QUESTION 1. (20 pts) With respect to free convection:

a. What is an extensive, quiescent fluid? (4 points) b. What are the two major physical considerations or forces for free convection? (4 points) c. What is the Grashof number in words (ratio of forces) and mathematically? (6 points) d. What is the Rayleigh number in words and mathematically? (6 points)

SOLUTION a. As defined on page 561 of the text (Incropera, et.al., 4/e), “an extensive medium is, in principle, an infinite medium. Since a quiescent fluid is one that is otherwise at rest”. In other words, an extensive, quiescent fluid is one that has infinite bulk properties (extensive) with no forced convection, and by all practical purposes, is stagnant (quiescent). b. The two major physical considerations or forces for free convection are buoyant forces and viscous forces. Buoyant forces arise from a fluid density gradient and a body force proportional to the density at any given instant; this proportional force is often gravity. Viscous forces are always present and actually provide a counter-force situation to the buoyant force. This is similar to Newton’s 3rd Law of Motion. More discussion of these forces is provided in Section 9.1 of the text. c. The Grashof number is the ratio of buoyant forces to viscous forces acting on a fluid.

d. The Rayleigh number is the product of the Grashof number and the Prandtl number, effectively being the ratio of buoyant forces to viscous forces multiplied by the ratio of momentum to thermal diffusivities.

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PROBLEM 9.10 KNOWN: Interior air and wall temperatures; wall height.

FIND: (a) Average heat transfer coefficient when T∞ = 20°C and Ts = 10°C, (b) Average heat

ransfer coefficient when T∞ = 27°C and Ts = 37°C. t SCHEMATIC:

A SSUMPTIONS: (a) Wall is at a uniform temperature, (b) Room air is quiescent.

PROPERTIES: Table A-4, Air (Tf = 288K, 1 atm): β = 1/Tf = 3.472 × 10-3 K-1, ν = 14.82 × 10

-6 m

2/s, k = 0.0253 W/m⋅K, α = 20.9 × 10-6 m2/s, Pr = 0.710; (Tf = 305K, 1 atm): β = 1/Tf = 3.279 ×

0-3 K-1, ν = 16.39 × 10

-6 m2/s, k = 0.0267 W/m⋅K, α = 23.2 × 10-6 m2/s, Pr = 0.706. 1

ANALYSIS: The appropriate correlation for the average heat transfer coefficient for free convection on a vertical wall is Eq. 9.26.

( )

20.1667hL 0.387 RaLNu 0.825L 0.296k 0.5631 0.492 / Pr

= = +

+

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎡ ⎤⎩ ⎣ ⎦ ⎭

w here RaL = g β ∆T L3/να, Eq. 9.25, with ∆T = Ts - T∞ or T∞ - Ts.

(a) Substituting numerical values typical of winter conditions gives

( ) ( )2 3 1 39.8 m/s 3.472 10 K 20 10 K 2.5m 10Ra 1.711 10L 6 2 -6 214.82 10 m / s 20.96 10 m / s

− −× × −= =

−× × ××

( )

( )

20.1667100.387 1.711 10Nu 0.825 299.6.L 0.2960.5631 0.492 / 0.710

×= + =

+

⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎡ ⎤⎪ ⎪⎣ ⎦⎩ ⎭

Hence, ( ) 2h Nu k/L 299.6 0.0253 W/m K / 2.5m 3.03 W/m K.L= = ⋅ = ⋅ < (b) Substituting numerical values typical of summer conditions gives

( ) ( )2 3 1 39.8 m/s 3.279 10 K 37 27 K 2.5 m 10Ra 1.320 10L 6 2 6 223.2 10 m / s 16.39 10 m / s

− −× × −= =

− −× × ××

( )

( )

20.1667100.387 1.320 10Nu 0.825 275.8.L 0.2960.5631 0.492 / 0.706

×= + =

+

⎧ ⎫⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎡ ⎤

⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

Hence, 2h Nu k/L 275.8 0.0267 W/m K/2.5m 2.94 W/m K.L= = × ⋅ = ⋅ < COMMENTS: There is a small influence due to Tf on h for these conditions. We should expect radiation effects to be important with such low values of h.

PROBLEM 9.78 KNOWN: Sphere with embedded electrical heater is maintained at a uniform surface temperature

hen suspended in various media. w F IND: Required electrical power for these media: (a) atmospheric air, (b) water, (c) ethylene glycol.

SCHEMATIC:

A

SSUMPTIONS: (1) Negligible surface radiation effects, (2) Extensive and quiescent media.

PROPERTIES: Evaluated at Tf = (Ts + T∞)/2 = 330K: ν⋅106, m2/s k⋅103, W/m⋅K α⋅106, m2/s Pr β⋅103, K-1

Table A-4, Air (1 atm) Table A-6, Water Table A-5, Ethylene glycol

18.91 0.497 5.15

28.5 650 260

26.9 0.158

0.0936

0.711 3.15 55.0

3.03 0.504 0.65

ANALYSIS: The electrical power (Pe) required to offset convection heat transfer is ( ) (2

conv s s sq h A T T h D T Tπ∞= − = − ).∞ (1) The free convection heat transfer coefficient for the sphere can be estimated from Eq. 9.35 using Eq. 9.25 to evaluate RaD.

( )D

1/ 4 3D

D4/ 99 /16 11D

Pr 0.70.589Ra g T DNu 2 Ra .

1 0.469 / Pr Ra 10

βνα

≥⎧⎪ Δ⎪= + =⎨⎪⎡ ⎤+ ≤⎪⎢ ⎥ ⎩⎣ ⎦

(2,3)

(a) For air

( )( ) ( )32 3 1

4D 6 2 6 2

9.8m / s 3.03 10 K 94 20 K 0.025mRa 6.750 10

18.91 10 m / s 26.9 10 m / s

− −

− −

× −= =

× × ××

( )( )

D

1/ 442

D 4/ 99 /16

0.589 6.750 10k 0.0285 W / m Kh Nu 2 10.6 W / m KD 0.025m

1 0.469 / 0.711

⎧ ⎫×⎪ ⎪⋅ ⎪ ⎪= = + =⎨ ⎬

⎪ ⎪⎡ ⎤+⎪ ⎪⎢ ⎥⎣ ⎦⎩ ⎭

⋅

( ) ( )22convq 10.6 W / m K 0.025m 94 20 K 1.55W.π= × ⋅ − =

Continued …..

PROBLEM 9.78 (Cont.) (b,c) Summary of the calculations above and for water and ethylene glycol:

Fluid RaD ( )2Dh W / m K⋅ q(W)

Air 6.750 × 104 10.6 1.55 < Water 7.273 × 107 1299 187 <

Ethylene glycol 15.82 × 106 393 57.0 < COMMENTS: Note large differences in the coefficients and heat rates for the fluids.

PROBLEM 9.90 K NOWN: Temperatures and dimensions of a window-storm window combination.

F IND: Rate of heat loss by free convection.

SCHEMATIC:

ASSUMPTIONS: (1) Both glass plates are of uniform temperature with insulated nterconnecting walls and (2) Negligible radiation exchange. i

PROPERTIES: Table A-4, Air (278K, 1 atm): ν = 13.93 × 10-6 m2/s, k = 0.0245 W/m⋅K, α 19.6 × 10-6 m2/s, Pr = 0.71, β = 0.00360 K-1. =

A NALYSIS: For the vertical cavity,

( ) ( )( )( )32 131 2

L 6 2 6 2

9.8m / s 0.00360K 30 C 0.06mg T T LRa

19.6 10 m / s 13.93 10 m / s

βαν

−

− −

°−= =

× × ×

5

LRa 8.37 10 .= × W ith (H/L) = 20, Eq. 9.52 may be used as a first approximation for Pr = 0.71,

( ) ( ) ( ) ( )L1/ 40.3 0.012 0.31/ 4 0.012 5

LNu 0.42 Ra Pr H / L 0.42 8.37 10 0.71 20− −= = × LNu 5.2=

L2k 0.0245W / m Kh Nu 5.2 2.1W / m K

L 0.06m.⋅

= = = ⋅ T he heat loss by free convection is then

( )1 2q h A T T= − < ( )( )2q 2.1W / m K 1.2m 0.8m 30 C 61W.= ⋅ × ° = COMMENTS: In such an application, radiation losses should also be considered, and infiltration effects could render heat loss by free convection significant.