# MECH 4003 Lecture Notes - Lecture 9: Reynolds Number, Junction Temperature, Thermocouple

by OC2206240

This

**preview**shows pages 1-3. to view the full**9 pages of the document.**PROBLEM 6.43

KNOWN: Ambient, interior and dewpoint temperatures. Vehicle speed and dimensions of

windshield. Heat transfer correlation for external flow.

FIND: Minimum value of convection coefficient needed to prevent condensation on interior surface

of windshield.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties.

PROPERTIES: Table A-3, glass: kg = 1.4 W/m⋅K. Prescribed, air: k = 0.023 W/m⋅K, ν = 12.5 ×

10-6 m2/s, Pr = 0.70.

ANALYSIS: From the prescribed thermal circuit, conservation of energy yields

,i s,i s,i ,o

igo

TT TT

1/h t/k 1/h

∞∞

−−

=+

where o

h may be obtained from the correlation

L0.8 1/3

oL

hL

Nu 0.030Re Pr

k

==

With V = (70 mph × 1585 m/mile)/3600 s/h = 30.8 m/s, ReD = (30.8 m/s × 0.800 m)/12.5 × 10-6 m2/s

= 1.97 × 106 and

(

)

()

0.8 1/3

62

o0.023W / m K

h 0.030 1.97 10 0.70 83.1W / m K

0.800 m

⋅

=×=⋅

From the energy balance, with Ts,i = Tdp = 10°C

()

()

1

s,i ,o

igo

,i s,i

TT t1

hkh

TT

−

∞

∞

⎛⎞

−

=+

⎜⎟

⎜⎟

−⎝⎠

()

()

1

i2

10 15 C 0.006 m 1

h50 10 C 1.4W / m K 83.1W / m K

−

⎛⎞

+°

⎜⎟=+

⎜⎟

−° ⋅ ⋅

⎝⎠

2

i

h 38.3W / m K=⋅ <

COMMENTS: The output of the fan in the automobile’s heater/defroster system must maintain a

velocity for flow over the inner surface that is large enough to provide the foregoing value of i

h. In

addition, the output of the heater must be sufficient to maintain the prescribed value of T∞,i at this

velocity.

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PROBLEM 6.51

KNOWN: Average frictional shear stress of s

τ

=

0.0625 N/m2 on upper surface of circuit board with

densely packed integrated circuits (ICs)

FIND: Allowable power dissipation from the upper surface of the board if the average surface

temperature of the ICs must not exceed a rise of 25°C above ambient air temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) The modified Reynolds analogy is applicable, (3)

Negligible heat transfer from bottom side of the circuit board, and (4) Thermophysical properties

required for the analysis evaluated at 300 K,

PROPERTIES: Table A-4, Air (Tf = 300 K, 1 atm): ρ = 1.161 kg/m3, cp = 1007 J/kg⋅K, Pr = 0.707.

ANALYSIS: The power dissipation from the circuit board can be calculated from the convection rate

equation assuming an excess temperature (Ts - T∞) = 25°C.

()

ss

qhAT T

∞

=− (1)

The average convection coefficient can be estimated from the Reynolds analogy and the measured

average frictional shear stress s.

τ

2/3 s

ff2p

Ch

St Pr C St

2Vc

V/2

τ

ρ

ρ

=== (2,3,4)

With V = u∞ and substituting numerical values, find h.

2/3

s2p

hPr

Vc

V

τ

ρ

ρ

=

sp 2/3

c

hPr

V

τ

−

=

()

22/3 2

0.0625 N / m 1007 J / kg K

h 0.707 39.7 W / m K

2m/s

−

×⋅

==⋅

Substituting this result into Eq. (1), the allowable power dissipation is

()

22

q 39.7 W / m K 0.120 0.120 m 25 K 14.3 W=⋅×××=

<

COMMENTS: For this analysis using the modified or Chilton-Colburn analogy, we found Cf =

0.0269 and St = 0.0170. Using the Reynolds analogy, the results are slightly different with

2

h31.5W/mK=⋅ and q = 11.3 W.

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