Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 2, Problem 87P

A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of 5 m/s. The two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is 0.9   N s / m 2 . The part of the plate immersed in oil at any given time is 2-m long and 0.5-m wide. If the plate moves through the mid-plane between the surfaces, determine the force required to maintain this motion. What would your response be if the plate was 1 cm from the bottom surface (h2) and 3 cm from the top surface

Chapter 2, Problem 87P, A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity
FIGURE P247

Expert Solution & Answer
Check Mark
To determine

The force required maintain motion at 5m/s

The force on the plate at 1cm from the bottom surface and 3cm from the top surface.

Answer to Problem 87P

The force required maintain motion at 5m/s is 450N.

The force on the plate at 1cm from the bottom surface and 3cm from the top surface is 600N.

Explanation of Solution

Given information:

The moving velocity of the thin plate is 5m/s, the distance between two stationary surface is 4cm, viscosity of the medium is 0.9Ns/m2, the length of the plate is 2m, and width of the plate is 0.5m, height of the plate from the bottom surface is 1cm, and form the top surface is 3cm.

Write the expression for area of the plate.

  A=l×w....... (I)

Here the length of the plate is l, width of the plate w, and area of the plate is A.

Write the expression for shear force acting on the upper side of the plate.

  Fupper=μAVh1....... (II)

Here, force acting on the upper side of the plate is Fupper, viscosity of the medium is μ, and velocity of the thin plate is V.

Write the expression for shear force acting on the lower side of the plate.

  Flower=μAVh2....... (III)

Here, shear force acting on the lower side of the plate is Flower.

Write the expression for force required to maintain the motion

  F=Fupper+Flower....... (IV)

Here force required to maintain the motion is F

Write the expression for shear force acting on the upper side of the plate.

  F(upper)=μAVh1....... (V)

Here, force acting on the upper side of the plate is F(upper).

Write the expression for shear force acting on the lower side of the plate.

  F(lower)=μAVh2....... (VI)

Write the expression for force required to maintain the motion

  F=F(upper)+F(lower)

Calculation:

Write the height of the plate from upper surface when the plate is between the mid of the surfaces.

  h1=42cm=(42cm)(0.01mcm)=0.02m

Here, height of the plate from upper surface is h1,

Write the height of the plate from bottom surface when the plate is between the mid of the surfaces.

  h2=42cm=42cm(0.01mcm)=0.02m

Here, height of the plate from bottom surface is h2.

Write the height of the plate from upper surface.

  h1=3cm=3cm(0.01mcm)=0.03m

Here, the height of the plate from upper surface is h1

Write the height of the plate from bottom surface.

  h2=1cm=1cm(0.01mcm)=0.01m

Here, the height of the plate from bottom surface is h2.

Substitute 2m for A and 0.5m for w in Equation (I).

  A=(2m)×(0.5m)=1m2

Substitute 0.9Ns/m2 for μ, 1m2 for A, 5m/s for V,and 0.02m for h1 in Equation (II).

  Fupper=(0.9Ns/m2)(1m2)(5m/s0.02m)=4.50.02N=225N

Substitute 0.9Ns/m2 for μ, 1m2 for A, 5m/s for V,and 0.02m for h2 in Equation (III).

  Flower=(0.9Ns/m2)(1m2)(5m/s0.02m)=4.50.02N=225N

Substitute 225N for Fupper and 225N for Flower in Equation (IV).

  F=225N+225N=450N

Substitute 0.9Ns/m2 for μ, 1m2 for A, 5m/s for V,and 0.03m for h1 in Equation (V).

  Fupper=(0.9Ns/m2)(1m2)(5m/s0.03m)=4.50.03N=150N

Substitute 0.9Ns/m2 for μ, 1m2 for A, 5m/s for V,and 0.01m for h2 in Equation (VI).

  Flower=(0.9Ns/m2)(1m2)(5m/s0.01m)=4.50.01N=450N

Substitute 150N for F(upper) and 450N for F(lower) in Equation (IV).

  F=150N+450N=600N

Conclusion:

The force required maintain motion at 5m/s is 450N.

The force on the plate at 1cm form the bottom surface and 3cm from the top surface is 600N.

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Chapter 2 Solutions

Fluid Mechanics: Fundamentals and Applications

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