The plate in Fig. 1–16 rests on top of the thin film of water, which is at a temperature of 25°C. If a pressure difference occurs between A and B, and a small force Fis applied to the plate, the velocity profile across the thickness of the water can be described as u = (40y – 800y²) m/s, where y is in meters. Determine the shear stress acting on the fixed surface and on the bottom of the plate.
The plate in Fig. 1–16 rests on top of the thin film of water, which is at a temperature of 25°C. If a pressure difference occurs between A and B, and a small force Fis applied to the plate, the velocity profile across the thickness of the water can be described as u = (40y – 800y²) m/s, where y is in meters. Determine the shear stress acting on the fixed surface and on the bottom of the plate.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![I want the same solution, but with the mention of
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solution.
EXAMPLE 1.5
The plate in Fig. 1-16 rests on top of the thin film of water, which is at
a temperature of 25 C. If a pressure difference occurs between A and B,
and a small force F is applied to the plate, the velocity profile across the
thickness of the water can be described as u = (40y – 800y² ) m/s,
where y is in meters. Determine the shear stress acting on the fixed
surface and on the bottom of the plate.
0.32 m/s
10 mm
Fig. 1-16
SOLUTION
Fluid Description. Water is a Newtonian fluid, and so Newton's law
of viscosity applies. The viscosity of water at 25°C is found from
Appendix A to be u = 0.897 ( 10-3) N s/m.
Analysis. Before applying Newton's law of viscosity, we must first
obtain the velocity gradient.
du
d
-(40y – 800y²) m/s = (40 – 1600y) s-1
%3D
dy
dy
Therefore, at the fixed surface, y = 0,
du
= [0.897(10 ³) N -s/m²] (40 – 0) s-1
dy
7 = 35.88(10 3) N/m² = 35.9 mPa
And, at the bottom of the moving plate, y = 0.01 m,
Ans.
du
[0.897(10 3) N-s/m²][40 – 1600(0.01)] s
dyly-0.01 m
7 = 21.5 mPa
Ans.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6dc890e0-088e-4fe8-9205-836d14849f06%2Fa9a58d95-55c7-462f-8a02-502ba773931e%2Fzyhjcmc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:I want the same solution, but with the mention of
questions data and a detailed mathematical
solution.
EXAMPLE 1.5
The plate in Fig. 1-16 rests on top of the thin film of water, which is at
a temperature of 25 C. If a pressure difference occurs between A and B,
and a small force F is applied to the plate, the velocity profile across the
thickness of the water can be described as u = (40y – 800y² ) m/s,
where y is in meters. Determine the shear stress acting on the fixed
surface and on the bottom of the plate.
0.32 m/s
10 mm
Fig. 1-16
SOLUTION
Fluid Description. Water is a Newtonian fluid, and so Newton's law
of viscosity applies. The viscosity of water at 25°C is found from
Appendix A to be u = 0.897 ( 10-3) N s/m.
Analysis. Before applying Newton's law of viscosity, we must first
obtain the velocity gradient.
du
d
-(40y – 800y²) m/s = (40 – 1600y) s-1
%3D
dy
dy
Therefore, at the fixed surface, y = 0,
du
= [0.897(10 ³) N -s/m²] (40 – 0) s-1
dy
7 = 35.88(10 3) N/m² = 35.9 mPa
And, at the bottom of the moving plate, y = 0.01 m,
Ans.
du
[0.897(10 3) N-s/m²][40 – 1600(0.01)] s
dyly-0.01 m
7 = 21.5 mPa
Ans.
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