Solucionário - Mecânica dos Fluidos - 4 Edição - Merle C Potter, David C Wiggert, Bassem H Ramadan

Prévia do material em texto

1
CHAPTER 1 
 
Basic Considerations 
 
 
1.1 Conservation of mass — Mass — density 
 Newton’s second law — Momentum — velocity 
 The first law of thermodynamics — internal energy — temperature 
 
1.2 a) density = mass/volume = M L/ 3 
 b) pressure = force/area = F L ML T L M LT/ / /2 2 2 2= = 
 c) power = force × velocity = F L T ML T L T ML T× = × =/ / / /2 2 3 
 d) energy = force × distance = ML T L ML T/ /2 2 2× = 
 e) mass flux = ρAV = M/L3 × L2 × L/T = M/T 
 f) flow rate = AV = L2 × L/T = L3/T 
 
1.3 a) density = 
M
L
FT L
L
FT L3
2
3
2 4/ /= 
 b) pressure = F/L2 
 c) power = F × velocity = F × L/T = FL/T 
 d) energy = F × L = FL 
 e) mass flux = 
M
T
FT L
T
FT L= =
2 /
/ 
 f) flow rate = AV = L2 × L/T = L3/T 
 
1.4 (C) m = F/a or kg = N/m/s2 = N.s2/m. 
 
1.5 (B) [µ] = [τ/du/dy] = (F/L2)/(L/T)/L = F.T/L2. 
 
1.6 a) L = [C] T2. ∴[C] = L/T2 
 b) F = [C]M. ∴[C] = F/M = ML/T2 M = L/T2 
 c) L3/T = [C] L2 L2/3 . ∴[C] = L T L L L T3 2 2 3 1 3/ / /⋅ ⋅ = 
 Note: the slope S0 has no dimensions. 
 
1.7 a) m = [C] s2. ∴[C] = m/s2 
 b) N = [C] kg. ∴[C] = N/kg = kg ⋅ m/s2⋅ kg = m/s2 
 c) m3/s = [C] m2 m2/3. ∴[C] = m3/s⋅m2⋅ m2/3 = m1/3/s 
 
1.8 a) pressure: N/m2 = kg ⋅ m/s2/m2 = kg/m⋅ s2 
 b) energy: N⋅ m = kg ⋅ m/s2 × m = kg⋅ m2/s2 
 c) power: N⋅ m/s = kg ⋅ m2/s3 
 d) viscosity: N⋅ s/m2 = kg m
s
s
1
m
kg / m s2 2
⋅ ⋅ = ⋅ 
 2
 e) heat flux: J/s = 
N m
s
kg m
s
m
s
kg m / s2
2 3⋅ = ⋅ ⋅ = ⋅ 
f) specific heat: 
J
kg K
N m
kg K
kg m
s
m
kg K
m / K s
2
2 2
⋅
= ⋅
⋅
= ⋅ ⋅
⋅
= ⋅ 
 
1.9 kg
m
s
m
s
m2 + + =c k f . Since all terms must have the same dimensions (units) we 
require: 
 [c] = kg/s, [k] = kg/s2 = N s / m s N / m,2 2⋅ ⋅ = [f] = kg m / s N.2⋅ = 
 Note: we could express the units on c as [c] = kg / s N s / m s N s / m2= ⋅ ⋅ = ⋅ 
 
1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3 
 e) 1.2 cm2 f) 76 mm3 
 
1.11 a) 1.25× 108 N b) 3.21× 10−5 s c) 6.7× 108 Pa 
 d) 5.6× 10−12 m3 e) 5.2 × 10−2 m2 f) 7.8 × 109 m3 
 
1.12 (A) 8 92.36 10 23.6 10 23.6 nPa.− −× = × = 
 
1.13 2 2 2
0.06854
0.225 0.738
0.00194 3.281
m m
d d
λ
ρ ρ
= =
×
 
where m is in slugs, ρ in slug/ft3 and d in feet. We used the conversions in the 
front cover. 
 
1.14 a) 20 cm/hr = 5
20
/3600 5.555 10 m/s
100
−= × 520 /3600 5.555 10 m/s
100
−= × 
 b) 2000 rev/min = 2000 2× π/60 = 209.4 rad/s 
 c) 50 Hp = 50 × 745.7 = 37 285 W 
 d) 100 ft3/min = 100 × 0.02832/60 = 0.0472 m3/s 
 e) 2000 kN/cm2 = 2× 106 N/cm2 × 1002 cm2/m2 = 2× 1010 N/m2 
 f) 4 slug/min = 4 × 14.59/60 = 0.9727 kg/s 
 g) 500 g/L = 500 × 10−3 kg/10−3 m3 = 500 kg/m3 
 h) 500 kWh = 500 × 1000× 3600 = 1.8× 109 J 
 
1.15 a) F = ma = 10 × 40 = 400 N. 
 b) F − W = ma. ∴ F = 10 × 40 + 10 × 9.81 = 498.1 N. 
 c) F − W sin 30° = ma. ∴ F = 10 × 40 + 9.81 × 0.5 = 449 N. 
 
1.16 (C) The mass is the same on earth and the moon: [4(8 )] 32 .
du
r r
dr
τ µ µ µ= = = 
 
1.17 The mass is the same on the earth and the moon: 
 3
 m = 
60
32 2
1863
.
. .= ∴ Wmoon = 1.863× 5.4 = 10.06 lb 
 
1.18 (C) shear sin 4200sin30 2100 N.F F θ= = =
o 
 shear 4
2100
= 84 kPa 
250 10
F
A
τ −= =×
 
1.19 a) λ
ρ
= = ×
× ×
= ×
−
−
−. .
.
. ( . )
.43225 225
4 8 10
184 3 7 10
102
26
10 2
6m
d
m or 0.00043 mm 
 b) λ
ρ
= = ×
× ×
= ×
−
−
−. .
.
. ( . )
.225 225
4 8 10
00103 3 7 10
7 7 102
26
10 2
5m
d
m or 0.077 mm 
 c) 
26
2 10 2
4.8 10
.225 .225 .0039m
.00002 (3.7 10 )
m
d
λ
ρ
−
−
×
= = =
× ×
or 3.9 mm 
 
1.20 Use the values from Table B.3 in the Appendix. 
 a) 52.3 + 101.3 = 153.6 kPa. 
 b) 52.3 + 89.85 = 142.2 kPa. 
 c) 52.3 + 54.4 = 106.7 kPa (use a straight- line interpolation). 
 d) 52.3 + 26.49 = 78.8 kPa. 
 e) 52.3 + 1.196 = 53.5 kPa. 
 
1.21 a) 101 − 31 = 70 kPa abs. b) 760 − 31
101
 × 760 = 527 mm of Hg abs. 
 c) 14.7 − 31
101
 × 14.7 = 10.2 psia. d) 34 − 31
101
 × 34 = 23.6 ft of H2O abs. 
 e) 30 − 31
101
 × 30 = 20.8 in. of Hg abs. 
 
1.22 p = po e−gz/RT = 101 e−9.81 × 4000/287 × (15 + 273) = 62.8 kPa 
 From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is 
 % error = 
62.8 61.6
61.6
−
 × 100 = 1.95 %. 
 
1.23 a) p = 973 + 
22,560 20,000
25,000 20,000
−
−
 (785 - 973) = 877 psf 
 T = −12.3 + 22,560 20,000
25,000 20,000
−
−
 (−30.1 + 12.3) = −21.4°F 
 b) p = 973 + .512 (785 − 973) + .512
2
 (−.488) (628 − 2 × 785 + 973) = 873 psf 
 T = −12.3 + .512 (−30.1 + 12.3) + .512
2
 (−.488) (−48 + 2 × 30.1 − 12.3) = −21.4°F 
Note: The results in (b) are more accurate than the results in (a). When we use a 
linear interpolation, we lose significant digits in the result. 
 4
 
1.24 T = −48 + 33,000 30,000
35,000 30,000
−
−
 (−65.8 + 48) = −59°F or (−59 − 32) 5
9
 = −50.6°C 
 
1.25 (B) 
1.26 p = n
F
A
 = 4
26.5 cos 42
152 10−×
o
 = 1296 MN/m2 = 1296 MPa. 
 
1.27 
4
n
4
t
(120000) .2 10 2 . 4 N
20 .2 10 .0004N
F
F
−
−
= × × = 

= × × = 
 F = 2 2n tF F+ = 2.400 N. 
θ = tan−1 .0004
2.4
 =.0095° 
 
1.28 ρ = m
V−
= 0 2
180 1728
.
/
 = 1.92 slug/ft3. τ = ρg = 1.92 × 32.2 = 61.8 lb/ft3. 
 
1.29 ρ = 1000 − (T − 4)2/180 = 1000 − (70 − 4)2/180 = 976 kg/m3 
 γ = 9800 − (T − 4)2/18 = 9800 − (70 − 4)2/180 = 9560 N/m3 
 % error for ρ = 976 978
978
−
 × 100 = −.20% 
 % error for γ = 9560 978 9.81
978 9.81
− ×
×
 × 100 = −.36% 
 
1.30 S = 13.6 − .0024T = 13.6 − .0024 × 50 = 13.48. 
% error = 
13.48 13.6
13.6
−
 × 100 = −.88% 
 
1.31 a) m = 
W V
g
γ=
612 400 500 10
9.81g
−× ×= = 0.632 kg 
 b) m = 
612 400 500 10
9.77
−× ×
 = 0.635 kg 
 c) m = 
612 400 500 10
9.83
−× ×
 = 0.631 kg 
 
1.32 S =
/
water
m Vρ
ρ
= 10/. 1.2
water
V
ρ
= .
1.94
 ∴ V = 4.30 ft3 
 
1.33 (D) 
2 2
3
water
( 4) (80 4)
1000 1000 968 kg/m
180 180
Tρ − −= − = − = 
 
 5
1.34 τ = µ 
du
dr
 = 1.92 × 10−5 
2
30(2 1/12)
(1/12)
 ×
 
 
 = 0.014 lb/ft2 
 
1.35 T = force × moment arm = τ 2πRL × R = µ 
du
dr
 2πR2L = µ 
2
0.4
1000
R
 +  
 2πR2L. 
 ∴µ = 
2 2
2
0.0026
0.4 0.41000 2 1000 2 .01 0.2
12
T
R L
R
π π
=
   + + × ×      
 = 0.414 N.s/m2. 
 
1.36 Use Eq.1.5.8: T =
32 R L
h
π ω µ
 = 
( )2 .5 / 12π π× × × × ×3 2000 2
60
4 006
01 12
.
. /
 = 2.74 ft- lb. 
 Hp = 
Tω
550
2 74 209
550
= ×. .4 = 1.04 Hp 
 
1.37 Fbelt = µ
du
dy
A = × −131 10 10
002
3.
.
(.6 × 4) = 15.7 N. 
Hp = 
F V× = ×
746
15 7 10
746
.
 = 0.210 Hp 
 
1.38 Assume a linear velocity so .
du r
dy h
ω= Due to the area 
 element shown, dT = dF × r = τdA × r = µ du
dy
 2πr dr × r. 
 
dr
r
τ
 
 
 T = 3
0
2R
r dr
h
µω π
∫ = 
2
4
2 36 10
400 2
60
3 12
2 08 12
4
5 4
πµω π
π
h
R =
× × × × ×
×
−. ( / )
. /
 = 91 × 10−5 ft- lb. 
 
1.39 
2
30(2 1/12)
(1/12)
 ×
 
 
2 2
0 0[32 / ] 32 / .
du
r r r r
dr
τ µ µ µ= = = ∴τr = 0 = 0, 
 
 τr=0.25 = 32 × 1 × 10−3 × 2
.25/100
(.5/100)
= 3.2 Pa, τr=0.5 = 32 × 1 × 10−3 × 2
.5/100
(.5/100)
= 6.4 Pa 
 
1.40 (A) 3[10 5000 ] 10 10 5000 0.02 1 Pa.
du
r
dr
τ µ µ −= = × = × × × = 
1.41 The velocity at a radius r is rω. The shear stress is τ µ= ∆
∆
u
y
. 
The torque is dT = τrdA on a differential element. We have 
 6
 
0.08
0
 = = 2
0.0002
r
T rdA rdx
ωτ µ π∫ ∫ , 
2000 2
209.4 rad/s
60
πω ×= = 
 where x is measured along the rotating surface. From the geometry 2x = r, so that 
 
0.08 0.08
2 3
0 0
209.4 / 2 329000
= 0.1 2 329000 (0.08 )
0.0002 32
x x
T dx x dxπ× = =∫ ∫ = 56.1 N . m 
 
1.42 If τ µ= du
dy
 = cons’t and µ = AeB/T = AeBy/K = AeCy, then 
 AeCy 
du
dy
 = cons’t. ∴ du
dy
 = De−Cy. 
 Finally, 
0 0
yu
Cydu De dy−=∫ ∫ or u(y) = 0
yCyD e
C
−− = E (e−Cy − 1) 
 where A, B, C, D, E, and K are constants. 
 
1.43 
µ = =
=



Ae Ae
Ae
B T B
B
/ /
/
.
.
001
000357
293
353 ∴A = 2.334 × 10
−6, B = 1776. 
 
 µ40 = 2.334 × 10−6 e1776/313 = 6.80 × 10−4 N.s/m2 
 
1.44 m = ρV . Then dm = ρd V + V dρ. Assume mass to be constant in a volume 
subjectedto a pressure increase; then dm = 0. ∴ρd V = − V dρ, or d V
V
.
d ρ
ρ
= − 
 
1.45 B =
V− p
V
∆
∆
2200 MPa.= V∴∆ V−= 2 10
2200
p
B
∆ − ×= = −0.00909 m3 or −9090 cm3 
1.46 Use c = 1450 m/s. L = c∆t = 1450 × 0.62 = 899 m 
 
1.47 =
B V
p
∆∆ −
V
= −2100 −13
20
.
 = 136.5 MPa 
 
1.48 a) 327,000 144/1.93c = × = 4670 fps b) 327,000 144/1.93c = × = 4940 fps 
 c) 308,000 144/1.87c = × = 4870 fps 
 
1.49 V∆ =3.8 × 10−4 × −20 × 1 = .0076 m3. 
∆p = −B V∆
V
.0076
2270
1
−= − = 17.25 MPa 
 
1.50 p = 
2 2 0741
5 10 6
σ
R
= ×
× −
.
 = 2.96 × 104 Pa or 29.6 kPa. Bubbles: p = 4σ/R = 59.3 kPa 
 
 7
1.51 Use Table B.1: σ = 0.00504 lb/ft. ∴p = 4 4 .00504
1/32 12R
σ ×=
×
 = 7.74 psf or 0.0538 psi 
 
1.52 See Example 1.4: h = 
4 cos 4 0.0736 0.866
0.130 m.
1000 9.81 0.0002gD
σ β
ρ
× ×= =
× ×
 
 
1.53 (D) 
6
4 cos 4 0.0736 1
3 m or 300 cm.
1000 9.81 10 10
h
gD
σ β
ρ −
× ×= = =
× × ×
 
 
1.54 See Example 1.4: h = 
4 cos 4 0.032cos130
1.94 13.6 32.2 0.8/12gD
σ β
ρ
×=
× × ×
o
 
 = −0.00145 ft or −0.0174 in 
 
1.55 force up = σ × L × 2 cosβ = force down = ρghtL. ∴h = 2σ β
ρ
cos
.
gt
 
1.56 Draw a free-body diagram: 
 The force must balance: 
 W = 2σL or π ρ σd L g L
2
4
2




= . 
 ∴ =d
g
8σ
πρ
 
 
 
 
W
σL σL
needle
 
 
1.57 From the free-body diagram in No. 1.47, a force balance yields: 
 Is 
π ρd g
2
4
< 2σ? π(. ) . .004
4
7850 9 81 2 0741
2
× < × 
 0.968 < 0.1482 ∴No 
 
1.58 Each surface tension force = σ × π D. There is a force on the outside 
 and one on the inside of the ring. 
 ∴F = 2σπD neglecting the weight of the ring. 
 
 
 
F
D
 
 
 
1.59 
 
h(x)
h
dW
σdl
 
 
 
From the infinitesimal free-body shown: 
cos .d gh xdxσ θ ρ α=l cosθ = dx
dl
. 
 
/d d x d
h
g xdx g x
σ σ
ρ α ρ α
∴ = =l l 
We assumed small α so that the element 
thickness is αx. 
 8
 
 
1.60 The absolute pressure is p = −80 + 92 = 12 kPa. At 50°C water has a vapor 
pressure of 12.2 kPa; so T = 50°C is a maximum temperature. The water would 
“boil” above this temperature. 
 
1.61 The engineer knew that water boils near the vapor pressure. At 82°C the vapor 
pressure from Table B.1 is 50.8 (by interpolation). From Table B.3, the elevation 
that has a pressure of 50.8 kPa is interpolated to be 5500 m. 
 
1.62 At 40°C the vapor pressure from Table B.1 is 7.4 kPa. This would be the 
minimum pressure that could be obtained since the water would vaporize below 
this pressure. 
 
1.63 The absolute pressure is 14.5 − 11.5 = 3.0 psia. If bubbles were observed to form 
at 3.0 psia (this is boiling), the temperature from Table B.1 is interpolated, using 
vapor pressure, to be 141°F. 
 
1.64 The inlet pressure to a pump cannot be less than 0 kPa absolute. Assuming 
atmospheric pressure to be 100 kPa, we have 
 10 000 + 100 = 600 x. ∴x = 16.83 km. 
 
1.65 (C) 
 
1.66 ρ= =
× +
=p
RT
1013
0 287 273 15
.
. ( )
1.226 kg/m3. γ = 1.226 × 9.81 = 12.03 N/m3 
 
1.67 3in
101.3
1.226 kg/m .
0.287 (15 273)
p
RT
ρ = = =
× +
 3out
85
1.19 kg/m .
0.287 248
ρ = =
×
 
Yes. The heavier air outside enters at the bottom and the lighter air inside exits at 
the top. A circulation is set up and the air moves from the outside in and the inside 
out: infiltration. This is the “chimney” effect. 
 
1.68 3
750 44
0.1339 slug/ft .
1716 470
p
RT
ρ ×= = =
×
 m Vρ= 0.1339 15 2.01 slug.= × = 
 
1.69 (C) 
p V
m = 800 4 59.95 kg
0.1886 (10 273)RT
×= =
× +
 
 
1.70 
p
W V
RT
= 100 (10 20 4) 9.81 9333 N.
0.287 293
g = × × × × =
×
 
 
 
 
 9
1.71 Assume that the steel belts and tire rigidity result in a constant volume so that m1 
= m2: 
V 1 V= 1 1 2 22
1 2
2
2 1
1
 or .
150 460
(35 14.7) 67.4 psia or 52.7 psi gage.
10 460
m RT m RT
p p
T
p p
T
=
+∴ = = + =
− +
 
 
1.72 The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have 
 
100000 1 9.81. 10200 kg.m m× = × ∴ = 
 Hence, 
 
p V
m =
3100 4 / 3
10200. 12.6 m or 25.2 m.
0.287 288
r
r d
RT
π×= = ∴ = =
×
 
 
1.73 2 2
1
0 ( 10). 20 32.2. 25.4 fps.
2
KE PE mV mg V V= ∆ + ∆ = + − ∴ = × ∴ = 
2 210 ( 20). 40 32.2. 35.9 fps.
2
mV mg V V= + − ∴ = × ∴ = 
 
1.74 2 21-2
1
. a) 200 0 5( 10 ). 19.15 m/s.
2 f f
W KE V V= ∆ × = × − ∴ = 
 b) 
10
2 2
0
1
20 15( 10 ).
2 f
sds V= × −∫ 
 
2
2 210 120 15( 10 ). 15.27 m/s.
2 2 f f
V V× = × − ∴ = 
 c) 
10
2 2
0
1
200cos 15( 10 ).
20 2 f
s
ds V
π = × −∫ 
2 220 1200sin 15( 10 ). 16.42 m/s.
2 2 f f
V V
π
π
× = × − ∴ = 
 
1.75 21 2 1 2 2 1
1
. 10 40 0.2 0 . 40000.
2
E E u u u u= × × + = + ∴ − =% % % % 
 
40000
. 55.8 C where comes from Table B.4.
717v v
u c T T c∆ = ∆ ∴∆ = = o% 
 The following shows that the units check: 
 
2 2 2 2 2
car
2 2 2
air
kg m / s m kg C m kg C
C
kg J/(kg C) N m s (kg m/s ) m s
m V
m c
 × ⋅ ⋅ ⋅ ⋅ ⋅= = = = 
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅  
o o
o
o
 
 where we used N = kg.m/s2 from Newton’s 2nd law. 
 
 
 
 10
1.76 
2
2
2 1 H O
1
. .
2
E E mV m c T= = ∆ 
 
2
61 100 10001500 1000 2000 10 4180 . 69.2 C.
2 3600
T T−
× × × = × × × ∆ ∴∆ =  
o 
 We used c = 4180 J/kg. Co from Table B.5. (See Problem 1.75 for a units check.) 
 
1.77 water . 0.2 40000 100 4.18 . 19.1 C.f fm h m c T T T= ∆ × = × ∆ ∴∆ =
o 
The specific heat c was found in Table B.5. Note: We used kJ on the left and kJ 
on the right. 
 
1.78. (B) ice water ice water water. 320 .E E m m c T∆ = ∆ × = × ∆ 
 6 35 (40 10 ) 1000 320 (2 10 ) 1000 4.18 . 7.66 C.T T− −× × × × = × × × ∆ ∴∆ = o 
We assumed the density of ice to be equal to that of water, namely 1000 
kg/m3. Ice is actually slightly lighter than water, but it is not necessary for 
such accuracy in this problem. 
 
1.79. W p d V= mRT
V
=∫ d V
d V
mRT=∫
V
ln
V
mRT=∫
2
V
2
1 1
ln
p
mRT
p
= 
 since, for the T = const process, 1p V 1 2p V= 2. Finally, 
 1-2
4 1
1716 530ln 78,310 ft-lb.
32.2 2
W = × × = − 
The 1st law states that 
 0. 78,310 ft-lb or 101 Btu.vQ W u mc T Q W− = ∆ = ∆ = ∴ = = − −% 
 
1.80 If the volume is fixed the reversible work is zero since the boundary does not 
 move. Also, since V 1 2
1 2
, 
mRT T T
p p p
= = so the temperature doubles if the 
 pressure doubles. Hence, using Table B.4 and Eq. 1.7.17, 
200 2
a) (1.004 0.287)(2 293 293) 999 kJ
0.287 293v
Q mc T
×= ∆ = − × − =
×
 
b) 
200 2
(1.004 0.287)(2 373 373) 999 kJ
0.287 373v
Q mc T
×= ∆ = − × − =
×
 
c) 
200 2
(1.004 0.287)(2 473 473) 999 kJ
0.287 473v
Q mc T
×= ∆ = − × − =
×
 
 
1.81 W p d V= (p V= 2 V− 11). If = const, 
T
p
V
2
1
T
V
= 2 1
2
 so if 2 ,T T=∫ 
 then V 2 2V= 1 and (2W p V= 1 V− 1) p V= 1 1.mRT= 
 a) 2 0.287 333 191 kJW = × × = 
 b) 2 0.287 423 243 kJW = × × = 
 11
 c) 2 0.287 473 272 kJW = × × = 
 
1.82 = 1.4 287 318 357 m/s. 357 8.32 2970 m.c kRT L c t= × × = = ∆ = × = 
 
1/ 0.4/1.4
2
2 1
1
500
(20 273) 151.8 K or 121.2 C
5000
k k
p
T T
p
−   = = + = −     
o 
 
1.83 We assume an isentropic process for the maximum pressure: 
/ 1 1.4/0.4
2
2 1
1
423
(150 100) 904 kPa abs or 804 kPa gage.
293
k k
T
p p
T
−   = = + =     
 
Note: We assumed patm = 100 kPa since it was not given. Also, a measured 
pressure is a gage pressure. 
 
1.84 
/ 1 1.4/0.4
2
2 1
1
473
100 534 kPa abs.
293
k k
T
p p
T
−   = = =     
 
 2 1( ) (1.004 0.287)(473 293) 129 kJ/kg.vw u c T T= − ∆ = − − = − − − = − 
 We used Eq. 1.7.17 for cv. 
 
1.85 a) 1.4 287 293 343.1 m/sc kRT= = × × = 
b) 1.4 188.9 293 266.9 m/sc kRT= = × × = 
c) 1.4 296.8 293 348.9 m/sc kRT= = × × = 
d) 1.4 4124 293 1301 m/sc kRT= = × × = 
e) 1.4 461.5 293 424.1 m/sc kRT= = × × = 
Note: We must use the units on R to be J/kg.K in the above equations. 
 
1.86 (D) For this high-frequency wave, 287 323 304 m/s.c RT= = × = 
 
1.87 At 10 000 m the speedof sound 1.4 287 223 299 m/s.c kRT= = × × = 
At sea level, 1.4 287 288 340 m/s.c kRT= = × × = 
 
340 299
% decrease 100 12.06 %.
340
−= × = 
 
1.88 a) = 1.4 287 253 319 m/s. 319 8.32 2654 m.c kRT L c t= × × = = ∆ = × = 
b) = 1.4 287 293 343 m/s. 343 8.32 2854 m.c kRT L c t= × × = = ∆ = × = 
c) = 1.4 287 318 357 m/s. 357 8.32 2970 m.c kRT L c t= × × = = ∆ = × = 
 12 
C HAPTER 2 
 
Fluid Statics 
 
2.1 Σ ∆ ∆
∆ ∆
F ma p z p s
y z
ay y y y= − =: sinα ρ 2
 
 Σ ∆ ∆
∆ ∆ ∆ ∆
F ma p y p s
y z
a g
y z
z z z z= − = +: cosα ρ ρ2 2
 
 Since ∆ ∆s ycosα = and ∆ ∆s z sin α = , we have 
 
 p p
y
ay y− = ρ
∆
2
 and ( )p p z a gz z− = +ρ
∆
2
 
 Let ∆y → 0 and ∆z → 0: 
 
p p
p p
y
z
− =
− =



0
0
 ∴ = =p p py z . 
 
2.2 p = γh. a) 9810 × 10 = 98 100 Pa or 98.1 kPa 
 b) (0.8 × 9810) × 10 = 78 480 Pa or 78.5 kPa 
 c) (13.6 × 9810) × 10 = 1 334 000 Pa or 1334 kPa 
 d) (1.59 × 9810) × 10 = 155 980 Pa or 156.0 kPa 
 e) (0.68 × 9810) × 10 = 66 710 Pa or 66.7 kPa 
 
2.3 h = p/γ. a) h = 250 000/9810 = 25.5 m 
 b) h = 250 000/(0.8 × 9810) = 31.9 m 
 c) h = 250 000/(13.6 × 9810) = 1.874 m 
 d) h = 250 000/(1.59 × 9810) = 16.0 m 
 e) h = 250 000/(0.68 × 9810) = 37.5 m 
 
2.4 (C) (13.6 9810) (28.5 0.0254) 96600 PaHgp hγ= = × × × = 
2.5 S = 
20 144
62.4 20
p
hγ
×
=
×
 = 2.31. ρ = 1.94 × 2.31 = 4.48 slug/ft3. 
 
2.6 a) p = γh = 0.76 × (13.6 × 9810) = 9810 h. ∴h = 10.34 m. 
 b) (13.6 × 9810) × 0.75 = 9810 h. ∴h = 10.2 m. 
 c) (13.6 × 9810) × 0.01 = 9810 h. ∴h = 0.136 m or 13.6 cm. 
 
2.7 a) p = γ1h1 + γ2h2 = 9810 × 0.2 + (13.6 × 9810) × 0.02 = 4630 Pa or 4.63 kPa. 
 b) 9810 × 0.052 + 15 630 × 0.026 = 916 Pa or 0.916 kPa. 
 c) 9016 × 3 + 9810 × 2 + (13.6 × 9810) × 0.1 = 60 010 Pa or 60.0 kPa. 
 
y
z
pz∆y
py∆z
p∆s
∆s
α
∆z
∆y
ρg∆V
 13 
2.8 ∆p = ρgh = 0.0024 × 32.2 (–10,000) = –773 psf or –5.37 psi. 
 
 
2.9 (D) 0 84000 1.00 9.81 4000 44760 Pap p ghρ= − = − × × = 
 
 
2.10 
 
inside
100 9.81
3 13.51 Pa
.287 253
100 9.81
3 11.67 Pa
.287 293
outside o
o
base
i
i
pg
p g h h
RT
p
pg
p g h h
RT
ρ
ρ
× ∆ = ∆ = ∆ = × = ×  ∴ ∆× ∆ = ∆ = ∆ = × =
× 
 = 1.84 Pa 
 If no wind is present this ∆pbase would produce a small infiltration since the higher 
 pressure outside would force outside air into the bottom region (through cracks). 
 
 
2.11 p = ρgdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and 
S(10) = 1.1. By definition ρ = 1000 S, where ρwater = 1000 kg/m3. 
Then dp = 1000 (1 + h/100) gdh. Integrate: 
 dp h gdh
p
= +∫∫ 1000 1 100
0
10
0
( / ) 
 p = × +
×
1000 9 81 10
10
2 100
2
. ( ) = 103 000 Pa or 103 kPa 
 Note: we could have used an average S: Savg = 1.05, so that ρavg = 1050 kg/m
3. 
 
2.12 
v
∇ = + +p
p
x
i
p
y
j
p
z
k
∂
∂
∂
∂
∂
∂
$ $ $ 
 = – ρa ix $ – ρα y j$ – ρα z k$ – ρgk$ = – ( )ρ a i a j a kx y z$ $ $+ + – ρgk$ 
 = – ρva – ρvg 
 ∴ ∇ = − +
v v vp a gρ( ) 
 
 
2.13 /0 0[( ) / ]
g R
atmp p T z T
αα= − 
 = 100 [(288 − 0.0065 × 300)/288]9.81/.0065 × 287 = 96.49 kPa 
 p p ghatm= − = − ×
× ×ρ 100
100
287 288
9 81 300 1000
.
. / = 96.44 kPa 
 % error = 
96 96
96
100
.44 .49
.49
−
× = −0.052% 
 The density variation can be ignored over heights of 300 m or less. 
 14 
2.14 
/
0
0
0
g R
atm atm
T z
p p p p p
T
α
α −
∆ = − = − 
 
 
 = 100 
288 0065 20
288
1
9 81 0065 287− ×


 −






×. . /.
 = −0.237 Pa or −0.000237 kPa 
 This change is very small and can most often be ignored. 
 
2.15 Eq. 1.5.11 gives 310,000 144 .
dp
d
ρ
ρ
× = But, dp = ρgdh. Therefore, 
 
74.464 10
gdh dρ ρ
ρ
×
= or 2 7
32.2
4.464 10
d
dh
ρ
ρ
=
×
 
 Integrate, using 0ρ = 2.00 slug/ft
3: 
 2 7
2 0
32.2
4.464 10
hd
dh
ρ ρ
ρ
=∫ ∫
×
. ∴
1 1
2ρ
 
− − 
 
= 7.21 × 10-7 h or 
7
2
1 14.42 10 h
ρ
−
=
− ×
 
 Now, 
 7
7 7
0 0
2 2
ln(1 14.42 10 )
1 14.42 10 14.42 10
h h g g
p gdh dh h
h
ρ −
− −
= = = − ×∫ ∫
− × − ×
 
 Assume ρ = const: 
 2.0 32.2 64.4p gh h hρ= = × × = 
 
a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf. 
96,600 96,700
% error 100 0.103 %
96,700
−
= × = − 
b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf. 
322,000 323,200
% error 100 0.371 %
323,200
−
= × = − 
 c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf. 
966,000 976,600
% error 100 1.085 %
976,600
−
= × = − 
 
2.16 Use the result of Example 2.2: p = 101 e−gz/RT. 
 
 a) p = 101 e−9.81 ×10 000/287 ×273 = 28.9 kPa. 
 b) p = 101 e−9.81 ×10 000/287 ×288 = 30.8 kPa. 
 c) p = 101 e−9.81 ×10 000/287 ×258 = 26.9 kPa. 
 
2.17 Use Eq. 2.4.8: p = 
9.81
0.0065 287101(1 0.0065 /288) .z ×− 
 a) z = 3000. ∴p = 69.9 kPa. b) z = 6000. ∴p = 47.0 kPa. 
 c) z = 9000. ∴p = 30.6 kPa. d) z = 11 000. ∴p = 22.5 kPa. 
 
 15 
2.18 Use the result of Example 2.2: /
0
= .gz RT
p
e
p
− 
 
0
ln .
p gz
p RT
= − 
0.001 32.2
ln .
14.7 1716 455
z
= −
×
 ∴z = 232,700 ft. 
 
2.19 p = γh = (13.6 × 9810) × 0.25 = 33 350 Pa or 33.35 kPa. 
 
2.20 a) p = γh. 450 000 = (13.6 × 9810) h. ∴h = 3.373 m 
 b) p + 11.78 × 1.5 = (13.6 × 9810) h. Use p = 450 000, then h = 3.373 m 
 % error is 0.000 %. 
 
2.21 Referring to Fig. 2.6a, the pressure in the pipe is p = ρgh. If p = 2400 Pa, then 
 2400 = ρgh = ρ × 9.81 h. 
 a) ρ =
×
2400
981 36. .
 = 680 kg/m3. ∴gasoline 
 b) ρ =
×
2400
981 272. .
 = 899 kg/m3. ∴benzene 
 c) ρ =
×
2400
981 245. .
 = 999 kg/m3. ∴water 
 d) ρ =
×
2400
9 81 154. .
 = 1589 kg/m3. ∴carbon tetrachloride 
 
2.22 Referring to Fig. 2.6a, the pressure is p = ρwgh = 2
1
.
2 a
Vρ Then V
ghw
a
2 2=
ρ
ρ
. 
 a) V 2
2 1000 9 81 06
123
=
× × ×. .
.
 = 957. ∴V = 30.9 m/s 
 b) V 2
2 194 32 2 3 12
00238
=
× × ×. . /
.
 = 13,124. ∴V = 115 ft/sec 
 c) V 2
2 1000 9 81 1
123
=
× × ×. .
.
 = 1595. ∴V = 39.9 m/s 
 d) V 2
2 194 32 2 5 12
00238
=
× × ×. . /
.
 = 21,870. ∴V = 148 ft/sec 
 
2.23 (C) 0 30000 0.3 9810 0.1 8020 Paw atm x x water wp p h hγ γ= + − = + × − × = 
 
2.24 See Fig. 2.6b: p1 = –γ1h + γ2H. 
 p1 = –0.86 × 62.4 × 
5
12
 + 13.6 × 62.4 × 
9 5
12
.
 = 649.5 psf or 4.51 psi. 
 
2.25 0 1 1 2 2 3 3 4 4p p gh gh gh ghρ ρ ρ ρ= + + + + 
 = 3200 + 917×9.81×0.2 + 1000×9.81×0.1 + 1258×9.81×0.15 + 1593×9.81×0.18 
 = 10 640 Pa or 10.64 kPa 
 16 
 
2.26 ( ) ( ) ( )p p p p p p p p1 4 1 2 2 3 3 4− = − + − + − (Use ∆ ∆p g h= ρ ) 
 40 000 – 16 000 = 1000×9.81(–.2) + 13 600×9.81×H + 920×9.81×.3. 
 ∴H = .1743 m or 17.43 cm 
 
2.27 ( ) ( ) ( )p p p p p p p p1 4 1 2 2 3 3 4− = − + − + − (Use ∆ ∆p g h= ρ ) 
 po – pw = 900×9.81(–.2) + 13 600×9.81(–.1) + 1000×9.81×.15 
= –12 300Pa or –12.3 kPa 
 
2.28 ( ) ( ) ( ) ( )p p p p p p p p p p1 5 1 2 2 3 3 4 4 5− = − + − + − + − 
 p1 = 9810(–.02) + 13 600×9.81(.–04) + 9810(–.02)+13 600×9.81×.16 
 = 15 620 Pa or 15.62 kPa 
 
2.29 pw + 9810 × .15 – 13.6 × 9810 × .1 – .68 × 9810 × .2 + .86 × 9810 × .15 = po. 
 ∴pw – po = 11 940 Pa or 11.94 kPa. 
 
2.30 pw – 9810 × .12 – .68 × 9810 × .1 + .86 × 9810 × .1 = po. 
 With pw = 15 000, po = 14 000 Pa or 14.0 kPa. 
 
2.31 a) p + 9810 × 2 = 13.6 × 9810 × .1. ∴p = –6278 Pa or –6.28 kPa. 
 b) p + 9810 × .8 = 13.6 × 9810 × .2. ∴p = 18 835 Pa or 18.84 kPa. 
 c) p + 62.4 × 6 = 13.6 × 62.4 × 4/12. ∴p = –91.5 psf or –0.635 psi. 
 d) p + 62.4 × 2 = 13.6 × 62.4 × 8/12. ∴p = 441 psf or 3.06 psi. 
 
2.32 p – 9810 × 4 + 13.6 × 9810 × .16 = 0. ∴p = 17 890 Pa or 17.89 kPa. 
 
2.33 (A) (13.6 9810) 0.16 21350 Pa.ap Hγ= − = − × × = − 
 , 21350 10000 11350 13.6 9810 . 0.0851 maafter after afterp H H= − + = − = × ∴ = 
 
2.34 8200 + 9810 × .25 = 1.59 × 9810 × H. ∴H = 0.683 m 
 Hnew = .683 + .273 = .956 m. ∆H = 
.273
2
 = .1365. 
 p + 9810 (.25 + .1365) = 1.59 × 9810 × .956. 
 ∴p = 11 120 Pa or 11.12 kPa. 
 
2.35 p + 9810 × .05 + 1.59 × 9810 × .07 – .8 × 9810× .1 = 13.6 × 9810 × .05. 
 ∴p = 5873 Pa or 5.87 kPa. 
 
 Note: In our solutions we usually retain 3 significant digits in the answers (if a number 
starts with “1” then 4 digits are retained. In most problems a material property is used, 
i.e., S = 1.59. This is only 3 sig. digits! ∴ only 3 are usually retained in the answer! 
 
H
∆H
∆H
 17 
2.36 Before pressure is applied the air column on the right is 48" high. After pressure is 
applied, it is (4 – H/2) ft high. For an isothermal process 1p V 1 2p V= 2 using 
absolute pressures. Thus, 
 14.7 × 144 × 4A = p2(4 – H / 2 )A or p2 = 
8467
4 2− H /
. 
 From a pressure balance on the manometer (pressures in psf): 
 30 × 144 + 14.7 × 144 = 13.6 × 62.4 H + 
8467
4 2− H /
, 
 or H2 – 15.59 H + 40.73 = 0. ∴H = 12.27 or 3.32 ft. 
 
2.37 a) ( ) ( ) ( ) ( )p p p p p p p p p p1 5 1 2 2 3 3 4 4 5− = − + − + − + − 
4000 = 9800(0.16–0.22) + 15 600(0.10–0.16) + 133 400H + 15 600(0.07–H). 
∴H = .0376 m or 3.76 cm 
b) 0.6×144 = 62.4(–2/12) + 99.5(–2/12) + 849H + 99.5(2.5/12 – H). 
∴H = .1236 ft or 1.483 in. 
 
2.38 a) 
2 2
2 2
1 1 2 3 2
2 /
2 2( ) /
H D d
p D dγ γ γ γ
∆
=
∆ − + + −
 
2
2
2(.1/.005)
9800 2 15 600 2(133 400 15 600)(.1/.005)
=
− + × + −
 = × −8 10 6.487 H 
 ∴ = × ×−∆H 8 10 4006.487 = 0.0034 m or 3.4 mm 
 b) ∆H =
− + × + −
×
2 4 2
62 2 99 5 2 99 5 4 2
06 144
2
2
( /. )
.4 . (849 . )( /. )
. = 0.01153 ft or 0.138 in. 
 
2.39 ( ) ( ) ( )p p p p p p p p1 4 1 2 2 3 3 4− = − + − + − (poil = 14.0 kPa from No. 2.30) 
 15 500 – 14 000 = 9800(0.12 + ∆z) + 680(0.1 – 2∆z) + 860(–0.1 – ∆z). 
∴∆z = 0.0451 m or 4.51 cm 
 
2.40 a) pair = –6250 + 625 = –5620 Pa. 
 –5620 + 9800(2 + ∆z) – 13 600 × 9.81(0.1 + 2∆z) = 0. ∴∆z = 0.0025. 
 ∴h = 0.1 + 2∆z = .15 m or 15 cm 
 b) pair = 18 800 + 1880 = 20 680 Pa. 
 20 680 + 9800(0.8 + ∆z) – 13 600 × 9.81(0.2 + 2∆z) = 0. ∴∆z = 0.00715 m 
 ∴h = .2+ 2∆z = .214 or 21.4 cm 
 c) pair = –91.5 + 9.15 = –82.4 psf. 
 –82.4 + 62.4(6 + ∆z) – 13.6 × 62.4(4/12 + 2∆z) = 0. ∴∆z = 0.00558 ft. 
 ∴h = 4/12 + 2 (0.00558) = 0.3445 ft or 4.13 in. 
 d) pair = 441 + 44.1 = 485 psf 
 485 + 62.4(2 + ∆z) – 13.6 × 62.4(8/12 + 2∆z) = 0. ∴∆z = 0.0267 ft. 
 ∴h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in. 
 
 18 
2.41 F hA= γ = 9810 × 10 × π × .32/4 = 6934 N. 
 
2.42 
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32670 N
3 3 3 3
P P× × = × × × × × × × ∴ = a) F = pc A = 9800 × 2 
× 42 = 313 600 N or 313.6 kN 
 b) 
2 2
9800 1 (2 4) 9800 2 9800 1 98000 N or 98.0 kN
3 3c
F p A= = × × × + × × + × × =
 c) F = pc A = 9800 × 1 × 2 × 4 × 2 = 110 900 N or 110.9 kN 
 d) F = pc A = 9800 × 1 × 2 × 4/.866 = 90 500 N or 90.5 kN 
 
2.43 For saturated ground, the force on the bottom tending to lift the vault is 
 F = pc A = 9800 × 1.5 × (2 × 1) = 29 400 N. 
 The weight of the vault is approximately 
 W gVρ= 2400 9.81walls = × [2(2×1.5×.1) + 2(2×1×.1) + 2(.8×1.3×.1)] = 28 400 N. 
 The vault will tend to rise out of the ground. 
 
2.44 F = pc A = 6660 × 2 × π × 22 = 167 400 N or 167.4 kN 
 Find γ in Table B.5 in the Appendix. 
 
2.45 a) F = pc A = 9800 (10 − 2.828/3) (2.828 × 2/2) = 251 000 N or 251 kN 
 where the height of the triangle is (32 − 12)1/2 = 2.828 m. 
 b) F = pc A = 9800 × 10 (2.828 × 2/2) = 277 100 N or 277.1 kN 
 c) F = pc A = 9800 (10 − 2.828 × .866/3) (2.828 × 2/2) = 254 500 N or 254.5 kN 
 
2.46 a) F hA= = × × =γ 62 27 33 24.4 . 40,930 lb. 
 y p = +
×
×
27 33
6 8 36
27 33 24
3
.
/
.
 = 27.46'. ∴y = 30 – 27.46 = 2.54'. 
 8/5.46 = 3/x. ∴x = 2.05’. (2.05, 2.54) ft. 
 
 
 b) F = 62.4 × 30 × 24 = 44,930 lb. The centroid is the center of pressure. 
 y = 2.667'. 8/5.333 = 3/x. ∴x = 2.000' (2.000, 2.667) ft. 
 
 c) F = 62.4 (30 – 2.667 × .707) × 24 = 42,100 lb. 
 y p = +
×
×
39 77
6 8 36
39 77 24
3
.
/
.
 = 39.86'. y = 42.43 – 39.86 = 2.57' 
 8/5.43 = 3/x. ∴x = 2.04'. (2.04, 2.57) ft. 
 
2.47 (B) The force acts 1/3 the distance from the hinge to the water line: 
5 1 5 5
(2 ) (2 ) [9800 1 3 (2 )]. 32670 N
3 3 3 3
P P× × = × × × × × × × ∴ = 
 
 
(x, y)
y
x
 19 
2.48 a) F hA= = × ×γ π9810 6 2 2 = 739 700 N or 739.7 kN. 
 y y
I
Ayp
= + = +
×
×
6
2 4
4 6
4π
π
/
 = 6.167 m. ∴(x, y)p = (0, –0.167) m 
 
 b) F hA= = × ×γ π9810 6 2 = 369 800 N or 369.8 kN. 
 y p = +
×
×
6
2 8
2 6
4π
π
/
 = 6.167 m. x2 + y2 = 4 
 x F
x
pdA x y xdy y y dyp = = − = − −∫ ∫∫
−−2 2
6
2
4 62
2
2
2
2γ γ
( ) ( )( ) . 
 ∴ × = − − + =
−
∫x y y y dypγ π
γ
γ6 2
2
24 4 6 322
2
2
3( ) . ∴xp = 0.8488 m 
 
 
 ∴(x, y)p = (0.8488, –0.167) m 
 
c) F = 9810 × (4 + 4/3) × 6 = 313 900 N or 313.9 kN. 
 y p = +
×
×
5333
3 4 36
5 333 6
3
.
/
.
 = 5.500 m. ∴y = –1.5 
 4/2.5 = 1.5/x. ∴x = 0.9375. ∴(x,y)p = (0.9375, –1.5) m 
 
 
d) F = × + ×9810 4
2
3
4( sin 36.9°) × 6 = 330 000 N 
 y p = +
×
×
5 6
5 2 36
6 5 6
3
.
.4 /
.
 = 5.657 m. ∴y = 0.343 m 
 
 
 3 cos 53.13° = 1.8, 2.5 – 1.8 = 0.7, 2.4/2.057 = . /7 1x . ∴ x1 = 0.6. 
 x = 1.8 + 0.6 = 2.4. ∴(x,y)p = (2.4, 0.343) m. 
 
2.49 F hA= = × × ×γ 62 11 6 10.4 ( ) = 41,180 lb. 
 y y
I
yAp
= + = +
×
×
11
6 10 12
11 60
3 /
 = 11.758'. 
 (16 – 11.758) 41,180 = 10P. ∴P = 17,470 lb. 
 
 2.50 F hA= = × ×γ 9810 6 20 = 1.777 × 106 N, or 1177 kN. 
 y y
I
Ayp
= + = +
×
×
7 5
4 5 12
7 5 20
3
.
/
.
 = 7.778 m. 
 (10 – 7.778) 1177 = 5 P. ∴P = 523 kN. 
 
 
y
x
(x, y)
dA
dy
x
y
3 4
53.13
o
yp
P
F
 20 
2.51 F hA= = × ×γ 9810 12 20 = 2.354 × 106 N, or 2354 kN. 
 y y
I
Ayp
= + = +
×
×
15
4 5 12
15 20
3 /
 = 15.139 m. 
 (17.5 – 15.139) 2354 = 5 P. ∴P = 1112 kN. 
 
2.52 y y
I
Ay
H bH
bH H
H H
Hp = + = + ×
= + =
2
12
2 2 6
2
3
3 /
/
. y p is measured from the surface. 
 ∴From the bottom, H y H H Hp− = − =
2
3
1
3
. 
 Note: this result is independent of the angle α, so it is true for a vertical area or a sloped 
 area. 
 
2.53 3
1
sin40 3 . ( 2) sin40 . 2( 2) .
2 3
l
F l l F l P l l Pγ γ= × × = + ∴ = +o o 
a) 9810× 23 = 2(2 + 2)P. ∴P = 9810 N 
b) 9810× 43 = 2(4 + 2)P. ∴P = 52 300 N 
c) 9810× 53 = 2(5 + 2)P. ∴P = 87 600 N 
 
2.54 h = −1 2 2 2. .4 = 1.1314 m. A = 1.2 × 1.1314 + .4 × 1.1314 = 1.8102 m2 
 Use 2 forces: F h Ac1 1 9800 5657 1 2 11314= = × × ×γ . ( . . ) = 7527 N 
 F h Ac2 2 9800
11314
3
11314= = × × ×γ
.
(.4 . ) = 1673 N 
 y p1
2
3
11314= ( . ). y y
I
A yp2
2
2
311314
3
11314 36
11314 2 11314 3
= + = +
×
× ×
. .4 . /
.4 . / . /
 = 0.5657 m 
 ΣMhinge = 0: 7527
11314
3
1673 11314 0 5657 11314× + × − −
.
( . . ) . P = 0. ∴P = 3346 N. 
 
2.55 To open, the resultant force must be just above the hinge, i.e., yp must be just less than h.
 Let yp = h, the condition when the gate is about to open: 
 y h H A h H I h H h H= + = + = + +( ) / , ( ) , [ ( )]( ) /3 2 362 3 
 ∴ =
+
+
+
+ +
=
+
+
+
=
+
y
h H h H
h H h H
h H h H h H
p 3
2 36
3 3 6 2
4
2
( ) /
( ) ( ) /
 
 a) h
h H
=
+
2
. ∴h = H = 0.9 m 
 b) h = H = 1.2 m 
 c) h = H = 1.5 m 
 
2.56 The gate is about to open when the center of pressure is at the hinge. 
 a) y H H
b
H bp
= + = + +
×
+
1 2 1 8 2
1 8 12
9 1 8
3
. ( . / )
. /
(. ) .
. ∴H = 0. 
 21 
 b) y H H
b
H bp
= + = + +
×
+
1 2 2 0 2
2 12
1 2
3
. ( . / )
/
( )
. ∴H = 0.6667 m. 
 c) y H H
b
H bp
= + = + +
×
+
1 2 2 2 2
2 2 12
11 2 2
3
. ( . / )
. /
( . ) .
. ∴H = 2.933 m. 
 
2.57 (A) The gate opens when the center of pressure in at the hinge: 
 
31.2 11.2 (1.2 ) /12
5. 5 1.2.
2 2 (1.2 ) (11.2 ) / 2p
h I h b h
y y y
Ay h b h
+ + +
= + = + = + = +
+ +
 
 This can be solved by trial-and –error, or we can simply substitute one of the 
 answers into the equation and check to see if it is correct. This yields h = 1.08 m. 
 
2.58 F
H
bH bH1
2
2
1
2
= × =γ γ 
 F H b b H2 = × =γ γl l 
 
1
2 3 2
2γ γbH
H
b H× = ×l
l
. ∴ =H 3l 
 a) H = ×3 2 =3.464 m b) H = 1.732 m c) H = 10.39' d) H = 5.196' 
 
 
 Assume 1 m deep 
2.59 The dam will topple if the moment about “O” of F1 and F3 exceeds 
 the restoring moment of W and F2. 
 a) W = × × + ×( .4 )( / )2 9810 6 50 24 50 2 = 21.19 × 106 N 
 dw =
× + ×
+
300 27 600 16
300 600
 = 19.67 m. 
 F2 = 9810 × 5 × 11.09 = 0.544 × 106 N. d2
11 09
3
=
.
 = 3.697 m. 
 
 F1 9810
45
2
45= × × = 9.933 × 106 N. d1 = 15 m. 
 F3 9810
45 10
2
30= ×
+
× = 8.093 × 106 N. d3
2 943 15 5150 20
2 943 5150
=
× + ×
+
. .
. .
 = 18.18 m. 
 
Wd F d
F d F d
w + = × ⋅
+ = × ⋅



2 2
6
1 1 3 3
6
418 8 10
2961 10
.
.
N m
N m
 ∴will not topple. 
 
 b) W = (2.4 × 9810) (6 × 65 + 65 × 12) = 27.55 × 106 N. 
 dw = 
390 27 780 16
390 780
× + ×
+
 = 19.67 m. 
 F2
60 54 10≅ ×. N. d2 3 70≅ . m. 
 F1 = 9810 × 30 × 60 = 17.66 × 106 N. d1 = 20 m. 
 F3 9810
60 10
2
30= ×
+
× = 10.3 × 106 N. d3
2 943 15 7358 20
2 943 7 358
=
× + ×
+
. .
. .
 = 18.57 m. 
F2
F1
H/3
l/2
F1 F2
F3
W
O
 22 
 
Wd F d
F d F d
w + = × ⋅
+ = × ⋅



2 2
6
1 1 3 3
6
543 9 10
544 5 10
.
.
N m
N m
 ∴it will topple. 
 
 c) Since it will topple for H = 60, it certainly will topple if H = 75 m. 
 assume 1 m deep 
2.60 The dam will topple if there is a net clockwise moment about “O.” 
a) W W W W= + = × × × ×1 2 1 6 43 1 62 2. ( ) .4 .4 = 38,640 lb. 
 W2 24 43 2 62 2= × × ×( / ) .4 .4 = 77,280 lb. 
 W3 40 22 33 2 62= × ×( . / ) .4 = 27,870 lb @ 20.89 ft. 
 F1 62 20 40 1= × × ×.4 ( ) = 49,920 lb @ 40/3 ft. 
 F2 62 5 10 1= × × ×.4 ( ) = 3120 lb @ 3.33 ft 
 
1
3
2
= 18,720 lb @ 15 ft
 
= 28,080 lb @ 20 ft
p
p
F
F
F
= 

 
 
 
 OMΣ : (49,920)(40/3) + (18,720)(15) + (28,080)(20) − (38,640)(3) 
 − (77,280)(14) − (27,870)(20.89) − (3120)(3.33) < 0. ∴won’t tip. 
 
 b) W1 = 6 × 63 × 62.4 × 2.4 = 56,610 lb. W2 = (24 × 63/2) × 62.4 × 2.4 = 113,220 lb. 
 F1 62 30 60= × ×.4 = 112,300 lb. 3 (60 22.86/2) 62.4W = × × = 42,790 lb. 
 F2 62 5 10= × ×.4 = 3120 lb 
 Fp1 62 10 30= × ×.4 = 18,720 lb. Fp2 62 50 30 2= × ×.4 / = 46,800 lb. 
 OMΣ : (112,300)(20) + (18,720)(15) + (46,800)(20) − (56,610)(3) 
 − (113,220)(14) − 42,790(21.24) = 799,000 > 0. ∴will tip. 
 
 c) Since it will topple for H = 60 ft., it will also topple for H = 80 ft. 
 
2.61 ΣMhinge = 0. 2.5P – dw × W – d1 × F1 = 0. 
 ∴ P = × × × +
×
× ×
×
×






1
2 5
2
3
9800 1 8
4 2
3
9800
2
4
4
2
. π
π
 = 62 700 N 
 Note: This calculation is simpler than that of Example 2.7. Actually, We could 
 have moved the horizontal force FH and a vertical force FV (equal to W) 
 simultaneously to the center of the circle and then 2.5P = 2FH.=2F1. This 
 was outlined at the end of Example 2.7. 
 
 
2.62 Since all infinitesimal pressure forces pass thru the center, we can place the resultant 
 forces at the center. Since the vertical components pass thru the bottom point, they 
 produce no moment about that point. Hence, consider only horizontal forces: 
 
( ) 9810 2 (4 10) 784 800N
( ) 0.86 9810 1 20 168 700N 
waterH
H oil
F
F
= × × × =
= × × × =
 
 ΣM P: . . . 2 784 8 2 168 7 2= × − × ∴P = 616.1 kN. 
 
F1 F2
F3
W
O
W3
F1
dw
d1
PW
 23 
2.63 Place the resultant force
v v
F FH V+ at the center of the circular arc. 
v
FH passes thru the 
 hinge showing that P FV= . 
 a) P FV= = × × + × =9810 6 2 4 4 594( )π 200 N or 594.2 kN. 
 b) P = FV = 62.4 (20 × 6 × 12 + 9π × 12) = 111,000 lb. 
 
2.64 (D) Place the force 
v v
F FH V+ at the center of the circular arc. FH passes through the 
hinge: 
 24 1.2 9800 ( 1.2 /4) 9800 300000. 5.16 m.VP F w w wπ∴ = = × × + × × = ∴ = 
 
2.65 Place the resultant 
v v
F FH V+ at the circular arc center. 
v
FH passes thru the hinge so that 
 P FV= . Use the water that could be contained above the gate; it produces the same 
 pressure distribution and hence the same FV . 
 P FV= = 9810 (6 × 3 × 4 + 9π) = 983 700 N or 983.7 kN. 
 
2.66 Place the resultant
v v
F FH V+ at the center. 
v
FV passes thru the hinge 
 2 × (9810 × 1 × 10) = 2.8 P. ∴P = 70 070 N or 70.07 kN. 
 
2.67 The incremental pressure forces on the circular quarter arc pass through the hinge so that 
 no moment is produced by such forces. Moments about the hinge gives: 
 3 P = 0.9 W = 0.9 × 400. ∴P = 120 N. 
 
 2.68 The resultant 
v v
F FH V+ of the unknown liquid acts thru the center of the circular arc. 
v
FV 
 passes thru the hinge. Thus we use only ( ) .FH oil Assume 1 m wide. 
 a) ΣM
R R
R
R
S
R
R
R
Rx: . 3
9810
2
4
3
9800
4 2
2



+





 =



π
π
γ ∴ γ x = 4580 N/m
3 
 b) ΣM
R R
R
R
S
R
R
R
Rx: . . . 3
62 4
2
4
3
62 4
4 2
2



+





 =



π
π
γ ∴ γ x = 29.1 lb/ft
3 
 
2.69 The force of the water is only vertical (FV)w, acting thru the center. The force of the oil 
 can also be positioned at the center: 
 a) P FH o= = × × ×( ) ( . ) . .0 8 9810 0 3 3 6 = 8476 N. 
 ΣF W F Fy V o V w= = + −0 ( ) ( ) 
 0 = S × 9810 π × .62 × 6 + .
.
36
36
4
−


π
× 6 × (.8 × 9810) – 9810 × π × .18 × 6 
 − × × × −9810 8 2 6 62. . . ∴ =S 0 955. . 
 
 b) g Vρ .W= = 1996 lb. 
 ΣF W F Fy V o V w= = + −0 ( ) ( ) 
 24 
 0 = S × 62.4 × π × 22 × 20 + 4
4
4
−


π
× 20 × .8 × 62.4 – 62.4 × π × 2 × 20 
 − × × × ×62 4 8 2 2 202. . . ∴ =S 0 955. . 
 
2.70 The pressure in the dome is 
 a) p = 60 000 – 9810 × 3 – 0.8 × 9810 × 2 = 14 870 Pa or 14.87 kPa. 
 The force is F = pAprojected = (π × 32) × 14.87 = 420.4 kN. 
 
 b) From a free-body diagram of the dome filled with oil: 
 Fweld + W = pA 
 Using the pressure from part (a): 
 Fweld = 14 870 × π × 32 – (.8 × 9810) × 
1
2
4
3
33π ×

 = –23 400 N 
 or –23.4 kN 
 
 
2.71 A free-body diagram of the gate and water is shown. 
 
H
F d W H Pw3
+ = × . 
 a) H = 2 m. F = 9810 × 1 × 4 = 39 240 N. 
 W xdy
y
dy= = =
×
∫∫9810 2 9810 2 2
2 9810
2
2
3 2
1 2
0
2
0
2 3 2/ /
/
 = 26 160 N. 
 d x
x
xdy
xdy
x dx
x dx
w = = = =




∫
∫
∫
∫
2
1
2
4
4
1
2
1 4
1 3
3
0
1
2
0
1
/
/
 = 0.375 m. 
 ∴ = × + ×P
1
3
39
0 375
2
26 160 240
.
 = 17 980 N or 17.98 kN. 
 
 b) H = 8'. F = 62.4 × 4 × 32 = 7,987 lb. 
 W xdy x dx= = × = × ×∫∫62 4 62 4 4 62 16 2 32
0
2
3.4 .4 .4 / = 2,662 lb. 
 d x
x dx
x dx
w = = =




∫
∫
1
2
4
4
1
2
16 4
8 3
3
0
2
2
0
2
/
/
 = 0.75'. 
1 8
7,987 0.75 2,662 2910 lb
8 3
P  = × + × = 
 
 
 
2.72 (A) W Vγ= 
 900 9.81 9810 0.01 15 . 6 mw w× = × × ∴ = 
W
pA
Fweld
dA=xdy
y
x
F
h/3
 25 
2.73 W = weight of displaced water. 
a) 20 000 + 250 000 = 9810 × 3 2(6 /2).d d+ ∴d2 + 12d – 18.35 = 0. ∴d = 1.372 m. 
 b) 270 000 = 1.03 × 9810 × 3 2(6 /2).d d+ d2 + 12d – 17.81 = 0. ∴d = 1.336 m. 
 
2.74 25 + FB = 100. ∴FB = 75 = 9810 −V . ∴ −V = 7.645 × 10−3 m3 
 γ × 7.645 × 10−3 = 100. or 7645 cm3 
 ∴γ = 13 080 N/m3. 
 
2.75 3000 × 60 = 25 × 300 ∆d × 62.4. ∴∆d = 0.3846' or 4.62". 
 
2.76 100 000 × 9.81 + 6 000 000 = (12 × 30 + 8h × 30) 9810 
 ∴h = 1.465 m. ∴distance from top = 2 – 1.465 = 0.535 m 
 
2.77 T + FB = W. (See Fig. 2.11 c.) 
 T = 40 000 – 1.59 × 9810 × 2 = 8804 N or 8.804 kN. 
 
 2.78 The forces acting on the balloon are its weight W, the buoyant force FB, and the weight of 
 the air in the balloonFa. Sum forces: 
 
 FB = W + Fa or 
4
3
1000
4
3
3 3π ρ π ρR g R ga= + 
 
4
3
5
100 9 81
287 293
1000
4
3
5
100 981
287
3 3π π×
×
×
= + ×
×.
.
.
.
.
Ta
 ∴Ta = 350.4 K or 77.4°C 
 
2.79 The forces acting on the blimp are the payload Fp, the weight of the blimp W, the buoyant 
 force FB, and the weight of the helium Fh: 
 FB = Fp + W + Fh 
 1500 150
100 9 81
287 288
2π × ×
×
×
.
.
 = Fp + 0.1 Fp + 1500 π × 1502 × 
100 981
2 077 288
×
×
.
.
 
 4o /64.I dπ= . Npeople = 
986 10
800
8. ×
 = 1.23 × 106 
 Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add 
 significant weight. 
 
2.80 Neglect the bouyant force of air. A force balance yields 
 FB = W + F 
 = 50 + 10 = 60 = 9800 −V . ∴ −V = .006122 m3 
 Density: g Vρ .W= 
 ρ × ×9 81 006122. . = 50. ∴ρ = 832.5 kg/m3 
 Specific wt: γ = ρg = 832.5 × 9.81 = 8167 N/m3 
 Specific gravity: S = ρ/ρwater = 832.5/1000 = 0.8325 
 
 26 
2.81 From a force balance FB = W + pA. 
 a) The buoyant force is found as follows (h > 16'): 
 cos ,θ =
− −h R
R
15
 Area = θR2 – (h – 15 – R) R sinθ 
 ∴FB = 10 × 62.4[πR2 − θR2 + (h – 15 – R) R sinθ]. 
 FB = 1500 + γhA. 
 The h that makes the above 2 FB’s equal is found by trial-and- 
 error: 
 h = 16.5: 1859 ? 1577 h = 16.8: 1866 ? 1858 
 h = 17.0: 1870 ? 1960 ∴h = 16.82 ft. 
 b) Assume h > 16
1
3
 ft. and use the above equations with R = 1.333': 
 h = 16.4: 1857 ? 1853 ∴h = 16.4 ft. 
 c) Assume h < 16
2
3
ft. With R = 1.667', 
 FB = 10 × 62.4[θR2 − (R – h + 15) R sinθ]. 
 FB = 1500 + γhA. cos θ =
− +R h
R
15
 
 Trial-and-error for h: 
 h = 16: 1849 ? 1374 h = 16.2: 1853 ? 1765 
 h = 16.4: 1857 ? 2170 ∴h = 16.25 ft. 
 
 
 
2.82 a) W FB= . [ ]0 01 136 1000 015 4 9 81 98102. . . / . .+ × × × × = −h Vπ 
 − =
×
× +
×
× = × −V
π π.
.
.
. . .
015
4
15
005
4
06 2 769 10
2 2
5 3m ∴h = 7.361 × 10−3 m 
 ∴ =mHg
213.6 1000 .015 / 4hπ× × × = 0.01769 kg 
 b) (.01 + .01769) 9.81 = 9810 
π π×
× +
×
×






.
.
.
. .
015
4
15
005
4
12
2 2
Sx ∴Sx = 0.959. 
 c) (.01 + .01769) 9.81 = 9810 
π ×
×
.
.
015
4
15
2
 Sx. ∴Sx = 1.045. 
 
2.83 (. ) .
.
.
.
. .01 9 81 9810
015
4
15
005
4
12
2 2
+ =
×
× +
×
×





mHg
π π
 ∴mHg = 0.01886. 
 a) (.01 + .01886) 9.81 = 9810 
π ×
×
.
.
015
4
15
2
Sx. ∴Sx = 1.089. 
 b) mHg = 0.01886 kg. 
 
h − 15
θ R
pA
FB
W
h − 15
θ R
 27 
2.84 a) 
4 4
o
(10/12)
64 64
d
I
π π ×
= = = 0.02367 ft4. 
 − = =
× × × ×
V
W
rH O2
8 62 5 12 12 12
62
2. .4 ( / ) /
.4
π
 = 0.4363. depth = 
.4363
( / )π 5 12 2
 = 0.8' 
 ∴ = − −GM . /.4363 (. .4)02367 5 = –0.0457'. ∴It will not float with ends horizontal. 
 
 b) Io = 0.02367 ft4, −V = 0.3636 ft3, depth = 0.6667' 
 GM = − −. /. ( ) /02367 3636 5 4 12 = –0.01823'. ∴It will not float as given. 
 
 c) −V = 0.2909, depth = 6.4", GM = 
.
.
.02367
2909
4 3 2
12
−
−
 = 0.0147. ∴It will float. 
 
2.85 With ends horizontal 4o /64.I dπ= The displaced volume is
 − = × = × −V d h dx xγ π γ
2 5 34 9800 8 014 10/ . since h = d. The depth the cylinder will 
 sink is 
 depth = 
−
= × = ×− −
V
A
d d dx x8 014 10 4 10 20 10
5 3 2 5. / / .γ π γ 
 The distance CG is CG
h
dx= − ×
−
2
10 2 10 25. /γ . Then 
 GM
d
d
d
d
x
x= ×
− + × >−
−π
γ
γ
4
5 3
564
8 014 10 2
10 2 10 2 0
/
.
. / . 
 This gives (divide by d and multiply by γx): 
 612.5 – .5 γx + 5.1 × 10-5 γ x
2 > 0. 
 Consequently, 
 γx > 8369 N/m3 or γx < 1435 N/m3 
 
2.86 
3
3.water
water water
S dW
V S d
γ
γ γ
− = = = 
3
3.water
water water
S dW
V S d
γ
γ γ
− = = = ∴h = Sd. 
 GM
d
Sd
d Sd d
S
S
= − − = − +
4
3
12
2 2
1
12
1
2 2
/
( / / ) ( ). 
 If GM = 0 the cube is neutral and 6S2 – 6S + 1 = 0. 
 ∴ =
± −
S
6 36 24
12
 = 0.7887, 0.2113. 
 
 The cube is unstable if 0.2113 < S < 0.7887. 
 Note: Try S = 0.8 and S = 0.1 to see if GM > 0. This indicates stability. 
 
2.87 As shown, y =
× + ×
+
16 9 16 4
16 16
 = 6.5 cm above the bottom edge.
 G
S
S
A
A
=
× + × + ×
× + × + ×
4 9 5 16 8 5 16 4
5 8 2 8 16
γ γ γ
γ γ γ
. .
.
 = 6.5 cm. 
C
G
h
 28 
∴130 + 104 SA = 174 + 64 SA. ∴ SA = 1.1. 
 
2.88 a) y =
× + × + ×
+ +
16 4 8 1 8 7
16 8 8
 = 4. x =
× + × + ×
+ +
16 1 8 4 8 4
16 8 8
 = 2.5. 
 For G: y =
× × + × × + × ×
× + × + ×
12 16 4 5 8 1 1 5 8 7
1 2 16 5 8 15 8
. . .
. . .
 = 4.682. 
 
 x =
× + × × + × ×
× + × + ×
12 16 5 8 4 15 8 4
12 16 5 8 15 8
. . .
. . .
 = 2.364. 
 
 G must be directly under C. 
 tan
.
.
.θ =
136
682
 ∴θ =11.3°. 
 
 b) y =
× + × + ×
+ +
4 2 2
1
2
2 3 5
4 2 2
.
 = 2. x =
× + × + ×
+ +
4
1
2
2 2 2 2
4 2 2
 = 1.25 
 For G:y =
× × + × + ×
× + × + ×
12 4 2 5 1 15 7
12 4 5 2 15 2
. . .
. . .
 = 2.34. x =
× + × + ×
× + × + ×
12 2 5 4 15 4
1 2 4 5 2 15 2
. . .
. . .
 = 1.182 
 ∆y = 0.34, ∆x = 0.068. tan
.
.
.θ =
068
34
 ∴θ = 11.3°. 
 
2.89 The centroid C is 1.5 m below the water surface. ∴ CG = 1.5 m. 
 Using Eq. 2.4.47: GM =
×
× ×
− = − = >
l
l
8 12
8 3
15 1777 15 0 277 0
3 /
. . . . . 
 ∴The barge is stable. 
 
 
2.90 y =
× + ×
+
8 3 16 97 1
8 16 97
.485 .414 .
.485 .
 = 1.8 m. ∴ CG = −1 8 1 5. . = 0.3 m. 
 Using Eq. 2.4.47: GM =
×
− = − =
l
l
8 12
34 97
3 1 3 116
3.485 /
.
. .46 . . . ∴Stable. 
 
2.91 (A) 
2
5
20000 20000 6660 (1.2 ) 24070 Pa
9.81
24070 0.02 30.25 N
plug
plug plug
p h
F p A
γ
π
= + = + × × =
= = × × =
. 
 
2.92 a) tan
.
.α = =
20
9 81 4
H
 ∴H = 8.155 m. pmax = 9810 (8.155 + 2) = 99 620 Pa 
 b) pmax = ρ(g + az) h = 1000 (9.81 + 20) × 2 = 59 620 Pa 
 c) pmax = 1.94 × 60 (–12) – 1.94 (32.2 + 60) (–6) = 2470 psf or 17.15 psi 
 d) pmax = 1.94 (32.2 + 60) (–6) = 1073 psf or 7.45 psi 
 
0.682
C
0.136
G
 29 
2.93 The air volume is the same before and after. 
 ∴ 0.5 × 8 = hb/2. tan
.
.α = =
10
9 81
h
b
 
 4
9 81
10
=
h
h
2
.
.
 ∴h = 2.856. ∴Use dotted line. 
 2 5
1
2
2 5 2 4. . .452 .w + × × = ∴w = 0.374 m. 
 a) pA = –1000 × 10 (0 – 7.626) – 1000 × 9.81 × 2.5 = 51 740 Pa or 51.74 kPa 
 b) pB = –1000 × 10 (0 – 7.626) = 76 260 Pa or 76.26 kPa 
 c) pC = 0. Air fills the space to the dotted line. 
 
2.94 Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure. 
 a) 60 000 = –1000 ax (0–8) – 1000 × 9.81 0 2 5
8
9 81
− −














.
.
ax 
 4 = 
h
ax
2 9 81
2
× .
 60 = 8 ax + 24.52 – 9.81 
8
9 81
ax
.
. ax – 4.435 = 1.1074 ax . 
 ax
2 – 10.1 ax + 19.67 = 0 ∴ax = 2.64, 7.46 m/s2 
 b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10) − +





2 5
8
9 81
.
.
.
ax 
 60 = 8 ax + 49.52 – 19.81 
8
19 81
ax
.
. ax – 1.31 = 1.574 ax . 
 ax
2 – 5.1 ax + 1.44 = 0 ∴ax = 0.25, 4.8 m/s2 
 c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 +
8
14 81
ax
.
). 
 60 = 8 ax + 37.0 – 14.81 
8
14 81
ax
.
. ax – 2.875 = 1.361 ax . 
 ax
2 – 7.6 ax + 8.266 = 0 ∴ax = 1.32, 6.28 m/s2 
 
2.95 a) ax = 20 × .866 = 17.32 m/s2, az = 10 m/s2. Use Eq. 2.5.2 with the peep hole as 
position 1. The x-axis is horizontal passing thru A. We have 
 pA = –1000 × 17.32 (0 – 1.232) – 1000 (9.81 + 10) (0 – 1.866) = 58 290 Pa 
 b) pA = –1000 × 8.66 (0 – 1.848) – 1000 (9.81 + 5) (0 – 2.799) = 57 460 Pa 
 c) The peep hole is located at (3.696, 5.598). Use Eq. 2.5.2: 
 pA = –1.94 × 51.96 (0 – 3.696) – 1.94 (32.2 + 30) (0 – 5.598) = 1048 psf 
 d) The peep hole is located at (4.928, 7.464). Use Eq. 2.5.2: 
 pA = –1.94 × 25.98 (–4.928) – 1.94 (32.2 + 15) (–7.464) = 932 psf 
 
2.96 a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2, 
 p(z) = –1000 × 10 (–7.626) – 1000 × 9.81(z) = 76 260 – 9810 z 
α
b
h
AB
z
1
xw
C
 30 
 ∴ = −∫F z dzAB ( )
.
76 260 9810 4
0
2 5
 = 640 000 N or 640 kN 
 b) The pressure on the bottom BC is 
 p(x) = –1000 × 10 (x – 7.626) = 76 260 – 10 000 x. 
 ∴ = −∫F x dxBC ( )
.
76 260 10 000 4
0
7 626
 = 1.163 × 106 N or 1163 kN 
 c) On the top p(x) = –1000 × 10 (x – 5.174) where position 1 is on the top surface: 
 ∴ = −∫F x dxtop ( )
.
51 740 10 000 4
0
5 174
 = 5.35 × 105 N or 535 kN 
2.97 a) The pressure at A is 58.29 kPa. At B it is 
 pB = –1000 × 17.32 (1.732–1.232) 
 – 1000 (19.81) (1–1.866) = 8495 Pa. 
 Since the pressure varies linearly over AB, we 
 can use an average pressure times the area: 
FAB =
+
× ×
58 290 8495
2
15 2
 
. = 100 200 N or 100.2 kN 
 
 
 b) pD = 0. pC = –1000 × 17.32 (–.5–1.232) − 1000 × 19.81(.866–1.866) = 49 810 Pa. 
 FCD = × × ×
1
2
49 810 15 2 . = 74 720 N or 74.72 kN. 
 c) pA = 58 290 Pa. pC = 49 810 Pa. ∴ =
+
×FAC
58 29 49 81
2
15
. .
. = 81.08 kN. 
 
 
2.98 Use Eq. 2.5.2 with position 1 at the open end: 
a) pA = 0 since z2 = z1. 
 pB = 1000 × 19.81 × 0.6 = 11 890 Pa. 
 pC = 11 890 Pa. 
 
 b) pA = –1000 × 10 (.9–0) = –9000 Pa. 
 pB = –000 × 10 (.9)–1000 × 9.81(-.6) = –3114 Pa 
 pC = –1000 × 9.81 × (–.6) = 5886 Pa. 
 
c) pA = –1000×20 (0.9) = –18 000 Pa. 
 pB = –1000 × 20 × 0.9–1000×19.81(−0.6) = –6110 Pa. pC = 11 890 Pa 
 d) pA = 0. pB = 1.94 × (32.2-60) 
25
12



 = −112 psf. pC = –112 psf. 
 e) pA = 1.94 × 60 −




37 5
12
.
 = −364 psf. 
 pB = 1.94 × 60 −




37 5
12
.
– 1.94 × 32.2 −


25
12
 = –234 psf. 
x
z
A
BC
x
z
1
 31 
 pC = –1.94 × 32.2 −




25
12
= 130 psf. 
 f) pA = 1.94 × 30 
37 5
12
.


 = 182 psf. 
 pB = –1.94(–30) 
37 5
12
.


 – 1.94 × 62.2 −




25
12
 = 433 psf. 
 pC = –1.94 × 62.2 × −




25
12
= 251 psf. 
 
2.99 Use Eq. 2.6.4 with position 1 at the open end: 
 ω
π
=
×50 2
60
 = 5.236 rad/s. 
 a) pA =
×
× ×
1000 5 236
2
6 15
2
2. (. . ) = 11 100 Pa. 
 pB =
1
2
× 1000 × 5.2362 × .92 + 9810 × .6 = 16 990 Pa. 
 pC = 9810 × .6 = 5886 Pa. 
 
 
 b) pA =
1
2
× 1000 × 5.2362 × 0.62 = 4935 Pa. 
 pB =
1
2
× 1000 × 5.2362 × 0.62 + 9810 × 0.4 = 8859 Pa. 
 pC = 9810 × 0.4 = 3924 Pa. 
 c) pA =
1
2
× 1.94 × 5.2362 × 
37 5
12
2.


 = 259.7 psf. 
 pB =
1
2
× 1.94 × 5.2362 × 
37 5
12
62
25
12
2.
.4

 + × = 389.7 psf. 
 pC =62
25
12
.4 × = 130 psf. 
 d) pA =
1
2
× 1.94 × 5.2362 × 
22 5
12
2.


 = 93.5 psf. 
 pB =
1
2
× 1.94 × 5.2362 × 
22 5
12
2.


 + 62
15
12
.4 × = 171.5 psf. 
 pC = 62
15
12
.4 × = 78 psf. 
 
A
BC
z
1
r
ω
 32 
2.100 Use Eq. 2.6.4 with position 1 at the open end. 
 a) pA =
1
2
× 1000 × 102 (0 – 0.92) = –40 500 Pa. 
 pB = –40 500 + 9810 × 0.6 = –34 600 Pa. 
 pC = 9810 × 0.6 = 5886 Pa. 
 b) pA =
1
2
× 1000 × 102 (0 – 0.62) = –18 000 Pa. 
 pB = –18 000 + 9810 × 0.4 = –14 080 Pa. 
 
 pC = 9810 × 0.4 = 3924 Pa. 
 c) pA =
1
2
× 1.94 × 102 0
37 5
144
2
−






.
 = –947 psf. 
 pB = -947 + 62.4 × 
25
12
 = –817 psf. pC = 62.4 × 
25
12
 = 130 psf. 
 d) pA =
1
2
× 1.94 × 102 −






22 5
12
2
2
.
 = –341 psf. 
 pB = –341 + 62.4 × 
15
12
 = –263 psf. pC = 62.4 × 
15
12
 = 78 psf. 
 
2.101.1Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0. 
 a) 0 = 
1
2
× 1000 ω2 (0 – 0.452) – 9810 (0 – 0.6). ∴ω = 7.62 rad/s. 
 b) 0 = 
1
2
× 1000 ω2 (0 – 0.32) – 9810 (0 – 0.4). ∴ω = 9.34 rad/s. 
 c) 0 = 
1
2
× 1.94 ω2 0
18 75
12
2
2−






.
– 62.4 −


25
12
. ∴ω = 7.41 rad/s. 
 d) 0 = 
1
2
× 1.94 ω2 −






11 25
12
2
2
.
– 62.4 −


15
12
. ∴ω = 9.57 rad/s. 
 
 
 
2.102 The air volume before and after is equal. 
 ∴ = × ×
1
2
6 20
2 2π πr h . . . ∴ r h0
2 = 0.144. 
a) Using Eq. 2.6.5: r0
2 25 2× / = 9.81 h 
 ∴h = 0.428 m 
 ∴pA = 
1
2
× 1000 × 52 × 0.62 – 9810 (–0.372) 
 = 8149 Pa. 
 
 
 b) r0
2 27 2× / = 9.81 h. ∴h = 0.6 m. 
∴pA =
1000
2
× 72 × 0.62 + 9810 × 0.2 = 10 780 Pa. 
A
BC
z
1
r
ω
z
1
r
ω
2
1
2
h
z
rA
r0
 33 
c) For ω = 10, part of the bottom is bared. 
 π π π× × = −. . .6 2
1
2
1
2
2
0
2
1
2
1r h r h 
 Using Eq. 2.6.5: 
 
ω 2 0
2
2
r
g
h= , 
ω 2 1
2
2
r
g
h= 1 . 
∴ = −0144
2 2
2
2
2 1
2.
g
h
g
h
ω ω
 or
 h h2 1
2
20 144 10
2 981
− =
×
×
.
.
. 
 
 
 Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. ∴h = 0.859 m, r1 = 0.108 m. 
 ∴pA = 
1
2
× 1000 × 102 (0.62 – 0.1082) = 17 400 Pa. 
 d) Following part (c): h h2 1
2
20144 20
2 9 81
− =
×
×
.
.
. 1.6h – .64 = 2.936.∴ h = 2.235 m. 
 ∴pA = 
1
2
× 1000 × 202 (0.62 – 0.2652) = 57 900 Pa r1 = 0.265 m 
 
2.103 The answers to Problem 2.102 are increased by 25 000 Pa. 
 a) 33 150 Pa b) 35 780 Pa c) 42 400 Pa d) 82 900 Pa 
 
2.104 p r r g h( ) [ (. )].= − − −
1
2
0 82 2ρω ρ 
 p r r h( ) (. )= + −500 9810 82 2ω if h < .8. 
 p r r r( ) ( )= −500 2 2 1
2ω if h > .8. 
a) F p rdr r r dr= = +∫∫ 2 2 12 500 36503
0
6
π π ( )
.
 = 6670 N. 
 (We used h = .428 m) 
 
 
 b) F p rdr r r dr= = +∫∫ 2 2 24 500 19623
0
6
π π ( )
.
 = 7210 N. (We used h = 0.6 m) 
 c) F p rdr r r dr= = −
−
∫∫ 2 2 50 000 1083
108
6
2π π ( ( . )
.
.
 = 9520 N. (We used r1 = 0.108 m) 
 d) F p rdr r r dr= = −∫∫ 2 2 200 000 2653
265
6
2π π ( ( . )
.
.
 = 26 400 N. (r1 = 0.265 m) 
 
 
 
 
1
h
z
rA
r0
h1
dr
dA = 2πrdr
 34 
CHAPTER 3 
 
Introduction to Fluids in Motion 
 
3.1 
 
3.2 Pathlines: Release several at an instant in time and take a time exposure of the 
 subsequent motions of the bulbs. 
 
 Sreakline: Continue to release the devises at a given location and after the last 
 one is released, take a snapshot of the “line” of bulbs. Repeat this 
 for several different release locations for additional streaklines. 
 
3.3 
 
3.4 
 
streakline
pathline
streamline
streakline
pathline hose
boytime tt = 0
streamlines
 t = 2 hr
pathline
 t = 2 hr
streakline at t = 3 hry
x
 35 
3.5 a)u dx
dt
t v
dy
dt
t= = + = =2 2 2 
 x t t c y t c= + + = +2 1
2
22 
 = +y y2 
 ∴ − + =x xy y y2 22 4 ∴parabola. 
 
 b) x t t c c c= + + ∴ = − = −2 1 1 22 8 4. , . and 
 = + + ± + −y y4 2 4 8( ) 
 ∴ − + + − =x xy y x y2 22 8 12 0. ∴parabola. 
 
3.6 
ˆˆ ˆ ( )
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ using , .
zV dr udy vdxV ui vj wk
dr dxi dy j dzk i j k j i k
 × = −= + + 

= + + × = × = −
vv v
v 
 
3.7 Lagrangian: Several college students would be hired to ride bikes around the 
 various roads, making notes of quantities of interest. 
 
 Eulerian: Several college students would be positioned at each intersection 
 and quantities would be recorded as a function of time. 
 
3.8 a) At t V= = =2 0 0 0 2 22 and m / s( , , ) . 
 At t V= − = + =2 1 2 0 3 2 3 6062 2 and m / s( , , ) . . 
 b) At t V= =2 0 0 0 0 and ( , , ) . 
 At t V= − = − + − =2 1 2 0 2 8 8 2462 2 and m / s( , , ) ( ) ( ) . . 
 c) At t V= = − =2 0 0 0 4 42 and m / s( , , ) ( ) . 
 At t V= − = + − + − =2 1 2 0 2 4 4 62 2 2 and m / s( , , ) ( ) ( ) . 
 
3.9 (D) 5 ˆ( 51.4 10 )j−− × 
A simultaneous solution yields 4/5 and 3/5.x yn n= = (They must 
both have the same sign. 
 
3.10 a) cos $ / ( ) / . . .α α= ⋅ = + + = ∴ =
v
oV i V 1 2 3 2 0 832 33 692 2 
 
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
y x
x x
⋅ = + ⋅ + =
+ =
+ =



∴
= −
+ =
$ . ( $ $) ( $ $) .0 3 2 0
3 2 0
1
3
2
9
4
1
2 2
2 2∴ = = − = −n n n i jx y
2
13
3
13
1
13
2 3, $ ( $ $). or 
39.8o
y
x
streamlines
 t = 5 s
(27, 21)
(35, 25)
 36 
 b) cos $ / / ( ) ( ) . .α α= ⋅ = − − + − = − ∴ =
v
oV i V 2 2 8 0 2425 1042 2 
 
v
V n i j n i n j
n n
n n
n n
n nx y
x y
x y
x y
y y
⋅ = − − ⋅ + =
− − =
+ =



∴
= −
+ =
$ . ( $ $) ( $ $) .0 2 8 0
2 8 0
1
4
16 12 2 2 2
 
 ∴ = = − = − +n n n i jy x
1
17
4
17
1
17
4, $ ( $ $). or 
 
 c) cos $ / / ( ) . . .α α= ⋅ = + − = ∴ = −
v
oV i V 5 5 8 0 6202 51 672 2 
 
v
V n i j n i n j
n n
n n
n n
n n
x y
x y
x y
x y
y y
⋅ = − ⋅ + =
− =
+ =



∴
=
+ =
$ . ( $ $) ( $ $) .0 5 8 0
5 8 0
1
8
5
64
25
1
2 2
2 2
 
 ∴ = = = +n n n i jy x
5
89
8
89
1
89
8 5, $ ( $ $). or 
 
3.11 a) [ ]v vV dr x i xtj dxi dyj× = + + × + =0 2 0. ( )$ $ ( $ $) . 
 ∴ + − =
+
=( ) .x dy xtdx t
xdx
x
dy2 0
2
 or 
 Integrate: [ ]t xdx
x
dy t x n x y C
+
= − + = +∫∫ 2 2 2. . l 
 2 1 2 3 2( ) .− = − + ∴ln C C = 0.8028. 
 [ ]t x n x y− + = +2 2 0 8028l . 
 
 b) [ ]v vV dr xyi y j dxi dyj× = − × + =0 2 02. $ $ ( $ $) . 
 ∴ + = = −xydy y dx dx
x
dy
y
2 0
22 or . 
 Integrate: 2 2 1 2l l l lnx n y C n n C= − = − −( / ). ( ) ( / ). 
 ∴ = − = − − ∴ = −C nx n y x y2 2 22 2. ( / ). . l l 
 
 c) [ ]v vV dr x i y tj dxi dyj× = + − × + =0 4 02 2. ( )$ $ ( $ $) . 
 ( ) .x dy y tdx tdx
x
dy
y
2 2
2 24 0 4
+ + =
+
= − or 
 Integrate: t x C
y
C
2 2
1 2
2
1
2
1
2
1 1tan . tan .− −+



= +



= − 
 ∴ = − −



=−C yt
x
0 9636
2
0 9636 21. . tan . 
 
 37 
3.12 (C) 2ˆ ˆ ˆ ˆ ˆ ˆ2 (2 ) (2 2 ) 16 8 16 .V V V Va u v w xy yi y xi yj i i j
t x y z
∂ ∂ ∂ ∂
= + + + = − − = − − +
∂ ∂ ∂ ∂
v v v v
v 
 2 28 16 17.89 m/sa∴ = + = 
 
3.13 a) DV
Dt
u
V
x
v
V
y
w
V
z
V
t
v v v v v
= + + +
∂
∂
∂
∂
∂
∂
∂
∂
=0. 
 b) u V
x
v
V
y
w
V
z
V
t
x i y j xi yj
∂
∂
∂
∂
∂
∂
∂
∂
v v v v
+ + + = + = +2 2 2 2 4 4( $) ( $) $ $ = 8 4$ $i j− 
 c) u V
x
v V
y
w V
z
V
t
x t xti ytj xyt xtj ztk x i xyj∂
∂
∂
∂
∂
∂
∂
∂
v v v v
+ + + = + + + + +2 22 2 2 2 2 2( $ $) ( $ $) $ 
 + = − −2 68 100 54yzk i j k$ $ $ 
 d) u V
x
v V
y
w V
z
V
t
x i yzj xyz xzj tz xyj tk zk∂
∂
∂
∂
∂
∂
∂
∂
v v v v
+ + + = − − − + − + +($ $) ( $) ( $ $) $2 2 2 2 
 = xi yz x yz xyzt j zt z k$ ( )$ ( ) $− − + + +2 4 22 2 2 
 = 2 114 15$ $ $i j k− + 
3.14 
v
Ω = −





 + −



+ −






1
2
1
2
1
2
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
w
y
v
z
i
u
z
w
x
j
v
x
u
y
k$ $ $. 
 a) 
v
Ω = − =
1
2
20
∂
∂
u
y
k yk$ $ = −20 $k 
 b) 
v
Ω = − + − + −
1
2
0 0
1
2
0 0
1
2
0 0( )$ ( )$ ( ) $i j k = 0 
 c) 
v
Ω = − + − + −
1
2
2 0
1
2
0 0
1
2
2 0( )$ ( ) $ ( ) $zt i j yt k = 6 2$ $i k− 
 d) 
v
Ω = + + − + − −
1
2
0 2
1
2
0 0
1
2
2 0( )$ ( )$ ( ) $xy i j yz k = − +2 3$ $i k 
 
3.15 The vorticity v
v
ω = 2Ω. Using the results of Problem 3.7: 
 a) vω = −40$i b) vω = 0 c) vω = 12 4$ $i k− d) vω = − +4 6$ $i k 
 
3.16 a) ε ∂
∂
ε
∂
∂
ε
∂
∂xx yy zz
u
x
v
y
w
z
= = = = = =0 0 0, , . 
 ε
∂
∂
∂
∂
ε
∂
∂
∂
∂xy xz
u
y
v
x
y
u
z
w
x
= +





 = − = = +



=
1
2
20 20
1
2
0, , 
 ε
∂
∂
∂
∂yz
v
z
w
y
= +





 = ∴ =










1
2
0
0 20 0
20 0 0
0 0 0
. rate - of strain 
 38 
 b) 
ε ε ε
ε ε ε
xx yy zz
xy xz yz
= = =
= = =
2 2 0
0 0 0
, , .
, , .
 
 
 rate-of strain = 
2 0 0
0 2 0
0 0 0










 
 
 c) ε ε εxx yy zzxt xt yt= = = = = = −2 8 2 8 2 4, , . 
 ε ε εxy xz yzyt zt= = − = = = =
1
2
2 2
1
2
0 0
1
2
2 6( ) , ( ) , ( ) . 
 rate-of strain = 
8 2 0
2 8 6
0 6 4
−
−
−










 
 d) ε ε εxx yy zzxz t= = − = − = =1 2 12 2, , . 
 ε ε εxy xz yzyz xy= − = = = = − =
1
2
2 3
1
2
0 0
1
2
2 2( ) , ( ) , ( ) . 
 rate-of strain = 
1 3 0
3 12 2
0 2 2
−










 
 
3.17 a) a
r r r r rr
= −







− +



−



−10
40 80
10
40
1
40
2 3 2 2cos cos
sin
( sin )θ θ
θ
θ 
 − +

 

 = − − −
1
10
40
10 2 1 125 1
2
2
2
r r
sin ( .5)( ) . ( )θ = 9.375 m/s2. 
 a
r r r r rθ
θ θ
θ
θ= −







+ +



+



10
40 80
10
40
10
40
2 3 2 2cos sin
sin
cos 
 − −



1
100
1600
4r r
sin cosθ θ = 0 since sin 180° = 0. 
 aφ = 0. 
 b) ω ω ω θ θθr z r r r r
= = = − +



− −



−0 0
1
10
40 1
10
40
2 2, , sin ( sin ) = 0. 
 At (4, 180°) vω = 0 since vω = 0 everywhere. 
 
3.18 a) a
r r r r rr
= −







− +



− −



10
80 240
10
80
10
80
3 4 3 3cos cos
sin
( sin )θ θ
θ
θ 
 − +

 

 = − −10
80
8 75 1 9375 1
3
2 2
r r
sin
. ( )(. )( )
θ = 8.203 m/s2 
 aθ = 0 since sin 180° = 0. aφ = 0 since v φ = 0. 
 b) ω r = 0, ω θ = 0, ω φ = 0, since sin 180° = 0. 
 
 39 
3.19 V Va u
t x
∂ ∂
∂ ∂
= +
v v
v v+
V
w
y
∂
+
∂
v
.̂
V u
i
z t
∂
∂
∂
=
∂
v
 For steady flow ∂ ∂u t a/ .= =0 0 so that v 
 
3.20 Assume u(r,x) and v(r,x) are not zero. Then, replacing z with x in the appropriate 
equations of Table 3.1 and recognizing that v θ ∂ ∂θ= =0 0 and / : 
 r x
v v u u
a v u a v u
r x r x
∂ ∂ ∂ ∂
∂ ∂ ∂ ∂
= + = + 
 
3.21 a) u e tt= − − = = ∞−2 1 0 1 10( )( ) ./ 2 m / s at 
 ( )a u
t
e tx
t= = − 



= =−
∂
∂
2 1 0
1
10
0 2 010( ) . ./ m / s at 2 
 b) u e tt= − − = = ∞−2 1 0 12 10( .5 )( ) ./ 1.875 m / s at 
 a e tx
t= − 



= =−2 1 0 2
1
10
0 0125 02 2 10( .5 / ) . ./ m / s at 2 
 c) u e tt= − − =−2 1 2 2 12 2 10( / )( ) ./ 0 for all 
 a e tx
t= − 



=−2 1 2 2
1
10
02 2 10( / ) ./ for all 
 
3.22 DT Tu
Dt x
∂
∂
= v+
T
w
y
∂
∂
+ 220(1 ) sin 0.5878
100 100 5
T T t
y
z t
∂ ∂ π π π
∂ ∂
 + = − − = − × 
 
 
 = −0.3693 °C/s. 
 
3.23 D
Dt
u
x
v
y
w
z t
e
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = − × − − ×
−
10 1 23 10 4 3000 10
4
( . ) = − ×
⋅
−9 11 10 4. .
kg
m s3
 
 
3.24 D
Dt
u
x
v
y
w
z t
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = −



10
1000
4
= −
⋅
2500
kg
m s3
. 
 
3.25 D
Dt
u
x
ρ ∂ρ
∂
= = ×4 01(. ) = 0.04 kg/m3⋅s 
 
3.26 (D) 22
10
[10(4 ) ]
(4 )
x
u u u u u
a u v w u x
t x y z x xx
−∂ ∂ ∂ ∂ ∂ ∂= + + + = = −
∂ ∂ ∂ ∂ ∂ ∂−
 
 3 22
10 10 1
10( 2)( 1)(4 ) 20 6.25 m/s .
4 8(4 )
x
x
−= − − − = × × =
−
 
3.27 D
Dt
V
t
= ⋅ ∇ +
v v ∂
∂
 observing that the dot product of two vectors
v
A A i A j A kx y z= + +$ $
$ 
 and 
v v v
B B i B j B k A B A B A B A Bx y z x x y y z z= + + ⋅ = + +$ $ $ . is 
 40 
 
3.28 
a
u
t
V u
a
v
t
V v
a
w
t
V w
a
V
t
V V
x
y
z
= + ⋅∇
= + ⋅∇
= + ⋅∇








∴ = + ⋅∇
∂
∂
∂
∂
∂
∂
∂
∂
v v
v v
v v
v
v
v v v
 ( ) 
 
3.29 Using Eq. 3.2.12: 
 a) 
v v
v v v v v v
v
v
A a
d s
dt
V r
d
dt
r= + + × + × × + ×
2
2 2Ω Ω Ω
Ω
( ) 
 = 2 20 4 20 20 1( $ $) $ ( $ .5$)k i k k i× + × × = 160 600$ $j i− m / s2 
 b) 
v v v v v v oA V r k j k k i= × + × × = × − + × ×2 2 20 20 30 20 20 3Ω Ω Ω( ) ( $ cos $) $ ( $ $) = −507$i 
3.30 
v
Ω =
× ×
= × −
2
24 60 60
7 272 10 5
π $ . $k k rad/s. 
 vv i k i k= − − = − −5 707 707 3 3( . $ . $) .535$ .535 $ m/s. 
 
v v v v v v
A V r= × + × ×2Ω Ω Ω( ) 
 = 2 7 272 10 3 3 7 272 105 5× × × − − + × ×− −. $ ( .535$ .535 $) . $k i k k 
 [ . $ ( . $ . $)]7 272 10 6 10 707 7075 6× × × − +− k i k = − × +−51 4 10 0 02245. $ . $j i m / s 2 . 
Note: We have neglected the acceleration of the earth relative to the sun since it is quite small 
(it is d s dt2 2v / ). The component 5 ˆ( 51.4 10 )j−− × is the Coriolis acceleration and causes air 
motions to move c.w. or c.c.w. in the two hemispheres. 
 
3.31 a) two-dimensional (r, z) b) two-dimensional (x, y) 
 c) two-dimensional (r, z)d) two-dimensional (r, z) 
 e) three-dimensional (x, y, z) f) three-dimensional (x, y, z) 
g) two-dimensional (r, z) h) one-dimensional (r) 
 
3.32 Steady: a, c, e, f, h Unsteady: b, d, g 
 
3.33 b. It is an unsteady plane flow. 
 
3.34 a) d) e) 
 
 
 
3.35 f, h 
 41 
 
3.36 a) inviscid. b) inviscid. c) inviscid. 
 d) viscous inside the boundary layer. 
e) viscous inside the boundary layers and separated regions. 
 f) viscous. g) viscous. h) viscous. 
 
3.37 d and e. Each flow possesses a stagnation point. 
 
3.38 
 
 
3.39 (C) The only velocity component is u(x). We have neglected v(x) since it is 
 quite small. If v(x) in not negligible, the flow would be two-dimensional. 
 
3.40 Re = V L / ν = 2 × .015/.77 × 10-6 = 39 000. ∴Turbulent. 
 
3.41 Re = =VL
ν
.2 × .8/1.4 × 10-5 = 11 400. ∴Turbulent. 
 
3.42 Re .
.
= =
×
× −
VL
ν
4 06
1 7 10 5
 = 14 100. ∴Turbulent. 
 Note: We used the smallest dimension to be safe! 
 
3.43 a) Re . .
.51
= =
×
×
=−
V D
ν
1 2 0 01
1 10
795.5 Always laminar. 
 
 b) Re .
.51
= =
×
×
=−
VD
ν
1 2 1
1 10
795 500. May not be laminar. 
 
3.44 Re = 3 × 105 = T
Vx
ν
. ν µ ρ= / where µ µ= ( ).T 
 a) T = 223 K or −50°C. ∴ = × ⋅−µ 1 10 5.5 N s / m .2 
 ∴ = ×
×
= ×
−
−ν
1 5 10
3376 1 23
2 5 10
5
5.
. .
. . m /s2 
 3 10 900 1000
3600 2 10
5
5× =
×
× × −
xT
.5
. ∴xT = 0.03 m or 3 cm 
 b) T = −48°F. ∴µ = 3.3 × 10−7 lb-sec/ft2. ν = × = ×
−
−3 3 10
00089
3 7 10
7
4.
.
. ft2/sec. 
 3 10 600 5280
3600 3 7 10
5
4× =
×
× × −
x T
.
. ∴xT = 0.13' or 1.5" 
 
 42 
3.45 Assume the flow is parallel to the leaf. Then 3 × 105 = Vx T / .ν 
 ∴ = × = × × × =−x VT 3 10 3 10 1 4 10 6 8 17
5 5 4ν / .5 . / . m . 
 The flow is expected to be laminar. 
 
3.46 a) M V
c
= =
× ×
=
100
14 287 236
0 325
.
. . For accurate calculations the flow is 
 compressible. Assume incompressible flow if an error of 4%, or so, is 
 acceptable. 
 b) M V
c
= =
× ×
=
80
14 287 288
0 235
.
. . ∴Assume incompressible. 
 c) M V
c
= =
× ×
=
100
14 287 373
0 258
.
. . ∴Assume incompressible. 
3.47 D
Dt
u
x
v
y
w
z t
ρ ∂ρ
∂
∂ρ
∂
∂ρ
∂
∂ρ
∂
= + + + = 0. For a steady, plane flow 
 ∂ρ ∂/ t = 0 and w = 0. Then 
 u
x
v
y
∂ρ
∂
∂ρ
∂
+ = 0. 
 
3.48 D u
Dt x
ρ ∂ρ
∂
= v
y
∂ρ
∂
+ w+
z t
∂ρ ∂ρ
∂ ∂
+ 0.= ∴incompressible. 
 
3.49 (B) 
2 9810 0.800
. 113 m/s.
2 1.23
water
air
hV p
V
γ
ρ ρ
×
= = = ∴ = 
 
3.50 V
p2
2
=
ρ
. Use ρ = 0.0021 slug/ft3. 
 a) v p= = × ×2 2 3 144 0021/ . / .ρ = 203 ft/sec. 
 b) v p= = × ×2 2 9 144 0021/ . / .ρ = 351 ft/sec. 
 c) v p= = × ×2 2 09 144 0021/ . / .ρ = 111 ft/sec. 
 
3.51 p V= = ×

 

ρ
2 2
2
123
120 1000
3600
2. / = 683 Pa. 
 ∴F = pA = 683 π × 0.0752 = 12.1 N. 
 
3.52 V
p2
2
0+ =
ρ
. ∴ =
−
=
×
V
p2 2 2000
123ρ .
 = 57.0 m/s 
 43 
3.53 (C) 
2 2 2
1 2 1. 0.200 0.600. 2 9.81 0.400 2.80 m/s.
2 2 2
V V Vp
V
g g gγ
+ = + = ∴ = × × = 
 
3.54 (B) The manometer reading h implies: 
2 2
21 1 2 2
2 2
2
 or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
ρ ρ
+ = + = − ∴ = The 
temperature (the viscosity of the water) and the diameter of the pipe are 
not needed. 
 
3.55 a) 
22
0
2 2
VV p
ρ
+ =
2
2( 10 ). . 50
2
o o
o
p px p
p p x ρ
ρ ρ ρ
−
+ + = ∴ = − 
 b) 
22
0
2 2
VV p
ρ
+ =
2
2(10 ). . 50
2
o o
o
p py p
p p y ρ
ρ ρ ρ
+ + = ∴ = − 
3.56 
22
2 2
U pV p
ρ ρ
∞ ∞+ = + . 
 a) v v U r rr cθ θ= = = − −∞0 180 1 1
2 2 and o: ( / )( ). 
 ( )∴ = − = − 










∞ ∞p U v U
r
r
r
rr
c cρ ρ
2 2
22 2 2
2
2
4
. 
 b) Let r r p Uc T= = ∞: 
ρ
2
2 
 c) ( ) [ ]v r r v U p v Ur c= = − ∴ = − = −∞ ∞ ∞0 2 2 2 1 4
2 2 2 and = U 2: sin . sinθ θθ
ρ ρ
θ 
 d) Let θ ρ= = − ∞90
3
290
2o : p U 
 
3.57 
22
2 2
U pV p
ρ ρ
∞ ∞+ = + . 
 a) v θ θ= =0 180 and 
o : ( )p U v U r
r
r
rr
c c= − = 



− 











∞ ∞
ρ ρ
2 2
22 2 2
3 6
. 
 b) Let r r p Uc T= = ∞: 
1
2
2ρ . 
 c) ( ) [ ]v r r p v Ur c= = − = −∞ ∞0 2 2 1 4
2 2 2 and = U 2: sin
ρ ρ
θθ 
 d) Let θ ρ= = − ∞90
3
290
2o : p U 
 
 44 
3.58 
22
2 2
U pV p
ρ ρ
∞ ∞+ = + . 
 a) ( )p U u
x x
= − = − +








 = − +









∞
ρ ρ π
π
ρ
2 2
10 10
20
2
50 1 1
12 2 2
2 2
 
 = − +



50
2 1
2ρ x x
 
 b) 10 when 1. 50 ( 2 1) 50u x p ρ ρ−= = − = − − + = 
 c) ( )
2 2
2 2 2 60 130 30 450 1 1
2 2 2
p U u
x x
ρ ρ π
ρ
π∞
      = − = − + = − +      
         
= − +



450
2 1
2ρ x x
 
 d) 10 when 1. 450 ( 2 1) 450u x p ρ ρ−= = − = − − + = 
 
3.59 
V p V p
V p p1
2
1 2
2
2
1 1 22 2
0 20+ = + = − =
ρ ρ
. and kPa. 
 ( )V p p V22 1 2 2
2 2
1000
20 6 32= − = ∴ =
ρ
( . 000) = 40. m / s 
 
3.60 Assume the velocity in the plenum is zero. Then 
 
2 2
21 1 2 2
2 2
2
 or (60 10.2). 9.39 m/s
2 2 1.13
V p V p
V V
ρ ρ
+ = + = − ∴ = 
 We found ρ = 113. kg / m 3 in Table B.2. 
 
3.61 Bernoulli from the stream to the pitot probe: p V pT = +ρ
2
2
. 
 Manometer: .T Hgp H H h p hγ γ γ γ+ − − = − 
 Then, 
2
2 Hg
V
p H H pρ γ γ+ + − = . 2 (2 )HgV H
γ γ
ρ
−
∴ = 
 
 a) V V2 13 6 1 9800
1000
2 0 04 3 14=
−
× ∴ =
( . )
( . ). . m / s 
 b) V V2 13 6 1 9800
1000
2 0 1 4 97=
−
× ∴ =
( . )
( . ). . m / s 
 c) V V2 13 6 1 62 4
1 94
2 2 12 11 62=
−
× ∴ =
( . ) .
.
( / ). . fps 
 d) V V2 13 6 1 62 4
1 94
2 4 12 16 44=
−
× ∴ =
( . ) .
.
( / ). . fps 
 
 45 
3.62 The pressure at 90° from Problem 3.56 is 290 3 /2.p Uρ ∞= − The pressure at the 
 stagnation point is 2 /2.Tp Uρ ∞= The manometer provides: p H pT − =γ 90 
 2 21 31.204 9800 0.04 1.204 . 12.76 m/s
2 2
U U U∞ ∞ ∞× − × = − × ∴ = 
3.63 The pressure at 90° from Problem 3.57 is 290 3 /2.p Uρ ∞= − The pressure at the 
 stagnation point is 2 /2.Tp Uρ ∞= The manometer provides: p H pT − =γ 90 
 2 21 31.204 9800 0.04 1.204 . 12.76 m/s
2 2
U U U∞ ∞ ∞× − × = − × ∴ = 
 
3.64 Assume an incompressible flow with point 1 outside in the room where p1 0= 
 and v 1 0= . The Bernoulli’s equation gives, with p hw2 2= γ , 
 
2
1
2
V 1p
ρ
+
2
2 2 .
2
V p
ρ
= + 
 a) 0
2
9800 0 02
1 204
18 042
2
2= +
− ×
∴ =
V
V
.
.
. . m / s 
 b) 0
2
9800 0 08
1 204
36 12
2
2= +
− ×
∴ =
V
V
.
.
. . m / s 
 c) 0
2
62 4 1 12
0 00233
66 82
2
2= +
− ×
∴ =
V
V
. /
.
. . fps 
 d) 0
2
62 4 4 12
0 00233
133 62
2
2= +
− ×
∴ =
V
V
. /
.
. . fps 
 
3.65 Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel
 where p V1 10 0= = and . Bernoulli’s equation gives 
 0
2
1
2
2
2
2
2 2
2= + ∴ = −
V p
p V
ρ
ρ. 
 a) ρ = =
×
= ∴ = − × × = −
p
RT
p
90
0 287 253
1 239
1
2
1 239 100 61952
2
.
. . . kg / m Pa3 
 b) ρ = =
×
= ∴ = − × × = −
p
RT
p
95
0 287 273
1 212
1
2
1 212 100 60602
2
.
. . . kg / m Pa3 
 c) ρ = =
×
= ∴ = − × × = −
p
RT
p
92
0 287 293
1 094
1
2
1 094 100 54702
2
.
. . . kg / m Pa3 
 d) ρ = =
×
= ∴ = − × × = −
p
RT
p
100
0 287 313
1 113
1
2
1113 100 55662
2
.
. . . kg / m Pa3 
 
3.66 (A) 
2
1
2
V
g
2
1 2 2
2
p V p
gγ γ
+ = +
2
2
2
800000
. . 40 m/s.
9810 2 9.81
V
V= ∴ =
×
 
 46 
 
3.67 a) p h V h hA A A= = × = = =γ 9800 4 39 0 2 200 Pa, Using . , 
 
2
2
AV
g
A
A
p
h
γ
+ +
2
2 2
22
V p
h
g γ
= + +
2
2
2. 2A
V
p p
g
γ= − 
 = −
×
× = −39
14
2 9 81
9800 58
2
 200 700 Pa
.
 
 
 b) 0 and 0.B Bp V= = Bernoulli’s eq. gives, with the datum through the pipe, 
 V
g
p
h
V
g
p
h pB B B
2
2
2
2
2 2
2
2 2
4
14
2 9 81
9800 58+ + = + + = −
×





 = −
γ γ
.
.
 700 Pa 
 
3.68 Bernoulli: 
2
2 2
2
V p
g γ
+
2
1 1
2
V p
g γ
= + 
 Manometer: 
2
2
1 22Hg
V
p z H H z p
g
γ γ γ γγ+ + − − = + 
 Substitute Bernoulli’s into the manometer equation: 
 ( )
2
1
1 1.2Hg
V
p H p
g
γ γ γ+ − = + 
 a) Use H = 0.01 m: 
V
V1
2
1
9800
2 9 81
13 6 1 9800 0 01 1
×
×
= − × ∴ =
.
( . ) . .572 m / s 
 Substitute into Bernoulli: 
 p
V V
g1
2
2
1
2 2 2
2
20 1
2 9 81
9800 198=
−
=
−
×
× =γ
.572
.
 600 Pa 
 b) Use H = 0.05 m: 
V
V1
2
1
9800
2 9 81
13 6 1 9800 0 05 3
×
×
= − × ∴ =
.
( . ) . .516 m / s 
 Substitute into Bernoulli: 
 p
V V
g1
2
2
1
2 2 2
2
20 3
2 9 81
9800 193=
−
=
−
×
× =γ
.516
.
 600 Pa 
 c) Use H = 0.1 m: 
V
V1
2
1
9800
2 9 81
13 6 1 9800 0 1 4 972
×
×
= − × ∴ =
.
( . ) . . m / s 
 Substitute into Bernoulli: 
 p
V V
g1
2
2
1
2 2 2
2
20 4 972
2 9 81
9800 187=
−
=
−
×
× =γ
.
.
 400 Pa 
 
 47 
3.69 Bernoulli across nozzle: 
2
1
2
V 21 2 2
2
p V p
ρ ρ
+ = + 2 1. 2 /V p ρ∴ = 
 Bernoulli to max. height: 
2
1
2
V
g
1
1
p
h
γ
+ +
2
2
2
V
g
= 2
p
γ
+ 2 2 1. / .h h p γ+ ∴ = 
 
a) V p2 12 2 700 1000 37 42= = × =/ / .ρ 000 m / s 
 h p2 1 700= =/ γ 000 / 9800 = 71.4 m 
 
b) V p2 12 2 1 1000 52 92= = × =/ / .ρ 400 000 m / s 
 h p2 1= =/ γ 1 400 000 / 9800 = 142.9 m 
 
 c) V p2 12 2 100 1 94 121 8= = × × =/ / . .ρ 144 fps 
 h p2 1= = ×/ γ 100 144 / 62.4 = 231 ft 
 
 d) V p2 12 2 200 1 94 172 3= = × × =/ / . .ρ 144 fps 
 h p2 1 200= = ×/ γ 144/ 62.4 = 462 ft 
 
3.70 a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream 
 flow: 
 
2
1
2
V
g
1p
γ
+
2
2 2
1 2
V p
h
g γ
+ = + 2 2. 2 ( )h V g H h+ ∴ = − 
 
 b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the 
 bottom of the downstream flow: 
 
2
1
2
V
g
2
1 2 2
1 2. 2
p V p
h h
gγ γ
+ + = + + 
 Using p H p h h h V g H h1 2 1 2 2 2= = = = −γ γ, , ( ) and 
 
3.71 
2
1
2
V 21 2 2 .
2
p V p
ρ ρ
+ = + p2 = −100 000 Pa, the lowest possible pressure. 
 a) 600
2
2
2 000
1000
100 000
1000
= −
V
. ∴ V2 = 37.4 m/s. 
 
 b) 300
2
2
2 000
1000
100 000
1000
= −
V
. ∴ V2 = 28.3 m/s. 
 48 
 c) 80 144
1.94
14.7 144
1.94
×
= −
×V2
2
2
. ∴ V2 = 118.6 ft/sec. 
 
 d) 40 144
1.94
14.7 144
1.94
×
= −
×V2
2
2
. ∴ V2 = 90.1 ft/sec. 
 
3.72 A water system must never have a negative pressure, since a leak could ingest 
 impurities. ∴ The least pressure is zero gage. 
 
V p
gz
V p
gz1
2
1
1
2
2
2
22 2
+ + = + +
ρ ρ
. V V1 2= . Let z1 0= , and p2 0= . 
 500 000
1000
 = 9 81 2. .z ∴ z2 = 51.0 m. 
 
3.73 a) ( ) ( )p V V1 22 12 2 22
1000
2
2 10= − = −
ρ = −48 000 Pa 
 b) ( ) ( )2 2 2 21 2 1 902 2 10 43300 Pa2 2p V V
ρ
= − = − = − 
 c) ( ) ( )2 2 2 21 2 1 680 2 10 32600 Pa2 2p V V
ρ
= − = − = − 
 d) ( ) ( )2 2 2 21 2 1 1.23 2 10 59.0 Pa2 2p V V
ρ
= − = − = − 
 
3.74 
V p V p1
2
1 2
2
2
2 2
+ = +
ρ ρ
. ( ) ( )2 2 2 21 2 1 1.23 2 82 2p V V
ρ
= − = − = −36.9 Pa 
 
3.75 (D) ( ) ( )2 2 2 21 2 1 902 30 15 304400 Pa2 2p V V
ρ
= − = − = 
 
3.76 Apply Bernoulli’s equation between the exit (point 2) where the radius is R and 
 a point 1 in between the exit and the center of the tube at a radius r less than R: 
 
V p V p
p
V V1
2
1 2
2
2
1
2
2
1
2
2 2 2
+ = + ∴ =
−
ρ ρ
ρ. . 
Since V V2 1< , we see that p1 is negative (a vacuum) so that the envelope would
 tend to rise due to the negative pressure over most of its area (except for a small 
area near the end of the tube). 
 
3.77 Re .= V D
ν
 For air ν ≅ × −1 10 5.5 . Use reasonable dimensions from your 
experience! 
 49 
 a) Re .
.5
.=
×
×
= ×−
20 0 03
1 10
4 105
4 ∴Separate 
 b) Re .
.5
.=
×
×
=−
20 0 005
1 10
67005 ∴Separate 
 c) Re
.5
. .=
×
×
= ×−
20 2
1 10
2 7 105
6 ∴Separate 
 d) Re .
.5
.=
×
×
=−
5 0 002
1 10
6705 ∴Separate 
 e) Re
.5
. .=
×
×
= ×−
20 2
1 10
2 7 105
6 ∴Separate 
 f) Re
.5
.=
×
×
= ×−
100 3
1 10
2 105
7 
∴It will tend to separate, except streamlining the components 
 eliminates separation. 
 
3.78 A burr downstream of the opening will create a region that 
 acts similar to a stagnation region thereby creating a high 
 pressure since the velocity will be relatively low in that region. 
 
 
3.79 ∆ ∆p V
R
n= = × =ρ
2 2
1000
10
0 05
0 02
.
. 40 000 Pa Along AB, we 
 expect V VA B> <10 10 m / s and m /s. 
 
 
3.80 The higher pressure at B will force the fluid toward the lower 
 pressure at A, especially in the wall region of slow moving 
 fluid, thereby causing a secondary flow normal to the pipe’s 
 axis. This results in a relatively high loss for an elbow. 
 
 
3.81 Refer to Bernoulli’s equation: 
V p V p1
2
1 2
2
2
2 2
+ = +
ρ ρ
 
 p pA B> since V VA B< 
 
 p pC D< since V VC D> 
 
 p pB D> since V VD B> 
stagnation
 region
A
B
VA
VB
 50 
CHAPTER 4 
 
The Integral Forms of the 
Fundamental Laws 
 
4.1 a) No net force may act on the system: Σ
v
F = 0. 
 b) The energy transferred to or from the system must be zero: Q - W = 0. 
c) If 3 3 2 ˆ ˆˆ 10 ( ) 0nV V n i j= ⋅ = ⋅ − =
v
is the same for all volume elements then 
Σ
v v
F
D
Dt
V dm= ∫ , or Σ
v v
F
D
Dt
mV= ( ). Since mass is constant for a system 
Σ
v
v
F m
DV
Dt
= . Since DV
Dt
a F ma
v
v v v
= =, . Σ 
 
4.2 Extensive properties: Mass, m; Momentum, mV
v
; kinetic energy, 1
2
mV 2 ; 
 potential energy, mgh; enthalpy, H. 
 Associated intensive properties (divide by the mass): unity, 1; velocity, 
v
V ;V2/2; 
 gh; H/m = h (specific enthalpy). 
 Intensive properties: Temperature, T; time, t; pressure, p; density, ρ; viscosity, µ. 
 
4.3 (B) 
 
4.4 
 
System ( )t V= 1
c.v.( )t V= 1
System ( )t t V+ ∆ = 1 V+ 2
c.v.( )t t V+ ∆ = 1
 
 
 
 
4.5 
 
System ( )t V= 1 V+ 2
c.v.( )t V= 1 V+ 2
System ( )t t V+ ∆ = 2 V+ 3
c.v.( )t t V+ ∆ = 1 V+ 2
 
 
1 2
1
2 3
pump
 51 
4.6 a) The energy equation (the 1st law of Thermo). 
 b) The conservation of mass. 
 c) Newton’s 2nd law. 
 d) The energy equation. 
 e) The energy equation. 
 
4.7 
 
 
 
 
4.8 
 
 
4.9 
 
4.10 $ $ $ . ($ $)n i j i j1
1
2
1
2
0 707= − − = − + . $ . $ .5 $n i j2 0 866 0= − . $ $n j3 = − . 
 1 1 1 ˆ ˆ ˆˆ 10 [ 0.707( )] 7.07 fpsnV V n i i j= ⋅ = ⋅ − + = −
v
 
 V V n i i jn2 2 2 10 0866 05 866= ⋅ = ⋅ − =
v
$ $ ( . $ . $) . fps 
 3 3 2 ˆ ˆˆ 10 ( ) 0nV V n i j= ⋅ = ⋅ − =
v
 
 
 
 
4.11 flux = ηρ $n VA⋅
v
 
 flux1 = ηρ ηρ[ . ($ $)] $ / .− + ⋅ = −0 707 10 0 707 10i j iA A 
 flux2 = ηρ ηρ( . $ .5$) $ / .0 866 0 10 0 866 10i j iA A− ⋅ = 
 flux3 = ηρ( $) $− ⋅ =j iA10 03 
 
 
n̂
v n̂
v
ωn̂ v n̂
v
n̂
v
n̂
v n̂n̂
n̂
v v
v
v
n̂n̂
n̂
n̂
v
v
v
v
n̂ n̂ v
v
 52 
4.12 ( $) ( .5$ . $) $( )
v
B n A i j j⋅ = + ⋅ × 15 0 0 866 10 12 
 = × × =15 0 866 120 1559. cm 3 
 Volume = 15 60 10 12 1559 sin cm 3o × × = 
 
4.13 The control volume must be independent of time. Since all space coordinates are 
integrated out on the left, only time remains; thus, we use an ordinary derivative 
to differentiate a function of time. But, on the right, we note that ρ and η may be 
functions of (x, y, z, t); hence, the partial derivative is used. 
 
4.14 
 
4.15 
 
 
4.16 
 
4.17 If fluid crosses the control surface only on areas A1 and A2, 
 ρ ρ ρ$ $ $
. .
n VdA n VdA n V dA
AAc s
⋅ = ⋅ + ⋅ =∫∫∫
v v v
0
21
 
 For uniform flow all quantities are constant over each area: 
 ρ ρ1 1 1 2 2 2 0
21
$ $n V dA n V dA
AA
⋅ + ⋅ =∫∫
v v
 
 Let A1 be the inlet so $n V V A1 1 1 2⋅ = −
v
 and be the outlet so $ .n V V2 2 2⋅ =
v
 Then 
 − + =ρ ρ1 1 1 2 2 2 0V A V A 
 or 
 ρ ρ2 2 2 1 1 1A V A V= 
1
2
1system (∆t) is in
volumes 1 and 2
c.v. (0) = c.v. (∆t)
 = volume 1
1
2
3
system (∆t) = V1 + V2 + V3
c.v. (∆t) = V1 + V2
system boundary
 at (t + ∆t)
 53 
 
4.18 Use