CHEE 3363: Fluid Mechanics for Chemical Engineers

Homework #2Due date: Wednesday, January 30, at start of class.

Recommended reading: Fox, sections 2.2–2.4.Recommended viewing: the NCFM movie Surface Tension.

1. The quantities viscosity µ, velocity v, and surface tension σ may be combined into a dimen-sionless group. Find the combination that is proportional to µ.

2. A 30-mm-diameter shaft is pulled through a cylindrical bearing as shown in the figure. Thelubricant filling the 0.4-mm gap between the shaft and bearing is an oil of kinematic viscosity7.0 × 10−4 m2/s and a specific gravity of 0.88. Determine the force P required to pull theshaft at a velocity of 4 m/s.

3. The belt in the figure moves at steady velocity V and skims the top of a tank of oil of viscosityµ. Assuming that the film of oil between the belt and the surface is thin, develop a simpleformula for the belt-drive power P required as a function of h, L, V,B, µ. Neglect airdrag.What power P in watts is required if the belt moves at 3 m/s over SAE 30W oil at 20◦C,with L = 3 m, b = 40 cm, and h = 2 cm?

Chapter 1 • Introduction 29

The total error is dominated by the 8% error in the estimate of clearance, (Ro – Ri). We might state the experimental result for viscosity as

exp 0.29 8.2% Ansµ ! ± = .kg0.29 0.024

m s±

"

1.52 The belt in Fig. P1.52 moves at steady velocity V and skims the top of a tank of oil of viscosity µ. Assuming a linear velocity profile, develop a simple formula for the belt-drive power P required as a function of (h, L, V, B, µ). 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:

oil belt beltV

P A V (bL)V .h

Ans# µ! "= ! =# $% &

2 LV b

hµ

(b) For SAE 30W oil, µ ! 0.29 kg/m " s. Then, for the given belt parameters, 2 2

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kg m 2.0 m kg mP V bL/h 0.29 2.5 (0.6 m) 73 . (b)

m s s 0.03 m sAnsµ ! " "! "= = ! =

# $# $" % &% &

73 W

1.53* 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

wr dr

d(Torque) r dA r 2 r ,h sin$# µ %

&! "! "= = # $# $% &% &

4. In the concentric cylinder viscometer shown in the figure, the viscometer is driven by a fallingmass M connected by a pulley and cord to the inner cylinder of radius R. The annular gapof width a and height H is filled with the liquid to be tested. The mass falls at a constantspeed vm.

(a) Derive an algebraic expression for the viscosity µ of the liquid in the viscometer as afunction of M , g, vm, r, R, a, and H.

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(b) Evaluate the viscosity of the liquid for the following: M = 0.2 kg, vm = 40 mm/s, r =23 mm, R = 55 mm, a = 0.15 mm, and H = 70 mm.Problem 2.55 [4]

Given: Data on the viscometer

Find: Time for viscometer to lose 99% of speed

Solution:

The given data is R 50 mm�� H 80 mm�� a 0.20 mm�� I 0.0273 kg� m2�� μ 0.1

N s�

m2��

The equation of motion for the slowing viscometer is I α� Torque� τ� A� R��

where � is the angular acceleration and � is the viscous stress, and A is the surface area of the viscometer

The stress is given by τ μ dudy�� μ V 0�

a��

μ V�a

�μ R� ω�

a�

where V and � are the instantaneous linear and angular velocities.

Hence I α� Idωdt

��μ R� ω�

a� A� R��

μ R2� A�

aω��

Separating variables dωω

μ R2� A�a I�

� dt��

Integrating and using IC � = �0 ω t( ) ω0 e

μ R2� A�a I�

� t���

The time to slow down by 99% is obtained from solving 0.01 ω0� ω0 e

μ R2� A�a I�

� t��� so t

a I�

μ R2� A�

� ln 0.01( )��

Note that A 2 π� R� H�� so ta I�

2 π� μ� R3� H�

� ln 0.01( )��

t0.0002 m� 0.0273� kg� m2

�2 π��

m2

0.1 N� s��

1

0.05 m�( )3�

10.08 m��

N s2�

kg m�� ln 0.01( )�� t 4.00s�

5. A pivot bearing used on the shaft of an electrical instrument is shown in the Figure. An oilwith a viscosity of µ = 0.02 lb·s/ft2 fills the 0.003-in gap between the rotating shaft and thestationary base. The cone angle is 30◦. Determine the frictional torque on the shaft when itrotates at 6,000 rpm.

µ = 0.02 lb·s/ft2

0.1 in

0.003 in

6,000 rpm

6. Conceptual problem: Describe an experimental setup to measure the surface tension of aliquid similar to water. (The NCFM movie Surface Tension may have useful ideas here.)What experimental precision could be expected?

7. Surface tension forces can be strong enough to allow a double-edge steel razor blade to “float”

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on water, but a single-edge blade will sink. Assume that the surface tension forces act at anangle of θ relative to the water surface as shown in the figure.

(a) The mass of the double-edge blade is 0.64 × 10−3 kg and the total length of its sidesis 206 mm. Determine the value of θ required to maintain equilibrium balance betweenthe blade weight and the resultant surface tension force.

(b) The mass of the single edge blade is 2.6 × 10−3 kg and the total length of its sides is154 mm. Explain why this blade sinks, and support your answer with calculations.

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