A arte da guerra

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PROBLEM 6.1 KNOWN: Variation of hx with x for laminar flow over a flat plate. FIND: Ratio of average coefficient, h x , to local coefficient, hx, at x. SCHEMATIC:

ANALYSIS: The average value of hx between 0 and x is hx = hx hx Hence, 1 x C x ∫ h x dx = ∫ x -1/2dx x 0 x 0 C 1/2 = 2x = 2Cx -1/2 x = 2h x . hx = 2. hx

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COMMENTS: Both the local and average coefficients decrease withincreasing distance x from the leading edge, as shown in the sketch below.

PROBLEM 6.2 KNOWN: Variation of local convection coefficient with x for free convection from a vertical heated plate. FIND: Ratio of average to local convection coefficient. SCHEMATIC:

ANALYSIS: The average coefficient from 0 to x is 1 x C x -1/4 h x = ∫ h x dx = ∫ x dx x 0 x 0 4 C 3/4 4 4 hx = x = C x -1/4 = h x . 3 x 3 3Hence, hx 4 = . hx 3

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The variations with distance of the local and average convection coefficients are shown in the sketch.

COMMENTS: Note that h x / h x = 4 / 3 is independent of x. Hence the average coefficient 4 for an entire plate of length L is h L = h L , where hL is the local coefficient at x = L. Note 3 also that the average exceeds the local. Why?

PROBLEM 6.3 KNOWN:Expression for the local heat transfer coefficient of a circular, hot gas jet at T∞ directed normal to a circular plate at Ts of radius ro. FIND: Heat transfer rate to the plate by convection. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Flow is axisymmetric about the plate, (3) For h(r), a and b are constants and n ≠ -2. ANALYSIS: The convective heat transfer rate to the plate follows fromNewton’s law of cooling q conv = ∫ dq conv = ∫ h ( r ) ⋅ dA ⋅ ( T∞ − Ts ).
A A

The local heat transfer coefficient is known to have the form, h ( r ) = a + br n and the differential area on the plate surface is
dA = 2π r dr.

Hence, the heat rate is q conv = ∫
ro 0

(a + brn ) ⋅ 2π r dr ⋅ (T∞ − Ts )
r

b n+2  o a q conv = 2π ( T∞ − Ts )  r 2 + r  n+2 2 0 b n+2  a 2 q conv =2π  ro + ro  ( T∞ − Ts ) . n+2 2 

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COMMENTS: Note the importance of the requirement, n ≠ -2. Typically, the radius of the jet is much smaller than that of the plate.

PROBLEM 6.4 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. SCHEMATIC:ANALYSIS: The average convection coefficient is 1 L 1 L h x dx = ∫ 0.7 + 13.6x − 3.4x 2 dx L ∫0 L 0 1 hL = 0.7L + 6.8L2 − 1.13L3 = 0.7 + 6.8L − 1.13L2 L hL =

(

(

)

)

h L = 0.7 + 6.8 (3) − 1.13 (9 ) = 10.9 W/m 2 ⋅ K. The local coefficient at x = 3m is h L = 0.7 + 13.6 (3) − 3.4 (9 ) = 10.9 W/m 2 ⋅ K. Hence, h L / h L = 1.0. COMMENTS: The result h L / h L = 1.0 is unique to x = 3m and isa consequence of the existence of a maximum for h x x . The maximum occurs at x = 2m, where

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(dh x / dx ) = 0 and

(d2h x / dx2 < 0.)

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PROBLEM 6.5
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,

h=−k ∂ T ∂ y y =0 k ( 70 × 600x ) =+ (Ts − T∞ ) (Ts − T∞ )

where Ts = 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 Ts - TW = 70°C = 70 K,

h=

0.0284 W m ⋅ K ( 42, 000x ) K m 70 K

= 17x W m 2 ⋅ K

(

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and the convection coefficient increaseslinearly with x.

The average coefficient over the range 0 ≤ x ≤ 5 m is 5 1 L 17 5 17 x 2 h = ∫ hdx = ∫ xdx = = 42.5 W m 2 ⋅ K 0 0 L 5 5 2 0

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PROBLEM 6.6
KNOWN: Variation of local convection coefficient with distance x from a heated plate with a uniform temperature Ts. FIND: (a) An expression for the average coefficient h12 for the section of length (x2 - x1) in terms of C, x1 and x2, and...
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