solution manual Fluid Mechanics 9th Edition by White Xue

Solution Manual Fluid Mechanics 9th Edition by White Xue – Updated 2024

Complete Solution Manual With Answers

Sample Chapter Is Below

 

 

Chapter 1  Introduction 1-1

Chapter 1  Introduction

P1.1 A gas at 20C may be rarefied if it contains less than 1012 molecules per mm3. If

Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent?

Solution: The mass of one molecule of air may be computed as

Molecular weight 28.97 mol m 

1

   



4.81E 23 g Avogadro’s number 6.023E23 molecules/g mol

Then the density of air containing 1012 molecules per mm3 is, in SI units,

  

12

molecules g 10 4.81E 23



       

3

mm

molecule

g kg 4.81E 11 4.81E 5

   

3 3

mm m

Finally, from the perfect gas law, Eq. (1.13), at 20C  293 K, we obtain the pressure:

 

2

    

   4.0Pa

  

kg m p RT 4.81E 5 287 (293 K) . m s K

 

ns

3 2

  



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McGraw-Hill Education.Chapter 1  Introduction 1-2

P1.2 Table A.6 lists the density of the standard atmosphere as a function of altitude. Use

these values to estimate, crudely, say, within a factor of 2, the number of molecules of air in

the entire atmosphere of the earth.

Solution: Make a plot of density  versus altitude z in the atmosphere, from Table A.6:

1.2255 kg/m3

Density in the Atmosphere



0 z 30,000 m

This writer’s approximation: The curve is approximately an exponential,    o exp(-bz), with

b approximately equal to 0.00011 per meter. Integrate this over the entire atmosphere, with the

radius of the earth equal to 6377 km:

b z m d vol R dz

e   

  

atmosphere o earth

 





2

( ) [ ](4 )

0

2 3 2

  

o earth

  

4 (1.2255 / )4 (6.377 6 ) 5.7 18

R kg m E m E kg

b m

0.00011 /

Dividing by the mass of one molecule  4.8E23 g (see Prob. 1.1 above), we obtain

the total number of molecules in the earth’s atmosphere:

m(atmosphere) 5.7E21 grams N

molecules Ans.    

1.2Ε44

molecules



m(one molecule) 4.8E 23 gm/molecule

This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere.

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McGraw-Hill Education.Chapter 1  Introduction 1-3

P1.3 For the triangular element in Fig.

P1.3, show that a tilted free liquid surface,

in contact with an atmosphere at pressure

pa, must undergo shear stress and hence

begin to flow.

Solution: Assume zero shear. Due to

element weight, the pressure along the lower

and right sides must vary linearly as shown,

to a higher value at point C. Vertical forces

are presumably in balance with ele-ment

weight included. But horizontal forces are

out of balance, with the unbalanced force

being to the left, due to the shaded excess-

pressure triangle on the right side BC. Thus

hydrostatic pressures cannot keep the

element in balance, and shear and flow result.

Fig. P1.3

P1.4 Sand, and other granular materials, definitely flow, that is, you can pour them

from a container or a hopper. There are whole textbooks on the “transport” of granular

materials [54]. Therefore, is sand a fluid? Explain.

Solution: Granular materials do indeed flow, at a rate that can be measured by

“flowmeters”. But they are not true fluids, because they can support a small shear stress

without flowing. They may rest at a finite angle without flowing, which is not possible for

liquids (see Prob. P1.3). The maximum such angle, above which sand begins to flow, is

called the angle of repose. A familiar example is sugar, which pours easily but forms a

significant angle of repose on a heaping spoonful. The physics of granular materials are

complicated by effects such as particle cohesion, clumping, vibration, and size

segregation. See Ref. 48 to learn more.

________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-4

P1.5 A formula for estimating the mean free path of a perfect gas is:

 

1.26 1.26

RT

  (1)



RT

p

where the latter form follows from the ideal-gas law,   pRT. What are the dimensions

of the constant “1.26”? Estimate the mean free path of air at 20C and 7 kPa. Is air rarefied

at this condition?

Solution: We know the dimensions of every term except “1.26”:

2

 

M M L { } {L} { } { } {R} {T} { } LT L T

                

 

3 2

     

Therefore the above formula (first form) may be written dimensionally as

{M/LT} {L} {1.26?} 

 

3 2 2

  

{1.26?}{L} {M/L} [{L /T }{ }]

Since we have {L} on both sides, {1.26}  {unity}, that is, the constant is dimensionless.

The formula is therefore dimensionally homogeneous and should hold for any unit system.

For air at 20C  293 K and 7000 Pa, the density is   pRT  (7000)/[(287)(293)]  0.0832

kgm3

. From Table A-2, its viscosity is 1.80E5 N  s/m2. Then the formula predicts

a mean free path of

1.80E 5

1.26

 9.4E 7 m

Ans.

(0.0832)[(287)(293)]

1/2

This is quite small. We would judge this gas to approximate a continuum if the physical

scales in the flow are greater than about 100 , that is, greater than about 94  m.

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McGraw-Hill Education.Chapter 1  Introduction 1-5

P1.6 Henri Darcy, a French engineer, proposed that the pressure drop Δp for flow at

velocity V through a tube of length L could be correlated in the form

p

 LV 2



If Darcy’s formulation is consistent, what are the dimensions of the coefficient α?

Solution: From Table 1.2, introduce the dimensions of each variable:

  

p



  



  

ML1T2

ML3

  



  

L2

T 2

  

 LV2

     L   L2

  

T 2

  

Solve for {α} = {L-1} Ans.

[The complete Darcy correlation is α = f /(2D), where D is the tube diameter, and f is a

dimensionless friction factor (Chap. 6).]

________________________________________________________________________

P1.7 Convert the following inappropriate quantities into SI units: (a) 2.283E7

U.S. gallons per day; (b) 4.48 furlongs per minute (racehorse speed); and (c)

72,800 avoirdupois ounces per acre.

Solution: (a) (2.283E7 gal/day) x (0.0037854 m3/gal) ÷ (86,400 s/day) =

1.0 m3/s Ans.(a)

(b) 1 furlong = (⅛)mile = 660 ft.

Then (4.48 furlongs/min)x(660 ft/furlong)x(0.3048 m/ft)÷(60 s/min) =

15 m/s Ans.(b)

(c) (72,800 oz/acre)÷(16 oz/lbf)x(4.4482 N/lbf)÷(4046.9 acre/m2) =

5.0 N/m2 = 5.0 Pa Ans.(c)

________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-6

P1.8 Suppose that bending stress  in a beam depends upon bending moment M and

beam area moment of inertia I and is proportional to the beam half-thickness y. Suppose

also that, for the particular case M  2900 inlbf, y  1.5 in, and I  0.4 in4, the predicted

stress is 75 MPa. Find the only possible dimensionally homogeneous formula for .

Solution: We are given that   y fcn(M,I) and we are not to study up on strength of

materials but only to use dimensional reasoning. For homogeneity, the right hand side

must have dimensions of stress, that is,

 

M { } {y}{fcn(M,I)}, or: {L}{fcn(M,I)} LT



     

2

or: the function must have dimensions  

M

{fcn(M,I)} L T

2 2

    

Therefore, to achieve dimensional homogeneity, we somehow must combine bending

moment, whose dimensions are {ML2T–2}, with area moment of inertia, {I}  {L4}, and

end up with {ML–2T–2}. Well, it is clear that {I} contains neither mass {M} nor time {T}

dimensions, but the bending moment contains both mass and time and in exactly the com-

bination we need, {MT–2}. Thus it must be that  is proportional to M also. Now we have

reduced the problem to:

 

2

 

M ML yM fcn(I), or {L} {fcn(I)}, or: {fcn(I)} LT T



2 2

4 {L }

          

We need just enough I’s to give dimensions of {L–4}: we need the formula to be exactly

inverse in I. The correct dimensionally homogeneous beam bending formula is thus:

My C , I



where {C} {unity} . Ans

The formula admits to an arbitrary dimensionless constant C whose value can only be

obtained from known data. Convert stress into English units:  

 (75 MPa)(6894.8) 

10880 lbfin2

. Substitute the given data into the proposed formula:



lbf My (2900 lbf in)(1.5 in) 10880 C C , or: I in 0.4 in



   C 1.00 

Ans.

2 4

The data show that C  1, or   My/I, our old friend from strength of materials.

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McGraw-Hill Education.Chapter 1  Introduction 1-7

P1.9 A hemispherical container, 26 inches in diameter, is filled with a liquid at 20C

and weighed. The liquid weight is found to be 1617 ounces. (a) What is the density of

the fluid, in kg/m3? (b) What fluid might this be? Assume standard gravity, g = 9.807

m/s2

.

Solution: First find the volume of the liquid in m3:

Hemisphere volume 

1





2

  

6

  D3



1





2

  

6

  26 in

 3



 4601in3  61024 in3



 

m3

   0.0754m3



Liquid mass  1617 oz 16 101lbm 0.45359 kg

  

lbm

   45.84 kg

Then the liquid density

45.84 kg

0.0754 m3

 607 kg

m3 Ans.(a)

From Appendix Table A.3, this could very well be ammonia. Ans.(b)

_______________________________________________________________________

P1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is:



9 F 3 DV V D

2 2

   

16

where D  sphere diameter,   viscosity, and   density. Is the formula homogeneous?

Solution: Write this formula in dimensional form, using Table 1-2:

9 {F} {3 }{ }{D}{V}  



  

     2

2 2

{ }{V} {D} ? 16

 

2

ML M L M L or: {1} {L} {1} {L } ? LT T T L T

                           

2 3 2

where, hoping for homogeneity, we have assumed that all constants (3, ,9,16) are pure,

i.e., {unity}. Well, yes indeed, all terms have dimensions {MLT2}! Therefore the Stokes-

Oseen formula (derived in fact from a theory) is dimensionally homogeneous.

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McGraw-Hill Education.Chapter 1  Introduction 1-8

P1.11 In English Engineering units, the specific heat cp of air at room temperature is

approximately 0.24 Btu/(lbm-

F). When working with kinetic energy relations, it is more

appropriate to express cp as a velocity-squared per absolute degree. Give the numerical

value, in this form, of cp for air in (a) SI units, and (b) BG units.

Solution: From Appendix C, Conversion Factors, 1 Btu = 1055.056 J (or N-m) = 778.17

ft-lbf, and 1 lbm = 0.4536 kg = (1/32.174) slug. Thus the conversions are:

_______________________________________________________________________

P1.12 For low-speed (laminar) flow in a tube of radius ro, the velocity u takes the form



p u B r r

 

 2 2

o



where  is viscosity and p the pressure drop. What are the dimensions of B?

Solution: Using Table 1-2, write this equation in dimensional form:

  

2 2

 

2 2 { p} L {M/LT } L {u} {B} {r }, or: {B?} {L } {B?} , { } T {M/LT} T 

          

or: {B}  {L–1} Ans.

The parameter B must have dimensions of inverse length. In fact, B is not a constant, it

hides one of the variables in pipe flow. The proper form of the pipe flow relation is



p u C r r

 

 2 2

o

L 

where L is the length of the pipe and C is a dimensionless constant which has the theoretical

laminar-flow value of (1/4)—see Sect. 6.4.

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McGraw-Hill Education.Chapter 1  Introduction 1-9

P1.13 The efficiency  of a pump is defined as







Q p

Input Power

where Q is volume flow and p the pressure rise produced by the pump. What is  if

p  35 psi, Q  40 Ls, and the input power is 16 horsepower?

Solution: The student should perhaps verify that Qp has units of power, so that  is a

dimensionless ratio. Then convert everything to consistent units, for example, BG:

2



L ft lbf lbf ft lbf Q 40 1.41 ; p 35 5040 ; Power 16(550) 8800

      

2 2

s s s in ft

 

3 2 (1.41 ft s)(5040 lbf ft ) 0.81 or



 

 

8800 ft lbf s

81%

Ans.

Similarly, one could convert to SI units: Q  0.04 m3/s, p  241300 Pa, and input power 

16(745.7)  11930 W, thus h  (0.04)(241300)/(11930)  0.81. Ans.

P1.14 The volume flow Q over a dam is proportional to dam width B and also varies with

gravity g and excess water height H upstream, as shown in Fig. P1.14. What is the only

possible dimensionally homo-geneous relation for this flow rate?

Solution: So far we know that

Q 

B fcn(H,g). Write this in dimensional

form:

3

 

L {Q} {B}{f(H,g)} {L}{f(H,g)}, T

      

2

 

L

or: {f(H,g)} T

    

Fig. P1.14

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McGraw-Hill Education.Chapter 1  Introduction 1-10

So the function fcn(H,g) must provide dimensions of {L2/T}, but only g contains time.

Therefore g must enter in the form g1/2 to accomplish this. The relation is now

Q  Bg1/2fcn(H), or: {L3/T}  {L}{L1/2/T}{fcn(H)}, or: {fcn(H)}  {L3/2}In

order for fcn(H) to provide dimensions of {L3/2}, the function must be a 3/2 power. Thus

the final desired homogeneous relation for dam flow is:

Q  C B g1/2 H3/2

, where C is a dimensionless constant Ans.

P1.15 The height H that fluid rises in a liquid barometer tube depends upon the liquid

density ρ, the barometric pressure p, and the acceleration of gravity g. (a) Arrange these

four variables into a single dimensionless group. (b) Can you deduce (or guess) the

numerical value of your group?

Solution: This is a problem in dimensional analysis, covered in detail in Chapter 5. Use

the symbols for dimensions suggested with Eq. (1.2): M for mass, L for length, T for

time, F for force,

{H}= {L}, {ρ} = {M/L3}, {g} = {L/T 2}, {p} = {F/L2} = {M/(LT 2)}

where the change in pressure dimensions uses Newton’s law, {F} = {ML/T2}. We see

that we can cancel mass by dividing density by pressure:

  



p

  



  

ML3

ML1T2

  



  

T 2

L2

  

We can eliminate time by multiplying by {g}: {(ρ/p)(g)} = {(T 2/L2)(L/T 2)} = {L–1}.

Finally, we can eliminate length by multiplying by the height {H}:

  



g

  

p

  (H )

  

 L1

  L   1   dimensionless

Thus the desired dimensionless group is ρgH/p, or its inverse, p/ρgH. Answer (a)

(b) You might remember from physics, or other study, that the barometer formula is

p ≈ ρgH. Thus this dimensionless group has a value of approximately 1.0, or unity.

Answer (b)

_______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-11

P1.16 Test the dimensional homogeneity of the boundary-layer x-momentum equation:

   

u u p u v g

   

x

      

x y x y

Solution: This equation, like all theoretical partial differential equations in mechanics,

is dimensionally homogeneous. Test each term in sequence:

      

2

     

M M

u u MLL/T p M/LT u v ; x y T L x L L 2 2 2 2

                        

    

3

LT LT 2

     

M M



M L M/LT { g } ; x L

                2 2 2 2

 

x 3 2

L T

L T L T

All terms have dimension {ML–2T–2}. This equation may use any consistent units.

P1.17 Investigate the consistency of the Hazen-Williams formula from hydraulics:

2.63 p

  

0.54

Q 61.9D L

    

What are the dimensions of the constant “61.9”? Can this equation be used with confidence

for a variety of liquids and gases?

Solution: Write out the dimensions of each side of the equation:

0.54

      

0.54 3 2 ? 2.63 2.63 L p M/LT {Q} {61.9}{D } {61.9}{L } T L L

              

The constant 61.9 has fractional dimensions: {61.9}  {L1.45T0.08M–0.54} Ans.

Clearly, the formula is extremely inconsistent and cannot be used with confidence

for any given fluid or condition or units. Actually, the Hazen-Williams formula, still in

common use in the watersupply industry, is valid only for water flow in smooth pipes

larger than 2-in. diameter and turbulent velocities less than 10 ft/s and (certain) English

units. This formula should be held at arm’s length and given a vote of “No Confidence.”

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McGraw-Hill Education.Chapter 1  Introduction 1-12

*P1.18 (“*” means “difficult”—not just a

plug-and-chug, that is) For small particles at

low velocities, the first (linear) term in Stokes’

drag law, Prob. 1.10, is dominant, hence

F  KV, where K is a constant. Suppose

a particle of mass m is constrained to move horizontally from the initial position x  0 with

initial velocity V  V0. Show (a) that its velocity will decrease exponentially with time;

and (b) that it will stop after travelling a distance x  mV0/K.

Solution: Set up and solve the differential equation for forces in the x-direction:

V t

dV dV m F Drag ma , or: KV m , integrate dt dt V K

       

x x

   

mV V V e x V dt 1 e

mt K mt K o

   

o

t

V 0

o

  Solve and (a,b) Ans.

K

0

o V

Thus, as asked, V drops off exponentially with time, and, as ,

t x K

 

m

P1.19 In his study of the circular hydraulic jump formed by a faucet flowing into a sink,

Watson [47] proposes a parameter combining volume flow rate Q, density  and viscosity

 of the fluid, and depth h of the water in the sink. He claims that the grouping is

dimensionless, with Q in the numerator. Can you verify this?

Solution: Check the dimensions of these four variables, from Table 1.2:

3 3 { } { / } ; { } { / } ; { } { / } ; { } { } Q L T M L M LT h L      

Can we make this dimensionless? First eliminate mass {M} by dividing density by viscosity,

that is,  / has units {T/L2}. (I am pretending that kinematic viscosity is unfamiliar to the

students in this introductory chapter.) Then combine   and Q to eliminate time: (  Q

has units {L}. Finally, divide that by a single depth h to form a dimensionless group:



3 3 { / }{ / } { } Q M L L T   

Ans



h M LT L

{1} dimensionless . Watson is correct. { / }{ }

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McGraw-Hill Education.Chapter 1  Introduction 1-13

P1.20 Books on porous media and atomization claim that the viscosity  and surface

tension  of a fluid can be combined with a characteristic velocity U to form an

important dimensionless parameter. (a) Verify that this is so. (b) Evaluate this parameter

for water at 20C and a velocity of 3.5 cm/s. NOTE: Extra credit if you know the name

of this parameter.

Solution: We know from Table 1.2 that { }= {ML-1T-1}, {U} = {LT-1}, and {  }= {FL-1} =

{MT-2}. To eliminate mass {M}, we must divide  by  , giving { /  } = {TL-1}.

Multiplying by the velocity will thus cancel all dimensions:





U Ans a

is dimensionless, as is its inverse, .( )





U

The grouping is called the Capillary Number. (b) For water at 20C and a velocity of 3.5

cm/s, use Table A.3 to find  = 0.001 kg/m-s and  = 0.0728 N/m. Evaluate



 

U kg m s m s Ans b

  

(0.001 / )(0.035 / ) , .( )

0.00048 2080



2

(0.0728 / )

kg s



U

_______________________________________________________________________

P1.21 Aeronautical engineers measure the pitching moment Mo of a wing and then write

it in the following form for use in other cases:

M

o

  V 2 AC

where V is the wing velocity, A the wing area, C the wing chord length, and ρ the air

density. What are the dimensions of the coefficient β?

Solution: Write out the dimensions of each term in the formula:

M

  FL

o

 

 



 

ML2

T 2



 

  V 2 AC

 



     L2



 

 

 

T 2

L2  L   M

 

  

L3

  



 



 

ML2

T 2



 

 

Thus {β} = {unity} or dimensionless. It is proportional to the moment coefficient in

aerodynamics.

_______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-14

P1.22 The Ekman number, Ek, arises in geophysical fluid dynamics. It is a

dimensionless parameter combining seawater density  , a characteristic length L,

seawater viscosity  , and the Coriolis frequency sin , where  is the rotation rate of

the earth and  is the latitude angle. Determine the correct form of Ek if the viscosity is

in the numerator.

Solution : First list the dimensions of the various quantities:

-3 -1 { } {ML }; { } {L} ; { } {ML T } ; { sin } {T } L       -1 -1  

Note that sin is itself dimensionless, so the Coriolis frequency has the dimensions of .

Only  and  contain mass {M}, so if  is in the numerator,  must be in the denominator.

That combination  / we know to be the kinematic viscosity, with units {L2T-1}. Of the

two remaining variables, only sin contains time {T-1}, so it must be in the denominator.

So far, we have the grouping  /( sin , which has the dimensions {L2}. So we put the

length-squared into the denominator and we are finished:





Ans

2 Dimensionless Ekman number: Ek .

L



sin

 

_______________________________________________________________________

P1.23 During World War II, Sir Geoffrey Taylor, a British fluid dynamicist, used dimensional

analysis to estimate the energy released by an atomic bomb explosion. He assumed that the

energy released, E, was a function of blast wave radius R, air density  , and time t. Arrange

these variables into a single dimensionless group, which we may term the blast wave number.

Solution: These variables have the dimensions {E} = {ML2/T2}, {R} = {L}, { } = {M/L3}, and

{t} = {T}. Multiplying E by t2 eliminates time, then dividing by   eliminates mass, leaving

{L5} in the numerator. It becomes dimensionless when we divide by R5. Thus

2

E

t

Bl ast

wave

number



5



R

____________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-15

P1.24 Air, assumed to be an ideal gas with k = 1.40, flows isentropically through a nozzle.

At section 1, conditions are sea level standard (see Table A.6). At section 2, the temperature is

–50C. Estimate (a) the pressure, and (b) the density of the air at section 2.

Solution: constant k,

From Table A.6, p1 = 101350 Pa, T1 = 288.16 K, and  1 = 1.2255 kg/m3

.

Convert to absolute temperature, T2 = -50°C = 223.26 K. Then, for a perfect gas with

p T

2 2

( )

k k

 

/( 1) 1.4/(1.4 1) 3.5

223.16 ( ) (0.7744) 0.4087 288.16

   

p T

1 1

 

Thus (0.4087)(101350 ) .( )

p Pa Pa Ans a

2

41,400



T

2 2

( )

k

 

1/( 1) 1/(1.4 1) 2.5

223.16 ( ) (0.7744) 0.5278 288.16

   



T

1 1

3

Thus (0.5278)(1.2255 / )



 

2

kg m k

0.647 3 / .( ) g m Ans b

Alternately, once p2 was known, we could have simply computed  2 from the ideal-gas law.

 2 = p2/RT2 = (41400)/[287(223.16)] = 0.647 kg/m3

P1.25 On a summer day in Narragansett, Rhode Island, the air temperature is 74ºF and

the barometric pressure is 14.5 lbf/in2. Estimate the air density in kg/m3

.

Solution: This is a problem in handling awkward units. Even if we use the BG system,

we have to convert. But, since the problem calls for a metric result, better we should

convert to SI units:

T = 74ºF + 460 = 534ºR x 0.5556 (inside front cover) = 297 K

p = 14.5 lbf/in2 x 6894.8 (inside front cover) = 100,700 Pa

The SI gas constant, from Eq. (1.12), is 287 m2/(s2∙K). Thus, from the ideal gas law, Eq.

(1.10),



p

RT

100,700 N/m2



287 m2



 1.18 Ns2

m4

 

s2

 

 K

  297K

This doesn’t look like a density unit, until we realize that 1 N ≡ 1 kg∙m/s2. Making this

substitution, we find that

ρ = 1.18 kg/m3 Answer

Notice that faithful use of SI units will lead to faithful SI results, without further

conversion.

______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-16

P1.26 the tire?

A tire has a volume of 3.0 ft3 and a ‘gage’ pressure (above atmospheric pressure)

of 32 psi at 75F. If the ambient pressure is sea-level standard, what is the weight of air in

Solution: Convert the temperature from 75F to 535R. Convert the pressure to psf:

2 2 2 2 From this compute the density of the air in the tire:

2

   

air

p 6724 lbf/ft 0.00732 slug/ft

3

  

RT (1717 ft lbf/slug R)(535 R)

Then the total weight of air in the tire is

3 2 3

   0.707 lbf

air W g (0.00732 slug/ft )(32.2 ft/s )(3.0 ft ) Ans.

2 p (32 lbf/in)(144 in/ft ) 2116 lbf/ft 4608 21 16 6724 lbf/ft     

800 900

v, ft3/lbm 0.7014 0.8464 0.9653 1.074 1.177

Does this data

P1.27 For steam at a pressure of 45 atm, some values of temperature and specific

volume are as follows, from Ref. 23:

T, ºF 500 600 700 Find an average value of the predicted gas constant R in m2/(s2∙K). reasonably approximate an ideal gas? If not, explain.

Solution: If ideal, the calculated gas constant would, from Table A.4, be about 461

m2/(s2∙K). Try this for the first value, at T = 500ºF = 960ºR. Change to SI units, using

the inside front cover conversions:

T = 500ºF = 960ºR ÷ 1.8 = 533 K; p = 45(101,350 Pa) = 4561 kPa

v = 0.7014 ft3/lbm ÷ 16.019 = 0.0438 m3/kg

pv

(4.561E6 Pa)(0.0438m3/kg)

2

pv  RT,or : R

T

533K 374 m

2

s

 K

This is about 19% lower than the recommended value of Rsteam = 461 m2/(s2∙K).

Continue by filling out the rest of the table:

T, ºF 500 600 700 These are all low, with an average of 418, nine per cent low. The temperature is too low

and the pressure too high. We are too near the saturation line for the ideal gas law to be

accurate.

800 900

R, m2/(s2∙K) 374 409 427 437 443

________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-17

P1.28 Wet air, at 100% relative humidity, is at 40C and 1 atm. Using Dalton’s law of

partial pressures, compute the density of this wet air and compare with dry air.

Solution: Change T from 40C to 313 K. Dalton’s law of partial pressures is

m m p 1 atm p p R T R T

     a w

tot air water a w

 

 

    a w

p p or: m m m for an ideal gas RT R T

tot a w

a w

where, from Table A-4, Rair 

287 and Rwater  461 m2/(s2K). Meanwhile, from Table A-5,

at 40C, the vapor pressure of saturated (100% humid) water is 7375 Pa, whence the partial

pressure of the air is pa  1 atm  pw  101350  7375  93975 Pa.

Solving for the mixture density, we obtain





m m p p 93975 7375 1.046 0.051

a w a w

kg

1.10

        3



R T R T 287(313) 461(313)

m

a w

Ans.

By comparison, the density of dry air for the same conditions is

   

p 101350 kg 1.13

dry air 3

RT 287(313) m

Thus, at 40°C, wet, 100% humidity, air is lighter than dry air, by about 2.7%.

P1.29 A compressed-air tank holds 5 ft3 of air at 120 lbf/in2 “gage,” that is, above

atmospheric pressure. Estimate the work input, in ft-lbf, required to compress this air from

the atmosphere, assuming a reversible isothermal process. (p0 = 14.7 psia).

Solution: Integrate the work of compression, assuming an ideal gas:

2 2

   

mRT p W p d d mRT ln p ln

1-2 2 2

 



2 2

  

              

 

1 1

p

1 1

where the latter form follows from the ideal gas law for isothermal changes. For the given

numerical data, we obtain the quantitative work done:

     

2

3



p lbf 134.7 W p ln 134.7 144 (5ft ) ln . p 14.7 ft

1-2 2 2 2

215,000 ftlbf

Ans

             

 

1

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McGraw-Hill Education.Chapter 1  Introduction 1-18

P1.30 Repeat Prob. 1.29 if the tank is filled with compressed water instead of air. Why is

the work input thousands of times less than that in Prob. 1.29?

Solution: First evaluate the density change of water. At 1 atm, 

o  1.94 slug/ft3. At 120

psi(gage)  134.7 psia, the density would rise slightly according to Eq. (1.22):

 

7

 

3

p 134.7 3001 3000, solve 1.940753 slug/ft , p 14.7 1.94

       

o

water Hence m 3

   

(1.940753)(5 ft ) 9.704 slug

The density change is extremely small. Now the work done, as in Prob. 1.29 above, is

  

m m d W pd pd p p m for a linear pressure rise

           2 2 2

 



1-2 avg 2 2

1 1 1

  

avg

  

3

 

21 ftlbf

Ans.

14.7 134.7 lbf 0.000753 ft Hence W 144 (9.704 slug) 2 slug ft 1.9404

      

1-2 2 2

   

[Exact integration of Eq. (1.22) would give the same numerical result.] Compressing water

(extremely small  ) takes ten thousand times less energy than compressing air, which is

why it is safe to test high-pressure systems with water but dangerous with air.

P1.31 One cubic foot of argon gas at 10C and 1 atm is compressed isentropically to a

new pressure of 600 kPa. (a) What will be its new density and temperature? (b) If

allowed to cool, at this new volume, back to 10C, what will be the final pressure?

Assume constant specific heats.

Solution: This is an exercise in having students recall their thermodynamics. From

Table A.4, for argon gas, R = 208 m2/(s2

-K) and k = 1.67. Note T1 = 283K. First

compute the initial density:

2

1

   

101350 / 1.72 /

p N m kg m

3

1 2 2

RT m s K K

1



(208 / )(283 )

For an isentropic process at constant k,

k

1.67

p Pa Ans a

2 2 2

    

  

3

600,000 5.92 ( ) ( ) , Solve .( ) 101,350 1.72

2

4.99 kg/m

p Pa



1 1

k k

/( 1) 1.67 / 0.67 2 2 2

p T T T C Ans a



    

2

578K

5.92 ( ) ( ) , Solve 305 .( ) 283

p T K

1 1

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McGraw-Hill Education.Chapter 1  Introduction 1-19

(b) 283K. Thus

Cooling at constant volume means  stays the same and the new temperature is

𝑝3 = 𝑅𝜌3𝑇3 = (208 m2 s2 ∙ K

⁄ )(4.99 kg m3

⁄ )(283K) = 293700 N/𝑚2 𝐴𝑛𝑠.(b)

2

k g

m

T R



p kPa 2 9 4

3 b

 

3

3

0 0 . 5 ( 2

2 0 8 ) (

K

)

2 8 3 ) (



Pa

0 0 0 , 2 9 4

An s

) . (

3

m

s

K

P1.32 A blimp is approximated by a prolate spheroid 90 m long and 30 m in diameter.

Estimate the weight of 20°C gas within the blimp for (a) helium at 1.1 atm; and (b) air at

1.0 atm. What might the difference between these two values represent (Chap. 2)?

Solution: Find a handbook. The volume of a prolate spheroid is, for our data,

2 2 3 2 2 LR (90 m)(15 m) 42412 m 3 3

     

Estimate, from the ideal-gas law, the respective densities of helium and air:

    He

p 1.1(101350) kg (a) 0.1832 ; R T 2077(293) m

helium 3

He

    air

p 101350 kg (b) 1.205 . R T 287(293) m

air 3

air

Then the respective gas weights are

  

3

kg m W g 0.1832 9.81 (42412 m ) (a)

 

  76000 N

Ans.

He He 3 2

m s

      

501000N

     air air W g (1.205)(9.81)(42412) (b) Ans.

The difference between these two, 425000 N, is the buoyancy, or lifting ability, of the

blimp. [See Section 2.8 for the principles of buoyancy.]

P1.33 the tank, in m3

.

A tank contains 9 kg of CO2 at 20ºC and 2.0 MPa. Estimate the volume of

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McGraw-Hill Education.Chapter 1  Introduction 1-20

Solution: All we have to do is find the density. For CO2, from Table A.4, R = 189

m2/(s2∙K). Then



p

RT

(2,000,000 Pa)

([189m2 /(s2

 K](20 + 273K)

 36.1 kg / m3

3 3 Thenthetankvolume / (9 )/(36.1 / ) 0.25 . m kg kg m m Ans    

________________________________________________________________________

P1.34 Consider steam at the following state near the saturation line: (p1, T1)  (1.31

MPa, 290°C). Using the ideal gas relation (Table A.4) and the Steam Tables,

respectively, calculate and compare (a) the density ρ1; and (b) the density ρ2 if the steam

expands isentropically to a new pressure of 414 kPa. Discuss your results.

Solution: K. Then,

From Table A.4, for steam, k  1.33, and R  461 m2/(s2K). Convert T1  563

p Pa kg Ans

1

   

1,310,000 . (a)

5.05

1 2 2 3

RT m s K K m

1



(461 )(563 )

1/ 1/1.33

2 2 2

   

  

k p kPa kg or Ans

   

414 0.421, : . (b)

2.12

      



5.05 1310

 

2 3

1 1

p kPa m

From the online Steam Tables, just look it up:

SpiraxSarcoTables:  1

 5.23 kg/m3 Ans. (a),  2

 2.16 kg/m3 Ans. (b)

The ideal-gas error is only about 3%, even though the expansion approached the saturation line.

P1.35 In Table A-4, most common gases (air, nitrogen, oxygen, hydrogen, CO, NO) have

a specific heat ratio k  1.40. Why do argon and helium have such high values? Why does

NH3 have such a low value? What is the lowest k for any gas that you know?

Solution: In elementary kinetic theory of gases [21], k is related to the number of

“degrees of freedom” of the gas: k  1  2/N, where N is the number of different modes of

translation, rotation, and vibration possible for the gas molecule.

Example: Monotomic gas, N  3 (translation only), thus k  5/3

This explains why helium and argon, which are monatomic gases, have k  1.67.

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McGraw-Hill Education.Chapter 1  Introduction 1-21

Example: Diatomic gas, N  5 (translation plus 2 rotations), thus k  7/5

This explains why air, nitrogen, oxygen, NO, CO and hydrogen have k  1.40.

But NH3 has four atoms and therefore more than 5 degrees of freedom, hence k will be

less than 1.40. Most tables list k = 1.32 for NH3 at about 20C, implying N  6.

The lowest k known to this writer is for uranium hexafluoride,

238UF6, which is a very

complex, heavy molecule with many degrees of freedom. The estimated value of k for this

heavy gas is k  1.06.

P1.36 50ºC, are listed as follows:

Experimental data [49] for the density of n-pentane liquid for high pressures, at

Pressure, kPa 100 10230 20700 34310

Density, kg/m3 586.3 604.1 617.8 632.8

(a) Fit this data to reasonably accurate values of B and n from Eq. (1.19).

(b) Evaluate ρ at 30 MPa.

Solution:

Eq. (1.19) is p/po ≈ (B+1)(ρ/ρo)n

– B. The first column is po = 100 kPa and ρo = 586.3 kg/m3

.

(a) The writer found it easiest to guess n, say, n = 7 for water, and then solve for B from

the data:



 



p

po







  

o

 

n



 



B



n



  

o

 

1

The value n = 7 is too low. Try n = 8,9,10,11 – yes, 11 works, with B ≈ 260 ± 2.

Equation (1.19), with B = 260 and n = 11, is accurate to a fraction of 1%. Ans.(a)

(b) Use our new formula for p = 30 MPA = 30,000 kPa.

p



po

30000



100 (261) 

  

11

586.1

 

 260, solve   628 kg

m3 Ans.(b)

________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-22

P1.37 A near-ideal gas has M  44 and cv  610 J/(kgK). At 100°C, what are (a) its

specific heat ratio, and (b) its speed of sound?

Solution: The gas constant is R  





 189 J/(kgK). Then

       v v 1.31

Ans.

With k and R known, the speed of sound at 100ºC  373 K is estimated by

2 2 a kRT 1.31[189 m/(s K)](373 K) 304 m/s Ans. (b)

   

2 c R/(k 1), or: k 1 R/c 1 189/610 (a) [It is pr obably NO]

P1.38 In Fig. P1.38, if the fluid is glycerin at 20°C and the width between plates is 6

mm, what shear stress (in Pa) is required to move the upper plate at V  5.5 m/s? What is

the flow Reynolds number if “L” is taken to be the distance between plates?

Fig. P1.38

(a) For glycerin at 20°C, from Table 1.4,   1.5 N · s/m2. The shear stress is

Solution: found from Eq. (1) of Ex. 1.8:







  

V (1.5 Pas)(5.5 m/s) 1380 Pa

h (0.006 m)

Ans

. (a)

The density of glycerin at 20°C is 1264 kg/m3. Then the Reynolds number is defined by

Eq. (1.24), with L  h, and is found to be decidedly laminar, Re < 1500:

VL (1264 kg/m )(5.5 m/s)(0.006 m) Re 3



  

28

L



Ans

. (b) 1.5 kg/m s



P1.39 Knowing   1.80E5 Pa · s for air at 20°C from Table 1-4, estimate its viscosity

at 500°C by (a) the power-law, (b) the Sutherland law, and (c) the Law of Corresponding

States, Fig. 1.5. Compare with the accepted value  (500°C)  3.58E5 Pa · s.

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McGraw-Hill Education.Chapter 1  Introduction 1-23

Solution: and from Eq. (1.30a),

First change T from 500°C to 773 K. (a) For the power-law for air, n  0.7,

n

 

0.7

kg

773 (T/T ) (1.80E 5) . (a) 293

 

       

3.55E 5

Ans

o o

ms



This is less than 1% low. (b) For the Sutherland law, for air, S  110 K, and from Eq. (1.30b),

1.5 1.5

   

(T/T ) (T S)  

o o

(773/293) (293 110)

 

o

(1.80E 5) (T S)           

 

(773 110)

kg



3.52E 5

ms

. (b) Ans



This is only 1.7% low. (c) Finally use Fig. 1.5. Critical values for air from Ref. 3 are:

c c Air: 1.93E 5 Pas T 132 K (“mixture” estimates)

    

At 773 K, the temperature ratio is T/Tc  773/132  5.9. From Fig. 1.5, read  µc  1.8.

Then our critical-point-correlation estimate of air viscosity is only 3% low:

kg

     

3.5E 5

c 1.8 (1.8)(1.93E 5) . (c) Ans

ms



P1.40 Glycerin at 20ºC fills the space between a hollow sleeve of diameter12 cm and a

fixed coaxial solid rod of diameter 11.8 cm. The outer sleeve is rotated at 120 rev/min.

Assuming no temperature change, estimate the torque required, in N∙m per meter of rod

length, to hold the inner rod fixed.

ω =120 rev/min

Solution: From Table A.3, the viscosity of glycerin is

1.49 kg/(m∙s). The clearance C is the difference in radii,

6 cm – 5.9 cm = 0.1 cm = 1 mm. The velocity of the sleeve

surface is V = ωro = [2π(120)/60](0.06) = 0.754 m/s.

The shear stressin the glycerin is approximately

τ = μV/C = (1.49)(0.754)/(0.001) = 1123 Pa.

6 cm

5.9 cm

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McGraw-Hill Education.Chapter 1  Introduction 1-24

The shear forces are all perpendicular to the radius

and thus have a total torque

i i Torque rLr N m m m m permeter Ans      2 (2 ) (1123 / )[2 (0.059 )(1 )](0.059 ) . 25Nm

 

________________________________________________________________________

P1.41 An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is

falling concentrically through a long vertical sleeve of diameter 6.04 cm. The clearance

is filled with SAE 50 oil at 20C. Estimate the terminal (zero acceleration) fall velocity.

Neglect air drag and assume a linear velocity distribution in the oil. [HINT: You are given

diameters, not radii.]

Solution: From Table A.3 for SAE 50 oil,  = 0.86 kg/m-s. The clearance is the

difference in radii: 3.02 – 3.0 cm = 0.02 cm = 0.0002 m. At terminal velocity, the

cylinder weight must balance the viscous drag on the cylinder surface:

V W A DL C r r

    

  

( )( ) , where clearance

wall wall sleeve cylinder

C

kg V N m m





or : 30 [0.86 )( )] (0.06 )(0.40 ) 0.0002



m s m

Solve for / .

 0.0925

V m s Ans

________________________________________________________________________

P1.42 Helium at 20ºC has a viscosity of 1.97E-5 kg/(m∙s). Use the power law

approximation and the data of Table A.4 to estimate the temperature, in ºC, at which

helium’s viscosity will double.

Solution: From Eq. (1.27), the power-law, μ/μo ≈ (T/To)n, has, in Table A.4, a value n =

0.67 for helium. With To = 20° = 293.16K, this formula requires that

μ/μo = 2.0 = (T/293.16)0.67, or: T = (2.814)(293.16) = 825 K = 552°C Ans.

______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-25

P1.43 For the flow between two parallel plates of Fig. 1.8, reanalyze for the case of slip

flow at both walls. Use the simple slip condition, uwall = l (du/dy)wall, where l is the mean

free path of the fluid. (a) Sketch the expected velocity profile and (b) find an expression

for the shear stress at each wall.

Solution: As in Fig. 1.8, the shear stress remains constant between the two plates. The

analysis is correct up to the relation u = a + b y . There would be equal slip velocities,

 u, at both walls, as shown in the following sketch

U

 u

y

h

u(y)

Ans. (a)

 u

x

Fixed plate

Because of slip at the walls, the boundary conditions are different for u = a + b y :

At 0 : ( )

    



du y u u a b dy

y



0

U

         



At y h u U u a bh U b b bh or b h

: , : 2



One way to write the final solution is

u  u  ( U

h  2l ) y , where u  ( l

h  2l )U ,

and wall  ( du

U

)wall 

dy

h  2l

Ans.(b)

______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-26

P1.44 One type of viscometer is simply a long capillary tube.. A commercial device is

shown in Prob. C1.10. One measures the volume flow rate Q and the pressure drop Δp

and, of course, the radius and length of the tube. The theoretical formula, which will be

discussed in Chap. 6, is   4 8 /( ) p QL R  

. For a capillary of diameter 4 mm and

length 254 mm, the test fluid flows at 0.9 m3/h when the pressure drop is 400 kPa. Find

the predicted viscosity in kg/m∙s.

Solution: the theoretical formula:

Convert: Q = 0.9/3600 = 0.00025 m3/s, R = D/2 = 0.002 m. Then apply

4 4 400,000 8 /( ) 8 (0.00025)(0.254)/[ (0.002)] p QL R       

Solve for μ = 0.0396 kg/m·s ≈ 0.040 kg/m·s Ans.

If the density is, say, 900 kg/m3, the Reynolds number ReD is about 1800, or below

transition.

P1.45 A block of weight W slides down

an inclined plane on a thin film of oil, as in

Fig. P1.45 at right. The film contact area is

A and its thickness h. Assuming a linear

velocity distribution in the film, derive an

analytic expression for the terminal velocity

V of the block. Fig. P1.45

Solution: Let “x” be down the incline, in the direction of V. By “terminal” velocity we

mean that there is no acceleration. Assume a linear viscous velocity distribution in the film

below the block. Then a force balance in the x direction gives:

                

V F W sin A W sin A ma 0, h

x x

hWsin



or: V . Ans



terminal



A

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McGraw-Hill Education.Chapter 1  Introduction 1-27

P1.46 A simple and popular model for two non-newtonian fluids in Fig. 1.9a is the power-law:



du

(

C )

n

dy

where C and n are constants fit to the fluid [16]. From Fig. 1.9a, deduce the values of the

exponent n for which the fluid is (a) newtonian; (b) dilatant; and (c) pseudoplastic. (d) Consider

the specific model constant C = 0.4 N-sn/m2, with the fluid being sheared between two parallel

plates as in Fig. 1.8. If the shear stress in the fluid is 1200 Pa, find the velocity V of the upper

plate for the cases (d) n = 1.0; (e) n = 1.2; and (f) n = 0.8.

Solution: By comparing the behavior of the model law with Fig. 1.9a, we see that

(a) Newtonian: n = 1 ; (b) Dilatant: n > 1 ; (c) Pseudoplastic: n < 1 Ans.(a,b,c)

From the discussion of Fig. 1.8, it is clear that the shear stress is constant in a fluid sheared

between two plates. The velocity profile remains a straight line (if the flow is laminar), and the

strain rate duIdy = V/h . Thus, for this flow, the model becomes  = C(V/h)n. For the three

given numerical cases, we calculate:

N

N



(

d

)

n



1

:





1200



C

(

V

/

h

)

n



4 . 0

(

2

m

m

1

s

2

001 . 0 V

)(

m

1

)

sol ve

V



m

3. 0

Ans

,

s

N

(

e

)

n



2 . 1

:





1200



C

(

V

/

h

)

n



4 . N

0 (



2

m

m

2 . 1

s

2

001 . 0 V

)(

m

2 . 1

)

m

sol ve

V



0. 79

,

s

N

(

f

)

n



8 . 0

:





1200



C

(

V

/

h

)

n



4 . N

0 (



2

m

2

m

8 . 0

s

001 . 0 V

)(

m

8 . 0

m

)

sol ve

V



22

,

s

A small change in the exponent n can sharply change the numerical values.

. (

d

)

Ans

Ans

. (

e

)

. (

f

)

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-28

P1.47 Data for the apparent viscosity of average human blood, at normal body

temperature of 37C, varies with shear strain rate, as shown in the following table.

Shear strain rate, s-1 1 10 100 1000

Apparent viscosity,

kg/(ms)

0.011 0.009 0.006 0.004

(a) Is blood a nonnewtonian fluid? (b) If so, what type of fluid is it? (c) How do these

viscosities compare with plain water at 37C?

Solution: (a) By definition, since viscosity varies with strain rate, blood is a

nonnewtonian fluid.

(b) Since the apparent viscosity decreases with strain rate, it must be a pseudoplastic

fluid, as in Fig. 1.9(a). The decrease is too slight to call this a “plastic” fluid. (c) These

viscosity values are from six to fifteen times the viscosity of pure water at 37C, which is

about 0.00070 kg/m-s. The viscosity of the liquid part of blood, called plasma, is about

1.8 times that of water. Then there is a sharp increase of blood viscosity due to

hematocrit, which is the percentage, by volume, of red cells and platelets in the blood.

For normal human beings, the hematocrit varies from 40% to 60%, which makes this

blood about six times the viscosity of plasma.

_______________________________________________________________________

P1.48 A thin moving plate is separated from two fixed plates by two fluids of unequal

viscosity and unequal spacing, as shown below. The contact area is A. Determine (a) the

force required, and (b) is there a necessary relation between the two viscosity values?

Solution: (a) Assuming a linear velocity distribution on each side of the plate, we obtain

 

1 2

 

  

V V A ( )

1 2 F A A . a Ans

     

h h

1 2

The formula is of course valid only for laminar (nonturbulent) steady viscous flow.

(b) Since the center plate separates the two fluids, they may have separate, unrelated shear

stresses, and there is no necessary relation between the two viscosities.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-29

P1.49 An amazing number of commercial and laboratory devices have been developed

to measure fluid viscosity, as described in Ref. 27. Consider a concentric shaft, fixed axially

and rotated inside the sleeve. Let the inner and outer cylinders have radii ri and ro,

respectively, with total sleeve length L. Let the rotational rate be (rad/s) and the applied

torque be M. Using these parameters, derive a theoretical relation for the viscosity  of the

fluid between the cylinders.

Solution: Assuming a linear velocity distribution in the annular clearance, the shear stress is



V



r

i

 

  

 

r r r

o i

This stress causes a force dF   dA   (ri d )L on each element of surface area of the inner

shaft. The moment of this force about the shaft axis is dM  ri dF. Put all this together:



r r L M r dF r r Ld

  

 2 3

i i i

   

2 i i



 

r r r r

0

o i o i

3

  Solve for the viscosity: . Ans

( )         i i r r r L

P1.50 A simple viscometer measures the time t for a solid sphere to fall a distance L

through a test fluid of density  . The fluid viscosity  is then given by

2

net W t DL





 

if t

3

DL

 

where D is the sphere diameter and Wnet is the sphere net weight in the fluid.

(a) Show that both of these formulas are dimensionally homogeneous.

(b) Suppose that a 2.5 mm diameter aluminum sphere (density 2700 kg/m3) falls in an oil of

density 875 kg/m3

. If the time to fall 50 cm is 32 s, estimate the oil viscosity and verify that the

inequality is valid.

Solution: (a) Test the dimensions of each term in the two equations:

2

 

     

t M ML T T M

net

W ( / )( ) { } and , dimensions OK. (3 ) (1)( )( )



Yes

               

LT DL L L LT



   

2 (1)( / )( )( ) { } { } and  

3



 

DL M L L L t T T Ans

Yes

       

   



M LT

{ } , dimensions OK. . (a) /

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-30

(b) Evaluate the two equations for the data. We need the net weight of the sphere in the fluid:

      

W gVol

3 2 3

net sphere fluid fluid ( ) ( ) (2700 875 kg/m)(9.81 m/s )( /6)(0.0025 m)



0.000146 N

Then 

W

nett

3DL

(0.000146 N )(32 s)

3(0.0025 m)(0.5 m) 0.40 kg

ms

Ans. (b)

3 2 DL 2(875 / )(0.0025 )(0.5 )



kg m m m

t s

Check 32 compared to  



0.40 /



kg m s



s t

5.5 OK, is greater

P1.51 An approximation for the boundary-layer

shape in Figs. 1.6b and P1.51 is the formula

y

U



y

u 0

(

y

)

U

 y

2 si n(

)



,







where U is the stream velocity far from the wall

and  is the boundary layer thickness, as in Fig. P.151.

If the fluid is helium at 20C and 1 atm, and if U =

u(y)

10.8 m/s and  = 3 mm, use the formula to (a) estimate

0

the wall shear stress  w in Pa; and (b) find the position

in the boundary layer where  is one-half of  w.

Fig. P1.51

Solution: From Table A.4, for helium, take R = 2077 m2/(s2

-K) and  = 1.97E-5 kg/m-s.

(a) Then the wall shear stress is calculated as



u



w







y





y

| 0

y



0





(

U

cos

)

2



2



y



U

 



2





97 . 1 (

E

: val ues Num er i cal





w



5

kg

/

m



8 . 10 ) (

s

m

/

s

)

Pa 0. 11

Ans

. (

a

)

003 . 0 ( 2

m

)

y = 

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McGraw-Hill Education.Chapter 1  Introduction 1-31

A very small shear stress, but it has a profound effect on the flow pattern.

(b) The variation of shear stress across the boundary layer is a cosine wave,  =  (du/dy):



U



y



y







2 δ



( b Ans

y w

y

)



)

2 cos(





)

2





2 cos(



when



: or

,

y

) . (

w 3



2

2



3

_______________________________________________________________________

P1.52 The belt in Fig. P1.52 moves at steady velocity V and skims the top of a tank of

oil of viscosity  . Assuming a linear velocity profile, develop a simple formula for the belt-

drive power P required as a function of (h, L, V, B,  ). Neglect air drag. What power P in

watts is required if the belt moves at 2.5 m/s over SAE 30W oil at 20C, with L  2 m, b

 60 cm, and h  3 cm?

Fig. P1.52

Solution: The power is the viscous resisting force times the belt velocity:

 

2 L

V b

Ans

V P A V (bL)V . h

 

oil belt belt



      

h

(b) For SAE 30W oil,   0.29 kg/m  s. Then, for the given belt parameters,

2

  

 

2 2

kg m 2.0 m kg m P V bL/h 0.29 2.5 (0.6 m) 73 . (b) m s s 0.03 m s

     



  

73 W

Ans

  

3

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McGraw-Hill Education.Chapter 1  Introduction 1-32

*P1.53 A solid cone of base ro and initial angular velocity  o is rotating inside a

conical seat. Neglect air drag and derive a formula for the cone’s angular velocity  (t) if

there is no applied torque.

Solution: At any radial position r  ro on the cone surface and instantaneous rate  ,

Fig. P1.53

  



r dr d(Torque) r dA r 2 r , h sin

       w

   

or: Torque M

ro

 

 2r3 dr 

hsin

0

4

  ro

2hsin

We may compute the cone’s slowing down from the angular momentum relation:



2

d 3 M I , where I (cone) mr , m cone mass dt 10

   

o o o

Separating the variables, we may integrate:

w t 4

2





r

 

5 r t



d dt, or: .

o

 

o

 

 

o

exp 3mhsin 

Ans



o

    

2hI sin



o 0

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-33

*P1.54 A disk of radius R rotates at

angular velocity  inside an oil container of

viscosity  , as in Fig. P1.54. Assuming a

linear velocity profile and neglecting shear

on the outer disk edges, derive an expres-

sion for the viscous torque on the disk. Fig. P1.54

Solution: At any r  R, the viscous shear    r/h on both sides of the disk. Thus,





   



r d(torque) dM 2r dA 2r 2 rdr, h

w

R





3

 

4 R

or: M 4 r dr

   



Ans

h

h

0

W

h

Wo

Fig. P1.55

P1.55 A block of weight W is being pulled

over a table by another weight Wo, as shown in

Fig. P1.55. Find an algebraic formula for the

steady velocity U of the block if it slides on an

oil film of thickness h and viscosity  . The block

bottom area A is in contact with the oil.

Neglect the cord weight and the pulley friction.

Solution: This problem is a lot easier to solve than to set up and sketch. For steady motion,

there is no acceleration, and the falling weight balances the viscous resistance of the oil film:

U



F

x

bl oc k



,

0





A



W

o



(



h

)

A



W

o

for Sol ve

U



Ans

.



A

The block weight W has no effect on steady horizontal motion except to smush the oil film.

W

o

h

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McGraw-Hill Education.Chapter 1  Introduction 1-34

*P1.56 For the cone-plate viscometer in

Fig. P1.56, the angle is very small, and the

gap is filled with test liquid  . Assuming a

linear velocity profile, derive a formula for

the viscosity  in terms of the torque M and

cone parameters.

Fig. P1.56

Solution: For any radius r  R, the liquid gap is h  r tan . Then

    

r dr d(Torque) dM dA r 2 r r, or rtan cos     

    

w

  

 

P1.57 Extend the steady flow between a fixed lower plate and a moving upper plate,

from Fig. 1.8, to the case of two immiscible liquids between the plates, as in Fig. P1.57.

V

h2

 2

y

h1

 1

x

Fixed

Fig. P1.57

(a) Sketch the expected no-slip velocity distribution u(y) between the plates. (b) Find an

analytic expression for the velocity U at the interface between the two liquid layers. (c)

What is the result if the viscosities and layer thicknesses are equal?

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McGraw-Hill Education.Chapter 1  Introduction 1-35

Solution: We begin with the hint, from Fig. 1.8, that the shear stress is constant

between the two plates. The velocity profile would be a straight line in each layer, with

different slopes:

V

(2)

y

U Ans. (a)

(1)

x

Fixed

Here we have drawn the case where  2 >  1, hence the upper profile slope is less. (b) Set

the two shear stresses equal, assuming no-slip at each wall:

U V U

 

( 0) ( )

  

   

1 1 2 2

h h

1 2

1 Solve for .( )



[ ]

U V Ans b h



1

1 2





h

2 1

(c) For equal viscosities and layer thicknesses, we get the simple result U = V/2. Ans.

______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-36

P1.58 The laminar-pipe-flow example of Prob. 1.14 leads to a capillary viscometer

[29], using the formula   r0 4p/(8LQ). Given r0 

2 mm and L 

25 cm. The data are

Q, m3/hr: 0.36 0.72 1.08 1.44 1.80

p, kPa: 159 318 477 1274 1851

Estimate the fluid viscosity. What is wrong with the last two data points?

Solution: Apply our formula, with consistent units, to the first data point:

r p (0.002 m) (159000 N/m ) N s p 159kPa: 4 4 2

 

 

o

    



3 2

0.040 8LQ 8(0.25 m)(0.36/3600 m /s) m

Do the same thing for all five data points:

p, kPa: 159 318 477 1274 1851

 , N·s/m2: 0.040 0.040 0.040 0.080(?) 0.093(?) Ans.

The last two estimates, though measured properly, are incorrect. The Reynolds number of the

capillary has risen above 2000 and the flow is turbulent, which requires a different formula.

P1.59 A solid cylinder of diameter D, length L, density  s falls due to gravity inside a tube

of diameter Do. The clearance, o (D D) D,   is filled with a film of viscous fluid (, ).

Derive a formula for terminal fall velocity and apply to SAE 30 oil at 20C for a steel cylinder

with D  2 cm, Do  2.04 cm, and L  15 cm. Neglect the effect of any air in the tube.

Solution: The geometry is similar to Prob. 1.47, only vertical instead of horizontal. At

terminal velocity, the cylinder weight should equal the viscous drag:



 

 2

V a 0: F W Drag g D L DL, 4 (D D)/2

z z s

        

  

  

o

s o gD(D D)





 or: V . Ans

8



For the particular numerical case given, ρsteel  7850 kg/m3. For SAE 30 oil at 20C,

  0.29 kg/m·s from Table 1.4. Then the formula predicts

3 2



 

gD(D D) (7850 kg/m )(9.81 m/s )(0.02 m)(0.0204 0.02 m)

V

s o

 

terminal

8 8(0.29 kg/m s)





. Ans 0.265 m/s



_________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-37

P1.60 Pipelines are cleaned by pushing through them a close-fitting cylinder called a pig.

The name comes from the squealing noise it makes sliding along. A new non-toxic pig,

driven by compressed air, is used for cleaning cosmetic and beverage pipes. Suppose the pig

diameter is 5-15/16 in and its length 26 in. It cleans a 6-in-diameter pipe at a speed of 1.2

m/s. If the clearance is filled with glycerin at 20C, what pressure difference, in pascals, is

needed to drive the pig? Assume a linear velocity profile in the oil and neglect air drag.

Solution: Since the problem calls for pascals, convert everything to SI units:

Find the shear stress in the oil, multiply that by the cylinder wall area to get the required force,

and divide the force by the area of the cylinder face to find the required pressure difference.

D

cyl

15



5 (

16

m

in

0254 .

)(

0

in

)



1508 . 0

m

;

L



(

26

m

in

0254 . 0

)(

in

)



6604 . 0

m

15

Clearance



C



(

D



D

pipe

cyl

)

/

2



(

6



5

16

m

in

0254 . 0

)(

in

)

/

2

000794 . 0



m

Table

A.3,

glycerin

:





1

49 .

kg

/

m



s

,





1260

kg

/

3

m

(



not

Shear

stress







V

/

C



49 . 1 (

kg

/

m



s

)(

2 . 1

m

/

000794 .

s

)

/

0

m



Shear

force

F





A

wall





(



D

cyl

L

)



(

2252

N

/

m

2

)[



1508 . 0

(

m

6604 . 0

)(

m

)]



705

N

F

705

N

Finally,

Δ

p





39500

Pa

Ans

A

cyl

2

face

(



/

)( 4

1508 . 0

m

)

needed

)

2252

N

/

2

m

.

Comment: The Reynolds number of the clearance flow, Re =  VC/ , is approximately 0.8.

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McGraw-Hill Education.Chapter 1  Introduction 1-38

P1.61 An air-hockey puck has m  50 g and D  9 cm. When placed on a 20C air table,

the blower forms a 0.12-mm-thick air film under the puck. The puck is struck with an

initial velocity of 10 m/s. How long will it take the puck to (a) slow down to 1 m/s; (b) stop

completely? Also (c) how far will the puck have travelled for case (a)?

Solution: For air at 20C take   1.8E5 kg/m·s. Let A be the bottom area of the

puck, A  D2/4. Let x be in the direction of travel. Then the only force acting in the

x direction is the air drag resisting the motion, assuming a linear velocity distribution in

the air:

V dV F A A m , where h airfilmthickness h dt

     

 x

Separate the variables and integrate to find the velocity of the decelerating puck:

V t

     

dV 

Kt

A K dt, or V V e , where K V o

mh

V 0

o

Integrate again to find the displacement of the puck:

   t

o



Kt

V x V dt [1 e ] K

0

Apply to the particular case given: air,   1.8E5 kgm·s, m  50 g, D  9 cm, h  0.12 mm,

Vo  10 m/s. First evaluate the time-constant K:

A (1.8E 5 kg/m s)[( /4)(0.09 m) ] K 2

 

 

1

  

0.0191 s

mh (0.050 kg)(0.00012 m)

(a) When the puck slows down to 1 m/s, we obtain the time:



1 Kt (0.0191 s )t

 

   

o V 1 m/s Ve (10 m/s)e , or t 121 s Ans. (a)

(b) The puck will stop completely only when e–Kt  0, or: t   Ans. (b)

(c) For part (a), the puck will have travelled, in 121 seconds,

o

 

     Kt (0.0191)(121)

Ans. 472 m

V 10 m/s x (1 e ) [1 e ] (c) K 0.0191 s



1

This may perhaps be a little unrealistic. But the air-hockey puck does decelerate slowly!

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McGraw-Hill Education.Chapter 1  Introduction 1-39

P1.62 The hydrogen bubbles in Fig. 1.13 have D  0.01 mm. Assume an “air-water”

interface at 30C. What is the excess pressure within the bubble?

Solution: At 30C the surface tension from Table A-1 is 0.0712 N/m. For a droplet or

bubble with one spherical surface, from Eq. (1.32),

   

2Y 2(0.0712 N/m) p R (5E 6 m)

28500Pa Ans.



P1.63 Derive Eq. (1.37) by making a force balance on the fluid interface in Fig. 1.9c.

Solution: The surface tension forces YdL1 and YdL2 have a slight vertical component.

Thus summation of forces in the vertical gives the result

 z 2 1

 



F 0 2YdL sin(d /2)



  

2YdL sin(d /2) pdA

1 2

Fig. 1.9c

But dA  dL1dL2 and sin(d /2)  d /2, so we may solve for the pressure difference:

       

1 1

2 1 1 2 1 2

Y

dL d dL d d d p Y Y dL dL dL dL 1 2

 Ans.

             

R R

1 2 1 2

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-40

P1.64 Pressure in a water container can be measured by an open vertical tube called a

piezometer – see Fig. P2.11 for a typical sketch. If the expected water rise is about 20

cm, what tube diameter is needed to ensure that the error due to capillarity will be less

than 3 per cent?

Solution: in Ex. 1.8:

For water on glass, take ϒ = 0.073 N/m and θ = 0º. We want the capillary rise

h to be less than 3% of 20 cm, or about 0.006 m. The appropriate formula was developed

2 cos 2(0.073 / )cos(0 ) 0.006 N m h m 

 

   

3 2



, 0.0025 (998 / )(9.81 / )

solve for R m gR kg m m s R

Thus the desired tube diameter D = 2R should be greater than 0.005 m = 5 mm Ans.

_________________________________________________________________

1.65 The system in Fig. P1.65 is used to

estimate the pressure p1 in the tank by

measuring the 15-cm height of liquid in the

1-mm-diameter tube. The fluid is at 60C.

Calculate the true fluid height in the tube

and the percent error due to capillarity if

the fluid is (a) water; and

(b) mercury.

Fig. P1.65

Solution: This is a somewhat more realistic variation of Ex. 1.9. Use values from that

example for contact angle  :

(a) Water at 60C:   9640 N/m3

,   0:

4Ycos 4(0.0662 N/m)cos(0 ) h 



  

3



0.0275 m, D (9640 N/m )(0.001 m)

or: htrue  15.0 – 2.75 cm  12.25 cm (+22% error) Ans. (a)

(b) Mercury at 60C:   132200 N/m3

,   130:

4Ycos 4(0.47 N/m)cos130 h 



  

3



0.0091 m, D (132200 N/m )(0.001 m)

    true or: h 15.0 0.91 15.91cm( 6%error) Ans. (b)

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McGraw-Hill Education.Chapter 1  Introduction 1-41

1.66 DuNouy Tensiometer is a ring-pull device. It lifts a thin wire ring from water. The

force is very small and may be measured by a calibrated soft-spring balance.

Platinumiridium is recommended for the ring, being noncorrosive and highly wetting to

most liquids. A thin wire ring, 3 cm in diameter, is lifted from a water surface at 20°C in

the device. Neglecting the wire weight, what is the lift force required?

Solution: In the literature this ring-pull device is called a DuNouy Tensiometer. The

forces are very small and may be measured by a calibrated soft-spring balance. Platinum-

iridium is recommended for the ring, being noncorrosive and highly wetting to most

liquids. There are two surfaces, inside and outside the ring, so the total force measured is

    F 2(Y D) 2Y D

This is crude—commercial devices recommend multiplying this relation by a correction

factor f  O(1) which accounts for wire diameter and the distorted surface shape.

For the given data, Y  0.0728 N/m (20C water/air) and the estimated pull force is

   F 2(0.0728 N/m)(0.03 m) . 00137N Ans.

For further details, see, e.g., F. Daniels et al., Experimental Physical Chemistry, 7th ed.,

McGraw-Hill Book Co., New York, 1970.

1.67 A vertical concentric annulus, with outer radius ro and inner radius ri, is lowered

into fluid of surface tension Y and contact angle   90. Derive an expression for the

capillary rise h in the annular gap, if the gap is very narrow.

Y

θ

h

ri

ro

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McGraw-Hill Education.Chapter 1  Introduction 1-42

Solution: For the figure above, the force balance on the annular fluid is

        2 2

o i o i cos (2 2 ) r r Y r r g h

 

Cancel where possible and the result is

h 2Y cos /{g(ro ri )} Ans.

*P1.68 Analyze the shape  (x) of the

water-air interface near a wall, as shown.

Assume small slope, R1  d2 /dx2. The

pressure difference across the interface is

p   g , with a contact angle  at x  0 and

a horizontal surface at x  . Find an

expression for the maximum height h.

Fig. P1.68

Solution: This is a two-dimensional surface-tension problem, with single curvature. The

surface tension rise is balanced by the weight of the film. Therefore the differential equation is

2

 

   

Y d d p g Y <<1 R dx dx

       

2

This is a second-order differential equation with the well-known solution,

      1 2 C exp[Kx] C exp[ Kx], K ( g/Y)

To keep  from going infinite as x  , it must be that C1  0. The constant C2 is found

from the maximum height at the wall:

     x 0 2 2 h C exp(0), hence C h

Meanwhile, the contact angle shown above must be such that,

  

x 0

d cot cot ) hK, thus h dx K

      

The complete (small-slope) solution to this problem is:

1/2 hexp[ ( g/Y) x],   

1/2 where h (Y/ g) cot . Ans



 

The formula clearly satisfies the requirement that   0 if x  . It requires “small slope”

and therefore the contact angle should be in the range 70    110

.

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McGraw-Hill Education.Chapter 1  Introduction 1-43

P1.69 A solid cylindrical needle of diameter

d, length L, and density  n may “float” on a

liquid surface. Neglect buoyancy and assume

a contact angle of 0. Calculate the maxi-

mum diameter needle able to float on the

surface. Fig. P1.69

Solution: The needle “dents” the surface downward and the surface tension forces are

upward, as shown. If these tensions are nearly vertical, a vertical force balance gives:



 2

   

z F 0 2YL g d L, or: 8Y

d

max





g

Ans

. (a) 4

(b) Calculate dmax for a steel needle (SG  7.84) in water at 20C. The formula becomes:

P1.70 Derive an expression for the

capillary-height change h, as shown, for a

fluid of surface tension Y and contact angle

 between two parallel plates W apart.

Evaluate h for water at 20C if W  0.5 mm.

Fig. P1.70

Solution: With b the width of the plates into the paper, the capillary forces on each wall

together balance the weight of water held above the reservoir free surface:

2Ycos



    gWhb 2(Ybcos ), or: h . Ans



gW

For water at 20C, Y  0.0728 N/m, ρg  9790 N/m3, and   0. Thus, for W  0.5 mm,



2(0.0728N/m)cos0 h 0.030 m

30mm

Ans.

   3

(9790N/m )(0.0005m)

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McGraw-Hill Education.Chapter 1  Introduction 1-44

*P1.71 A soap bubble of diameter D1 coalesces with another bubble of diameter D2 to

form a single bubble D3 with the same amount of air. For an isothermal process, express

D3 as a function of D1, D2, patm, and surface tension Y.

Solution: The masses remain the same for an isothermal process of an ideal gas:

    

m m m ,

  

1 2 1 1 2 2 3 3 3

p 4Y/r p 4Y/r p 4Y/r

  

 

3 3 3 a 1 a 2 a 3

                

or: D D D

          

1 2 3

         

RT 6 RT 6 RT 6

The temperature cancels out, and we may clean up and rearrange as follows:

3 2 3 2 3 2

a 3 3 a 2 2 a 1 1 pD 8YD pD 8YD pD 8YD     

   . Ans

This is a cubic polynomial with a known right hand side, to be solved for D3.

P1.72 Early mountaineers boiled water to estimate their altitude. If they reach the top and

find that water boils at 84C, approximately how high is the mountain?

Solution: From Table A-5 at 84C, vapor pressure pv  55.4 kPa. We may use this value

to interpolate for pressure in the standard altitude, Table A-6, to estimate

 z Ans.

4800 m

P1.73 A small submersible moves at velocity V in 20C water at 2-m depth, where

ambient pressure is 131 kPa. Its critical cavitation number is Ca  0.25. At what velocity

will cavitation bubbles form? Will the body cavitate if V  30 m/s and the water is cold

(5C)?

Solution: From Table A-5 at 20C read pv  2.337 kPa. By definition,

2(p p ) 2(131000 2337) Ca 0.25  

     a v

crit crit 2 3 2



, solve V 32.1 m/s

Ans



a V (998 kg/m )V

If we decrease water temperature to 5C, the vapor pressure reduces to 863 Pa, and the

density changes slightly, to 1000 kg/m3. For this condition, if V  30 m/s, we compute:



2(131000 863) Ca 0.289

 

(1000)(30)

2

This is greater than 0.25, therefore the body will not cavitate for these conditions. Ans. (b)

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-45

P1.74 Oil, with a vapor pressure of 20 kPa, is delivered through a pipeline by equally-

spaced pumps, each of which increases the oil pressure by 1.3 MPa. Friction losses in the

pipe are 150 Pa per meter of pipe. What is the maximum possible pump spacing to avoid

cavitation of the oil?

Solution: The absolute maximum length L occurs when the pump inlet pressure is

slightly greater than 20 kPa. The pump increases this by 1.3 MPa and friction drops the

pressure over a distance L until it again reaches 20 kPa. In other words, quite simply,

max 1.3 MPa 1,300,000 Pa (150 Pa/m)L, or L   

8660m

Ans.

It makes more sense to have the pump inlet at 1 atm, not 20 kPa, dropping L to about 8 km.

P1.75 An airplane flies at 555 mi/h. At what altitude in the standard atmosphere will

the airplane’s Mach number be exactly 0.8?

Solution: First convert V = 555 mi/h x 0.44704 = 248.1 m/s. Then the speed of sound is

V m s

a m s   

248.1 / 310 /

Ma 0.8

Reading in Table A.6, we estimate the altitude to be approximately 7500 m. Ans.

_____________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-46

P1.76 Derive a formula (a) for the bulk modulus β of an ideal gas, with constant

specific heats, and (b) calculate it for steam at 300ºC and 200 kPa. (c) Compare your

result to experimental data.

Solution: If an ideal gas is isentropic, p = Cρk, where k = cp/cv. Evaluate the bulk

modulus:





1 ( ) ( ) . k k k

d       

    

p d C Ck Ck k p Ans



s

 

From Table A.4, for H2O, k = 1.33, hence βsteam = (1.33)(200,000) = 266,000 Pa

Ans.(b)

(c) The steam tables do not list a bulk modulus, but we can use the identity β ≡ ρ a2

,

where a is the speed of sound. We can enter, for example, SpiraxSarco.com, at p = 200

kPa and T = 300ºC, to obtain ρ = 0.7598 kg/m3 and a = 585.88 m/s. Thus, for this

temperature and pressure,

βsteam ≈ (0.7598)(585.88)2 = 260,000 Pa Ans.(c)

These estimates are reasonably close.

P1.77 Assume that the n-pentane data of Prob. P1.36 represents isentropic conditions.

Estimate the value of the speed of sound at a pressure of 30 MPa. [Hint: The data

approximately fit Eq. (1.19) with B = 260 and n = 11.]

Solution: We have to differentiate Eq. (1.19) to find dp/dρ, using the given data po = 100

kPa, ρo = 586.3 kg/m3

, B = 260, and n = 11:

( 1) [ ( 1)( / ) ], : ( )n 

p B n





1 2

n o

dp dp d p B B or a d

    

 

o o

  

o o

To finish, we have to find ρ for p = 30 MPa = 30,000 kPa:

11 3

      

p solve kg m

30,000 (260 1)( ) 260 , 628.4 / 100 586.3

p

o

2 10

a a Ans

   990 m / s

100,000(261)11 628.4 then [ ]( ) 980,200 , solve . 586.3 586.3

____________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-47

P1.78 Sir Isaac Newton measured sound speed by timing the difference between seeing

a cannon’s puff of smoke and hearing its boom. If the cannon is on a mountain 5.2 miles

away, estimate the air temperature in C if the time difference is (a) 24.2 s; (b) 25.1 s.

Solution: Cannon booms are finite (shock) waves and travel slightly faster than sound

waves, but what the heck, assume it’s close enough to sound speed:

(a) 

x 5.2(5280)(0.3048) m a 345.8 1.4(287)T, T 298 K . (a) t 24.2 s

     

25C 

Ans



(b) 

x 5.2(5280)(0.3048) m a 333.4 1.4(287)T, T 277 K . (b) t 25.1 s

     

4C 

Ans



P1.79 From Table A.3, the density of glycerin at standard conditions is about 1260

kg/m3. At a very high pressure of 8000 lb/in2, its density increases to approximately 1275

kg/m3. Use this data to estimate the speed of sound of glycerin, in ft/s.

Solution: For a liquid, we simplify Eq. (1.38) to a pressure-density ratio, without

knowing if the process is isentropic or not. This should give satisfactory accuracy:

2



lb in Pa psi p m a E

(8000 15 / )(6895 / ) | 2 2



glycerin

  

3.67 6

 

3 2



(1275 1260) /

kg m s

   6300

3.67 6 1920 / / .

Hence a E m s ft s Ans

The accepted value, in Table 9.1, is 6100 ft/s. This accuracy (3%) is very good,

considering the small change in density (1.2%).

___________________________________________________________________

P1.80 In Problem P1.24, for the given data, the air velocity at section 2 is 360 m/s.

What is the Mach number at that section?

Solution: With T2 given as 223 K, evaluate the speed of sound at section 2, assuming, as

stated, an ideal gas:

a2  kRT2  1.4(287 m

2 / s

2

 K)(223K) 299 m / s

V2

360m / s

Then Ma2 



 1.20 Ans.

a2

299m / s

This kind of calculation will be a large part of the material in Chap. 9, Compressible Flow.

_______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-48

P1.81 Use Eq. (1.39) to find and sketch the streamlines of the following flow field:

; ; 0 , where isa constant u K x v K y w K   

Solution: Introduce these velocities into Eq. (1.39):

dx dy dz

  

( )



Kx Ky

0 indicates flowinthe x y plane

Cancel the K’s and integrate:

dx dy or x y C or C

      

, : ln( ) ln( ) ln( ) , : xy = Ans

.

x y

The streamlines are hyperbolas, depending upon the value of C. When C = 0, the

streamlines are the x and y axes. The complete pattern looks like this:

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McGraw-Hill Education.Chapter 1  Introduction 1-49

P1.82 Solution: A velocity field is given by u  V cos , v  V sin , and w  0, where V and  are

constants. Find an expression for the streamlines of this flow.

Equation (1.44) may be used to find the streamlines:

dx dy dx dy dy , or: tan u v Vcos Vsin dx

   



 

Solution: (tan )    y x constant Ans.

The streamlines are straight parallel lines which make an angle  with the x axis. In

other words, this velocity field represents a uniform stream V moving upward at angle

.

*P1.83 Use Eq. (1.39) to find and sketch the streamlines of the following flow field:

2 2 ( ) ; 2 ; 0 , where isaconstant u K x y v K x y w K    

Hint: This is a first-order exact differential equation.

Solution: The asterisk * denotes this as a relatively difficult problem. From Eq. (1.39),

2 2

dx dx dy dy or xy dx x y dy

     

2 2 , : 2 ( ) 0

 

( ) 2

u K x y v K x y

The latter form is suitable for determining ‘exactness’. The equation M dx + N dy = 0 is

exact if

/ / M y N x      . Check this:

2 2 (2 ) / 2 ( ) , Yes,the equationis exact. xy y x x y      

To finish, find the function F for which 2 2

/ 2 / F x M xy and F y x y        

Integrating each of these yields the final function F = x2y – y3/3 = constant, which is the

equation of the streamlines for this problem. The figure shows the flow patterns for

various values of the constant. It is as if the flow were trapped between six 60º walls.

The arrows are determined by checking the directions of u and v in each of the six

segments.

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McGraw-Hill Education.Chapter 1  Introduction 1-50

P1.84 In the early 1900’s, the British chemist Sir Cyril Hinshelwood quipped that fluid

dynamics was divided into ”workers who observed things they could not explain and workers

who explained things they could not observe”. To what historic situation was he referring?

Solution: He was referring to the split between hydraulics engineers, who performed

experiments but had no theory for what they were observing, and theoretical

hydrodynamicists, who found numerous mathematical solutions, for inviscid flow, that were

not verified by experiment.

________________________________________________________________________

P1.85a Report to the class on the achievements of Evangelista Torricelli.

Solution: Torricelli’s biography is taken from a goldmine of information which I did not

put in the references, preferring to let the students find it themselves: C. C. Gillespie (ed.),

Dictionary of Scientific Biography, 15 vols., Charles Scribner’s Sons, New York, 1976.

Torricelli (1608–1647) was born in Faenza, Italy, to poor parents who recognized his

genius and arranged through Jesuit priests to have him study mathematics, philosophy, and

(later) hydraulic engineering under Benedetto Castelli. His work on dynamics of projectiles

attracted the attention of Galileo himself, who took on Torricelli as an assistant in 1641.

Galileo died one year later, and Torricelli was appointed in his place as “mathematician

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McGraw-Hill Education.Chapter 1  Introduction 1-51

and philosopher” by Duke Ferdinando II of Tuscany. He then took up residence in

Florence, where he spent his five happiest years, until his death in 1647. In 1644 he

published his only known printed work, Opera Geometrica, which made him famous as a

mathematician and geometer.

In addition to many contributions to geometry and calculus, Torricelli was the first to

show that a zero-drag projectile formed a parabolic trajectory. His tables of trajectories for

various angles and initial velocities were used by Italian artillerymen. He was an excellent

machinist and constructed—and sold—the very finest telescope lenses in Italy.

Torricelli’s hydraulic studies were brief but stunning, leading Ernst Mach to proclaim

him the ‘founder of hydrodynamics.’ He deduced his theorem that the velocity of efflux

from a hole in a tank was equal to (2gh), where h is the height of the free surface above

the hole. He also showed that the efflux jet was parabolic and even commented on water-

droplet breakup and the effect of air resistance. By experimenting with various liquids in

closed tubes—including mercury (from mines in Tuscany)—he thereby invented the

barometer. From barometric pressure (about 30 feet of water) he was able to explain why

siphons did not work if the elevation change was too large. He also was the first to explain

that winds were produced by temperature and density differences in the atmo-sphere and

not by “evaporation.”

P1.85b Report to the class on the achievements of Henri de Pitot.

Solution: The following notes are abstracted from the Dictionary of Scientific Biography

(see Prob. 1.85a).

Pitot (1695–1771) was born in Aramon, France, to patrician parents. He hated to study

and entered the military instead, but only for a short time. Chance reading of a textbook

obtained in Grenoble led him back to academic studies of mathematics, astronomy, and

engineering. In 1723 he became assistant to Réamur at the French Academy of Sciences and

in 1740 became a civil engineer upon his appointment as a director of public works in

Languedoc Province. He retired in 1756 and returned to Aramon until his death in 1771.

Pitot’s research was apparently mediocre, described as “competent solutions to minor

problems without lasting significance”not a good recommendation for tenure

nowadays! His lasting contribution was the invention, in 1735, of the instrument which

bears his name: a glass tube bent at right angles and inserted into a moving stream with

the opening facing upstream. The water level in the tube rises a distance h above the

surface, and Pitot correctly deduced that the stream velocity  (2gh). This is still a basic

instrument in fluid mechanics.

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McGraw-Hill Education.Chapter 1  Introduction 1-52

P1.85c Report to the class on the achievements of Antoine Chézy.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Chézy (1718–1798) was born in Châlons-sur-Marne, France, studied engineering at the Ecole

des Ponts et Chaussées and then spent his entire career working for this school, finally being

appointed Director one year before his death. His chief contribution was to study the flow in open

channels and rivers, resulting in a famous formula, used even today, for the average velocity:

 V const AS/P

where A is the cross-section area, S the bottom slope, and P the wetted perimeter, i.e., the

length of the bottom and sides of the cross-section. The “constant” depends primarily on

the roughness of the channel bottom and sides. [See Chap. 10 for further details.]

P1.85d Report to the class on the achievements of Gotthilf Heinrich Ludwig Hagen.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Hagen (1884) was born in Königsberg, East Prussia, and studied there, having among

his teachers the famous mathematician Bessel. He became an engineer, teacher, and writer

and published a handbook on hydraulic engineering in 1841. He is best known for his study

in 1839 of pipe-flow resistance, for water flow at heads of 0.7 to 40 cm, diameters of 2.5

to 6 mm, and lengths of 47 to 110 cm. The measurements indicated that the pressure drop

was proportional to Q at low heads and proportional (approximately) to Q2 at higher heads,

where “strong movements” occurred—turbulence. He also showed that p was

approximately proportional to D4

.

Later, in an 1854 paper, Hagen noted that the difference between laminar and turbulent

flow was clearly visible in the efflux jet, which was either “smooth or fluctuating,” and in

glass tubes, where sawdust particles either “moved axially” or, at higher Q, “came into

whirling motion.” Thus Hagen was a true pioneer in fluid mechanics experimentation.

Unfortunately, his achievements were somewhat overshadowed by the more widely

publicized 1840 tube-flow studies of J. L. M. Poiseuille, the French physician.

P1.85e Report to the class on the achievements of Julius Weisbach.

Solution: The following notes are abstracted from the Dictionary of Scientific Biography

(see Prob. 1.85a) and also from Rouse and Ince [Ref. 12].

Weisbach (1806–1871) was born near Annaberg, Germany, the 8th of nine children of

working-class parents. He studied mathematics, physics, and mechanics at Göttingen and

Vienna and in 1931 became instructor of mathematics at Freiberg Gymnasium. In 1835 he

was promoted to full professor at the Bergakademie in Freiberg. He published 15 books

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McGraw-Hill Education.Chapter 1  Introduction 1-53

and 59 papers, primarily on hydraulics. He was a skilled laboratory worker and summarized

his results in Experimental-Hydraulik (Freiberg, 1855) and in the Lehrbuch der Ingenieur-

und Maschinen-Mechanik (Brunswick, 1845), which was still in print 60 years later.

There were 13 chapters on hydraulics in this latter treatise. Weisbach modernized the

subject of fluid mechanics, and his discussions and drawings of flow patterns would be

welcome in any 20th century textbook—see Rouse and Ince [23] for examples.

Weisbach was the first to write the pipe-resistance head-loss formula in modern form:

hf(pipe)  f(L/D)(V2/2g), where f was the dimensionless ‘friction factor,’ which Weisbach

noted was not a constant but related to the pipe flow parameters [see Sect. 6.4]. He was also

the first to derive the “weir equation” for volume flow rate Q over a dam of crest length L:

Q 

2

3





Cw (2g)1/2 H 

 

 



V2



3/2

2g

 



V2



3/2





 

2g

 

 





2

3

Cw (2g)1/2 H3/2

where H is the upstream water head level above the dam crest and Cw is a dimensionless

weir coefficient  O(unity). [see Sect. 10.7] In 1860 Weisbach received the first Honorary

Membership awarded by the German engineering society, the Verein Deutscher

Ingenieure.

P1.85f Report to the class on the achievements of George Gabriel Stokes.

Solution: The following notes are abstracted from the Dictionary of Scientific Biography

(see Prob. 1.85a).

Stokes (1819–1903) was born in Skreen, County Sligo, Ireland, to a clergical family

associated for generations with the Church of Ireland. He attended Bristol College and

Cambridge University and, upon graduation in 1841, was elected Fellow of Pembroke

College, Cambridge. In 1849, he became Lucasian Professor at Cambridge, a post once

held by Isaac Newton. His 60-year career was spent primarily at Cambridge and resulted

in many honors: President of the Cambridge Philosophical Society (1859), secretary (1854)

and president (1885) of the Royal Society of London, member of Parliament (1887–1891),

knighthood (1889), the Copley Medal (1893), and Master of Pembroke College (1902). A

true ‘natural philosopher,’ Stokes systematically explored hydro-dynamics, elasticity,

wave mechanics, diffraction, gravity, acoustics, heat, meteorology, and chemistry. His

primary research output was from 1840–1860, for he later became tied down with

administrative duties.

In hydrodynamics, Stokes has several formulas and fields named after him:

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McGraw-Hill Education.Chapter 1  Introduction 1-54

(1) The equations of motion of a linear viscous fluid: the Navier-Stokes equations.

(2) The motion of nonlinear deep-water surface waves: Stokes waves.

(3) The drag on a sphere at low Reynolds number: Stokes’ formula, F  3  VD.

(4) Flow over immersed bodies for Re << 1: Stokes flow.

(5) A metric (CGS) unit of kinematic viscosity, : 1 cm2/s  1 stoke.

(6) A relation between the 1st and 2nd coefficients of viscosity: Stokes’ hypothesis.

(7) A stream function for axisymmetric flow: Stokes’ stream function [see Chap. 8].

Although Navier, Poisson, and Saint-Venant had made derivations of the equations of

motion of a viscous fluid in the 1820’s and 1830’s, Stokes was quite unfamiliar with the

French literature. He published a completely independent derivation in 1845 of the Navier-

Stokes equations [see Sect. 4.3], using a ‘continuum-calculus’ rather than a ‘molecular’

viewpoint, and showed that these equations were directly analogous to the motion of elastic

solids. Although not really new, Stokes’ equations were notable for being the first to

replace the mysterious French ‘molecular coefficient’  by the coefficient of absolute

viscosity, .

P1.85g Report to the class on the achievements of Moritz Weber.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Weber (1871–1951) was professor of naval mechanics at the Polytechnic Institute of

Berlin. He clarified the principles of similitude (dimensional analysis) in the form used

today. It was he who named the Froude number and the Reynolds number in honor of those

workers. In a 1919 paper, he developed a dimensionless surface-tension (capillarity)

parameter [see Sect. 5.4] which was later named the Weber number in his honor.

P1.85h Report to the class on the achievements of Theodor von Kármán.

Solution: The following notes are abstracted from the Dictionary of Scientific Biography

(see Prob. 1.85a). Another good reference is his ghost-written (by Lee Edson) auto-

biography, The Wind and Beyond, Little-Brown, Boston, 1967.

Kármán (1881–1963) was born in Budapest, Hungary, to distinguished and well-

educated parents. He attended the Technical University of Budapest and in 1906 received

a fellowship to Göttingen, where he worked for six years with Ludwig Prandtl, who had

just developed boundary layer theory. He received a doctorate in 1912 from Göttingen and

was then appointed director of aeronautics at the Polytechnic Institute of Aachen. He

remained at Aachen until 1929, when he was named director of the newly formed

Guggenheim Aeronautical Laboratory at the California Institute of Technology. Kármán

developed CalTech into a premier research center for aeronautics. His leadership spurred

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McGraw-Hill Education.Chapter 1  Introduction 1-55

the growth of the aerospace industry in southern California. He helped found the Jet

Propulsion Laboratory and the Aerojet General Corporation. After World War II, Kármán

founded a research arm for NATO, the Advisory Group for Aeronautical Research and

Development, whose renowned educational institute in Brussels is now called the Von

Kármán Center.

Kármán was uniquely skilled in integrating physics, mathematics, and fluid mechanics

into a variety of phenomena. His most famous paper was written in 1912 to explain the

puzzling alternating vortices shed behind cylinders in a steady-flow experiment conducted

by K. Hiemenz, one of Kármán’s students—these are now called Kármán vortex streets [see

Fig. 5.2a]. Shed vortices are thought to have caused the destruction by winds of the Tacoma

Narrows Bridge in 1940 in Washington State.

Kármán wrote 171 articles and 5 books and his methods had a profound influence on

fluid mechanics education in the 20th century.

P1.85i Report to the class on the achievements of Paul Richard Heinrich Blasius.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Blasius (1883–1970) was Ludwig Prandtl’s first graduate student at Göttingen. His

1908 dissertation gave the analytic solution for the laminar boundary layer on a flat plate

[see Sect. 7.4]. Then, in two papers in 1911 and 1913, he gave the first demonstration that

pipe-flow resistance could be nondimensionalized as a plot of friction factor versus

1/4



Reynolds number—the first “Moody-type” chart. His correlation, 

d f 0.316 Re , is still

is use today. He later worked on analytical solutions of boundary layers with variable

pressure gradients.

P1.85j Report to the class on the achievements of Ludwig Prandtl.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Ludwig Prandtl (1875–1953) is described by Rouse and Ince [23] as the father of modern

fluid mechanics. Born in Munich, the son of a professor, Prandtl studied engineering and

received a doctorate in elasticity. But his first job as an engineer made him aware of the lack

of correlation between theory and experiment in fluid mechanics. He conducted research from

1901–1904 at the Polytechnic Institute of Hanover and presented a seminal paper in 1904,

outlining the new concept of “boundary layer theory.” He was promptly hired as professor and

director of applied mechanics at the University of Gottingen, where he remained throughout

his career. He, and his dozens of famous students, started a new “engineering science” of fluid

mechanics, emphasizing (1) mathematical analysis based upon by physical reasoning; (2) new

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McGraw-Hill Education.Chapter 1  Introduction 1-56

experimental techniques; and (3) new and inspired flow-visualization schemes which greatly

increased our understanding of flow phenomena.

In addition to boundary-layer theory, Prandtl made important contributions to

(1) wing theory; (2) turbulence modeling; (3) supersonic flow; (4) dimensional analysis; and

(5) instability and transition of laminar flow. He was a legendary engineering professor.

P1.85k Report to the class on the achievements of Osborne Reynolds.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Osborne Reynolds (1842–1912) was born in Belfast, Ireland, to a clerical family and

studied mathematics at Cambridge University. In 1868 he was appointed chair of

engineering at a college which is now known as the University of Manchester Institute

of Science and Technology (UMIST). He wrote on wide-ranging topics—mechanics,

electricity, navigation—and developed a new hydraulics laboratory at UMIST. He was the

first person to demonstrate cavitation, that is, formation of vapor bubbles due to high

velocity and low pressure. His most famous experiment, still performed in the

undergraduate laboratory at UMIST (see Fig. 6.5 in the text) demonstrated transition of

laminar pipe flow into turbulence. He also showed in this experiment that the viscosity was

very important and led him to the dimensionless stability parameter  VD/ now called the

Reynolds number in his honor. Perhaps his most important paper, in 1894, extended the

Navier-Stokes equations (see Eqs. 4.38 of the text) to time-averaged randomly fluctuating

turbulent flow, with a result now called the Reynolds equations of turbulence. Reynolds

also contributed to the concept of the control volume which forms the basis of integral

analysis of flow (Chap. 3).

P1.85l Report to the class on the achievements of John William Strutt, Lord Rayleigh.

Solution: The following notes are from Rouse and Ince [Ref. 12].

John William Strutt (1842–1919) was born in Essex, England, and inherited the title

Lord Rayleigh. He studied at Cambridge University and was a traditional hydro-dynamicist

in the spirit of Euler and Stokes. He taught at Cambridge most of his life and also served

as president of the Royal Society. He is most famous for his work (and his textbook) on

the theory of sound. In 1904 he won the Nobel Prize for the discovery of argon gas. He

made at least five important contributions to hydrodynamics: (1) the equations of bubble

dynamics in liquids, now known as Rayleigh-Plesset theory; (2) the theory of nonlinear

surface waves; (3) the capillary (surface tension) instability of jets; (4) the “heat-transfer

analogy” to laminar flow; and (5) dimensional similarity, especially related to viscosity

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McGraw-Hill Education.Chapter 1  Introduction 1-57

data for argon gas and later generalized into group theory which previewed Buckingham’s

Pi Theorem. He ended his career as president, in 1909, of the first British committee on

aeronautics.

P1.85m Report to the class on the achievements of Daniel Bernoulli.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Daniel Bernoulli (1700–1782) was born in Groningen, Holland, his father, Johann,

being a Dutch professor. He studied at the University of Basel, Switzerland, and taught

mathematics for a few years at St. Petersburg, Russia. There he wrote, and published in

1738, his famous treatise Hydrodynamica, for which he is best known. This text contained

numerous ingenious drawings illustrating various flow phenomena. Bernoulli used energy

concepts to establish proportional relations between kinetic and potential energy, with

pressure work added only in the abstract. Thus he never actually derived the famous

equation now bearing his name (Eq. 3.77 of the text), later derived in 1755 by his friend

Leonhard Euler. Daniel Bernoulli never married and thus never contributed additional

members to his famous family of mathematicians.

P1.85n Report to the class on the achievements of Leonhard Euler.

Solution: The following notes are from Rouse and Ince [Ref. 12].

Leonhard Euler (1707–1783) was born in Basel, Switzerland, and studied mathematics

under Johann Bernoulli, Daniel’s father. He succeeded Daniel Bernoulli as professor of

mathematics at the St. Petersburg Academy, leaving there in 1741 to join the faculty of

Berlin University. He lost his sight in 1766 but continued to work, aided by a prodigious

memory, and produced a vast output of scientific papers, dealing with mathematics, optics,

mechanics, hydrodynamics, and celestial mechanics (for which he is most famous today).

His famous paper of 1755 on fluid flow derived the full inviscid equations of fluid motion

(Eqs. 4.36 of the text) now called Euler’s equations. He used a fixed coordinate system,

now called the Eulerian frame of reference. The paper also presented, for the first time, the

correct form of Bernoulli’s equation (Eq. 3.77 of the text). Separately, in 1754 he produced

a seminal paper on the theory of reaction turbines, leading to Euler’s turbine equation (Eq.

11.11 of the text).

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McGraw-Hill Education.Chapter 1  Introduction 1-58

FUNDAMENTALS OF ENGINEERING EXAM PROBLEMS: Answers

FE-1.1 The absolute viscosity  of a fluid is primarily a function of

(a) density (b) temperature (c) pressure (d) velocity (e) surface tension

FE-1.2 Carbon dioxide, at 20C and 1 atm, is compressed isentropically to 4 atm.

Assume CO2 is an ideal gas. The final temperature would be

(a) 130C , (b) 162C , (c) 171C , (d) 237C , (e) 313C

FE-1.3 Helium has a molecular weight of 4.003. What is the weight of 2 cubic meters of

helium at 1 atmosphere and 20C?

(a) 3.3 N (b) 6.5 N (c) 11.8 N (d) 23.5 N (e) 94.2 N

FE-1.4 An oil has a kinematic viscosity of 1.25E–4 m2/s and a specific gravity of 0.80.

What is its dynamic (absolute) viscosity in kg/(m · s)?

(a) 0.08 (b) 0.10 (c) 0.125 (d) 1.0 (e) 1.25

FE-1.5 Consider a soap bubble of diameter 3 mm. If the surface tension coefficient is

0.072 N/m and external pressure is 0 Pa gage, what is the bubble’s internal gage pressure?

(a) 24 Pa (b) 48 Pa (c) 96 Pa (d) 192 Pa (e) 192 Pa

FE-1.6 The only possible dimensionless group which combines velocity V, body size L,

fluid density  , and surface tension coefficient  is:

(a) L  /V (b)  VL2/ (c)   V2/L (d) LV2/ (e) LV2/

FE-1.7 Two parallel plates, one moving at 4 m/s and the other fixed, are separated by

a 5-mm-thick layer of oil of specific gravity 0.80 and kinematic viscosity 1.25E4 m2/s.

What is the average shear stress in the oil?

(a) 80 Pa (b) 100 Pa (c) 125 Pa (d) 160 Pa (e) 200 Pa

FE-1.8 Carbon dioxide has a specific heat ratio of 1.30 and a gas constant of 189 J/(kg·C).

If its temperature rises from 20C to 45C, what is its internal energy rise?

(a) 12.6 kJ/kg (b) 15.8 kJ/kg (c) 17.6 kJ/kg (d) 20.5 kJ/kg (e) 25.1 kJ/kg

FE-1.9 A certain water flow at 20C has a critical cavitation number, where bubbles

form, Ca  0.25, where Ca  2(pa 

pvap)/( V2). If pa  1 atm and the vapor pressure is

0.34 psia, for what water velocity will bubbles form?

(a) 12 mi/hr (b) 28 mi/hr (c) 36 mi/hr (d) 55 mi/hr (e) 63 mi/hr

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McGraw-Hill Education.Chapter 1  Introduction 1-59

FE-1.10 Example 1.10 gave an analysis which predicted that the viscous moment on a

rotating disk was M =    R4/(2h). If the uncertainty of each of the four variables (,  , R, h) is

1.0 %, what is the estimated overall uncertainty of the moment M?

(a) 4.0 % , (b) 4.4 % , (c) 5.0 % , (d) 6.0 % , (e) 7.0 %

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McGraw-Hill Education.Chapter 1  Introduction 1-60

COMPREHENSIVE PROBLEMS

C1.1 Sometimes equations can be developed and practical problems solved by knowing

nothing more than the dimensions of the key parameters. For example, consider the heat loss

through a window in a building. Window efficiency is rated in terms of “R value,” which has

units of ft2·hr·F/Btu. A certain manufacturer offers a double-pane window with R  2.5

and also a triple-pane window with R  3.4. Both windows are 3 ft by 5 ft. On a given

winter day, the temperature difference between inside and outside is 45F. (a) Develop

and equation for window heat loss Q, in time period t, as a function of window area A, R

value, and temperature difference T. How much heat is lost through the above (a) double-

pane window, or (b) triple-pane window, in 24 hours? (c) Suppose the building is heated

with propane gas, at $3.25 per gallon, burning at 80% efficiency. Propane has 90,000 Btu of

available energy per gallon. In a 24-hour period, how much money would a homeowner save,

per window, by installing a triple-pane rather than a double-pane window? (d) Finally,

suppose the homeowner buys 20 such triple-pane windows for the house. A typical winter

equals about 120 heating days at T  45F. Each triple-pane window costs $85 more than

the double-pane window. Ignoring interest and inflation, how many years will it take the

homeowner to make up the additional cost of the triple-pane windows from heating bill

savings?

Solution: (a) The function Q  fcn(t, R, A, T) must have units of Btu. The only

combination of units which accomplishes this is:

Q

tT A

R

Ans. Thus Qlost

(24 hr)(45F)(3 ft5 ft)

2.5 ft2

hr

F/Btu

 6480 Btu Ans. (a)

(b) Triple-pane window: use R  3.4 instead of 2.5 to obtain Q3-pane  4760 Btu Ans. (b)

(c) The savings, using propane, for one triple-pane window for one 24-hour period is:

/ 2 5 . 3 $

g a l

1



Co s t





6 4 8 0 (

4 7 6 0

Bt u

)



0 7 8 . 0 $

An s

c e nt s 7 . 8

) . (

c

9 0 0 0 0

/

Bt u

g a l

8 0 . 0

e f f i c i e n c y

(d) Extrapolate to 20 windows, 120 cold days per year, and $85 extra cost per window:

wi ndow

t i me bac k





/ 85 $ d

9

Pay year s

Ans

. (

)

/ $ 078 . 0 (

/

day wi ndow

120 ) (

/

)

y ear day s

Not a very good investment. We are using ‘$’ and ‘windows’ as “units” in our equations!



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McGraw-Hill Education.Chapter 1  Introduction 1-61

C1.2 When a person ice-skates, the ice surface actually melts beneath the blades, so

that he or she skates on a thin film of water between the blade and the ice. (a) Find an

expression for total friction force F on the bottom of the blade as a function of skater

velocity V, blade length L, water film thickness h, water viscosity  , and blade width W.

(b) Suppose a skater of mass m, moving at constant speed Vo, suddenly stands stiffly

with skates pointed directly forward and allows herself to coast to a stop. Neglecting air

resistance, how far will she travel (on two blades) before she stops? Give the answer X

as a function of (Vo, m, L, h,  , W). (c) Compute X for the case Vo  4 m/s, m  100 kg,

L  30 cm, W  5 mm, and h  0.1 mm. Do you think our assumption of negligible air

resistance was a good one?

Solution: (a) The skate bottom and the melted ice are like two parallel plates:



      , (a) V VLW F A Ans.

h h

(b) Use F  ma to find the stopping distance:

VLW dV F F ma m

 2



     

x x

h dt

(the ‘2’ is for two blades)

Separate and integrate once to find the

velocity, once again to find the distance

traveled:

 



LW



t

mh V mh

dV LW dt or V V e X V dt Ans.

       2

o

V mh LW

2 , : , (b) 2

o

0



(c) Apply our specific numerical values to a 100-kg (!) person:

(4.0 / )(100 )(0.0001 )

ms kg m

X

 

7460m  Ans. (c)

 

2(1.788 3 / )(0.3 )(0.005 )

E kgm s m m

We could coast to the next town on ice skates! It appears that our assumption of negligible

air drag was grossly incorrect.

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McGraw-Hill Education.Chapter 1  Introduction 1-62

C1.3 Two thin flat plates are tilted at an angle  and placed in a tank of known surface

tension Y and contact angle  , as shown. At the free surface of the liquid in the tank, the

two plates are a distance L apart, and of width b into the paper. (a) What is the total

z-directed force, due to surface tension, acting on the liquid column between plates? (b) If

the liquid density is  , find an expression for Y in terms of the other variables.

Solution: (a) Considering the right side of

the liquid column, the surface tension acts

tangent to the local surface, that is, along the

dashed line at right. This force has magnitude

F  Yb, as shown. Its vertical component is

F cos( 



 ), as shown. There are two plates.

Therefore, the total z-directed force on the

liquid column is

Fvertical  2Yb cos( 

–  ) Ans. (a)

(b) The vertical force in (a) above holds up the entire weight of the liquid column between

plates, which is W   g{bh(L 

h tan )}. Set W equal to F and solve for

 

[gbh(L 

h tan )]/[2 cos(   )] Ans. (b)

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-63

C1.4 Oil of viscosity  and density 

drains steadily down the side of a tall, wide

vertical plate, as shown. The film is fully

developed, that is, its thickness  and

velocity profile w(x) are independent of

distance z down the plate. Assume that the

atmosphere offers no shear resistance to the

film surface.

(a) Sketch the approximate shape of the

velocity profile w(x), keeping in mind the

boundary conditions.

(b) Suppose film thickness d is measured, along with the slope of the velocity profile at the

wall, (dw/dx)wall, with a laser-Doppler anemometer (Chap. 6). Find an expression for  as

a function of ,  , (dw/dx)wall, and g. Note that both w and (dw/dx)wall will be negative as

shown.

Solution: (a) The velocity profile must be

such that there is no slip (w  0) at the wall

and no shear (dw/dx  0) at the film surface.

This is shown at right. Ans. (a)

(b) Consider a freebody of any vertical

length H of film, as at right. Since there is

no acceleration (fully developed film), the

weight of the film must exactly balance the

shear force on the wall:

dw

        

W gH b Hb dx

( ) ( ), wall wall wall

Solve this equality for the fluid viscosity:



 

g





Ans. (b)

dw dx

( / )

wall

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-64

C1.5 Viscosity can be measured by flow through a thin-bore or capillary tube if the

flow rate is low. For length L, (small) diameter ,

D L  pressure drop p, and (low)

volume flow rate Q, the formula for viscosity is  = D4p/(CLQ), where C is a constant.

(a) Verify that C is dimensionless. The following data are for water flowing through a

2-mm-diameter tube which is 1 meter long. The pressure drop is held constant at p = 5

kPa.

T, °C: 10.0 40.0 70.0

Q, L/min: 0.091 0.179 0.292

(b) Using proper SI units, determine an average value of C by accounting for the variation

with temperature of the viscosity of water.

Solution: (a) Check the dimensions of the formula and solve for {C}:

        

 

4 4 1 2

M D p L ML T M

( ) { } , { }



                 

3

LT CLQ LT C C L L T

{ }( )( / )

   

therefore {C}  {1} Dimensionless Ans. (a)

(b) Use the given data, with values of  water from Table A.1, to evaluate C, with L  1 m,

D  0.002 m, and p  5000 Pa. Convert the flow rate from L/min to m3/s.

T, °C: 10.0 40.0 70.0

Q, m3/s: 1.52E6 2.98E6 4.87E6

 water, kg/m-s: 1.307E3 0.657E3 0.405E3

C 

D4p/( LQ): 40.3 40.9 40.6

The estimated value of C  40.6 0.3. The theoretical value (Chap. 4) is C  128/  40.74.

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of

McGraw-Hill Education.Chapter 1  Introduction 1-65

C1.6 The rotating–cylinder viscometer in Fig. C1.6 shears the fluid in a narrow clearance,

r, as shown. Assume a linear velocity distribution in the gaps. If the driving torque M is

measured, find an expression for  by (a) neglecting, and (b) including the bottom friction.

Solution: 1.49:

(a) The fluid in the annular region has the same shear stress analysis as Prob.



R R L M RdF R dA R RL d

  2 3

       

        

( )( ) 2 ,

R R

0



M R





: (a)

or Ans. 3

2





R L

(b) Now add in the moment of the (variable) shear stresses on the bottom of the cylinder:

Fig. C1.6

_______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-66

C1.7 Make an analytical study of the transient behavior of the sliding block in Prob.

1.45. (a) Solve for V(t) if the block starts from rest, V = 0 at t = 0. (b) Calculate the time

t1 when the block has reached 98% of its terminal velocity.

Solution: Let x go down the slope, and write Newton’s law

W dV V

V, x



F W A

x

sin , where

g dt h        

W h

μ



Rearrange: sin , where dV A g g K V K dt hW

    . This is a first-order linear

differential equation, with the solution g V C C constant

  

sin



Kt

e

,

K

If V = 0 at t = 0, we find that C = – g sinθ/K. Introduce K for the final solution:



sin ( )[1 exp{ / ( )} ] hW V Ag hW t A

   Ans.(a)





As t →∞, V = Vterminal = (hWsinθ/μA) as in Prob. P1.45. We are 98% there if exp{-

(μAg/hW) t} = 0.02, or (μAg/hW) t1 = -ln(0.02) ≈ 3.91. Thus

1 98% 3.91 .( ) hW t Ans b





Ag

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McGraw-Hill Education.Chapter 1  Introduction 1-67

C1.8 A mechanical device, which uses the rotating cylinder of Fig. C1.6, is the Stormer

viscometer [Ref. 29 of Chap. 1]. Instead of being driven at constant , a cord is wrapped

around the shaft and attached to a falling weight W. The time t to turn the shaft a given number

of revolutions (usually 5) is measured and correlated with viscosity. The Stormer formula is

/ t A W B   

( )

where A and B are constants which are determined by calibrating the device with a known

fluid. Here are calibration data for a Stormer viscometer tested in glycerol, using a weight of

50 N:

 , kg/m·s: 0.23 0.34 0.57 0.84 1.15

t, sec: 15 23 38 56 77

(a) Find reasonable values of A and B to fit this calibration data. [Hint: The data are not

very sensitive to the value of B.] (b) A more viscous fluid is tested with a 100-N weight

and the measured time is 44 s. Estimate the viscosity of this fluid.

Solution: t, to the values

(a) The data fit well, with a standard deviation of about 0.17 s in the value of

A  3000 and B  3.5 Ans. (a)

(b) With a new fluid and a new weight, the values of A and B should nevertheless be the

same:

A

    

t s

3000 44 , 100 3.5

new fluid

W B N solve for  

 

kg

1.42 Ans. (b)



m s

_______________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-68

L1

L2

W

V1

pivot

Cylinder, diameter D,

length L, in an oil film of

thickness  R

C1.9 The lever in Fig. P1.101 has a weight W

at one end and is tied to a cylinder at the other end.

The cylinder has negligible weight and buoyancy

and slides upward through a film of heavy oil of viscosity  . (a) If there is no acceleration (uniform

lever rotation) derive a formula for the rate of

fall V2 of the weight. Neglect the lever weight.

Assume a linear velocity profile in the oil film.

(b) Estimate the fall velocity of the weight if W = 20 N, L1 = 75 cm, L2 = 50 cm, D = 100 cm, L

= 22 cm,  R = 1 mm, and the oil is glycerin at 20C.

V2?

Fig. P1.101

Solution: pivot:

(a) If the motion is uniform, no acceleration, then the moments balance about the

V





0 1



W L



F

L

wher e

F





A

M w

1

1

,

1



w

(



) (

2 DL

p i v o t 

)



R

Since the lever is rigid, the endpoint velocities vary according to their lengths from the pivot:

 

L

 

V

L



V

/

L

V

V

1 /

1

; L

2

2

1

1

2

2

Combine these two relations to obtain the desired solution for a highly viscous fluid:

W



R

V



V

2 a

1

( 2

L

2

/

L

1

)



(

L

2

/

L

1

)

An s

. (

)

DL

 

(b) For glycerin at 20C, from Table A.3,  = 1.49 kg/m-s. The formula above yields

____________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-69

C1.10 A popular gravity-driven instrument is the Cannon-Ubbelohde viscometer, shown

in Fig. C1.10. The test liquid is drawn up above the bulb on the right side and allowed to

drain by gravity through the capillary tube below the bulb. The time t for the meniscus to

pass from upper to lower timing marks is recorded. The kinematic viscosity is computed

by the simple formula  = Ct, where C is a calibration constant. For   in the range of

100-500 mm2/s, the recommended constant is C = 0.50 mm2/s2, with an accuracy less

than 0.5%.

upper timing mark

bulb of known volume

Fig. C1.10

The Cannon-

Ubbelohde

viscometer.

lower timing mark

capillary tube

reservoir

(a) What liquids from Table A.3 are in this viscosity range? (b) Is the calibration formula

dimensionally consistent? (c) What system properties might the constant C depend upon?

(d) What problem in this chapter hints at a formula for estimating the viscosity?

Solution: (a) Very hard to tell, because values of  are not listed – sorry, I’ll add these

values if I remember. It turns out that only three of these 17 liquids are in the

100-500 mm2/s range: SAE 10W, 10W30, and 30W oils, at 120, 194, and 326 mm2/s

respectively.

(b) No, the formula is dimensionally inconsistent because C has units; thus C = 0.5 is

appropriate only for a mm2/s result.

(c) Hidden in C are the length and diameter of the capillary tube and the acceleration of

gravity – see Eq. (6.12) later.

(d) Problem P1.58 gives a formula for flow through a capillary tube. If the flow is

vertical and gravity driven, p =  gL and Q = (/4)d2V, leaving a new formula as

follows:   gd2/(32V).

____________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-70

C1.11 Mott [Ref. 49, p. 38] discusses a simple falling-ball viscometer, which we can

analyze later in Chapter 7. A small ball of diameter D and density  b falls through a tube

of test liquid.

The fall velocity V is calculated by the time to fall a measured distance. calculating the viscosity of the fluid is



The formula for



( 2



b





)

g

D

18

V

This result is limited by the requirement that the Reynolds number ( VD/ ) be less than 1.0.

Suppose a steel ball (SG = 7.87) of diameter 2.2 mm falls in SAE 25W oil (SG = 0.88) at 20C.

The measured fall velocity is 8.4 cm/s. (a) What is the viscosity of the oil, in kg/m-s?

(b) Is the Reynolds number small enough for a valid estimate?

Solution: oil density is 0.88(1000) = 880 kg/m3. Relating SG to water, Eq. (1.7), the steel density is 7.87(1000) = 7870 kg/m3 and the

Using SI units, the formula predicts

3

2

7870 ( 2



880

kg

/

m

81 . 9 ) (

m

/

s

0022 . 0 ) (

m

)



oi l



18

084 . 0 (

m

/

s

)



VD

) 0022 . 0 ) ( 084 . 0 ) ( 880 (

Check

Re









22 . 0

kg



0. 22

Ans

.

m



s

74 . 0



0 . 1

OK

As mentioned, we shall analyze this falling sphere problem in Chapter 7.

_____________________________________________________________________________

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McGraw-Hill Education.Chapter 1  Introduction 1-71

C1.12 A solid aluminum disk (SG = 2.7) is 2 inches in diameter and 3/16 inch thick. It

slides steadily down a 14 incline that is coated with a castor oil (SG = 0.96) film one

hundredth of an inch thick. The steady slide velocity is 2 cm/s. Using Figure A.1 and a

linear oil velocity profile assumption, estimate the temperature of the castor oil.

Solution: This problem reviews complicated units, volume and weight, shear stress, and

viscosity. It fits the sketch in Fig. P1.45. The writer converts to SI units.

3 2 2 (2700 / )(9.81 / ){ [(1/12)(0.3048)] }(3/16/12 )(0.3048) 0.256

     alum W gAh kg m m s N

The weight component along the incline balances the shear stress, in the castor oil, times

the bottom flat area of the disk.

 

0.02 / sin , or: (0.256 )sin(14 ) [ m s

V W A N m

   

2

h m

] [(1/12)(0.3048) ] (1/100/12)(0.3048)

Solve for 0.39 /( ) oil



 

kg m s

Looking on Fig. A.1 for castor oil, this viscosity corresponds approximately to 30C The specific gravity of the castor oil was a red herring and is not needed. Note also that,

since W is proportional to disk bottom area A, that area cancels out and is not needed.

Ans.

_______________________________________________________________________