*P1.55 The device in Fig. P1.54 is called a rotating disk viscometer[27]. Suppose that R 5 cm and h 1 mm. If the torquerequired to rotate the disk at 900 r/min is 0.537 N m,what is the viscosity of the fluid? If the uncertainty in eachparameter (M, R, h, ) is 1 percent, what is the overalluncertainty in the viscosity?

*P1.56 The device in Fig. P1.56 is called a cone-plate viscometer[27]. The angle of the cone is very small, so that sin , and the gap is filled with the test liquid. The torque Mto rotate the cone at a rate is measured. Assuming a lin-ear velocity profile in the fluid film, derive an expressionfor fluid viscosity as a function of (M, R, , ).

8rL04

Qp

Pipe end effects are neglected [27]. Suppose our capillaryhas r0 2 mm and L 25 cm. The following flow rateand pressure drop data are obtained for a certain fluid:

Q, m3/h 0.36 0.72 1.08 1.44 1.80

p, kPa 159 318 477 1274 1851

What is the viscosity of the fluid? Note: Only the first threepoints give the proper viscosity. What is peculiar about thelast two points, which were measured accurately?

P1.59 A solid cylinder of diameter D, length L, and density s

falls due to gravity inside a tube of diameter D0. The clear-ance, D0 D D, is filled with fluid of density andviscosity . Neglect the air above and below the cylinder.Derive a formula for the terminal fall velocity of the cylin-der. Apply your formula to the case of a steel cylinder,D 2 cm, D0 2.04 cm, L 15 cm, with a film of SAE30 oil at 20°C.

P1.60 For Prob. 1.52 suppose that P 0.1 hp when V 6 ft/s,L 4.5 ft, b 22 in, and h 7/8 in. Estimate the vis-cosity of the oil, in kg/(m s). If the uncertainty in eachparameter (P, L, b, h, V) is 1 percent, what is the over-all uncertainty in the viscosity?

*P1.61 An air-hockey puck has a mass of 50 g and is 9 cm in di-ameter. When placed on the air table, a 20°C air film, of0.12-mm thickness, forms under the puck. The puck isstruck with an initial velocity of 10 m/s. Assuming a lin-ear velocity distribution in the air film, how long will ittake the puck to (a) slow down to 1 m/s and (b) stop com-pletely? Also, (c) how far along this extremely long tablewill the puck have traveled for condition (a)?

P1.62 The hydrogen bubbles which produced the velocity pro-files in Fig. 1.13 are quite small, D 0.01 mm. If the hy-drogen-water interface is comparable to air-water and thewater temperature is 30°C estimate the excess pressurewithin the bubble.

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

P1.64 At 60°C the surface tension of mercury and water is 0.47and 0.0662 N/m, respectively. What capillary heightchanges will occur in these two fluids when they are incontact with air in a clean glass tube of diameter 0.4 mm?

P1.65 The system in Fig. P1.65 is used to calculate the pressurep1 in the tank by measuring the 15-cm height of liquid inthe 1-mm-diameter tube. The fluid is at 60°C (see Prob.1.64). Calculate the true fluid height in the tube and thepercent error due to capillarity if the fluid is (a) water and(b) mercury.

Problems 51

R R

Clearance h

Oil

Ω

P1.54

Ω

FluidR

θ θ

*P1.57 For the cone-plate viscometer of Fig. P1.56, suppose thatR 6 cm and 3°. If the torque required to rotate thecone at 600 r/min is 0.157 N m, what is the viscosity ofthe fluid? If the uncertainty in each parameter (M, R, ,) is 1 percent, what is the overall uncertainty in the vis-cosity?

*P1.58 The laminar-pipe-flow example of Prob. 1.12 can be usedto design a capillary viscometer [27]. If Q is the volumeflow rate, L is the pipe length, and p is the pressure dropfrom entrance to exit, the theory of Chap. 6 yields a for-mula for viscosity:

P1.56

30 Solutions Manual • Fluid Mechanics, Fifth Edition

πµωµω π

θ θ= =

or 43 o

0

ror: Torque M 2 r dr

h sin 2hsin

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

2o o o

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

dt 10

ω= − = =

Separating the variables, we may integrate:

o

w t4o

o 0

rddt, or: .

2hI sinAns

ω

πµωω θ

= −

πµω ωθ

= −2o

o5 r t

exp3mh sin

1.54* A disk of radius R rotates at angular velocity Ω inside an oil container of viscosity µ, 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 τ ≈ µΩr/h on both sides of the disk. Thus,

w

R3

0

rd(torque) dM 2r dA 2r 2 r dr,

h

or: M 4 r drh

Ans

µτ

µπ

Ω= = = π

Ω= = .4R

hΩπµ

1.55 Apply the rotating-disk viscometer of Prob. 1.54, to the particular case R = 5 cm, h = 1 mm, rotation rate 900 rev/min, measured torque M = 0.537 N·m. What is the fluid viscosity? If each parameter (M,R,h,Ω) has uncertainty of ±1%, what is the overall uncertainty of the measured viscosity?

Solution: The analytical formula M = πµΩR4/h was derived in Prob. 1.54. Convert the rotation rate to rad/s: Ω = (900 rev/min)(2π rad/rev ÷ 60 s/min) = 94.25 rad/s. Then,

4 4

hM (0.001 m)(0.537 N m) kg or

m sR (94.25 rad/s)(0.05 m)Ansµ

π π⋅ = = = . ⋅Ω 2

N s0.29

m

⋅

Chapter 1 • Introduction 31

For uncertainty, looking at the formula for µ, we have first powers in h, M, and Ω and a fourth power in R. The overall uncertainty estimate [see Eq. (1.44) and Ref. 31] would be

1/ 22 2 2 2h M R

2 2 2 2 1/ 2

S S S S (4S )

[(0.01) (0.01) (0.01) 4(0.01) ] 0.044 or: Ans

µ Ω ≈ + + +

≈ + + + ≈ .4.4%±

The uncertainty is dominated by the 4% error due to radius measurement. We might report the measured viscosity as µ ≈ 0.29 ± 4.4% kg/m·s or 0.29 ± 0.013 kg/m·s.

1.56* For the cone-plate viscometer in Fig. P1.56, the angle is very small, and the gap is filled with test liquid µ. Assuming a linear velocity profile, derive a formula for the viscosity µ in terms of the torque M and cone parameters.

Fig. P1.56

Solution: For any radius r ≤ R, the liquid gap is h = r tanθ. Then

wr dr

d(Torque) dM dA r 2 r r, orr tan cos

τ µ πθ θ

Ω = = =

R 32

0

2 2 RM r dr , or: .

sin 3sinAns

π µ π µ µθ θ

Ω Ω= = = 3

3M sin

2 R

θπΩ

1.57 Apply the cone-plate viscometer of Prob. 1.56 above to the special case R = 6 cm, θ = 3°, M = 0.157 N ⋅ m, and a rotation rate of 600 rev/min. What is the fluid viscosity? If each parameter (M,R,Ω,θ) has an uncertainty of ±1%, what is the uncertainty of µ?

Solution: We derived a suitable linear-velocity-profile formula in Prob. 1.56. Convert the rotation rate to rad/s: Ω = (600 rev/min)(2π rad/rev ÷ 60 s/min) = 62.83 rad/s. Then,

3 3

3Msin 3(0.157 N m)sin(3 ) kg or .

m s2 R 2 (62.83 rad/s)(0.06 m)Ans

θµπ π

⋅ ° = = = ⋅Ω 2

N s0.29

m

⋅

For uncertainty, looking at the formula for µ, we have first powers in θ, M, and Ω and a third power in R. The overall uncertainty estimate [see Eq. (1.44) and Ref. 31] would be

1/ 22 2 2 2M R

2 2 2 2 1/ 2

S S S S (3S )

[(0.01) (0.01) (0.01) 3(0.01) ] 0.035, or: Ans

µ θ Ω = + + +

≈ + + + = . .3 5± %

The uncertainty is dominated by the 3% error due to radius measurement. We might report the measured viscosity as µ ≈ 0.29 ± 3.5% kg/m·s or 0.29 ± 0.01 kg/m·s.

with manometer fluid m. One side of the manometer is opento the air, while the other is connected to new tubing whichextends to pressure measurement location 1, some height Hhigher in elevation than the surface of the manometer liquid.For consistency, let a be the density of the air in the room,t be the density of the gas inside the tube, m be the den-sity of the manometer liquid, and h be the height differencebetween the two sides of the manometer. See Fig. P2.38.(a) Find an expression for the gage pressure at the mea-surement point. Note: When calculating gage pressure, usethe local atmospheric pressure at the elevation of the mea-surement point. You may assume that h H; i.e., assumethe gas in the entire left side of the manometer is of den-sity t. (b) Write an expression for the error caused by as-suming that the gas inside the tubing has the same densityas that of the surrounding air. (c) How much error (in Pa)is caused by ignoring this density difference for the fol-lowing conditions: m 860 kg/m3, a 1.20 kg/m3,t 1.50 kg/m3, H 1.32 m, and h 0.58 cm? (d) Canyou think of a simple way to avoid this error?

is very large. If the inclined arm is fitted with graduations1 in apart, what should the angle be if each graduationcorresponds to 1 lbf/ft2 gage pressure for pA?

Problems 107

P2.38 An interesting article appeared in the AIAA Journal (vol. 30,no. 1, January 1992, pp. 279–280). The authors explain thatthe air inside fresh plastic tubing can be up to 25 percentmore dense than that of the surroundings, due to outgassingor other contaminants introduced at the time of manufacture.Most researchers, however, assume that the tubing is filledwith room air at standard air density, which can lead to sig-nificant errors when using this kind of tubing to measurepressures. To illustrate this, consider a U-tube manometer

h

(1)

(2)

30

2 mP2.35

P2.39 An 8-cm-diameter piston compresses manometer oil intoan inclined 7-mm-diameter tube, as shown in Fig. P2.39.When a weight W is added to the top of the piston, the oilrises an additional distance of 10 cm up the tube, as shown.How large is the weight, in N?

P2.40 A pump slowly introduces mercury into the bottom of theclosed tank in Fig. P2.40. At the instant shown, the airpressure pB 80 kPa. The pump stops when the air pres-sure rises to 110 kPa. All fluids remain at 20°C. What willbe the manometer reading h at that time, in cm, if it is con-nected to standard sea-level ambient air patm?

50 cm

50 cm

OilSG = 0.8

WaterSG = 1.0

L

P2.36

1 in

Reservoir

θ D = 5

16inpA

P2.37

h

H

1

U-tube manometer m

t (tubing gas)

a (air)

pa at location 1p1

P2.38

P2.44 Water flows downward in a pipe at 45°, as shown in Fig.P2.44. The pressure drop p1 p2 is partly due to gravityand partly due to friction. The mercury manometer readsa 6-in height difference. What is the total pressure dropp1 p2 in lbf/in2? What is the pressure drop due to fric-tion only between 1 and 2 in lbf/in2? Does the manome-ter reading correspond only to friction drop? Why?

108 Chapter 2 Pressure Distribution in a Fluid

P2.41 The system in Fig. P2.41 is at 20°C. Compute the pres-sure at point A in lbf/ft2 absolute.

D = 8 cm

d = 7 mm

Meriam redoil, SG = 0.827

10 cm

15˚

Piston

W

8 cm

9 cm

Air: pB

Water

Mercury

Pump

patm

h

2 cm

Hg10 cm

P2.39

P2.40

Water

Water

5 in

10 in6 in

Mercury

A

Oil, SG = 0.85pa = 14.7 lbf/in2

P2.41

h1

pA

1

pB

1

h

2

h1

ρ ρ

ρ

P2.42

5 ft

Flow

1

2

45˚

6 in

Mercury

Water

P2.44

P2.42 Very small pressure differences pA pB can be measuredaccurately by the two-fluid differential manometer in Fig.P2.42. Density 2 is only slightly larger than that of theupper fluid 1. Derive an expression for the proportional-ity between h and pA pB if the reservoirs are very large.

*P2.43 A mercury manometer, similar to Fig. P2.35, records h 1.2, 4.9, and 11.0 mm when the water velocities in the pipeare V 1.0, 2.0, and 3.0 m/s, respectively. Determine ifthese data can be correlated in the form p1 p2 CfV2,where Cf is dimensionless.

P2.45 In Fig. P2.45, determine the gage pressure at point A inPa. Is it higher or lower than atmospheric?

P2.46 In Fig. P2.46 both ends of the manometer are open to theatmosphere. Estimate the specific gravity of fluid X.

P2.47 The cylindrical tank in Fig. P2.47 is being filled with wa-ter at 20°C by a pump developing an exit pressure of 175kPa. At the instant shown, the air pressure is 110 kPa andH 35 cm. The pump stops when it can no longer raisethe water pressure. For isothermal air compression, esti-mate H at that time.

P2.48 Conduct the following experiment to illustrate air pres-sure. Find a thin wooden ruler (approximately 1 ft in

EES

P3.119 Revisit the turbine cascade system of Prob. 3.78, and de-rive a formula for the power P delivered, using the angular-momentum theorem of Eq. (3.55).

P3.120 A centrifugal pump impeller delivers 4000 gal/min of wa-ter at 20°C with a shaft rotation rate of 1750 r/min. Ne-glect losses. If r1 6 in, r2 14 in, b1 b2 1.75 in,Vt1 10 ft/s, and Vt2 110 ft/s, compute the absolute ve-locities (a) V1 and (b) V2 and (c) the horsepower required.(d) Compare with the ideal horsepower required.

P3.121 The pipe bend of Fig. P3.121 has D1 27 cm and D2 13 cm. When water at 20°C flows through the pipe at 4000gal/min, p1 194 kPa (gage). Compute the torque re-quired at point B to hold the bend stationary.

P3.124 A rotating dishwasher arm delivers at 60°C to six nozzles,as in Fig. P3.124. The total flow rate is 3.0 gal/min. Eachnozzle has a diameter of 1

36 in. If the nozzle flows are equal

and friction is neglected, estimate the steady rotation rateof the arm, in r/min.

Problems 201

P3.121

P3.122

P3.123

P3.124

P3.125

1

2

V2, p2 = pa

B

50 cm

50 cm

V1, p1

C

, V

V

h1

h2

V

FnL

h3

ρ

75˚

4 ft

Ω

150 ft/s

5 in

40˚

5 in 6 in

Q

x

y

0

d <R, L

Vw

Closedvalve

R L

<

*P3.122 Extend Prob. 3.46 to the problem of computing the centerof pressure L of the normal face Fn, as in Fig. P3.122. (Atthe center of pressure, no moments are required to holdthe plate at rest.) Neglect friction. Express your result interms of the sheet thickness h1 and the angle betweenthe plate and the oncoming jet 1.

P3.123 The waterwheel in Fig. P3.123 is being driven at 200 r/minby a 150-ft/s jet of water at 20°C. The jet diameter is 2.5in. Assuming no losses, what is the horsepower developedby the wheel? For what speed r/min will the horsepowerdeveloped be a maximum? Assume that there are manybuckets on the waterwheel.

*P3.125 A liquid of density flows through a 90° bend as shownin Fig. P3.125 and issues vertically from a uniformlyporous section of length L. Neglecting pipe and liquidweight, derive an expression for the torque M at point 0required to hold the pipe stationary.

Chapter 3 • Integral Relations for a Control Volume 213

3.121 The pipe bend of Fig. P3.121 has D1 = 27 cm and D2 = 13 cm. When water at 20°C flows through the pipe at 4000 gal/ min, p1 = 194 kPa (gage). Compute the torque required at point B to hold the bend stationary.

Solution: First convert Q = 4000 gal/ min = 0.252 m3/s. We need the exit velocity:

Fig. P3.121

2 2 1 12

0.252 m mV Q/A 19.0 Meanwhile, V Q/A 4.4

s s( /4)(0.13)π= = = = =

We don’t really need V1, because it passes through B and has no angular momentum. The angular momentum theorem is then applied to point B:

B B 1 1 2 2p A p A ( ) m( ) = + × + × − = 1 2 2 2 1 1M T r j r i r V r V× − ×

But r1 and p2 are zero,

Bhence m( ) Q[(0.5 0.5 ) (19.0 )]ρ= = + 2 2T r V i j i× ×

Thus, finally, TB = (998)(0.252)(0.5)(19.0)(−k) ≈ −2400 k N · m (clockwise) Ans.

3.122 Extend Prob. 3.46 to the problem of computing the center of pressure L of the normal face Fn, as in Fig. P3.122. (At the center of pressure, no moments are required to hold the plate at rest.) Neglect friction. Express your result in terms of the sheet thickness h1 and the angle θ between the plate and the oncoming jet 1.

Fig. P3.122

Solution: Recall that in Prob. 3.46 of this Manual, we found h2 = (h1/2)(1 + cosθ) and that h3 = (h1/2)(1 − cosθ). The force on the plate was Fn = ρQVsinθ. Take clockwise moments about O and use the angular momentum theorem:

( )ρ ρ ρ

= − = + −

= + − − = −

| | | | | | o n 2 z 3 z 1 z

2 2 22 2 3 3 2 3

M F L m m m

Vh (h V/2) Vh ( h V/2) 0 (1/2) V h h

2O 2 3O 3 1O 1r V r V r V× × ×

214 Solutions Manual • Fluid Mechanics, Fifth Edition

( ) ( )2 2 2 2 22 3 2 3

211

(1/2) V h h h hThus L .

2h sinV h sinAns

ρθρ θ

− −= − = = − 1

1h cot

2θ−

The latter result follows from the (h1, h2, h3) relations in 3.46. The C.P. is below point O.

3.123 The waterwheel in Fig. P3.123 is being driven at 200 r/min by a 150-ft/s jet of water at 20°C. The jet diameter is 2.5 in. Assuming no losses, what is the horse-power developed by the wheel? For what speed Ω r/min will the horsepower developed be a maximum? Assume that there are many buckets on the waterwheel.

Solution: First convert Ω = 200 rpm = 20.9 rad/s. The bucket velocity = Vb = ΩR = (20.9)(4) = 83.8 ft/s. From Prob. 3.51 of this Manual, if there are many buckets, the entire (absolute) jet mass flow does the work:

Fig. P3.123

ρ

π

= − − ° = −

= − 4

⋅= ≈

jet b jet b jet jet b jet b

2

P m V (V V )(1 cos165 ) A V V (V V )(1.966)

2.5(1.94) (150)(83.8)(150 83.8)(1.966)

12

ft lbf108200 .

sAns197 hp

Prob. 3.51: Max. power is for Vb = Vjet/2 = 75 ft/s, or Ω = 18.75 rad/s = 179 rpm Ans.

3.124 A rotating dishwasher arm delivers at 60°C to six nozzles, as in Fig. P3.124. The total flow rate is 3.0 gal/min. Each nozzle has a diameter of 3

16 in. If the nozzle flows are equal and friction is neglected, estimate the steady rotation rate of the arm, in r/min.

Fig. P3.124

xx a c

yy a c

zz a c

xy yx 12

a c

xz zx 12

a c

yz zy 12

a c

where a and c are constants of the fluid. Make all the sameassumptions as in the derivation of Eq. (4.140). (a) Find thevelocity profile u(y). (b) How does the velocity profile forthis case compare to that of a newtonian fluid?

wy

z

wx

uz

x

uy

wz

y

ux

Fundamentals of Engineering Exam Problems 273

2 cm

1.5 m

5 cm

SAE 10 oil

2 m/s

Fixed cylinder

Moving rod

2 cm

P4.90

Word Problems

W4.1 The total acceleration of a fluid particle is given by Eq.(4.2) in the eulerian system, where V is a known functionof space and time. Explain how we might evaluate parti-cle acceleration in the lagrangian frame, where particle po-sition r is a known function of time and initial position,r fcn(r0, t). Can you give an illustrative example?

W4.2 Is it true that the continuity relation, Eq. (4.6), is valid forboth viscous and inviscid, newtonian and nonnewtonian,compressible and incompressible flow? If so, are there anylimitations on this equation?

W4.3 Consider a CD compact disk rotating at angular velocity. Does it have vorticity in the sense of this chapter? Ifso, how much vorticity?

W4.4 How much acceleration can fluids endure? Are fluids likeastronauts, who feel that 5g is severe? Perhaps use the flowpattern of Example 4.8, at r R, to make some estimatesof fluid-acceleration magnitudes.

W4.5 State the conditions (there are more than one) under whichthe analysis of temperature distribution in a flow field canbe completely uncoupled, so that a separate analysis forvelocity and pressure is possible. Can we do this for bothlaminar and turbulent flow?

W4.6 Consider liquid flow over a dam or weir. How might theboundary conditions and the flow pattern change when wecompare water flow over a large prototype to SAE 30 oilflow over a tiny scale model?

W4.7 What is the difference between the stream function andour method of finding the streamlines from Sec. 1.9? Orare they essentially the same?

W4.8 Under what conditions do both the stream function andthe velocity potential $ exist for a flow field? When doesone exist but not the other?

W4.9 How might the remarkable three-dimensional Taylor in-stability of Fig. 4.18 be predicted? Discuss a general pro-cedure for examining the stability of a given flow pattern.

W4.10 Consider an irrotational, incompressible, axisymmetric(/ 0) flow in (r, z) coordinates. Does a stream func-tion exist? If so, does it satisfy Laplace’s equation? Arelines of constant equal to the flow streamlines? Does avelocity potential exist? If so, does it satisfy Laplace’sequation? Are lines of constant $ everywhere perpendicu-lar to the lines?

Fundamentals of Engineering Exam ProblemsThis chapter is not a favorite of the people who prepare the FEExam. Probably not a single problem from this chapter will appearon the exam, but if some did, they might be like these.FE4.1 Given the steady, incompressible velocity distribution V

3xi Cyj 0k, where C is a constant, if conservation ofmass is satisfied, the value of C should be(a) 3, (b) 3/2, (c) 0, (d) 3/2, (e) 3

FE4.2 Given the steady velocity distribution V 3xi 0j Cyk,

where C is a constant, if the flow is irrotational, the valueof C should be(a) 3, (b) 3/2, (c) 0, (d) 3/2, (e) 3

FE4.3 Given the steady, incompressible velocity distribution V 3xi Cyj 0k, where C is a constant, the shear stress xy

at the point (x, y, z) is given by(a) 3 , (b) (3x Cy) , (c) 0, (d) C ,(e) (3 C)