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.
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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.
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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