Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 1, Problem 1.9CP
<
To determine

(a)

Formula for the rate of fall V2 of the weight.

<
Expert Solution
Check Mark

Answer to Problem 1.9CP

Formula for the rate of fall V2 of the weight is V2=WRμπDL(L2L1)2

Explanation of Solution

Given:

The cylinder weight and buoyancy is negligible and slides upward by a film of heavy oil of viscosity.

Fluid Mechanics, Chapter 1, Problem 1.9CP

Concept Used:

Newton’s law of viscosity;

shere forceshere area =μVR

Calculation:

W=20N, L=75cm=75×102m

L2=50cm=50×102m,D=10cm = 10×102m

R =1mm = 1×103m,L = 22cm = 22×102m

Viscosity of glycerin at 20C is μ=0.950N-S/m2

Derive formula for the rate of fall ( V2 )

Apply ever rate at the pivot:

F×L1=W×L2

Force on left end, F=W L2L1

As per Newton’s law of viscosity;

shere forceshere area =μV1RFπDL=μV1R

F=μV1×πDLR,WL2L1=μV1πDLRV1=W L2RL1 μπDL

Also, assuming efficiency of ever as 100 %

(power) left end = (power) right-end

F×V1=W×V2V2=FV1W=(WL2/L1)(WL2RL1μπDL)×1WV2=WRμπDL( L2 L1 )2

Conclusion:

Therefore, formula for the rate of fall V2 of the weight is V2=WRμπDL(L2L1)2.

<
To determine

(b)

Fall velocity of the weight.

<
Expert Solution
Check Mark

Answer to Problem 1.9CP

The fall velocity of the weight V2=0.13537m/s

Explanation of Solution

Given:

The cylinder weight and buoyancy is negligible and slides upward by a film of heavy oil of viscosity.

W=20 N, L1=75 cm, L2=50 cm, D=10 cm, L=22 cm, ΔR=1 mm

Oil is glycerin at 20°C.

Concept Used:

Using the formula

V2=WRμπDL(L2L1)2

Calculation:

Estimation for V2

V2=20×1×1030.95×π×0.1×0.22×(0.50.75)2=0.3046×0.9949V2=0.13537m/s

Conclusion:

Thus, the fall velocity of the weight V2=0.13537m/s.

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Chapter 1 Solutions

Fluid Mechanics

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