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capítulo 6 ao 9
Tipologia: Exercícios
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KNOWN: Variation of hx with x for laminar flow over a flat plate.
FIND: Ratio of average coefficient, hx , to local coefficient, hx , at x.
ANALYSIS: The average value of hx between 0 and x is
x x
0 0
-1/ x x
1/2 -1/ x
x x
h h dx x dx x x C h 2x 2Cx x h 2h.
Hence,
x
x
h
h
COMMENTS: Both the local and average coefficients decrease with increasing distance x
from the leading edge, as shown in the sketch below.
KNOWN: Variation of local convection coefficient with x for free convection from a vertical
heated plate.
FIND: Ratio of average to local convection coefficient.
ANALYSIS: The average coefficient from 0 to x is
x x
0 0
-1/ x x
3/4 -1/ x x
h h dx x dx x x 4 C 4 4 h x C x h. 3 x 3 3
Hence, x
x
h 4 . h 3
The variations with distance of the local and average convection coefficients are shown in the
sketch.
COMMENTS: Note that h (^) x / h (^) x = 4 / 3is independent of x. Hence the average coefficient
for an entire plate of length L is (^) L L
h h 3
= , where hL is the local coefficient at x = L. Note
also that the average exceeds the local. Why?
KNOWN: Distribution of local convection coefficient for obstructed parallel flow over a flat
plate.
FIND: Average heat transfer coefficient and ratio of average to local at the trailing edge.
ANALYSIS: The average convection coefficient is
( )
( )
L L
0 0
2 L x
2 3 2 L
h h dx 0.7 13.6x 3.4x dx L L 1 h 0.7L 6.8L 1.13L 0.7 6.8L 1.13L L
2
The local coefficient at x = 3m is
2 h (^) L= 0.7 +13.6 3 − 3.4 9 = 10.9 W/m ⋅K.
Hence,
COMMENTS: The result hL / h (^) L= 1.0is unique to x = 3m and is a consequence of the
existence of a maximum for h (^) x1 6x. The maximum occurs at x = 2m, where
( ) (^) ( )
2 2 dh (^) x / dx (^) = 0 and d h (^) x/ dx (^) <0.
KNOWN: Temperature distribution in boundary layer for air flow over a flat plate.
FIND: Variation of local convection coefficient along the plate and value of average coefficient.
SCHEMATIC:
ANALYSIS: From Eq. 6.17,
y 0
s s
k T y (^) k 70 600x h T T T T
∞ ∞
where T (^) s = T(x,0) = 90°C. Evaluating k at the arithmetic mean of the freestream and surface
temperatures, T = (20 + 90)°C/2 = 55°C = 328 K, Table A.4 yields k = 0.0284 W/m⋅K. Hence, with
T (^) s - T = 70°C = 70 K,
( )
0.0284 W m K 42, 000x K m 2 h 17x W m K 70 K
= = ⋅ <
and the convection coefficient increases linearly with x.
The average coefficient over the range 0 ≤ x ≤ 5 m is
5 2 L (^5 )
0 0 0
1 17 17 x h hdx xdx 42.5 W m K L 5 5 2
= (^) ∫ = (^) ∫ = = ⋅ <
KNOWN: Radial distribution of local convection coefficient for flow normal to a circular
disk.
FIND: Expression for average Nusselt number.
ASSUMPTIONS: Constant properties
ANALYSIS: The average convection coefficient is
As ro
0
o
s s
n 2 o^ o o r 2 n+ o 3 n o o (^0)
h hdA A
1 k h Nu 1 a r/r 2 rdr r D
kNu r ar h r 2 n 2 r
π π
where Nu (^) o is the Nusselt number at the stagnation point (r = 0). Hence,
o
D
D
r 2 n 2 o o o 0
o
hD r/r a r Nu 2Nu k 2 n+2 r
Nu Nu 1 2a/ n 2
1/2 0.
COMMENTS: The increase in h(r) with r may be explained in terms of the sharp turn which
the boundary layer flow must make around the edge of the disk. The boundary layer
accelerates and its thickness decreases as it makes the turn, causing the local convection
coefficient to increase.
KNOWN: Convection correlation and temperature of an impinging air jet. Dimensions and initial
temperature of a heated copper disk. Properties of the air and copper.
FIND: Effect of jet velocity on temperature decay of disk following jet impingement.
SCHEMATIC:
ASSUMPTIONS: (1) Validity of lumped capacitance analysis, (2) Negligible heat transfer from sides
and bottom of disk, (3) Constant properties.
ANALYSIS: Performing an energy balance on the disk, it follows that
dt ρcL
where, (^) ( )( )
2 2 h (^) r = εσ T + Tsur T + Tsur and, from the solution to Problem 6.7,
1/ 2 0. D (^) D
k k 2a h Nu 1 0.814 Re Pr D D n 2
With a = 0.30 and n = 2, it follows that
1/ 2 0. D
h = k D 0.936 Re Pr
where Re (^) D = VD/ν. Using the Lumped Capacitance Model of IHT, the following temperature histories
were determined.
Continued …..
KNOWN: Local convection coefficient on rotating disk. Radius and surface temperature of disk.
Temperature of stagnant air.
FIND: Local heat flux and total heat rate. Nature of boundary layer.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible heat transfer from back surface and edge of disk.
ANALYSIS: If the local convection coefficient is independent of radius, the local heat flux at every
point on the disk is
2 2
Since h is independent of location,
2 h (^) = h (^) = 20 W / m (^) ⋅ Kand the total power requirement is
2 Pelec = q = hA (^) s Ts − T∞ = h πro Ts −T∞
( ) (^ ) (^ )
2 2
If the convection coefficient is independent of radius, the boundary layer must be of uniform
thickness δ. Within the boundary layer, air flow is principally in the circumferential direction. The
circumferential velocity component uθ corresponds to the rotational velocity of the disk at the surface
(y = 0) and increases with increasing r (uθ = Ωr). The velocity decreases with increasing distance y
from the surface, approaching zero at the outer edge of the boundary layer (y → δ).
KNOWN: Form of the velocity and temperature profiles for flow over a surface.
FIND: Expressions for the friction and convection coefficients.
ANALYSIS: The shear stress at the wall is
2 s y=0 y=
u A 2By 3Cy A. y
∂ τ μ μ μ ∂
Hence, the friction coefficient has the form,
s f (^2 )
u / 2 u
τ μ
ρ (^) ∞ ρ ∞
f (^2)
u
ν
∞
The convection coefficient is
2 f f (^) y=0 y=
s
k E 2Fy 3Gy k T/ y h T T D T
∂ ∂
∞ ∞
k E f h. D T∞
COMMENTS: It is a simple matter to obtain the important surface parameters from
knowledge of the corresponding boundary layer profiles. However, it is rarely a simple matter
to determine the form of the profile.
KNOWN: Boundary layer temperature distribution.
FIND: Surface heat flux.
PROPERTIES: Table A-4 , Air (T (^) s = 300K): k = 0.0263 W/m⋅K.
ANALYSIS: Applying Fourier’s law at y = 0, the heat flux is
y=
s s y=
s s
s
T u u y q k k T T Pr exp Pr y
u q k T T Pr
q 0.0263 W/m K 100K 0.7 5000 1/m.
∂
∂ ν ν
ν
∞ ∞ ∞
∞ ∞
2
COMMENTS: (1) Negative flux implies convection heat transfer to the surface.
(2) Note use of k at Ts to evaluate qs′′^ from Fourier’s law.
KNOWN : Air flow over a flat plate of length L = 1 m under conditions for which transition from
laminar to turbulent flow occurs at x (^) c = 0.5m based upon the critical Reynolds number, Rex,c = 5× 10
5 .
Forms for the local convection coefficients in the laminar and turbulent regions.
FIND : (a) Velocity of the air flow using thermophysical properties evaluated at 350 K, (b) An expression
from the leading edge, x, for the turbulent region, x (^) c ≤ x ≤ L, and (d) Compute and plot the local and
average convection coefficients, h (^) x and h (^) x, respectively, as a function of x for 0 ≤ x ≤ L.
SCHEMATIC :
ASSUMPTIONS : (1) Forms for the local coefficients in the laminar and turbulent regions, h (^) lam =
Clamx
-0. and h (^) tirb = Cturb x
-0. where Clam = 8.845 W/m
3/ ⋅K, Cturb = 49.75 W/m
2 ⋅K
, and x has units (m).
PROPERTIES: Table A.4 , Air (T = 350 K): k = 0.030 W/m⋅K, ν = 20.92 × 10
2 /s, Pr = 0.700.
ANALYSIS : (a) Using air properties evaluated at 350 K with x (^) c = 0.5 m,
c 5 x,c
u x Re 5 10 ν
∞ = = ×
5 5 6 2 u 5 10 ν x (^) c 5 10 20.92 10 m s 0.5 m 20.9 m s
− ∞ =^ ×^ =^ ×^ ×^ ×^ =^ <
(b) From Eq. 6.5, the average coefficient in the laminar region, 0 ≤ x ≤ x (^) c , is
x x (^) 0.5 0. lam 0 lam lam (^) o lam
1 1 1 h x h x dx C x dx C x x x x
− = (^) ∫ = (^) ∫ = ( )
2Clam x 2hlam x
− = = (1) <
(c) The average coefficient in the turbulent region, x (^) c ≤ x ≤ L, is
c c
c c
x x 0.5 0. x x turb 0 lam (^) x turb lam turb
0 x
1 x x h x h x dx h x dx C C x 0.5 0.
= + = +
∫ ∫
( ) (^) ( )
0.5 0.8 0. turb lam c turb c
1 h x 2C x 1.25C x x x
(2) <
(d) The local and average coefficients, Eqs. (1) and (2) are plotted below as a function of x for the range
0 ≤ x ≤ L.
0 0.5 1 Distance from leading edge, x (m)
0
50
100
150
Convection coefficient (W/m^2.K)
Local - laminar, x <= xc Local - turbulent, x => xcAverage - laminar, x <= xc Average - turbulent, x => xc
KNOWN: Transition Reynolds number. Velocity and temperature of atmospheric air, water,
engine oil and mercury flow over a flat plate.
FIND: Distance from leading edge at which transition occurs for each fluid.
ASSUMPTIONS: Transition Reynolds number is Re (^) x,c = 5 × 10.
5
PROPERTIES: For the fluids at T = 300K;
Fluid Table v(m
2 /s)
Air (1 atm) A-4 15.89 × 10
Water A-6 0.858 × 10
Engine Oil A-5 550 × 10
Mercury A-5 0.113 × 10
ANALYSIS: The point of transition is
5
c x,c
x Re. u 1 m/s
ν ν ∞
Substituting appropriate viscosities, find
Air 7.
Water 0.
Oil 275
Mercury 0.
COMMENTS: Due to the effect which viscous forces have on attenuating the instabilities
which bring about transition, the distance required to achieve transition increases with
increasing ν.
KNOWN: Two-dimensional flow conditions for which v = 0 and T = T(y).
FIND: (a) Verify that u = u(y), (b) Derive the x-momentum equation, (c) Derive the energy equation.
SCHEMATIC:
Pressure & shear forces Energy fluxes
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)
Negligible body forces, (4) v = 0, (5) T = T(y) or ∂T/∂x = 0, (6) Thermal energy generation occurs
only by viscous dissipation.
ANALYSIS: (a) From the mass continuity equation, it follows from the prescribed conditions that
∂u/∂x = 0. Hence u = u(y).
(b) From Newton’s second law of motion, ΣF (^) x= (Rate of increase of fluid momentum) (^) x,
p p p dx dy 1 dy dx 1 u u u u dx dy 1 u u dy 1 x y x
∂ ∂ τ ∂ τ τ ρ ρ ρ ∂ ∂ ∂
− + ⋅ + − + + ⋅ = + ⋅ − ⋅
2
2
p p u u u 0. x y x x (^) y
∂ ∂ τ ∂ ∂ ∂ ρ μ ∂ ∂ ∂ ∂ (^) ∂
(c) From the conservation of energy requirement and the prescribed conditions, it follows that
Ein (^) − Eout (^) =0, or
pu u e u / 2 dy 1 k u dy dx 1 y y
∂^ ∂^ τ ρ τ ∂ ∂
pu pu dx u e u 2 / 2 u e u 2 / 2 dx dy 1 u k T^ k T dy dx 1 0 x x y y y
∂ ∂ ∂ ∂ ∂ ρ ρ τ ∂ ∂ ∂ ∂ ∂
− + + + + + ⋅ − − + − ⋅ = ^ ^ ^ ^
or,
u (^2) T pu u e u / 2 k 0 y x x y y
∂ τ (^) ∂ ∂ ∂ ∂ ρ ∂ ∂ ∂ ∂ ∂
2
2
u p T u u k 0. y y x (^) y
∂ ∂ τ ∂ ∂ τ ∂ ∂ ∂ (^) ∂
Noting that the second and third terms cancel from the momentum equation,
(^2 )
2
u T k 0. y (^) y
∂ ∂ μ ∂ (^) ∂
KNOWN: Diameter, clearance, rotational speed and fluid properties of a lightly loaded journal
bearing. Temperature of bearing.
FIND: (a) Temperature distribution in the fluid, (b) Rate of heat transfer from bearing and operating
power.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Incompressible fluid with constant properties, (3)
Couette flow.
PROPERTIES: Oil (Given): ρ = 800 kg/m
3 , (^) ν = 10
2 /s, k = 0.13 W/m⋅K; μ = ρν = 8 × 10
kg/s⋅m.
ANALYSIS: (a) For Couette flow, the velocity distribution is linear, u(y) = U(y/L), and the energy
equation and general form of the temperature distribution are
2 2 2 2 (^2 ) 2 2
d T du U U C k T y y C. dy dy^ L^ 2k^ L^ k
μ μ μ
Considering the boundary conditions dT/dy) (^) y=L = 0 and T(0) = T 0 , find C 2 = T 0 and C 1 = μU
2 /L.
Hence,
( ) (^ )^ (^ )
2 2 T T 0 μ U / k y/L 1/ 2 y/L.
(b) Applying Fourier’s law at y = 0, the rate of heat transfer per unit length to the bearing is
( ) ( ) (^) ( )
2 3 2 3 3 y=
dT U^8 10 kg/s m^ 14.14 m/s q k D D 75 10 m 1507.5 W/m dy L (^) 0.25 10 m
μ π π π
− − −
× ⋅ ′ = − = − = − × × = − ×
where the velocity is determined as
The journal power requirement is
( y=L)^ s y=L( )
P ′ (^) = F′ U (^) = τ (^) ⋅π D U⋅
( )
2 -3 3
where the shear stress at y = L is
3 2 s y=L (^) y=L (^) -
U 14.14 m/s u/ y 8 10 kg/s m 452.5 kg/s m. L (^) 0.25 10 m
τ μ ∂ ∂ μ
COMMENTS: Note that q ′ (^) = P ,′ which is consistent with the energy conservation requirement.
KNOWN: Conditions associated with the Couette flow of air or water.
FIND: (a) Force and power requirements per unit surface area, (b) Viscous dissipation, (c) Maximum
fluid temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Fully-developed Couette flow, (2) Incompressible fluid with constant
properties.
PROPERTIES: Table A-4 , Air (300K): μ = 184.6 × 10
2 , k = 26.3 × 10
Water (300K): μ = 855 × 10
2 , k = 0.613 W/m⋅K.
ANALYSIS: (a) The force per unit area is associated with the shear stress. Hence, with the linear
Air:
7 2 2 air
200 m/s 184.6 10 N s/m 0.738 N/m 0.005 m
τ
−
Water:
6 2 2 water
200 m/s 855 10 N s/m 34.2 N/m. 0.005 m
τ
− = × ⋅ × =
With the required power given by P/A = τ ⋅ U,
Air: (^) ( ) (^) ( )
2 2 air
Water: (^) ( ) (^) ( )
2 2 water
P/A (^) = 34.2 N/m 200 m/s (^) =6840 W/m.
2 2 μ Φ = μ du/dy^ = μ U/L^. Hence,
2 7 4 3 air (^2)
N s 200 m/s 184.6 10 2.95 10 W/m m 0.005 m
μ
2 6 6 3 water (^2)
N s 200 m/s 855 10 1.37 10 W/m. m 0.005 m
μ
(c) From the solution to Part 4 of the text Example, the location of the maximum temperature
corresponds to ymax = L/2. Hence,
2 Tmax = T 0 + μU / 8k and
-7 2 2
max air
184.6 10 N s/m 200 m/s T 27 C 30.5 C 8 0.0263 W/m K
$ $
-6 2 2
max water
855 10 N s/m 200 m/s T 27 C 34.0 C. 8 0.613 W/m K
$ $
equal the power, P. (2) Although 1 μ Φ 6 1 μΦ 6 water air water^ air
, k >>k. Hence,
Tmax,water ≈Tmax,air.
