Question 6 A liquid flows down an inclined plane surface in a steady, fully developed laminar film of thickness h. Simplify the continuity and Navier-Stokes equations to model this flow field. Obtain expressions for the liquid velocity profile, the shear stress distribution, the volume flow rate, and the average velocity. Relate the liquid film thickness to the volume flow rate per unit depth of surface normal to the flow. Calculate the volume flow rate in a film of water h = 1 mm thick, flowing on a surface b = 1 mm wide, inclined at 0 = 15° to the horizontal. 0 15° h = 1 mm Width b = 1 m Answer: upg Q = V sine μ Txy = pgsine (h - y) bpgsine h³ μl 3 = Q bh v-22²) = hy bpgsinፀ h3 μ 3 h = 3μQ bpgsine Q = 0.846 L/s 1/3

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Question 6
A liquid flows down an inclined plane surface in a steady, fully developed laminar film of thickness h. Simplify the
continuity and Navier-Stokes equations to model this flow field. Obtain expressions for the liquid velocity profile,
the shear stress distribution, the volume flow rate, and the average velocity. Relate the liquid film thickness to the
volume flow rate per unit depth of surface normal to the flow. Calculate the volume flow rate in a film of water
h = 1 mm thick, flowing on a surface b = 1 mm wide, inclined at 0 15° to the horizontal.
=
18
0 = 15⁰
h = 1 mm
Width b = 1 m
Too
g
Answer:
u = pg
Q
V
sine
μ
=
Txy = pgsin0(h — y)
bpgsine h³
μ 3
Q
bh
hy
=
12
2
bpgsine h³
μ 3
3μQ
bpgsine
Q = 0.846 L/s
h =
1/3
Transcribed Image Text:Question 6 A liquid flows down an inclined plane surface in a steady, fully developed laminar film of thickness h. Simplify the continuity and Navier-Stokes equations to model this flow field. Obtain expressions for the liquid velocity profile, the shear stress distribution, the volume flow rate, and the average velocity. Relate the liquid film thickness to the volume flow rate per unit depth of surface normal to the flow. Calculate the volume flow rate in a film of water h = 1 mm thick, flowing on a surface b = 1 mm wide, inclined at 0 15° to the horizontal. = 18 0 = 15⁰ h = 1 mm Width b = 1 m Too g Answer: u = pg Q V sine μ = Txy = pgsin0(h — y) bpgsine h³ μ 3 Q bh hy = 12 2 bpgsine h³ μ 3 3μQ bpgsine Q = 0.846 L/s h = 1/3
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