Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 1, Problem 1.7CP
<
To determine

(a)

Discussion of the expression for V(t) for the block.

<
Expert Solution
Check Mark

Answer to Problem 1.7CP

  Thus, the expression for V(t) is =18.18118.181e0.13965t

Explanation of Solution

Given:

Fluid Mechanics, Chapter 1, Problem 1.7CP , additional homework tip  1

θ=15°,

h=1mm,

m=6kg,

A=35cm2

Dynamic viscosity of SAE 30 oil at 20°C, μ=0.2394pa.s

Concept Used:

Using F.B.D. and from Newton’s second law along inclined plane.

Calculation:

Force on block due to fluid can be given as

F=μVR×A{forlinearvelocityprofile}

F.B.D. is shown. From Newton’s second law along inclined plane.

Fluid Mechanics, Chapter 1, Problem 1.7CP , additional homework tip  2

mgsinθF=mamgsinθμVh×A=mdvdtmdvdt+(μAh)v=mgsinθdvdt+(μAmR)v=gsinθ...(1)

At terminal velocity υ=v, a=0, dvdt=0

From equation (1)

+μAmhV=gsinθV=9.81×sin15°×6×0.0010.2394×35×104V=18.181m/s(terminalvelocity)

from equation (1)

dVdt+(μAmR)V=gsinθ

Solution of this differential equation can be given by

V(t)=μ(t).gsinθ+Cμ(t)

Where μ(t) is integrating factor

μ(t)=e( μA mR )dt

μ(t)=e(μAmht)

So V(t)=e μA mh t.gsinθdt+CeμAmht

V(t)=(1 μA mhe μA mht.gsinθ)+CeμAmht

V(t)=6×0.001×9.81×sin15°0.2394×35×104+CeμAmht

V(t)=18.181+Ce( 0.2394×35×104 6×0.001)tV(t)=18.181+Ce0.13965t

at t=0, V(t)=0

o=18.181+C×e0.13965×0

C=18.181

So V(t)=18.18118.181e0.13965t

Conclusion:

  Thus, the expression for V(t) is =18.18118.181e0.13965t.

<
To determine

(b)

To calculate:

The time t1 when the block has reached 98% of its terminal velocity.

<
Expert Solution
Check Mark

Answer to Problem 1.7CP

The time t1 when the block has reached 98% of its terminal velocity =28.01sec

Explanation of Solution

Given:

Fluid Mechanics, Chapter 1, Problem 1.7CP , additional homework tip  3

θ=15°,

h=1mm,

m=6kg,

A=35cm2

Dynamic viscosity of SAE 30 oil at 20°C, μ=0.2394pa.s

Concept Used:

Formula used is V(t)=18.18118.181e0.13965t

Calculation:

At V=0.98V, (98% terminal velocity)

0.98×18.181=18.181(1e0.13965t1)

0.98=1e0.13965t1

1e0.13965t1=0.02{takelnonbothsides}

lne0.13965t1=ln0.02

0.13965t1=3.912{sincelnea=a}

t1=+3.912+0.13965t1=28.01sec

Conclusion:

Thus, the time t1 when the block has reached 98% of its terminal velocity =28.01sec.

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Chapter 1 Solutions

Fluid Mechanics

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