Fundamentals of heat and mass transfer frank p incropera david p dewitt solution manual ch06

PROBLEM 6.1

KNOWN: Variation of hx with x for laminar flow over a flat plate.

FIND: Ratio of average coefficient, xh , to local coefficient, hx, at x.

SCHEMATIC:

ANALYSIS: The average value of hx between 0 and x is

x x

0 0

-1/2

x x

1/2 -1/2

x

x x

1 C h h dx x dx

x x

C h 2x 2Cx

x

h 2h .

=∫ =∫

= =

=

Hence, x

x

h 2.

h = <

COMMENTS: Both the local and average coefficients decrease with increasing 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

x x

0 0

-1/4

x x

3/4 -1/4

x x

1 C h h dx x dx

x x

4C 4 4 h x C x h .

3x 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 /h 4/3 x x = is independent of x. Hence the average coefficient

for an entire plate of length L is L L

4 h h

3 = , where hL is the local coefficient at x = L. Note

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 from Newton’s law of

cooling

() ( ) A A

conv conv s q dq h r dA T T . = = ⋅⋅ − ∫ ∫ ∞

The local heat transfer coefficient is known to have the form,

( ) n h r a br = +

and the differential area on the plate surface is

dA 2 r dr. = π

Hence, the heat rate is

( ) ( )

( )

ro

0

o

n

conv s

r

2 n+2

conv s

0

q a br 2 r dr T T

a b q 2T T r r 2 n2

π

π

∞

∞

= + ⋅ ⋅−

  =− +     +

∫

( ) 2 n+2

conv o o s

a b q 2 r r T T. 2 n2

π ∞   =+ −     +

<

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

( )

( )

L L

0 0

2

L x

23 2

L

1 1 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 h 0.7 6.8 3 1.13 9 10.9 W/m K. L =+ − = ⋅ <

The local coefficient at x = 3m is

() () 2 h 0.7 13.6 3 3.4 9 10.9 W/m K. L =+ − = ⋅

Hence,

h / h 1.0. L L = <

COMMENTS: The result h / h 1.0 L L = is unique to x = 3m and is a consequence of the

existence of a maximum for h x x
. The maximum occurs at x = 2m, where

( ) ( ) 2 2

x x dh / dx 0 and d h / dx 0. = <

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,

( )

( )

( )

y 0

s s

kT y k 70 600x

h

TT TT

∂ ∂ =

∞ ∞

×

=− =+ − −

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 - T‡ = 70°C = 70 K,

( ) ( ) 2 0.0284 W m K 42,000x K m h 17x W m K

70K

⋅ = =⋅ <

and the convection coefficient increases linearly with x.

The average coefficient over the range 0 ≤ x ≤ 5 m is

5 2 L 5 2

0 0

0

1 17 17 x h hdx xdx 42.5 W m K

L 5 52 = = == ⋅ ∫ ∫ <

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 12 h for the section of length (x2 - x1) in terms of

C, x1 and x2, and (b) An expression for 12 h in terms of x1 and x2, and the average coefficients 1h and

2h , corresponding to lengths x1 and x2, respectively.

SCHEMATIC:

ASSUMPTIONS: (1) Laminar flow over a plate with uniform surface temperature, Ts, and (2)

Spatial variation of local coefficient is of the form 1/2

x h Cx− = , where C is a constant.

ANALYSIS: (a) The heat transfer rate per unit width from a longitudinal section, x2 - x1, can be

expressed as

q h x xTT 12 12 2 1 s ′ = −− ( )( ) ∞ (1)

where 12 h is the average coefficient for the section of length (x2 - x1). The heat rate can also be

written in terms of the local coefficient, Eq. (6.3), as

( )( ) 2 2

1 1

x x

12 x s s x x x

q h dx T T T T h dx ′ = − =− ∫ ∫ ∞ ∞ (2)

Combining Eq. (1) and (2),

( )

2

1

x

12 x x 2 1

1 h h dx

x x = − ∫ (3)

and substituting for the form of the local coefficient, 1/2

x h Cx− = , find that

( )

2

2

1

1

x 1/2 1/2 1/2

x 1/2 2 1 12 x 21 21 21

x

1 Cx x x

h Cx dx 2C

x x x x 1/2 x x

−   − = ==   −−−     ∫ (4)<

(b) The heat rate, given as Eq. (1), can also be expressed as

q h x T T hx T T 12 2 2 s 1 1 s ′ = −− − ( ) ∞ ∞ ( ) (5)

which is the difference between the heat rate for the plate over the section (0 - x2) and over the section

(0 - x1). Combining Eqs. (1) and (5), find,

2 2 11

12

2 1

h x hx h

x x

− = − (6)<

COMMENTS: (1) Note that, from Eq. 6.6,

x x 1/2 1/2

x x 0 0

1 1 h h dx Cx dx 2Cx

2 x

− − == = ∫ ∫ (7)

or xh = 2hx. Substituting Eq. (7) into Eq. (6), see that the result is the same as Eq. (4).

PROBLEM 6.7

KNOWN: Radial distribution of local convection coefficient for flow normal to a circular

disk.

FIND: Expression for average Nusselt number.

SCHEMATIC:

ASSUMPTIONS: Constant properties

ANALYSIS: The average convection coefficient is

( )

( )

As

r

o

0

o

s

s

n

o o 2

o

r 2 n+2

o

3 n

o o 0

1 h hdA

A

1 k h Nu 1 a r/r 2 rdr

r D

kNu r ar h

r n 2r 2

π

π

=

  = +    

 

= +  

  +  

∫

∫

where Nuo is the Nusselt number at the stagnation point (r = 0). Hence,

( )

( )

( )

o

D

D

r 2 n 2

o

o

o

0

o

hD a r r/r

Nu 2Nu

k 2 n+2 r

Nu Nu 1 2a/ n 2

  +   == +        

= ++    

( ) 1/2 0.36

D D Nu =+ +   1 2a/ n 2 0.814Re Pr .   <

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.

PROBLEM 6.8

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

E VcdT dt A q q st s conv rad = =− + ρ ( ) ′′ ′′
. Hence, with V = AsL,

( )( ) r sur dT hT T h T T

dt cL ρ

−+ − ∞ = −

where, ( )( ) 2 2

r sur sur h TT T T =+ + εσ and, from the solution to Problem 6.7,

1/ 2 0.36 D D

k k 2a h Nu 1 0.814 Re Pr

D D n2

  = =+     +

With a = 0.30 and n = 2, it follows that

( ) 1/ 2 0.36

D h k D 0.936Re Pr =

where ReD = VD/ν. Using the Lumped Capacitance Model of IHT, the following temperature histories

were determined.

Continued …..

PROBLEM 6.8 (Cont.)

0 500 1000 1500 2000 2500 3000

Time, t(s)

300

400

500

600

700

800

900

1000

Temperature, T(K)

V = 4 m/s

V = 20 m/s

V = 50 m/s

The temperature decay becomes more pronounced with increasing V, and a final temperature of 400 K is

reached at t = 2760, 1455 and 976s for V = 4, 20 and 50 m/s, respectively.

COMMENTS: The maximum Biot number, Bi = ( )r Cu h h Lk + , is associated with V = 50 m/s

(maximum h of 169 W/m2

⋅K) and t = 0 (maximum hr of 64 W/m2

⋅K), in which case the maximum Biot

number is Bi = (233 W/m2

⋅K)(0.025 m)/(386 W/m⋅K) = 0.015 < 0.1. Hence, the lumped capacitance

approximation is valid.