
50 1000. 20 000 . 1440 N. ( 13.89 m/s)
3600
×= = × ∴ = =W FV F F V
2 22 1 2 2
13.89( ). 1440 1.23 1 ( 13.89). 30.6 m/s.
2π
+= − = × × − ∴ =
V F m V V V V
2 33 330.6 13.89
1 69.9 m /s
2
π +
∴ = = × × =Q A V
1
3
13.890.625 or 62.5%
22.24η = = = p
V
V
4.152
Fix the reference frame to the boat so that V 1 2088
6029 33= × = . fps.
2
8840 58.67 fps.
60= × =V
2
2 1
10 29.33 58.67 ( ) 1.94 (58.67 29.33)
12 2π
+⎛ ⎞∴ = − = × × −
⎜ ⎟⎝ ⎠
F m V V =
5460 lb.
1
ft-lb5460 29.33 160,000 or 291 hp.
sec= × = × =W F V
210 29.33 58.67
1.94 186.2 slug/sec12 2
π +⎛ ⎞
= × × × =⎜ ⎟⎝ ⎠
m
4.154
1 1 1 1 1 max 10.2 .2 1.0. 1 m/s. 2 m/s. ( ) 20(0.1 ).= = × × ∴ = ∴ = ∴ = −V A V V V V y y
flux in =
0.1 0.1 32 2 2
0 0
0.12 2 1000 20 (0.1 ) 800 000 267 N.
3
ρ = × − = =
∫ ∫V dy y dy
The slope at section 1 is −20. 2( ) 20 .∴ = − +V y y A
Continuity: 1 1 2 2 2 1. 2 2 m/s.= ∴ = = A V A V V V
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Chapter 4 / The Integral Forms of the Fundamental Laws
55© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
22
2
(0)1/2.
(0.05) 1
= ⎫∴ = −⎬
= − ⎭
V AV A
V A
2
12 . 2.5. ( ) 2.5 20 .
2= − ∴ = ∴ = − A V y y
flux out =
0.050.05 32
0 0
( 0.125) 800 0002 1000(2.5 20 ) 800 000 [0.00153]
3 3
⎡ ⎤−− = =⎢ ⎥
⎢ ⎥⎣ ⎦∫
y y dy
= 408 3. N. ∴change = 408 − 267 = 141 N.
4.156
From the c.v. shown: 21 2 0 0( ) 2 .π τ π − = w p p r r L
5
.
2
0.03 144 .75 / 12
2 30 2.36 10
ow
w
w
p r du
L dr
du
dr
τ μ
−
Δ∴ = =
× ×
∴ = × × ×
191 ft/sec/ft=
4.158
1 1 2 ( ) ρ ρ = − ∫ topm A V V y dA
22
0
1.23 2 10 32 (28 )10 65.6 kg/s⎡ ⎤
= × × − + =⎢ ⎥⎢ ⎥⎣ ⎦
∫ y dy
22 2 2 2
1 1 10
1.23 (28 ) 10 65.6 32 1.23 20 322
ρ − = + − = + + × − × ×
∫ ∫
top
FV dA m V m V y dy
∴ =F 3780 N .
Momentum and Energy
4.160
a) Energy:V
g z
V
g z h L
1
2
12
2
22 2
+ = + + . See Problem 4.125(a).
8
2 9 81
0 61912
2 9 81
22 2
×
+ =
×
+ +.
..
.
.51 .h L ∴ =h L 1166. m.
1 1losses = 9810 (0.6 1) 8 1.166 54 900 W/m of width.γ ∴ = × × × × =LV h
p1 A1 p2 A2
τw2 π r o L
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Chapter 4 / The Integral Forms of the Fundamental Laws
56© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Moment of Momentum
4.164
2
419.89 m/s. Velocity in arm = .
1000 4 0.004 ρ π = = =
× × ×
ee
mV V
A
(2 ) ρ = × ×M r Ω VI d V 0.3
c.v. 0
ˆ ˆˆ4 ( 2 ) ρ = × − Ω ×∫ ∫ i k ir V Adr
0.3
0
ˆ ˆ8 0.36 ρ ρ = Ω = − Ω∫k k AV rdr AV
and ( ) ρ = ×M 0 r Vd
d V dt
c.v.
0=∑ ∫
( )c.s.ˆ ˆ ˆˆ( ) 0.3 0.707 0.707 ρ ρ × ⋅ = × +
∫r V V n i j k e e e edA V V V A
The z-component of 2
. .
ˆ( ) 0.3 0.707 ρ ρ × ⋅ = ×∫ r V V n e ec s
dA V A
Finally, 2( ) 0.36 4 0.3 0.707 . Using , ρ ρ − = Ω = × × =I z e e e e M AV V A AV A V
0.36 4 0.3 0.707 19.89.Ω = × × × ∴ =Ω 46 9. .rad / s
4.166
20 010 1000 0.01 . 31.8 m/s. ρ π = = = × ∴ =m AV V V
Continuity: 2 20 0.01 0.01 0.006( 0.05).π π × = × + × −eV V V r
0
20.01 0.006 .15. 11.1 m/s.π × = × × ∴ =e eV V V
0 19.1( 0.05) 42.4 212 .∴ = − − = −eV V r V r
0.05 0.2
0
0 0.05
ˆ ˆ ˆ ˆˆ ˆ2 ( 2 ) 2 [ 2 (42.4 212 ) ] ρ ρ = × + Ω × + × + Ω × −∫ ∫M i k i i k iI r V Adr r r Adr
0.05 0.22
0
0 0.05
ˆ ˆ4 4 (42.4 212 ) ρ ρ = Ω + Ω −∫ ∫k k V A rdr A r r dr
= 2 2 3 342.4 212 ˆ(0.2 0.05 ) (0.2 0.05 )
2 3
⎡ ⎤− − −⎢ ⎥⎣ ⎦
k
ˆ ˆ(0.05 0.3 ) 0.35 .= Ω + Ω = Ωk k
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Chapter 4 / The Integral Forms of the Fundamental Laws
57
0.2 0.22
0.05 0.05
ˆ ˆ ˆ ˆ( ) 0.006 11.1 1000 0.006 13.86 ρ × − × = − × × = −∫ ∫i j k ke er V V dr rdr
∴ − = −0 35 13 86. . .Ω 39.6 rad/s.∴Ω =
4.168
See Problem 4.165.2
2
0.00819.89 m/s. 19.89 3.18 m/s.
0.02= = × =eV V
0.32
0
ˆ ˆ ˆˆ ˆ4 ( 2 ) . 0.01 , 0.004 ρ π π ⎡ Ω ⎤⎛ ⎞
= × − Ω × + − × = × = ×⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
∫M i k i k iI ed
r V r Adr A Adt
0.3 0.32
0 0
ˆ ˆ8 4
ˆ ˆ360 36
ρ ρ Ω
= − Ω −
Ω= − Ω −
∫ ∫k k
k k
dV rdr A r dr
dt
d AV A
dt
2
c.s.
ˆˆ( ) ( ) 212 ρ × ⋅ =∫ r V V n k z e edA V A
Thus, 2360 36 212 or 31.8 373.Ω Ω
Ω + = + Ω =e ed d
AV A V Adt dt
The solution is 31.8 11.73.−Ω = +tCe The initial condition is Ω( ) .0 0=
∴ = −C 1173. .
Finally, 31.811.73(1 ) rad/s.−Ω = − te