Problem 1.9 [2]

1.9 Consider again the small particle of Problem 1.8. Express the distance required to reach 95 percent of its

terminal speed in terms of g, k, and W.

Given: Small particle accelerating from rest in a fluid. Net weight is W, resisting force is FD = kV, where

V is speed.

Find: Distance required to reach 95 percent of terminal speed, Vt.

Solution: Consider the particle to be a system. Apply Newton's second law.

Basic equation: ∑Fy = may

Assumptions:

1. W is net weight.

2. Resisting force acts opposite to V.

Then, dV W dVdt g dyF W kV = ma m Vy y= − = =∑ or V dVk

W g dy1 V− =

At terminal speed, ay = 0 and Wt kV V= = . Then

g

V dV1V g dy1 V− =

Separating variables t

1V

V dV g dy1 V

=−

Integrating, noting that velocity is zero initially

[ ]

0.950.95 2

00

2 2 2

2 2

22

2

ln 111

0.95 ln (1 0.95) ln (1)

0.95 ln 0.05 2.05

2.05 2.05

t

t

VV

t tt

t

t t t

t t

t

V dV Vgy VV VVV

V

gy V V V

gy V V

Wy Vg gt

⎡ ⎤⎛ ⎞= = − − −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦−

= − − − −

= − + =

∴ = =

∫

Problem 1.22 [1]

Given: Quantities in SI (or other) units.

Find: Quantities in BG units.

Solution: Use Table G.2.

(a) 50 m2⋅ 50 m2

⋅1 in⋅

0.0254 m⋅1 ft⋅

12 in⋅×⎛⎜

⎝⎞⎟⎠

2×= 538 ft2⋅=

(b) 250 cc⋅ 250 cm3⋅

1 m⋅100 cm⋅

1 in⋅0.0254 m⋅

×1 ft⋅

12 in⋅×⎛⎜

⎝⎞⎟⎠

3×= 8.83 10 3−

× ft3⋅=

(c) 100 kW⋅ 100 kW⋅1000 W⋅

1 kW⋅×

1 hp⋅746 W⋅

×= 134 hp⋅=

(d) 5lbf s⋅

ft2⋅ is already in BG units

Problem 1.28 [3]

Given: Information on canal geometry.

Find: Flow speed using the Manning equation, correctly and incorrectly!

Solution: Use Table G.2 and other sources (e.g., Google) as needed.

The Manning equation is VRh

23 S0

12

⋅

n= which assumes Rh in meters and V in m/s.

The given data is Rh 7.5 m⋅= S0110

= n 0.014=

Hence V7.5

23 1

10⎛⎜⎝

⎞⎟⎠

12

⋅

0.014= V 86.5

ms

⋅= (Note that we don't cancel units; we just write m/snext to the answer! Note also this is a very highspeed due to the extreme slope S0.)

Using the equation incorrectly: Rh 7.5 m⋅1 in⋅

0.0254 m⋅×

1 ft⋅12 in⋅

×= Rh 24.6 ft⋅=

Hence V24.6

23 1

10⎛⎜⎝

⎞⎟⎠

12

⋅

0.014= V 191

fts

⋅= (Note that we again don't cancel units; we justwrite ft/s next to the answer!)

This incorrect use does not provide the correct answer V 191fts

⋅12 in⋅1 ft⋅

×0.0254 m⋅

1 in⋅×= V 58.2

ms

= which is wrong!

This demonstrates that for this "engineering" equation we must be careful in its use!

To generate a Manning equation valid for Rh in ft and V in ft/s, we need to do the following:

Vfts

⎛⎜⎝

⎞⎟⎠

Vms

⎛⎜⎝

⎞⎟⎠

1 in⋅0.0254 m⋅

×1 ft⋅

12 in⋅×=

Rh m( )

23 S0

12

⋅

n1 in⋅

0.0254 m⋅1 ft⋅

12 in⋅×⎛⎜

⎝⎞⎟⎠

×=

Vfts

⎛⎜⎝

⎞⎟⎠

Rh ft( )

23 S0

12

⋅

n1 in⋅

0.0254 m⋅1 ft⋅

12 in⋅×⎛⎜

⎝⎞⎟⎠

23

−

×1 in⋅

0.0254 m⋅1 ft⋅

12 in⋅×⎛⎜

⎝⎞⎟⎠

×=Rh ft( )

23 S0

12

⋅

n1 in⋅

0.0254 m⋅1 ft⋅

12 in⋅×⎛⎜

⎝⎞⎟⎠

13

×=

In using this equation, we ignore the units and just evaluate the conversion factor 1.0254

112⋅⎛⎜

⎝⎞⎟⎠

13

1.49=

Hence Vfts

⎛⎜⎝

⎞⎟⎠

1.49 Rh ft( )

23

⋅ S0

12

⋅

n=

Handbooks sometimes provide this form of the Manning equation for direct use with BG units. In our case we are askedto instead define a new value for n:

nBGn

1.49= nBG 0.0094= where V

fts

⎛⎜⎝

⎞⎟⎠

Rh ft( )

23 S0

12

⋅

nBG=

Using this equation with Rh = 24.6 ft: V24.6

23 1

10⎛⎜⎝

⎞⎟⎠

12

⋅

0.0094= V 284

fts

=

Converting to m/s V 284fts

⋅12 in⋅1 ft⋅

×0.0254 m⋅

1 in⋅×= V 86.6

ms

= which is the correct answer!

Problem 1.32 [1]

Given: Equation for mean free path of a molecule.

Find: Dimensions of C for a diemsionally consistent equation.

Solution: Use the mean free path equation. Then "solve" for C and use dimensions.

The mean free path equation is λ Cm

ρ d2⋅

⋅=

"Solving" for C, and using dimensions Cλ ρ⋅ d2

⋅m

=

C

LM

L3× L2

×

M= 0=

The drag constant C is dimensionless.

Problem 2.17 [3]

Problem 2.24 [3] Part 1/2

Problem 2.24 [3] Part 2/2

Problem 2.48 [3]

NOTE: Figure is wrong - length is 0.85 m

Given: Data on double pipe heat exchanger

Find: Whether no-slip is satisfied; net viscous force on inner pipe

Solution:For the oil, the velocity profile is uz r( ) umax 1

rRii

⎛⎜⎝

⎞⎟⎠

2−

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅= where umaxRii

2Δp⋅

4 μ⋅ L⋅=

Check the no-slip condition. When r Rii= uz Rii( ) umax 1RiiRii

⎛⎜⎝

⎞⎟⎠

2

−⎡⎢⎢⎣

⎤⎥⎥⎦

⋅= 0=

For the water, the velocity profile is uz r( )1

4 μ⋅ΔpL

⋅ Rio2 r2−

Roi2 Rio

2−

lnRioRoi

⎛⎜⎝

⎞⎟⎠

lnr

Rio⎛⎜⎝

⎞⎟⎠

⋅−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅=

Check the no-slip condition. When r Roi= uz Roi( ) 14 μ⋅

ΔpL

⋅ Rio2 Roi

2−

Roi2 Rio

2−

lnRioRoi

⎛⎜⎝

⎞⎟⎠

lnRoiRio

⎛⎜⎝

⎞⎟⎠

⋅−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅=

uz Roi( ) 14 μ⋅

ΔpL

⋅ Rio2 Roi

2− Roi

2 Rio2

−⎛⎝

⎞⎠+⎡

⎣⎤⎦⋅= 0=

When r Rio= uz Rio( ) 14 μ⋅

ΔpL

⋅ Rio2 Rio

2−

Roi2 Rio

2−

lnRioRoi

⎛⎜⎝

⎞⎟⎠

lnRioRio

⎛⎜⎝

⎞⎟⎠

⋅−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅= 0=

The no-slip condition holds on all three surfaces.

The given data is Rii7.5 cm⋅

23 mm⋅−= Rii 3.45 cm⋅= Rio

7.5 cm⋅2

= Rio 3.75 cm⋅= Roi11 cm⋅

23 mm⋅−= Roi 5.2 cm⋅=

Δpw 2.5 Pa⋅= Δpoil 8 Pa⋅= L 0.85 m⋅=

The viscosity of water at 10oC is (Fig. A.2) μw 1.25 10 3−×

N s⋅

m2⋅=

The viscosity of SAE 10-30 oil at 100oC is (Fig. A.2) μoil 1 10 2−×

N s⋅

m2⋅=

For each, shear stress is given by τrx μdudr⋅=

For water τrx μ

duz r( )

dr⋅= μw

ddr⋅

14 μw⋅

ΔpwL

⋅ Rio2 r2−

Roi2 Rio

2−

lnRioRoi

⎛⎜⎝

⎞⎟⎠

lnr

Rio⎛⎜⎝

⎞⎟⎠

⋅−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

=

τrx14

ΔpwL

⋅ 2− r⋅Roi

2 Rio2

−

lnRioRoi

⎛⎜⎝

⎞⎟⎠

r⋅

−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅=

so on the pipe surface Fw τrx A⋅=14

ΔpwL

⋅ 2− Rio⋅Roi

2 Rio2

−

lnRioRoi

⎛⎜⎝

⎞⎟⎠

Rio⋅

−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅ 2⋅ π⋅ Rio⋅ L⋅=

Fw Δpw π⋅ Rio2

−Roi

2 Rio2

−

2 lnRioRoi

⎛⎜⎝

⎞⎟⎠

⋅

−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅=

HenceFw 2.5

N

m2⋅ π× 3.75 cm⋅

1 m⋅100 cm⋅

×⎛⎜⎝

⎞⎟⎠

2−

5.2 cm⋅( )2 3.75 cm⋅( )2−⎡⎣ ⎤⎦

1 m⋅100 cm⋅⎛⎜⎝

⎞⎟⎠

2×

2 ln3.755.2

⎛⎜⎝

⎞⎟⎠

⋅−

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

×=

Fw 0.00454N=

This is the force on the r-negative surface of the fluid; on the outer pipe itself we also have Fw 0.00454N=

For oil τrx μ

duz r( )

dr⋅= μoil

ddr⋅ umax 1

rRii

⎛⎜⎝

⎞⎟⎠

2−

⎡⎢⎢⎣

⎤⎥⎥⎦

⋅=2 μoil⋅ umax⋅ r⋅

Rii2

−=Δpoil r⋅

2 L⋅−=

so on the pipe surface Foil τrx A⋅=Δpoil Rii⋅

2 L⋅− 2⋅ π⋅ Rii⋅ L⋅= Δpoil− π⋅ Rii

2⋅=

This should not be a surprise: the pressure drop just balances the friction!

Hence Foil 8−N

m2⋅ π× 3.45 cm⋅

1 m⋅100 cm⋅

×⎛⎜⎝

⎞⎟⎠

2×= Foil 0.0299− N=

This is the force on the r-positive surface of the fluid; on the pipe it is equal and opposite Foil 0.0299N=

The total force is F Fw Foil+= F 0.0345N=

Note we didn't need the viscosities because all quantities depend on the Δp's!

Problem 2.65 [5]

Problem 2.72 [2]

Given: Data on size of various needles

Find: Which needles, if any, will float

Solution:For a steel needle of length L, diameter D, density ρs, to float in water with surface tension σ and contact angle θ, thevertical force due to surface tension must equal or exceed the weight

2 L⋅ σ⋅ cos θ( )⋅ W≥ m g⋅=π D2⋅4

ρs⋅ L⋅ g⋅= or D8 σ⋅ cos θ( )⋅

π ρs⋅ g⋅≤

From Table A.4 σ 72.8 10 3−×

Nm⋅= θ 0 deg⋅= and for water ρ 1000

kg

m3⋅=

From Table A.1, for steel SG 7.83=

Hence8 σ⋅ cos θ( )⋅π SG⋅ ρ⋅ g⋅

8π 7.83⋅

72.8× 10 3−×

Nm⋅

m3

999 kg⋅×

s2

9.81 m⋅×

kg m⋅

N s2⋅

×= 1.55 10 3−× m⋅= 1.55 mm⋅=

Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant)

Problem 3.14 [3]

Problem 3.21 [2]

Problem 3.50 [2]

FR dy

a = 1.25 ft

SG = 2.5

y

b = 1 ft

y’

w

Given: Geometry of access port

Find: Resultant force and location

Solution:

Basic equation FR Ap⌠⎮⎮⌡

d=dpdy

ρ g⋅= ΣMs y' FR⋅= FRy⌠⎮⎮⌡

d= Ay p⋅⌠⎮⎮⌡

d=

or, use computing equations FR pc A⋅= y' ycIxx

A yc⋅+=

We will show both methods

Assumptions: static fluid; ρ = constant; patm on other side

FR Ap⌠⎮⎮⌡

d= ASG ρ⋅ g⋅ y⋅⌠⎮⎮⌡

d= but dA w dy⋅= and wb

ya

= wba

y⋅=

Hence FR0

a

ySG ρ⋅ g⋅ y⋅ba⋅ y⋅

⌠⎮⎮⌡

d=

0

a

ySG ρ⋅ g⋅ba⋅ y2⋅

⌠⎮⎮⌡

d=SG ρ⋅ g⋅ b⋅ a2

⋅3

=

Alternatively FR pc A⋅= and pc SG ρ⋅ g⋅ yc⋅= SG ρ⋅ g⋅23⋅ a⋅= with A

12

a⋅ b⋅=

Hence FRSG ρ⋅ g⋅ b⋅ a2

⋅3

=

For y' y' FR⋅ Ay p⋅⌠⎮⎮⌡

d=

0

a

ySG ρ⋅ g⋅ba⋅ y3⋅

⌠⎮⎮⌡

d=SG ρ⋅ g⋅ b⋅ a3

⋅4

= y'SG ρ⋅ g⋅ b⋅ a3

⋅4 FR⋅

=34

a⋅=

Alternatively y' ycIxx

A yc⋅+= and Ixx

b a3⋅36

= (Google it!)

y'23

a⋅b a3⋅36

2a b⋅⋅

32 a⋅⋅+=

34

a⋅=

Using given data, and SG = 2.5 (Table A.1) FR2.53

1.94⋅slug

ft3⋅ 32.2×

ft

s2⋅ 1× ft⋅ 1.25 ft⋅( )2

×lbf s2

⋅slug ft⋅

×= FR 81.3 lbf⋅=

and y'34

a⋅= y' 0.938 ft⋅=

Problem 3.65 [3]

Problem 4.11 [3]

Given: Geometry of 3D surface

Find: Volume flow rate and momentum flux through area

kdxdyjdxdzAd ˆˆ +=r

jbyiaxV ˆˆ −=r

jyixV ˆˆ −=r

We will need the equation of the surface: yz213−= or zy 26 −=

a) Volume flow rate

( ) ( )( )

( )

sft90

sft90180

1060261010

ˆˆˆˆ

3

3

3

0

23

0

3

0

10

0

3

0

−=

+−=

+−=−−=−=−=

+⋅−=⋅=

∫∫∫ ∫

∫∫

Q

Q

zzdzzydzdxydz

kdxdyjdxdzjyixdAVQAA

r

Solution:

b) Momentum flux

( ) ( )( )

( )

( ) ( )

( )( ) ( )( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ⋅+−=

+−+−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−+⎟

⎠⎞⎜

⎝⎛ −⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

−+−−=

+−=

−−=⋅

∫∫∫

∫∫ ∫

∫∫

3

3

0

323

02

10

0

2

3

0

23

0

10

0

3

0

210

0

3

0

ftslug in is if

ss

ftslugˆ360ˆ450

ˆ3610810810ˆ91850

ˆ34123610ˆ6

2

ˆ2610ˆ26

ˆ10ˆ

ˆˆ

ρρρ

ρρ

ρρ

ρρ

ρρ

ρρ

ji

ji

jzzzizzx

jdzzidzzdxx

jdzyidxdzxy

ydxdzjyixAdVVAA

rrr

Problem 4.18 [3]

Given: Data on flow into and out of tank

Find: Time at which exit pump is switched on; time at which drain is opened; flow rate into drain

Solution:

Basic equationtMCV

∂

∂CS

ρ V→⋅ A→⋅( )∑+ 0=

Assumptions: 1) Uniform flow 2) Incompressible flow

After inlet pump is ontMCV

∂

∂CS

ρ V→⋅ A→⋅( )∑+ t

Mtank∂

∂ρ Vin⋅ Ain⋅−= 0=

tMtank

∂

∂ρ Atank⋅

dhdt⋅= ρ Vin⋅ Ain⋅= where h is the

level of waterin the tank

dhdt

VinAin

Atank⋅= Vin

DinDtank

⎛⎜⎝

⎞⎟⎠

2

⋅=

Hence the time to reach hexit = 0.7 m is texithexitdhdt

=hexitVin

DtankDin

⎛⎜⎝

⎞⎟⎠

2

⋅= texit 0.7 m⋅15

×sm⋅

3 m⋅0.1 m⋅⎛⎜⎝

⎞⎟⎠

2×= texit 126s=

After exit pump is ontMCV

∂

∂CS

ρ V→⋅ A→⋅( )∑+ t

Mtank∂

∂ρ Vin⋅ Ain⋅− ρ Vexit⋅ Aexit⋅+= 0= Atank

dhdt

⋅ Vin Ain⋅ Vexit Aexit⋅−=

dhdt

VinAin

Atank⋅ Vexit

AexitAtank⋅−= Vin

DinDtank

⎛⎜⎝

⎞⎟⎠

2

⋅ VexitDexitDtank

⎛⎜⎝

⎞⎟⎠

2

⋅−=

Hence the time to reach hdrain = 2 m is tdrain texithdrain hexit−( )

dhdt

+=hdrain hexit−( )

VinDin

Dtank

⎛⎜⎝

⎞⎟⎠

2

⋅ VexitDexitDtank

⎛⎜⎝

⎞⎟⎠

2

⋅−

=

tdrain 126 s⋅ 2 0.7−( ) m⋅1

5ms

⋅0.1 m⋅3 m⋅

⎛⎜⎝

⎞⎟⎠

2× 3

ms

⋅0.08 m⋅

3 m⋅⎛⎜⎝

⎞⎟⎠

2×−

×+= tdrain 506s=

The flow rate into the drain is equal to the net inflow (the level in the tank is now constant)

Qdrain Vinπ Din

2⋅

4⋅ Vexit

π Dexit2

⋅

4⋅−= Qdrain 5

ms

⋅π

4× 0.1 m⋅( )2

× 3ms

⋅π

4× 0.08 m⋅( )2

×−= Qdrain 0.0242m3

s=

Problem 4.25

Problem 4.34 [2]

Problem 4.41 Problem 4.49 [3]

P4.48.

Problem 4.42

Problem 4.50 [4]

Problem 4.66 [3]

Rx

V

yx

CS

Given: Water tank attached to mass

Find: Whether tank starts moving

Solution:Basic equation: Momentum flux in x direction for the tank

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow

Hence Rx V cos θ( )⋅ ρ⋅ V A⋅( )⋅= ρ V2⋅

π D2⋅4

⋅ cos θ( )⋅=

We need to find V. We could use the Bernoulli equation, but here it is known that V 2 g⋅ h⋅= where h = 4 m is theheight of fluid in the tank

V 2 9.81×m

s2⋅ 4× m⋅= V 8.86

ms

=

Hence Rx 1000kg

m3⋅ 8.86

ms

⋅⎛⎜⎝

⎞⎟⎠

2×

π

4× 0.04 m⋅( )2

× cos 60 deg⋅( )×= Rx 49.3N=

This force is equal to the tension T in the wire T Rx= T 49.3N=

For the block, the maximum friction force a mass of M = 9 kg can generate is Fmax M g⋅ μ⋅= where μ is static friction

Fmax 9 kg⋅ 9.81×m

s2⋅ 0.5×

N s2⋅

kg m⋅×= Fmax 44.1N=

Hence the tension T created by the water jet is larger than the maximum friction Fmax; the tank starts to move

Problem 4.55

Problem 4.68 [2]

Problem 4.81 [3]

Problem 4.73

Problem 4.87 [3]

Problem *4.107 [2]

CS

xy

Rx

V, A

Given: Water jet striking surface

Find: Force on surface

Solution:Basic equations: Momentum flux in x direction

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow

Hence Rx u1 ρ− u1⋅ A1⋅( )⋅= ρ− V2⋅ A⋅= ρ−

QA

⎛⎜⎝

⎞⎟⎠

2⋅ A⋅=

ρ Q2⋅A

−=4 ρ⋅ Q2

⋅

π D2⋅

−= where Q is the flow rate

The force of the jet on the surface is then F Rx−=4 ρ⋅ Q2

⋅

π D2⋅

=

For a fixed flow rate Q, the force of a jet varies as 1

D2: A smaller diameter leads to a larger force. This is because as

the diameter decreases the speed increases, and the impact force varies as the square of the speed, but linearly with area

For a force of F = 650 N

Qπ D2⋅ F⋅4 ρ⋅

= Qπ

46

1000m⋅⎛⎜

⎝⎞⎟⎠

2× 650× N⋅

m3

1000 kg⋅×

kg m⋅

s2 N⋅×

1 L⋅

10 3− m3⋅

×60 s⋅1 min⋅

×= Q 257L

min⋅=

Problem 5.13 [3]

Given: Data on boundary layer

Find: y component of velocity ratio; location of maximum value; plot velocity profiles; evaluate at particular point

Solution:

u x y, ( ) U32

yδ x( )

⎛⎜⎝

⎞⎟⎠

⋅12

yδ x( )

⎛⎜⎝

⎞⎟⎠

3⋅−

⎡⎢⎣

⎤⎥⎦

⋅= and δ x( ) c x⋅=

so u x y, ( ) U32

y

c x⋅⎛⎜⎝

⎞⎟⎠

⋅12

y

c x⋅⎛⎜⎝

⎞⎟⎠

3⋅−

⎡⎢⎣

⎤⎥⎦

⋅=

For incompressible flowx

u∂

∂ yv∂

∂+ 0=

Hence v x y, ( ) yx

u x y, ( )dd

⌠⎮⎮⎮⌡

d−= and dudx

34

U⋅y3

c3 x

52

⋅

y

c x

32

⋅

−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅=

so v x y, ( ) y34

U⋅y3

c3x5

2⋅

yc

x3

2⋅−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅

⌠⎮⎮⎮⌡

d−=

v x y, ( )38

U⋅y2

c x

32

⋅

y4

2 c3⋅ x

52

⋅

−⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅= v x y, ( )38

U⋅δ

x⋅

yδ

⎛⎜⎝

⎞⎟⎠

2 12

yδ

⎛⎜⎝

⎞⎟⎠

4⋅−

⎡⎢⎣

⎤⎥⎦

⋅=

The maximum occurs at y δ= as seen in the corresponding Excel workbook

vmax38

U⋅δ

x⋅ 1

12

1⋅−⎛⎜⎝

⎞⎟⎠

⋅=

At δ 5 mm⋅= and x 0.5 m⋅= , the maximum vertical velocity isvmax

U0.00188=

To find when v /U is maximum, use Solver

v /U y /d

0.00188 1.0

v /U y /d

0.000000 0.00.000037 0.10.000147 0.20.000322 0.30.000552 0.40.00082 0.50.00111 0.60.00139 0.70.00163 0.80.00181 0.90.00188 1.0

Vertical Velocity Distribution In Boundary layer

0.0

0.2

0.4

0.6

0.8

1.0

0.0000 0.0005 0.0010 0.0015 0.0020

v /U

y/ δ

Problem 5.43 [2]

Problem 5.49 [2]

Problem 5.59 [3]

Problem 5.83 [3]

Problem 6.11 [2]

Problem 6.22 [4] Part 1/2

Problem 6.19 cont'd

Problem 6.22 [4] Part 2/2

Problem 6.27 [2]

Problem 6.65 [3]

Given: Flow through reducing elbow

Find: Mass flow rate in terms of Δp, T1 and D1 and D2

Solution:

Basic equations: pρ

V2

2+ g z⋅+ const= Q V A⋅=

Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Ignore elevation change 6) p2 = patm

Available data: Q 20 gpm⋅= Q 0.0446ft3

s= D 1.5 in⋅= d 0.5 in⋅= ρ 1.94

slug

ft3⋅=

From contnuity V1Q

π D2⋅4

⎛⎜⎝

⎞⎟⎠

= V1 3.63fts

= V2Q

π d2⋅4

⎛⎜⎝

⎞⎟⎠

= V2 32.7fts

=

Hence, applying Bernoulli between the inlet (1) and exit (2) p1ρ

V12

2+

p2ρ

V22

2+=

or, in gage pressures p1gρ

2V2

2 V12

−⎛⎝

⎞⎠⋅= p1g 7.11psi=

From x-momentum Rx p1g A1⋅+ u1 mrate−( )⋅ u2 mrate( )⋅+= mrate− V1⋅= ρ− Q⋅ V1⋅= because u1 V1= u2 0=

Rx p1g−π D2⋅4

⋅ ρ Q⋅ V1⋅−= Rx 12.9− lbf=

The force on the supply pipe is then Kx Rx−= Kx 12.9 lbf= on the pipe to the right

Problem 6.77 [4]

Given: Water flow out of tube

Find: Pressure indicated by gage; force to hold body in place

Solution:Basic equations: Bernoulli, and momentum flux in x direction

pρ

V2

2+ g z⋅+ constant= Q V A⋅=

Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (gx = 0)

Applying Bernoulli between jet exit and stagnation point

p1ρ

V12

2+

p2ρ

V22

2+=

V22

2= where we work in gage pressure

p1ρ

2V2

2 V12

−⎛⎝

⎞⎠⋅=

But from continuity Q V1 A1⋅= V2 A2⋅= V2 V1A1A2⋅= V1

D2

D2 d2−

⋅= where D = 2 in and d = 1.5 in

V2 20fts

⋅22

22 1.52−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅= V2 45.7fts

=

Hence p112

1.94×slug

ft3⋅ 45.72 202

−( )×fts

⎛⎜⎝

⎞⎟⎠

2⋅

lbf s2⋅

slugft⋅×= p1 1638

lbf

ft2= p1 11.4psi= (gage)

The x mometum is F− p1 A1⋅+ p2 A2⋅− u1 ρ− V1⋅ A1⋅( )⋅ u2 ρ V2⋅ A2⋅( )⋅+=

F p1 A1⋅ ρ V12 A1⋅ V2

2 A2⋅−⎛⎝

⎞⎠⋅+= using gage pressures

F 11.4lbf

in2⋅

π 2 in⋅( )2⋅

4× 1.94

slug

ft3⋅ 20

fts

⋅⎛⎜⎝

⎞⎟⎠

2π 2 in⋅( )2⋅

4× 45.7

fts

⋅⎛⎜⎝

⎞⎟⎠

2π 2 in⋅( )2 1.5 in⋅( )2

−⎡⎣ ⎤⎦⋅4

×−⎡⎢⎣

⎤⎥⎦

×1 ft⋅

12 in⋅⎛⎜⎝

⎞⎟⎠

2×

lbf s2⋅

slugft⋅×+=

F 14.1 lbf= in the direction shown

Problem 7.4 [2]

Problem 7.26 [2]

Problem 7.33 (In Excel) [3]

Given: Bubble size depends on viscosity, density, surface tension, geometry and pressure

Find: Π groups

Solution:We will use the workbook of Example 7.1, modified for the current problem

The number of parameters is: n = 6The number of primary dimensions is: r = 3The number of repeat parameters is: m = r = 3The number of Π groups is: n - m = 3

Enter the dimensions (M, L, t) ofthe repeating parameters, and of up tofour other parameters (for up to four Π groups).The spreadsheet will compute the exponents a , b , and c for each.

REPEATING PARAMETERS:Choose ρ, Δp , D

M L tρ 1 -3Δp 1 -1 -2D 1

Π GROUPS:

M L t M L td 0 1 0 μ 1 -1 -1

Π1: a = 0 Π2: a = -0.5b = 0 b = -0.5c = -1 c = -1

M L t M L tσ 1 0 -2 0 0 0

Π3: a = 0 Π4: a = 0b = -1 b = 0c = -1 c = 0

Note that the Π1 group can be obtained by inspection

Hence Dd

=Π1 2

2

21

212

pDDp

Δ→

Δ

=Πρμ

ρ

μpDΔ

=Πσ

3

Problem 7.39 [3]

Problem 7.50 [3]

Problem 7.59 [3]

Problem 7.74 [2]

Problem 8.3 [3]

Given: Air entering pipe system

Find: Flow rate for turbulence in each section; Which become fully developed

Solution:From Table A.9 ν 1.62 10 4−

×ft2

s⋅=

The given data is L 5 ft⋅= D1 1 in⋅= D212

in⋅= D314

in⋅=

The critical Reynolds number is Recrit 2300=

Writing the Reynolds number as a function of flow rate

ReV D⋅

ν

=Q

π

4D2⋅

Dν

⋅= or QRe π⋅ ν⋅ D⋅

4=

Then the flow rates for turbulence to begin in each section of pipe are

Q1Recrit π⋅ ν⋅ D1⋅

4= Q1 2300

π

4× 1.62× 10 4−

×ft2

s⋅

112

× ft⋅= Q1 0.0244ft3

s=

Q2Recrit π⋅ ν⋅ D2⋅

4= Q2 0.0122

ft3

s= Q3

Recrit π⋅ ν⋅ D3⋅

4= Q3 0.00610

ft3

s=

Hence, smallest pipe becomes turbulent first, then second, then the largest.

For the smallest pipe transitioning to turbulence (Q3)

For pipe 3 Re3 2300= Llaminar 0.06 Re3⋅ D3⋅= Llaminar 2.87 ft= Llaminar < L: Not fully developed

or, for turbulent, Lmin 25 D3⋅= Lmin 0.521 ft= Lmax 40 D3⋅= Lmax 0.833 ft= Lmax/min < L: Not fully developed

For pipes 1 and 2 Llaminar 0.064 Q3⋅

π ν⋅ D1⋅

⎛⎜⎝

⎞⎟⎠

⋅ D1⋅= Llaminar 2.87 ft= Llaminar < L: Not fully developed

Llaminar 0.064 Q3⋅

π ν⋅ D2⋅

⎛⎜⎝

⎞⎟⎠

⋅ D2⋅= Llaminar 2.87 ft= Llaminar < L: Not fully developed

For the middle pipe transitioning to turbulence (Q2)

For pipe 2 Re2 2300= Llaminar 0.06 Re2⋅ D2⋅= Llaminar 5.75 ft=

Llaminar > L: Fully developed

or, for turbulent, Lmin 25 D2⋅= Lmin 1.04 ft= Lmax 40 D2⋅= Lmax 1.67 ft=

Lmax/min < L: Not fully developed

For pipes 1 and 3 L1 0.064 Q2⋅

π ν⋅ D1⋅

⎛⎜⎝

⎞⎟⎠

⋅ D1⋅= L1 5.75 ft=

L3min 25 D3⋅= L3min 0.521 ft= L3max 40 D3⋅= L3max 0.833 ft=

Lmax/min < L: Not fully developed

For the large pipe transitioning to turbulence (Q1)

For pipe 1 Re1 2300= Llaminar 0.06 Re1⋅ D1⋅= Llaminar 11.5 ft=

Llaminar > L: Fully developed

or, for turbulent, Lmin 25 D1⋅= Lmin 2.08 ft= Lmax 40 D1⋅= Lmax 3.33 ft=

Lmax/min < L: Not fully developed

For pipes 2 and 3 L2min 25 D2⋅= L2min 1.04 ft= L2max 40 D2⋅= L2max 1.67 ft=

Lmax/min < L: Not fully developed

L3min 25 D3⋅= L3min 0.521 ft= L3max 40 D3⋅= L3max 0.833 ft=

Lmax/min < L: Not fully developed

Problem 8.22 [2]

x y d

U2

U1

Given: Laminar flow between moving plates

Find: Expression for velocity; Volume flow rate per depth

Solution:Using the analysis of Section 8-2, the sum of forces in the x direction is

τ

yτ

∂

∂

dy2

⋅+ τ

yτ

∂

∂

dy2

⋅−⎛⎜⎝

⎞⎟⎠

−⎡⎢⎣

⎤⎥⎦

b⋅ dx⋅ px

p∂

∂

dx2

⋅− p−x

p∂

∂

dx2

⋅+⎛⎜⎝

⎞⎟⎠

b⋅ dy⋅+ 0=

Simplifying dτ

dydpdx

= 0= or μ

d2u

dy2⋅ 0=

Integrating twice u c1 y⋅ c2+=

Boundary conditions: u 0( ) U1−= c2 U1−= u y d=( ) U2= c1U1 U2+

d=

Hence u y( ) U1 U2+( ) yd⋅ U1−= u y( ) 75 y⋅ 0.25−= (u in m/s, y in m)

The volume flow rate is Q Au⌠⎮⎮⌡

d= b yu⌠⎮⎮⌡

d⋅= Q b

0

d

xU1 U2+( ) yd⋅ U1−⎡⎢

⎣⎤⎥⎦

⌠⎮⎮⌡

d⋅=

Q b d⋅U2 U1−( )

2⋅=

Qb

10 mm⋅1 m⋅

1000 mm⋅×

12

× 0.5 0.25−( )×ms

×= Q 0.00125

m3

sm

=

Problem 8.84 [2]

Problem 8.108 [2]

Problem 9.25 [2]

Given: Data on wind tunnel and boundary layers

Find: Pressure change between points 1 and 2

Solution:

Basic equations (4.12) pρ

V2

2+ g z⋅+ const=

Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal

For this flow ρ U⋅ A⋅ const=

The given data is U0 100fts

⋅= U1 U0= h 3 in⋅= A1 h2= A1 9 in2

⋅=

We also have δdisp2 0.035 in⋅=

Hence at the Point 2 A2 h 2 δdisp2⋅−( )2= A2 8.58 in2⋅=

Applying mass conservation between Points 1 and 2

ρ− U1⋅ A1⋅( ) ρ U2⋅ A2⋅( )+ 0= or U2 U1A1A2⋅= U2 105

fts

⋅=

The pressure change is found from Bernoullip1ρ

U12

2+

p2ρ

U22

2+= with ρ 0.00234

slug

ft3⋅=

Hence Δpρ

2U1

2 U22

−⎛⎝

⎞⎠⋅= Δp 8.05− 10 3−

× psi⋅= Δp 1.16−lbf

ft2⋅=

The pressure drops by a small amount as the air accelerates

Problem 9.113 [3]

Problem 9.121 [1]

Given: Data on car antenna

Find: Bending moment

Solution:

Basic equation: FD12

ρ⋅ A⋅ V2⋅ CD⋅=

The given or available data is V 120kmhr

⋅= V 33.3ms

⋅= L 1.8 m⋅= D 10 mm⋅=

A L D⋅= A 0.018m2=

ρ 1.225kg

m3⋅= ν 1.50 10 5−

×m2

s⋅= (Table A.10, 20oC)

For a cylinder, check Re ReV D⋅

ν= Re 2.22 104

×=

From Fig. 9.13 CD 1.0= FD12

ρ⋅ A⋅ V2⋅ CD⋅= FD 12.3N=

The bending moment is then M FDL2⋅= M 11.0 N m⋅⋅=