Cap. 03 Mecânica dos Fluídos - Fox - McDonald (8ª Edição) Solution PDF

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Given: Data on nitrogen tankFind: Pressure of nitrogen; minimum required wall thicknessAssumption: Ideal gas behaviorSolution:Ideal gas equation of state: pV⋅ =MR⋅ ⋅Twhere, from Table A, for nitrogen R 55.ft lbf⋅lbm R⋅= ⋅Then the pressure of nitrogen is pMR⋅ ⋅TV= MR⋅ ⋅T6πD3⋅⎛⎜⎝⎞⎟⎠= ⋅p 140 lbm⋅ ×55.f...

Description

Problem 3.1

[Difficulty: 2]

Given:

Data on nitrogen tank

Find:

Pressure of nitrogen; minimum required wall thickness

Assumption:

Ideal gas behavior

Solution: Ideal gas equation of state:

p ⋅V = M⋅R⋅T

where, from Table A.6, for nitrogen

R = 55.16⋅

Then the pressure of nitrogen is

p =

ft ⋅ lbf lbm⋅ R

M⋅ R⋅ T

= M⋅ R⋅ T⋅ ⎛⎜

3⎟ ⎝ π⋅ D ⎠

V

p = 140⋅ lbm × 55.16 ⋅

p = 3520 ⋅

6 ⎞

ft ⋅ lbf × ( 77 + 460)⋅ R × ⎡⎢ lbm⋅ R

⎤ × ⎛ ft ⎞ 3⎥ ⎝ 12⋅ in ⎠ ⎣ π × (2.5⋅ ft ) ⎦ 6

lbf 2

in

To determine wall thickness, consider a free body diagram for one hemisphere: π⋅D

ΣF = 0 = p ⋅

4

2

− σc ⋅ π ⋅ D ⋅ t

pπ D2/4

where σc is the circumferential stress in the container Then

t=

p⋅ π⋅ D

2

4 ⋅ π ⋅ D ⋅ σc

t = 3520 ⋅

lbf 2

in t = 0.0733⋅ ft

=

σc πDt

p⋅ D 4 ⋅ σc 2

×

in 2.5 ⋅ ft × 3 4 30 × 10 ⋅ lbf t = 0.880⋅ in

2

Problem 3.2

[Difficulty: 2]

Given: Pure water on a standard day Find:

Boiling temperature at (a) 1000 m and (b) 2000 m, and compare with sea level value.

Solution: We can determine the atmospheric pressure at the given altitudes from table A.3, Appendix A

The data are

Elevation (m) 0 1000 2000

p/p o

p (kPa)

1.000 0.887 0.785

101.3 89.9 79.5

We can also consult steam tables for the variation of saturation temperature with pressure: p (kPa) 70 80 90 101.3

T sat (°C) 90.0 93.5 96.7 100.0

We can interpolate the data from the steam tables to correlate saturation temperature with altitude: Elevation (m) 0 1000 2000

p (kPa) T sat (°C)

p/p o 1.000 0.887 0.785

101.3 89.9 79.5

The data are plotted here. They show that the saturation temperature drops approximately 3.4°C/1000 m.

100.0 96.7 93.3

Saturation Temperature (°C)

Variation of Saturation Temperature with Pressure Sea Level 100

1000 m

98 96

2000 m

94 92 90 88 70

75

80

85

90

95

Absolute Pressure (kPa)

100

105

Problem 3.3

[Difficulty: 2]

Given:

Data on flight of airplane

Find:

Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop."

Solution: Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 kg ρSL = 1.225 ⋅ 3 m

ρair = 0.7423 ⋅ ρSL

kg ρair = 0.909 3 m

We also have from the manometer equation, Eq. 3.7 and also

∆p = −ρair ⋅ g ⋅ ∆z Combining

∆hHg =

ρair ρHg

∆hHg =

⋅ ∆z =

ρair ⋅ ∆z SGHg ⋅ ρH2O

0.909 × 100 ⋅ m 13.55 × 999

∆p = − ρHg ⋅ g ⋅ ∆hHg SGHg = 13.55 from Table A.2

∆hHg = 6.72⋅ mm

For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m. From table A.3

ρair = 0.4292 ⋅ ρSL

kg ρair = 0.526 3 m

We also have from the manometer equation ρair8000 ⋅ g ⋅ ∆z8000 = ρair3000 ⋅ g ⋅ ∆z3000 where the numerical subscripts refer to conditions at 3000m and 8000m. Hence ∆z8000 =

ρair3000 ⋅ g ρair3000 ⋅ ∆z3000 = ⋅ ∆z3000 ρair8000 ⋅ g ρair8000

∆z8000 =

0.909 × 100 ⋅ m 0.526

∆z8000 = 173 m

Problem 3.4

[Difficulty: 3]

Given: Boiling points of water at different elevations Find: Change in elevation Solution: From the steam tables, we have the following data for the boiling point (saturation temperature) of water Tsat (oF) 195 185

p (psia) 10.39 8.39

The sea level pressure, from Table A.3, is pSL =

14.696

psia

Hence

Altitude vs Atmospheric Pressure p/pSL

195 185

0.707 0.571

15000 12500

Altitude (ft)

Tsat (oF)

From Table A.3 p/pSL

Altitude (m)

Altitude (ft)

0.7372 0.6920 0.6492 0.6085 0.5700

2500 3000 3500 4000 4500

8203 9843 11484 13124 14765

Data

10000

Linear Trendline

7500

z = -39217(p/pSL) + 37029 5000

R2 = 0.999

2500 0.55

0.60

0.65

0.70

p/pSL

Then, any one of a number ofExcel functions can be used to interpolate (Here we useE xcel 's Trendline analysis) p/pSL

Altitude (ft)

0.707 0.571

9303 14640

Current altitude is approximately

The change in altitude is then 5337 ft Alternatively, we can interpolate for each altitude by using a linear regression between adjacent data points p/pSL For

0.7372 0.6920

Altitude (m) 2500 3000

Then

0.7070

2834

Altitude (ft) 8203 9843 9299

The change in altitude is then 5338 ft

p/pSL 0.6085 0.5700 0.5730

Altitude (m) 4000 4500 4461

Altitude (ft) 13124 14765 14637

9303 ft

0.75

Problem 3.5

Given:

Data on system

Find:

Force on bottom of cube; tension in tether

[Difficulty: 2]

Solution: dp

Basic equation

dy

= − ρ⋅ g

∆p = ρ⋅ g⋅ h

or, for constant ρ

where h is measured downwards

The absolute pressure at the interface is

pinterface = patm + SGoil⋅ ρ⋅ g⋅ hoil

Then the pressure on the lower surface is

pL = pinterface + ρ⋅ g ⋅ hL = patm + ρ⋅ g⋅ SGoil⋅ hoil + hL

For the cube

(

V = 125⋅ mL

V = 1.25 × 10

−4

)

3

⋅m

1 3

Then the size of the cube is

d = V

d = 0.05 m

Hence

hL = hU + d

hL = 0.35 m

The force on the lower surface is

FL = pL⋅ A

where

(

and the depth in water to the upper surface is hU = 0.3⋅ m where hL is the depth in water to the lower surface 2

A = d

2

A = 0.0025 m

)

FL = ⎡patm + ρ⋅ g⋅ SGoil ⋅ hoil + hL ⎤ ⋅ A ⎣ ⎦

⎡ N⋅ s ⎥⎤ m kg 3 N 2 + 1000 ⋅ FL = ⎢101 × 10 ⋅ × 0.0025⋅ m × 9.81⋅ × (0.8 × 0.5⋅ m + 0.35⋅ m) × 2 3 2 ⎢ ⎥ kg ⋅ m s m m ⎣ ⎦ 2

FL = 270.894 N For the tension in the tether, an FBD gives

Note: Extra decimals needed for computing T later!

ΣFy = 0

FL − FU − W − T = 0

(

)

where FU = ⎡patm + ρ⋅ g ⋅ SGoil ⋅hoil + hU ⎤⋅ A ⎣ ⎦

or

T = FL − FU − W

Note that we could instead compute Using FU

(

)

∆F = FL − FU = ρ⋅ g ⋅SGoil ⋅ hL − hU ⋅ A

T = ∆F − W

⎡ N⋅ s ⎤⎥ m kg 3 N 2 + 1000⋅ FU = ⎢101 × 10 ⋅ × 0.0025 ⋅m × 9.81⋅ × (0.8 × 0.5⋅ m + 0.3⋅ m) × 2 3 2 ⎢ ⎥ kg ⋅ m s m m ⎣ ⎦ 2

FU = 269.668 N For the oak block (Table A.1)

and

Note: Extra decimals needed for computing T later!

SGoak = 0.77

W = 0.77 × 1000 ⋅

W = SG oak⋅ ρ⋅ g⋅ V

so

kg 3

m T = F L − FU − W

× 9.81⋅

m s

2

× 1.25 × 10

T = 0.282 N

−4

3

⋅m ×

2

N ⋅s kg⋅ m

W = 0.944 N

Problem 3.6

Given:

Data on system before and after applied force

Find:

Applied force

[Difficulty: 2]

Solution: Basic equation

dp dy

(

p = patm − ρ⋅ g ⋅ y − y0

or, for constant ρ

= − ρ⋅ g

For initial state

p1 = patm + ρ⋅ g⋅ h

For the initial FBD

ΣFy = 0

For final state

p2 = patm + ρ⋅ g⋅ H

For the final FBD

ΣFy = 0

and

)

F1 = p1⋅ A = ρ⋅ g⋅ h⋅ A

F1 − W = 0

( )

with

p y0 = patm

(Gage; F1 is hydrostatic upwards force)

W = F1 = ρ⋅ g⋅ h⋅ A

and

F2 = p2⋅ A = ρ⋅ g⋅ H⋅ A

F2 − W − F = 0

(Gage; F2 is hydrostatic upwards force)

F = F 2 − W = ρ⋅ g⋅ H⋅ A − ρ⋅ g⋅ h ⋅ A = ρ⋅ g⋅ A⋅ (H − h )

2

π ⋅D F = ρH2O⋅ SG⋅ g ⋅ ⋅ ( H − h) 4

From Fig. A.1

SG = 13.54

F = 1000 ⋅

kg 3

m F = 45.6 N

× 13.54 × 9.81 ⋅

m s

2

×

π 4

2

× (0.05⋅ m) × ( 0.2 − 0.025) ⋅ m ×

2

N ⋅s kg⋅ m

Problem 3.7

[Difficulty: 1]

Given: Pressure and temperature data from balloon Find: Plot density change as a function of elevation Assumption: Ideal gas behavior Solution:

Density Distribution 0.078

Using the ideal gas equation, ρ = p/RT

p (psia) 14.71 14.62 14.53 14.45 14.36 14.27 14.18 14.10 14.01 13.92 13.84

T (oF)

ρ (lbm/ft3)

53.6 52.0 50.9 50.4 50.2 50.0 50.5 51.4 52.9 54.0 53.8

0.07736 0.07715 0.07685 0.07647 0.07604 0.07560 0.07506 0.07447 0.07380 0.07319 0.07276

Density (lbm/ft3)

0.077 0.076 0.075 0.074 0.073 0.072 0

1

2

3

4

5

6

Elevation Point

7

8

9

10

Problem 3.8

[Difficulty: 2]

Given:

Properties of a cube floating at an interface

Find:

The pressures difference between the upper and lower surfaces; average cube density

Solution: The pressure difference is obtained from two applications of Eq. 3.7 pU = p0 + ρSAE10⋅ g⋅ (H − 0.1⋅ d)

pL = p0 + ρSAE10⋅ g⋅ H + ρH2O⋅ g⋅ 0.9⋅ d

where pU and pL are the upper and lower pressures, p0 is the oil free surface pressure, H is the depth of the interface, and d is the cube size Hence the pressure difference is

(

∆p = p L − pU = ρH2O ⋅ g⋅ 0.9⋅ d + ρSAE10 ⋅ g ⋅ 0.1⋅ d From Table A.2

∆p = ρH2O ⋅ g⋅ d ⋅ 0.9 + SGSAE10 ⋅ 0.1

SGSAE10 = 0.92 ∆p = 999⋅

kg 3

× 9.81⋅

m s

m

2

2

× 0.1⋅ m × (0.9 + 0.92 × 0.1) ×

N⋅s kg ⋅ m

∆p = 972 Pa

For the cube density, set up a free body force balance for the cube ΣF = 0 = ∆p ⋅ A − W Hence

W = ∆p⋅ A = ∆p ⋅d

2 2

m ∆p ⋅ d ∆p W = = = ρcube = 3 3 3 d⋅ g d ⋅g d d ⋅g ρcube = 972⋅

N 2

m

2

×

s kg ⋅ m 1 × × 2 0.1⋅ m 9.81⋅ m N⋅s

ρcube = 991

kg 3

m

)

Problem 3.9

Given:

Data on tire at 3500 m and at sea level

Find:

Absolute pressure at 3500 m; pressure at sea level

[Difficulty: 2]

Solution: At an elevation of 3500 m, from Table A.3: pSL = 101⋅ kPa

patm = 0.6492 ⋅ pSL

patm = 65.6⋅ kPa

and we have

pg = 0.25⋅ MPa

pg = 250⋅ kPa

p = p g + patm

At sea level

patm = 101 ⋅ kPa

p = 316⋅ kPa

Meanwhile, the tire has warmed up, from the ambient temperature at 3500 m, to 25oC. At an elevation of 3500 m, from Table A.3

Tcold = 265.4 ⋅ K

and

Thot = ( 25 + 273) ⋅ K

Thot = 298 K

Hence, assuming ideal gas behavior, pV = mRT, and that the tire is approximately a rigid container, the absolute pressure of the hot tire is Thot phot = ⋅p Tcold

phot = 354⋅ kPa

Then the gage pressure is pg = phot − patm

pg = 253⋅ kPa

Problem 3.10

Given:

Data on air bubble

Find:

Bubble diameter as it reaches surface

[Difficulty: 2]

Solution: dp = − ρsea⋅ g dy

Basic equation

and the ideal gas equation

p = ρ⋅ R⋅ T =

M ⋅ R⋅ T V

We assume the temperature is constant, and the density of sea water is constant

For constant sea water density

p = patm + SGsea⋅ ρ⋅ g⋅ h

Then the pressure at the initial depth is

p1 = patm + SGsea⋅ ρ⋅ g⋅ h1

The pressure as it reaches the surface is

p2 = patm

M⋅ R⋅ T

For the bubble

p =

Hence

p1 ⋅V1 = p2 ⋅V2

where p is the pressure at any depth h

but M and T are constant

V

V2 = V1⋅

or

M⋅ R⋅ T = const = p⋅ V

P1 p2

1

Then the size of the bubble at the surface is

From Table A.2

⎛ p1 ⎞ D2 = D1⋅ ⎜ ⎟ p ⎝ 2⎠

SGsea = 1.025

⎡ ⎢ ⎣

3

1

p2

1

⎡( patm + ρsea⋅ g⋅ h1)⎤ ⎛ ρsea⋅ g ⋅h1 ⎞ = D1 ⋅ ⎢ ⎥ = D1⋅ ⎜ 1 + ⎟ patm patm ⎣ ⎦ ⎝ ⎠ 3

3

(This is at 68oF)

D2 = 0.3⋅ in × ⎢1 + 1.025 × 1.94⋅

D2 = 0.477⋅ in

3 p1

3

D2 = D1 ⋅

or

slug ft

3

× 32.2 ×

ft s

2

2

× 100⋅ ft ×

in

14.7⋅ lbf

lbf⋅ s ⎥⎤ 1⋅ ft ⎞ × ⋅ ⎥ ⎝ 12⋅ in ⎠ slugft

×⎛

2

2

⎦

1 3

Problem 3.11

[Difficulty: 2]

Given:

Properties of a cube suspended by a wire in a fluid

Find:

The fluid specific gravity; the gage pressures on the upper and lower surfaces

Solution: From a free body analysis of the cube:

(

)

2

ΣF = 0 = T + pL − pU ⋅ d − M⋅ g

where M and d are the cube mass and size and pL and pU are the pressures on the lower and upper surfaces For each pressure we can use Eq. 3.7

p = p0 + ρ⋅ g⋅ h

Hence

pL − pU = ⎡p0 + ρ⋅ g⋅ ( H + d)⎤ − p0 + ρ⋅ g⋅ H = ρ⋅ g⋅ d = SG ⋅ρH2O ⋅ d ⎣ ⎦

(

)

where H is the depth of the upper surface

2 ⋅ slug× 32.2⋅ M⋅ g − T

SG =

Hence the force balance gives

s

SG =

3

ρH2O ⋅ g ⋅ d

ft 2

×

lbf ⋅ s

2

− 50.7 ⋅ lbf

slug ⋅ ft

SG = 1.75

2

lbf⋅ s 3 × (0.5 ⋅ ft ) 1.94 ⋅ × 32.2⋅ × 3 2 slug ⋅ ft ft s slug

ft

From Table A.1, the fluid is Meriam blue. The individual pressures are computed from Eq 3.7 p = p0 + ρ⋅ g⋅ h

For the upper surface

For the lower surface

pg = ρ⋅ g⋅ h = SG ⋅ρH2O ⋅ h

or

pg = 1.754 × 1.94⋅

slug

pg = 1.754 × 1.94⋅

slug

ft

ft

3

3

× 32.2⋅

ft s

× 32.2⋅

2

2

×

lbf ⋅ s 2 1⋅ ft ⎞ ⋅ ft × ×⎛ slug ⋅ ft ⎝ 12⋅ in ⎠ 3

2

2

lbf ⋅ s 2 1 1⋅ ft ⎞ × ⎛ + ⎞ ⋅ ft × ×⎛ 2 ⎝3 2 ⎠ slug ⋅ ft 12 ⋅ in⎠ ⎝ s ft

pg = 0.507⋅ psi 2

pg = 0.888⋅ psi

Note that the SG calculation can also be performed using a buoyancy approach (discussed later in the chapter): Consider a free body diagram of the cube:

ΣF = 0 = T + FB − M ⋅ g

where M is the cube mass and FB is the buoyancy force

FB = SG ⋅ ρH2O ⋅ L ⋅ g

Hence

3

T + SG ⋅ ρH2O ⋅ L ⋅ g − M ⋅ g = 0

3

or

SG =

M⋅ g − T ρH2O ⋅ g ⋅ L

3

as before

SG = 1.75

Problem 3.12

[Difficulty: 4]

Given:

Model behavior of seawater by assuming constant bulk modulus

Find:

(a) Expression for density as a function of depth h. (b) Show that result may be written as ρ = ρo + bh (c) Evaluate the constant b (d) Use results of (b) to obtain equation for p(h) (e) Determine depth at which error in predicted pressure is 0.01%

Solution:

SGo = 1.025

From Table A.2, App. A:

dp

Governing Equations:

= ρ⋅ g

dh

Ev =

Then

dρ or dp = ρ⋅ g⋅ dh = Ev ⋅ ρ

After integrating:

Now for

ρo⋅ g⋅ h Ev

ρ − ρo ρ ⋅ρ o

=

g⋅ h Ev

dρ 2

ρ

=

5

Ev = 2.42⋅ GPa = 3.51 × 10 ⋅ psi (Hydrostatic Pressure - h is positive downwards)

dp

(Definition of Bulk Modulus)

dρ ρ h

ρ

g Ev

dh

⌠ ⌠ g 1 ⎮ = ⎮ dh d ρ ⎮ 2 ⎮ Ev ρ ⎮ ⌡0 ⌡ρo

Now if we integrate:

Therefore: ρ =

Ev⋅ ρo Ev − g⋅ h⋅ ρo

ρ

and

ρo

1

= 1−

1)

Problem *3.113

[Difficulty: 4]

Open-Ended Problem Statement: A proposed ocean salvage scheme involves pumping air into “bags” placed within and around a wrecked vessel on the sea bottom. Comment on the practicality of this plan, supporting your conclusions with analyses.

Discussion: This plan has several problems that render it impractical. First, pressures at the sea bottom are very high. For example, Titanic was found in about 12,000 ft of seawater. The corresponding pressure is nearly 6,000 psi. Compressing air to this pressure is possible, but would require a multi-stage compressor and very high power. Second, it would be necessary to manage the buoyancy force after the bag and object are broken loose from the sea bed and begin to rise toward the surface. Ambient pressure would decrease as the bag and artifact rise toward the surface. The air would tend to expand as the pressure decreases, thereby tending to increase the volume of the bag. The buoyancy force acting on the bag is directly proportional to the bag volume, so it would increase as the assembly rises. The bag and artifact thus would tend to accelerate as they approach the sea surface. The assembly could broach the water surface with the possibility of damaging the artifact or the assembly. If the bag were of constant volume, the pressure inside the bag would remain esse...