PROBLEM 7.1
KNOWN: Temperature and velocity of fluids in parallel flow over a flat plate.
FIND: (a) Velocity and thermal boundary layer thicknesses at a prescribed distance from the leading
edge, and(b) For each fluid plot the boundary layer thicknesses as a function of distance.
SCHEMATIC:

ASSUMPTIONS: (1) Transition Reynolds number is 5 × 105.
PROPERTIES: Table A.4, Air (300 K, 1 atm): ν =15.89 × 10-6 m2/s, Pr = 0.707; Table A.6, Water
(300 K): ν = µ/ρ = 855 × 10-6 N⋅s/m2/997 kg/m3 = 0.858 × 10-6 m2/s, Pr = 5.83; Table A.5, Engine Oil
(300 K): ν = 550 × 10-6 m2/s, Pr = 6400; Table A.5,Mercury (300 K): ν = 0.113 × 10-6 m2/s, Pr =
0.0248.
ANALYSIS: (a) If the flow is laminar, the following expressions may be used to compute δ and δt,
respectively,

δ=

5x

δt =

Re1/ 2
x

δ

RexFluid

Pr1/ 3

Air
Water
Oil
Mercury

where
u x 1m s ( 0.04 m ) 0.04 m 2 s
Re x = ∞ =
=
ν
ν
ν

δ (mm)

δt (mm)

2517
4.66 × 104
72.7
3.54 × 105

3.99
0.93
23.5
0.34

4.48
0.52
1.27
1.17

(b) Using IHTwith the foregoing equations, the boundary layer thicknesses are plotted as a function of
distance from the leading edge, x.
10

5

BL thickness, deltat (mm)

BL thickness, delta (mm)

8
6
4
2
0

4
3
21
0

0

10

20

30

Distance from leading edge, x (mm)
Air
Water
Oil
Mercury

40

0

10

20

30

40

Distance from leading edge, x (mm)
Air
Water
Oil
Mercury

COMMENTS: (1) Note that δ ≈ δt for air, δ> δt for water, δ >> δt for oil, and δ < δt for mercury. As
expected, the boundary layer thicknesses increase with increasing distance from the leading edge.
(2) The value of δt for mercury should beviewed as a rough approximation since the expression for δ/δt
was derived subject to the approximation that Pr > 0.6.

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PROBLEM 7.2
KNOWN: Temperature and velocity of engine oil. Temperature andlength of flat plate.
FIND: (a) Velocity and thermal boundary layer thickness at trailing edge, (b) Heat flux and surface
shear stress at trailing edge, (c) Total drag force and heat transfer per unit... [continua]

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