Chapter 4 / The Integral Forms of the Fundamental Laws
CHAPTER 4
The Integral Forms of the
Fundamental Laws
FE-type Exam Review Problems: Problems 4-1 to 4-16
4.1
(B)
4.2
(D)
m   AV 
4.3
(A)
Refer to the circle of Problem 4.27:
75.7  2
Q  AV  (  0.42 
 0.10  0.40  sin 75.5 )  3  0.516 m3 /s.
360
4.4
(D)
p
200
AV 
  0.042  70  0.837 kg/s .
RT
0.287  293
WP V22  V12
p  p1

 2
.
Q
2g

WP  40 kW
V22  V12
and energy req'd =
p2  p1
4.5
(A)
0
4.6
(C)
Manometer:  H  p1   g
2g
Energy: K

WP
1200  200

.
  0.040


. 0
40
 47.1 kW.
0.85
p
1202
 2 .  p2  7 200 000 Pa.
2  9.8 9810
V22
V2
 p2 or 9810  0.02  p1   g 2 .
2g
2g
100 000
7.962

.
2  9.81
9810
 K  3.15.
Combine the equations: 9810  0.02  1.2 
4.7
(B)
hL  K
V 2 p

.
2g

V12
.
2
V1  18.1 m/s.
Q
0.040

 7.96 m/s.
A   0.042
100 000
7.962
K

.
 K  3.15.
2  9.81
9810
V 
29
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.8
(C)
WP V22  V12 p


.
Q
2g

WP  Qp  0.040  400  16 kW.
4.9
(D)
4.10
(A)
WP


16
 18.0 kW.
0.89
pB
4.582
7.162
36.0  15 

 3.2
.  pB  416 000 Pa
2  9.81 9810
2  9.81
In the above energy equation we used
V2
Q
0.2
hL  K
with V  
 4.42 m/s.
2g
A   0.22
V 
Q
0.1

 19.89 m / s.
A  .04 2
V 22 p 2
V 22

 z2  K
.
Energy —surface to entrance: H P 
2g 
2g
19.89 2 180 000
19.89 2
HP 

 50  5.6
 201.4 m.
2  9.81
9810
2  9.81
W P  QH P /  P  9810  0.1  201.4 / 0.75  263 000 W.
4.11
(A)
4.12
(C)
After the pressure is found, that pressure is multiplied by the area of the
window. The pressure is relatively constant over the area.
V12 p1 V22
p


 2.
2g 
2g

p1  9810 
(6.252  1)  12.732
 3 085 000 Pa.
2  9.81
p1 A1  F   Q(V2  V1 ). 3 085 000    0.052  F  1000  0.1 12.73(6.25  1)
 F  17 500 N.
4.13
(D)
 Fx  m(V2 x  V1x )  1000  0.01  0.2  50(50cos 60  50)  2500 N.
4.14
(A)
Fx  m(Vr 2 x  Vr1x )  1000    0.022  60  (40cos 45  40)  884 N.
Power  Fx  VB  884  20  17 700 W.
4.15
(A)
Let the vehicle move to the right. The scoop then diverts the water to the
right. Then
F  m(V2 x  V1x )  1000  0.05  2  60  [60  (60)]  720 000 N.
30
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Chapter 4 / The Integral Forms of the Fundamental Laws
Chapter 4 Problems: 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
1 ˆ 1 ˆ
i
j = 0.707(ˆi  ˆj) .
nˆ 2  0.866ˆi  0.5ˆj . nˆ 3  ˆj .
2
2
V1n  V1  nˆ 1  10ˆi   0.707(ˆi  ˆj)   7.07 fps
V2n  V2  nˆ 2  10ˆi  (0.866ˆi  0.5ˆj)  8.66 fps .
V3n  V3  nˆ 2  10ˆi  (ˆj)  0
4.24
nˆ 1  
4.26
(B  nˆ ) A  15(0.5ˆi  0.866ˆj)  ˆj (10 12)  15  0.866  120  1559 cm 3
Volume = 15 sin 60 10 12  1559 cm3
Conservation of Mass
4.32
Use Eq. 4.4.2 with mV representing the mass in the volume:
dmV
dmV
0
   nˆ  VdA 
  A2V2   A1V1
dt c.s.
dt

Finally,
dmV
 Q  m.
dt
dmV
 m   Q.
dt
1.25 2
2.5 2
 60 =  
V2.
V2 = 15 ft/sec.
144
144
1.25 2
1.25 2
m  A V  1.94
 60 = 3.968 slug/sec. Q = AV = 
 60 = 2.045 ft 3 /sec.
144
144
4.34
A1V1 = A2V2.  
4.36
m in = A1V1 + A2V2. 200 = 1000   0.0252  25 + 1000 Q2. Q2 = 0.1509 m3 /s.
4.38
1A1V1   2 A2V2 . 1 
p1
500
kg
1246
kg

 4.433 3 .  2 
 8.317 3
RT 0.287  393
0.287  522
m
m
4.433   0.052  600 = 8.317   0.052 V2. V2 = 319.8 m/s.
m   1 A1V1 = 20.89 kg/s.
Q 1  A1V1 = 4.712 m3 /s.
Q2 = 2.512 m3 /s.
31
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.40
d 22 2
 . d2 = 4.478 m
b) (2  1.5 + 1.5  1.5) 3 = 
4 2
cos = 1/2
 = 60o

4.42
R
a) Since the area is rectangular, V = 5 m/s.
m   AV  1000  0.08  0.8  5 = 320 kg/s.
Q=

m

= 0.32 m 3 /s.
c) V  0.08 = 10  0.04 + 5  0.02 + 5  0.02.
V = 7.5 m/s.
m
m   AV  1000  0.08  0.8  7.5 = 480 kg/s. Q 
= 0.48 m3 /s.

4.44
If dm / dt  0, then  1 A 1V1   2 A 2V2   3 A 3V3 . In terms of m 2 and Q 3 this
becomes, letting  1   2   3 ,
1000    0.02 2  12  m 2  1000  0.01.
4.46
 m2  5.08 kg/s.
 0.1

min  mout  m.   0.2  2 10     10(20 y  100 y 2 )2dy    0.1 2 10   m.
 0

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:
4

4      2    m .
3

4.48
m   VdA 
 m  0.6667  = 0.82 kg/s.
1/ 3
 2.21.3545y (6y  9y
2
)2  5dy
0
1/ 3


 22  6 y  2.127 y 2  9 y 2  3.19 y 3 dy = 4.528 slug/sec.
0
2
2
4
u max   2  fps. (See Prob. 4.43b).
3
3
3
4 
1
2.2  1.94
= 2.07 slug/ft3.  V A  2.07    5   = 4.6 slug/sec.

3 
3
2
Thus, V A  m since  = (y) and V = V(y) so that V  V .
V 
4.50
m3 of H 2O
4
m3 of air
2000    0.00153 3
 9000  5
= 1.5  (1.5h).
3
s
m of air
h = 0.565 m.
32
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.52
 in  m
2 m
 3.
m
V1 = 20 m/s (see Prob. 4.43c).
20 1000  0.022  10  1000  0.022  V3.
4.54
The control surface is close to the
interface at the instant shown.
Vi = interface velocity.
 e A eV e   i A i V i .
1.5    0.152  300 
Ve
 V3 = 12.04 m/s.
n̂
Vi
n̂
8000
 12  Vi .
0.287  673
Vi = 0.244 m/s.
4.56
For an incompressible flow (low speed air flow)
 udA  A
0. 2
2
V2 .
 20 y
1/ 5
 0.8dy    0.15 2 V 2 .
0
A1
5
20  0.8 0.2 6/ 5    0.15 2 V 2 .
6
 V2  27.3 m/s
4.58
Draw a control volume around the entire set-up:
dm tissue
0
 V 2 A 2  V1 A 1
dt
 d2  d2  
2
 m tissue    2
 h2   ( h1 tan  ) h1
 4 
or
 d 2  d 22 

 tissue   
m
h2  h12 h 1 tan 2  .
 4

4.60
0
4.62
dm
 A 2V 2  A 1V1
dt
  1000(  0.003 2  0.02  10  10 6 / 60).
m
0
m  3.99 104 kg/s
dm
  3 Q 3   1 A 1V1  m 2 where m  A h.
dt
a) 0  1000  0.6 2 h  1000  0.6 / 60  1000  0.02 2  10  10.
h  0.0111 m/s
or
11.1 mm/s
33
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.64 Choose the control volume to consist of the air volume inside the tank. The conservation of
mass equation is
d
0
  dV  CS  V  n dA
dt CV
Since the volume of the tank is constant, and for no flow into the tank, the equation is
0V
d
 eVe Ae
dt
p
d
1 dp
. At the instant of interest,
.

RT
dt RT dt
Substituting in the conservation of mass equation, we get
Assuming air behaves as an ideal gas,  


1.8 kg/m3  200 m/s    0.015 m 
eVe Ae
dp

( RT )  
dt
V
1.5 m3
dp
  14.5 kPa/s
dt
2


kJ
0.287 kg  K  298 K 


Energy Equation
4.66
W  T  pAV  
du
Abelt
dy
 20  500  2 /60  400  0.4  0.5 10  1.81105 100  0.5  0.8
 1047  800  0.000724  1847 W
4.68 80% of the power is used to increase the pressure while 20% increases the internal energy
(Q  0 because of the insulation). Hence,
mu  0.2W
1000  0.02  4.18T  0.2  500.

W
T
 40  0.89.

mg
T  0.836C
b) WT  40  0.89  (90 000/60)  9.81  523 900 W
4.70

4.72
V 12 p1
V 22 p 2

 z1 

 z2.
2g 
2g 
3 ft
V
h2
1
V2
12 2
36 2
6 

h
.
2
2
2  32.2
64.4 h2
20.1
Continuity: 3  12 = h2 V2. This can be solved by trial-and-error:
8.236  2  h2 .
h2
h2  8' :
8.24 ? 8.31
h2  1.8' : 8.24 ? 8.00
h2  7.9 ' : 8.24 ? 8.22
h2  7.93' .
h2  1.75' : 8.24 ? 8.31
h2  1.76 '.
34
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.74
Manometer: Position the datum at the top of the right mercury level.
9810.4  9810 z 2  p 2 
V 22
 1000  (9810  13.6).4  9810  2  p1
2
p
V 22
.4  z 2 

 13.6.4  2  1 .

2g

p2
Divide by   9810:
V 12 p1
V 22 p 2

 z1 

 z2.
2g 
2g 
Energy:
(2)
V 12
 12.6 .4.
2g
Subtract (1) from (2): With z1 = 2 m,
(1)
V1 = 9.94 m/s
2
4.76
 1
Q = 120  0.002228 =     V 1 .
 12 
2
V1 = 12.25 fps.
2
Continuity:
1
1.5
    V 1      V 2 .
 12 
 12 
Energy:
V 12 p1 V 22 p 2
V2



 0.37 1 .
2g 
2g 
2g
V2 = 5.44 fps.

12.25 2 5.44 2 
= 8702.9 psf or 60.44 psi
 p 2  60  144  62.4 0.63

64.4
64.4 

4.78
V1  Q / A1 
Energy:
0.08
= 28.29 m/s.
 .03 2
V2 = 9V1 = 254.6 m/s.
V12 p1 V22 p2
V2



.2 1 .
2g 
2g 
2g
 254.6 2
28.29 2 
6
 p1  9810 
 0.8
 = 32.1  10 Pa.
2

9
.
81
2

9
.
81


4.80 a) Energy:
V 02 p 0
V2 p

 z 0  2  2  z 2 .  V 2  2 gz 0  2  9.81  2.4 = 6.862 m/s.
2g 
2g 
Q = AV = 0.8  1  6.862 = 5.49 m3 /s.
For the second geometry the pressure on the surface is zero but it increases
with depth. The elevation of the surface is 0.8 m:
35
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Chapter 4 / The Integral Forms of the Fundamental Laws
V22
 z0 
 h.
 V2  2g( z 0  h)  2  9.81  2 = 6.264 m/s.
2g
Q = 0.8  6.264 = 5.01 m3 /s.
Note: z0 is measured from the channel bottom in the 2nd geometry.
z0 = H + h.
4.82
V 02 p 0
V2 p

 z0  2  2  z2.
2g 
2g 
V 22
80 000
4
.
9810
2  9.81
V2 = 19.04 m/s.
b) Q  A2V2    0.092 19.04 = 0.485 m 3 /s.
4.84
Manometer:
 H   z  p1  13.6 H   z  p2 .
p1

 12.6 H 

.
V 12 p 2 V 22



.

2g

2g
V 22  V 12
12.6 H 
.
2g
Combine energy and manometer:
Continuity:
V2 
d12
d22
 d4 
 V12  12.6H  2 g  14  1 .
d

 2 
V1 .
1/2
d 2   12.6H  2 g 
 Q  V1 1   4 4

4 4  d1 / d2  1 
a) Energy from surface to outlet:
Energy from constriction to outlet:
Continuity:
V1  4V2 .
d12
 H
 12.35d12d22  4 4
 d d
 1
2
V 22
 H.
2g
p1


1/2



 V 22  2 gH .
V 12 p 2 V 22


.
2g

2g
With p1 = pv = 2450 Pa and p2 = 100 000 Pa,
2450
16
100 000
1

 2 gH 

 2 gH .
9810 2  9.81
9810
2  9.81
4.88
p2
p1
Energy:
4.86

Energy: surface to surface: z 0  z 2  hL .  30  20  2
H = 0.663 m.
V 22
.
2g
 V 22 = 10 g.
V1 = 4V2.  V 12 = 160 g.
160 g (94 000)
Energy: surface to constriction: 30 

 z1
2g
9810
z1 = 40.4 m. H = 40.4 + 20 = 60.4 m.
Continuity:
36
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.90
Velocity at exit = V e . Velocity in constriction = V1 . Velocity in pipe = V2 .
Energy: surface to exit:
V e2
 H.
2g
 V e2  2 gH .
D2
Continuity across nozzle: V 2  2 V e . Also, V1  4V2 .
d
Energy: surface to constriction: H 
a) 5 
V12 pv

.
2g 
 97 550
1 
D4
.
 16  4  2 g  5 
2g 
.2
9810

 D  0.131 m
2
4.92
 1
m  A V  1.94       120  5.079 slug / sec.
 12 
 302  1202 120 144 
ft-lb
WP  5.079  32.2 

or 23.5 hp.
 / 0.85  12,950
62.4 
sec
 2  32.2
4.94
 0  10.2 2 600 000 
W T  2  1000  9.81

 0.87.
9810 
 2  9.81
We used V2 
4.96
W T  1.304  10 6 W.
Q
2

 10.2 m/s
A2   0.252
 T  0.924
 V 2  V12 p2 p1

c
a) Q  WS  mg  2
   z2  z1  v (T2  T1 )  .
 2 1
g
 2 g

The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13.
1 
p1g
85  9.81

 9.92 N/m3.
RT1 0.287  293
2 
600  9.81 20 500

.
0.287 T2
T2
600 000T2 85 000 716.5
 200 2

 (1 500 000) = 5  9.81



(T2  293).
20 500
9.92
9.81
 2  9.81

 T2  572 K
or
299 C .
Be careful of units. p2  600 000 Pa,
cv  716.5 J/kg K
37
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.98
V22
V22 
   mg

Energy: surface to exit: W

20

4
.
5

.
T T
2g 
 2g
15
  Q  15  9810  147 150 N / s.
V2 
 13.26 m / s.
mg
 .62
 13.26 2
13.26 2 

 WT  0.8  147 150
 20  4.5
.
2  9.81
 2  9.81
 WT  5390 kW.
4.100 Choose a control volume that consists of the entire system and apply conservation of
energy:
HP 
p1


V12
p V2
 z1  HT  2  2  z2  hL
2g
 2g
Carburetor
0.5 m
Section (1)
Section (2)
Pump
We recognize that HT = 0, V1 is negligible and hL = 210 V 2/2g where V = Q/A and
V2  Q / A2 . Rearranging we get:
HP 

p2  p1

V22

 z2  z1  hL
2g

6.3 106 m3 /s
Q
(0.321) 2
V 
 0.321 m/s  hL  210
 1.1 m
A  2.5 103 m 2
2  9.81
V2 


Q 6.3 106 m3 /s

 12.53 m/s
A2  4 104 m 2


Substituting the given values we get:
 95  100  kN/m2  0.5m  12.53 m/s 

2
HP

6.660 kN/m3
2 9.81 m/s2

 1.1 m  8.85 m
The power input to the pump is:



WP   QHP /P  6660N/m3 6.3 106 m3 /s 8.85m  / 0.75  0.5 W
38
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.102 Energy: across the nozzle:

V12 p2 V22



.
 2g 
2g
p1
V12
6.252 V12
400 000


.
9810
2  9.81 2  9.81
V2 
52
22
V1  6.25V1.
V1  4.58 m/s , VA  7.16 m/s , V2  28.6 m/s.
Energy: surface to exit:
28.62
4.582
7.162
H P  15 
 1.5
 3.2
.
2  9.81
2  9.81
2  9.81
 H P  36.8 m.
 QHP 9810  (  0.012 )  28.6  36.8
 WP 

 3820 W.
P
0.85
Energy: surface to “A”:
pA
7.162
7.162
15 

 3.2
.
2  9.81 9810
2  9.81
 p A  39 400 Pa
Energy: surface to “B”:
36.0  15 
p
4.582
7.162
 B  3.2
.
2  9.81 9810
2  9.81
 pB  416 000 Pa
4.104 Depth on raised section = y 2 . Continuity: 3  3  V 2 y 2 .
Energy (see Eq. 4.5.21):
 3.059 
Trial-and-error:
V2
32
 3  2  (0.4  y 2 ).
2g
2g
92
 y2 ,
2 g y 22
y2  2.0 :
y2  1.8 :
y2  2.1:
y2  2.3 :
or
 0.11 ? 0 

 0.05 ? 0 
 0.1 ? 0 

 0.1 ? 0 
y 23  3.059 y 22  4.128  0
 y2  1.85 m.
 y2  2.22 m.
The depth that actually occurs depends on the downstream conditions. We cannot select a
“correct” answer between the two.
39
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.106 The average velocity at section 2 is also 8 m/s. The kinetic-energy-correction factor for a
parabola is 2 (see Example 4.9). The energy equation is:
V 12 p1
V2 p

  2 2  2  hL .
2g 
2g 
82
150 000
82
110 000

2

 hL .
2  9.81
9810
2  9.81
9810
 hL  0.815 m .
4.108
1
1
a) V   VdA 
A
  0.012


1
AV 3
V
3
dA 

r2 
20  0.012
0.014 
10 1 
2

rdr




  5 m/s
 0.012 
0.012  2
4  0.012 


0.01

0

  0.012  53
0
3
r2 
10 1 
2 rdr
 0.012 


0.01
1
3
2000  0.01
3  0.01
3  0.016
0.018 




  2.00
0.012  5 3  2
4  0.012 6  0.014 8  0.016 
2
4
 V 2  V12

4.110 Engine power = FD  V  m  2
 u2  u1 


2


 V 2  V12

m f q f  FDV  m  2
 cv (T2  T1 ) 
2


4.112 0   2
V22 p2
V2 p
 LV

 z2  1  1  z1  32
2g 
2g 
gD 2
02
V2
106 180V
 0.35  32 
.
2  9.81
9.81 0.022
V 2  14.4V  3.434  0.
 V  0.235 m/s
4.114 Energy from surface to surface:
a) H P  40  5
and
Q  7.37 105 m3 /s
V22 p 2
V12 p 1
V2
HP 

 z2 

 z1  K
.
2g 
2g 
2g
Q2
 40  50.7 Q 2
2
  0.04  2  9.81
Try Q  0.25:
Try Q  0.30:
H P  43.2 (energy).
H P  44.6 (energy).
H P  58 (curve)
H P  48 (curve)
Solution: Q  0.32 m3 /s
40
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Chapter 4 / The Integral Forms of the Fundamental Laws
Momentum Equation
4.116
V 12 p1 V 22 p 2


 .
2g 
2g 
b)
d)
V2 
d2
(d / 2)2
V1  4 V1.
V12
400 000 16 V12
 V1  7.303 m/s.


.
2  9.81
9810
2  9.81
400 000 .03 2  F  1000 .03 2  7.303(4  7.303  7.303).  F  679 N .
V12
30 144 16 V12


.
 V1  17.24 fps.
2  32.2
62.4
2  32.2
30    1.52  F  1.94    (1.5 /12)2  17.242 (4  1).
V12 p1 V22 p2
4.118



.
2g 
2g 
V0  0.012  Ve  0.006  0.15.
 F  127 lb.
 Ve  11.1 m/s.
Fx  m(V2 x  V1x )
a) V 2 
V 12
10 2
400 000 2.441 V 12
V

1
.562
V
.


.  V1  23.56 m/s.
1
1
82
2  9.81
9810
2  9.81
 p1 A 1  F  m (V 2  V1 ).
400 000  0.052  F  1000  0.052  23.56(0.562  23.56) .  F  692 N .
c) V 2 
10 2
V 1  6.25 V 1 .
42
V 12 400 000 39.06 V 12


.  V1  4.585 m/s.
2g
9810
2g
400 000  0.052  F  1000  0.052  4.585(5.25  4.585).
4.120 V 2  4 V 1 .
V 12 p1 V 22 p 2


 .
2g 
2g 

 F  2275 N .
15 V 12 p1
 .
2g

2  9.81
 400 000  53.33.  V1  7.30 m/s, V2  29.2 m/s.
15  9810
p1 A 1  Fx  m (V2 x  V1x ).  Fx  400 000 .04 2  1000 .04 2  7.3 2  2280 N .
b) V12 
Fy  m (V 2 y  V1y ) = 1000 .04 2  7.3  (29.2)  1071 N .
4.122 A 1V1  A 2V2 .
 V2  11.11 m/s.
  0.0252  4   (0.0252  0.022 )V2 .
p1


V 12 p 2 V 22


.
2g

2g
p1A1
F
41
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Chapter 4 / The Integral Forms of the Fundamental Laws
 11.112  42 
p1  9810 
 53 700 Pa.
 2  9.81 


p1 A1  F  m(V2  V1 ).
 F  53 700  0.0252  1000  0.0252  4(11.11  4)
 49.6 N.
4.124 Continuity:
6 V1  0.2 V2 .
 V2  30 V1.
Energy (along bottom streamline):
F
F1
V 12 p1
V2 p

 z1  2  2  z 2
2g 
2g 
F2
V22 / 900
V22
6
 0.2.
2  9.81
2  9.81
 V2  10.67, V1  0.36 m/s.
Momentum:
F1  F2  F  m (V2  V1 )
9810  3(6  4)  9810  0.1(0.2  4)  F  1000  ( o.2  4) 10.67(10.67  0.36)
 F  618 000 N .
4.126 Continuity:
V2 y2  V1y1  4V2 y1.
(F acts to the right on the gate.)
 y2  4 y1.
1/2 
Use the result of Example 4.12:


1
8
y2    y1   y12  y1V12 
2
g






a) y2  4  0.8  3.2 m.
1
8


3.2   0.8   0.82 
 0.8  V12 
2 
9.81


1/2 
4.128 Refer to Example 4.12:

y
60 
 1 y1w    3  6w    6w 10 10   .
2
y1 

.

 V1  8.86 m/s.
(V1y1  6 10).
 y 6

1200
 ( y12  36)  600   1
 37.27.  y1  3.8 ft, V1  15.8 fps.
 or ( y1  6) y1 
2
32.2
 y1 
42
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.130 V1A1  2V2 A2 .
 p1  9810
V2  15
  0.052
 30 m/s.
2  0.0252
302  152
 337 500 Pa .
2  9.81
p1


V12 p2 V22


2g 
2g
Fx  m  V2x  V1x  .
p1A1  F  m(V1 ).
 F  p1A1  mV1
 337 500  0.052  1000  0.052 152  4420 N.
4.132 a)  Fx  m(V2 x  V1x ),
V1 
 F  mV1.
V2
m
300

 38.2 m/s
 A1 1000    0.052
F
V1
 F  300  38.2  11 460 N .
b)  F  m r (V1  VB )(cos   1).
28.2
 F  300 
(38.2  10)  6250 N .
38.2
4.134 b)  F  m r (V1  VB )(cos   1). 700  1000 .04 2 (V1  8) 2 (.866  1). V1  40.24 m/s
 m  A 1V1  1000 .04 2  40.24  202 kg / s.
VB  R  0.5  30  15 m / s.
4.136
R x  m (V1  VB )(cos   1)  1000 .025 2  40  25(.5  1).  R x  982 N.
W  10R x VB  10  982  15  147 300 W.
4.138  Fx  m (V1  VB )(cos 120  1)  4 .02 2  (400  180) 2 (.5  1).
VB  1.2  150  180 m / s.
 R x  365 N.
W  15  365  180  986 000 W.
The y-component force does no work.
4.140 b)
100sin 30  Vr1 sin 1 

 1  47 , Vr1  Vr 2  68.35 m/s
100 cos 30  40  Vr1 cos 1 



 V2  38.9 m/s,  2  29.5
V2 cos 60  68.35cos  2  40 

V2 sin 60  68.35sin  2
 Rx  m(V2 x  V1x )  1000  0.0152 100(38.9cos 60  100cos 30 ).  Rx  7500 N
 W  12VB Rx  12  40  7500  3.60  106 W
43
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.142 To find F, sum forces normal to the plate: Fn  m  Vout n  V1n  .




a)  F  1000.02.4  40 (40 sin 60 )  11 080 N . (We have neglected friction)
Ft  0  m2V2  m3 (V3 )  m1  40sin 30 .
Bernoulli: V1  V 2  V 3 .
0  m2  m3  0.5m1   m2  .75m1  0.75  320  240 kg/s.

Continuity: m1  m2  m3 
m3  80 kg/s.

4.144 F  mr (V1r )n  1000  0.02  0.4(40  VB ) 2 sin 60 .
Fx  8(40  VB2 )sin 2 60 .
W  VB Fx  8VB (40  VB )2  0.75  6(1600VB  80VB2  VB3 ).
dW
 6(1600  160VB  3VB2 )  0.
dVB
 VB  13.33 m/s.
4.146 F  mr (V1  VB )(cos  1)  90  .8  2.5  13.89  (13.89)(1)  34 700 N.
50 1000


 13.89 m/s   W  34 700 13.89  482 000 W or
 VB 
3600


647 hp
4.148 To solve this problem, choose a control volume attached to the reverse thruster vanes, as
shown below. The momentum equation is applied to a free body diagram:
Vr2
Momentum:
 Rx  0.5m  (Vr 2 ) x  (Vr 3 ) x   m   Vr1 x
Assume the pressure in the gases equals the
atmospheric pressure and that Vr1 = Vr2 = Vr3.
Hence,
Rx
Vr1
(Vr1 ) x  Vr1  800 m/s , and
(Vr 2 ) x  (Vr 3 ) x  Vr1  sin 
where = 20 . Then, momentum is
 Rx  0.5m  2Vr1 sin    m  Vr1
Vr3
 mVr1  sin   1
Momentum:
Rx  0.5m  2Vr1 sin    m  Vr1  mVr1  sin   1
44
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Chapter 4 / The Integral Forms of the Fundamental Laws
The mass flow rate of the exhaust gases is
m  mair  mfuel  mair 1  1 40   100 1.025  102.5 kg/s
Substituting the given values we calculate the reverse thrust:


Rx  102.5 kg/s 800 m/s  1  sin 20  110 kN
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 observing the flow while
standing on the boat.
W  FV1.
20 000  F 
F  m(V2  V1 ).
1440  1.23 12 
Q  A3V3   12 
p 
50 1000
.  F  1440 N.
3600
V2  13.89
(V2  13.89).
2
 V2  30.6 m/s.
30.6  13.89
 69.9 m3 /s
2
V1 13.89

 0.625 or
V3 22.24
62.5%
4.152 Fix the reference frame to the boat so that V1  20 
V2  40 
(V1  13.89 m/s)
88
 29.33 fps.
60
88
 58.67 fps.
60
2
 10  29.33  58.67
 F  m(V2  V1 )  1.94    
(58.67  29.33)  5460 lb.
2
 12 
W  F  V1  5460  29.33  160,000
ft-lb
or 291 hp.
sec
2
 10  29.33  58.67
m  1.94      
 186.2 slug/sec
2
 12 
45
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.154 0.2  V1A1  V1  .2 1.0. V1  1 m/s. V1 max  2 m/s.
0.1
flux in = 2 
0
V1( y)  20(0.1  y).
0.1
0.13
 V dy  2  1000  20 (0.1  y) dy  800 000
 267 N.
3
0
2
2
2
The slope at section 1 is 20.  V2 ( y)  20 y  A.
Continuity: A1V1  A2V2. V2  2V1  2 m/s.
V2 (0)  A

  V2  A  1/ 2.
V2 (0.05)  A  1
1
2  A  .  A  2.5.  V2 ( y)  2.5  20 y.
2
0.05
flux out = 2

0
0.05
 ( y  0.125)3 
1000(2.5  20 y) dy  800 000 

3

 0
2
 408.3 N.

800 000
[0.00153]
3
change = 408  267 = 141 N.
4.156 From the c.v. shown: ( p1  p2 ) r02   w 2 r0 L.
 w 
p ro
du

.
2L
dr w

du
dr

w
w2roL
0.03  144.75 / 12
2  30  2.36  10 5
p 1A 1
p 2A 2
 191 ft/sec/ft
4.158 mtop   A1V1    V2 ( y)dA
2


 1.23  2 10  32   (28  y 2 )10dy   65.6 kg/s


0
2

F
   V 2dA  mtop V1  m1V1  1.23 (28  y 2 ) 210dy  65.6  32  1.23  20  32 2
2
0
 F  3780 N .
4.160 a) Energy:
V 12
V2
 z 1  2  z 2  hL . See Problem 4.125(a).
2g
2g
82
1.912 2
 0.6 
 2.51  hL .
2  9.81
2  9.81
 hL  1166
.
m.
 losses =  A1V1hL  9810  (0.6 1)  8 1.166  54 900 W/m of width.
46
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.164 Ve 
m
4

 19.89 m/s.
 Ae 1000  4    0.0042
 r  (2Ω  V)  d V
MI 
Velocity in arm = V .
0.3
 4  rˆi  (2kˆ  V ˆi )  Adr
c.v.
0
 8  AV kˆ
0.3
 rdr  0.36 AV kˆ
0
M  0
and
d
 (r  V)  d V  0
dt c.v.
 (r  V)V  nˆ  dA  0.3ˆi   0.707Ve ˆj  0.707Vekˆ  Ve  Ae
c.s.
The z-component of
 r  V(V  nˆ )  dA  0.3  0.707Ve Ae 
2
c.s.
Finally, ( MI ) z  0.36  AV   4  0.3  0.707Ve2 Ae  . Using AV  Ae Ve ,
   46.9 rad / s.
0.36  4  0.3 0.707 19.89.
4.166 m  10   AV  1000  0.012 V0 .
 V0  31.8 m/s.
V0  0.012  V   0.012  Ve  0.006(r  0.05).
Continuity:
V0  0.012  Ve  0.006  .15.
 Ve  11.1 m/s.
 V  V0  19.1(r  0.05)Ve  42.4  212r.
0.05
MI 

2rˆi  (2kˆ  V0ˆi )  Adr 
0
 4V0  Akˆ
0.2

2rˆi  [2kˆ  (42.4  212r )ˆi ] Adr
0.05
0.05

0.2
rdr  4 Akˆ
0

(42.4r  212r 2 )dr
0.05
212
 42.4

(0.22  0.052 ) 
(0.23  0.053)  kˆ
= 
3
 2

 (0.05  0.3)kˆ  0.35kˆ .
0.2

rˆi  (Ve ˆj)Ve   0.006dr  11.12  1000  0.006
0.05
0.2

rdr kˆ  13.86 kˆ
0.05
  39.6 rad/s.
 0.35  13.86.
47
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.168 See Problem 4.165. Ve  19.89 m/s.
V
0.0082
0.022
 19.89  3.18 m/s.
0.3

 d ˆ  ˆ 
M I  4  rˆi  (2kˆ  V ˆi )   
k   ri  Adr. A    0.012 , Ae    0.0042
dt




0
 8  AV kˆ
0.3

rdr  4  A
0
d ˆ
k
dt
0.3
r
2
dr
0
d ˆ
 360 AV kˆ  36 A
k
dt
 (r  V)z (V  nˆ )  dA  212Ve Aekˆ
2
c.s.
Thus, 360 AV   36 A
d
d
 212Ve2 Ae or
 31.8  373.
dt
dt
The solution is   Ce 31.8t  11.73. The initial condition is (0)  0.  C  1173
. .
Finally,
  11.73(1  e31.8t ) rad/s.
48
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