Problem 2 Fomts) By applying 5 N force, glycerin is sheared between two parallel plates, as shown in Fig. 2. The gap between the plates (h) is 1.5 mm. The plates measure 10 cm in length and 5 cm in width. The bottom plate is stationary. Calculate: a) the velocity of the upper plate ( b) the Reynolds number (1 c) the power required to pull the plate Fig. 2 Flow between parallel plates Viscosity of Fluids Fluid Hydrogen Air F = 5 N Gasoline Water Ethyl alcohol Mercury SAE 30 oil P. kg/(ms) 8.8 E-6 1.8 E-5 2.9 E-4 1.0 E-3 1.2 E-3 1.5 E-3 0.29 1.5

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Problem 2 Fomts)
By applying 5 N force, glycerin is sheared between two parallel plates, as
shown in Fig. 2. The gap between the plates (h) is 1.5 mm. The plates
measure 10 cm in length and 5 cm in width. The bottom plate is stationary.
Calculate:
a) the velocity of the upper plate
b) the Reynolds number (1
c) the power required to pull the plate
0:45
Fig. 2 Flow between parallel plates
Viscosity of Fluids
Fluid
Hydrogen
Air
F =5 N
Gasoline
Water
Ethyl alcohol
Mercury
SAE 30 oil
Glycerin
P.
kg/(ms)
8.8 E-6
1.8 E-5
2.9 E-4
1.0 E-3
1.2 E-3
1.5 E-3
0.29
1.5
Transcribed Image Text:Problem 2 Fomts) By applying 5 N force, glycerin is sheared between two parallel plates, as shown in Fig. 2. The gap between the plates (h) is 1.5 mm. The plates measure 10 cm in length and 5 cm in width. The bottom plate is stationary. Calculate: a) the velocity of the upper plate b) the Reynolds number (1 c) the power required to pull the plate 0:45 Fig. 2 Flow between parallel plates Viscosity of Fluids Fluid Hydrogen Air F =5 N Gasoline Water Ethyl alcohol Mercury SAE 30 oil Glycerin P. kg/(ms) 8.8 E-6 1.8 E-5 2.9 E-4 1.0 E-3 1.2 E-3 1.5 E-3 0.29 1.5
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For part b, can you clarify the values used to calculate the Reynolds number? On my screen it is appearing as:

Re = 1000×1×0.11.5= 66.67

What is the 0.11.5?

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