Fenomenosdostransportes

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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 IHT with 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 and length 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 unitplate width, and (d) Plot the
boundary layer thickness and local values of the shear stress, convection coefficient, and heat flux as a
function of x for 0 ≤ x ≤ 1 m.
SCHEMATIC:

ASSUMPTIONS: (1) Critical Reynolds number is 5 × 105, (2) Flow over top and bottom surfaces.
PROPERTIES: Table A.5, Engine Oil (Tf = 333 K): ρ = 864 kg/m3, ν = 86.1 × 10-6 m2/s, k = 0.140
W/m⋅K, Pr = 1081.
ANALYSIS: (a)Calculate the Reynolds number to determine nature of the flow,

uL
0.1m s × 1m
ReL = ∞ =
= 1161
ν
86.1×10−6 m 2 s
Hence the flow is laminar at x = L, from Eqs. 7.19 and 7.24, and
−1/ 2

δ = 5L ReL1/ 2 = 5 (1m )(1161)
= 0.147 m
−1/ 3
δ t = δ Pr −1/ 3 = 0.147 m (1081)
= 0.0143m

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<

(b) The local convection coefficient, Eq. 7.23, and heat flux at x = L are

hL =

k
0.140 W m ⋅ K
1/ 2
1/ 3
0.332 Re1/2 Pr1/ 3 =
0.332 (1161) (1081) = 16.25 W m 2 ⋅ K
L
L
1m

q′′ = h L ( Ts − T∞ ) = 16.25 W m 2 ⋅ K ( 20 − 100 ) C = −1300 W m 2
x


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Also, the local shear stress is, from Eq. 7.20,
ρu2
864 kg m3
τ s,L = ∞ 0.664 Re−1/ 2 =
(0.1m s )2 0.664 (1161)−1/ 2
L

2

2

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τ s,L = 0.0842 kg m ⋅ s 2 = 0.0842 N m 2
(c) With the drag force per unit width given by D′ = 2Lτ s,L where the factor of 2 is includedto account
for both sides of the plate, it follows that

(

)

−1/ 2

2

D′ = 2L ρ u ∞ 2 1.328 Re L1/ 2 = (1m ) 864 kg m3 (0.1m s ) / 2 1.328 (1161)
2

= 0.337 N m

For laminar flow, the average value h L over the distance 0 to L is twice the local value, hL,
h L = 2h L = 32.5 W m 2 ⋅ K
The total heat transfer rate per unit width of the plate is

q′ = 2Lh L ( Ts − T∞ ) = 2 (1m ) 32.5 W m 2 ⋅ K (20 − 100 ) C = −5200 W m

<

<
Continued...

PROBLEM 7.2 (Cont.)
(c) Using IHT with the foregoing equations, the boundary layer thickness, and local values of the
convection coefficient and heat flux were calculated and plotted as a function of x.

deltax*10, hx*100, -q''x

5000

4000

3000

2000

1000

0
0

0.2

0.4

0.6

0.8

1

Distance from leading edge, x (m)
BL thickness, deltax * 10...
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