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KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, h (^) w and ha, respectively.
SCHEMATIC:
ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the direction normal to flow.
ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has the form
q = h′ ( π D ) ( Ts − T∞)
and solving for the heat transfer convection coefficient, find
q h =.
πD Ts T
Substituting numerical values for the water and air situations:
Water
28 10 W/m h = = 4,570 W/m K 0.030m 90-25 C
w
π
×
× $^
<
Air
400 W/m h = 65 W/m K. 0.030m 90-25 C
a
π
× $^
<
COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 70 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases. See Table 1.1.
KNOWN: Chip width and maximum allowable temperature. Coolant conditions.
FIND: Maximum allowable chip power for air and liquid coolants.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by radiation in air.
ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to the coolant. Hence,
P = q
and from Newton’s law of cooling,
P = hA(T - T∞) = h W 2 (T - T∞).
In air ,
P (^) max = 200 W/m 2 ⋅K(0.005 m) 2 (85 - 15) ° C = 0.35 W. <
In the dielectric liquid
P (^) max = 3000 W/m 2 ⋅K(0.005 m) 2 (85-15) ° C = 5.25 W. <
COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can dissipate far less energy than in the dielectric liquid.
KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient and emissivity of a person in the room.
FIND: Basis for difference in comfort level between summer and winter.
SCHEMATIC:
ASSUMPTIONS: (1) Person may be approximated as a small object in a large enclosure.
ANALYSIS: Thermal comfort is linked to heat loss from the human body, and a chilled feeling is associated with excessive heat loss. Because the temperature of the room air is fixed, the different summer and winter comfort levels can not be attributed to convection heat transfer from the body. In both cases, the heat flux is
Summer and Winter : q conv′′^ = h T( s − T∞ ) = 2 W/m^2 ⋅ K × 12 $C =24 W/m^2
However, the heat flux due to radiation will differ, with values of
Summer : q rad′′^ = εσ ( Ts^4 − Tsur 4 ) = 0.9 × 5.67 × 10 −^8 W/m 2 ⋅ K 4 ( 3054 − 300 4 )K 4 =28.3 W/m^2
Winter : q rad′′^ = εσ ( Ts^4 − Tsur 4 ) = 0.9 × 5.67 × 10 −^8 W/m 2 ⋅ K 4 ( 3054 − 2874 )K^4 =95.4 W/m^2
There is a significant difference between winter and summer radiation fluxes, and the chilled condition is attributable to the effect of the colder walls on radiation.
COMMENTS: For a representative surface area of A = 1.5 m 2 , the heat losses are q (^) conv =
36 W, q (^) rad(summer) = 42.5 W and qrad(winter) = 143.1 W. The winter time radiation loss is significant and if maintained over a 24 h period would amount to 2,950 kcal.
