Electronics Cooling: Mechanical Power Engineering Dept
Electronics Cooling: Mechanical Power Engineering Dept
Electronics Cooling: Mechanical Power Engineering Dept
MPE 635
• Fourier’s law
dT
q x k
dx
T2 T1
qx k
L
Thermal convection
• The heat transfer by convection is described by the
Newton's law of cooling:
q hA(TW T )
Air movement due to temperature difference
Forced fan
Air
Free convection
- gases 2-25
- liquids 50-1000
Forced convection
- gases 25-250
- liquids 50-20,000
Convection with two phase
- boiling or condensation 2500-100,000
Thermal convection
• Example 3.1: An electric current is passed through a wire 1mm
diameter and 10 cm long. This wire is submerged in liquid water at
atmospheric pressure, and the current is increased until the water
boils. For this situation h = 5000 W/m2.oC. And the water will be 100
oC. How much electric power must be supplied to the wire to maintain
Electric wire
Thermal convection
• Solution:
• The total convection loss from the wire is given
by.
q hA(TW T )
• For this problem the surface area of the wire is
A= π d L = π (1 x 10-3) (10 x 10-2) = 3.142 x10-4 m2
• The heat transfer is therefore
4
q 5000 3.142 10 (114 100) 21.99W
• And this is equal to the electric power which
must be applied.
Thermal radiation
• The mechanism of heat transfer by radiation
depends on the transfer of energy between
surfaces by electromagnetic waves in the wave
length interval between 0.1 to 100 μm.
• Radiation heat transfer can travel in vacuum
such as solar energy.
• Radiation heat transfer depends on the surface
properties such as colors, surface orientation
and fourth power of the absolute temperature
(T4) of the surface.
• The basic equation for radiation heat transfer
between two gray surfaces is given by
q fA(T T )
1
4
2
4
Thermal radiation
T
qcond kA qcond Rt ,cond T
L
L
Rt ,cond
kA E2 E 1 L
Re
qconv hAT qconv Rt ,convT I A
1
Rt ,conv
ha
Series Circuits:
E E1 E2
i
Re Re,1 Re, 2 Re,3
• By analogy
Toverall
q
Rt
T1 T 2
q
Rt ,conv Rt ,cond Rt ,conv
T1 T 2
q
1 L 1
h1 A kA h2 A
Parallel Circuit:
L1, k1, A1
T T
T1 L2, k2, A2
qi ki Ai L3, k3, A3 T2
Li Rt ,i qtot
qtot L4, k4, A4
1 1 1 1 1 1 1 L6, k6, A6
qtot T
L7, k7, A7
Rt ,1 Rt , 2 Rt , 3 Rt , 4 Rt , 5 Rt , 6 Rt , 7
T q1
qtot q2 R2
q2
Rt ,tot q3 q3
R3
1 1
qtot q4 R4 q4 qtot
q5 q5
Rt ,tot Rt ,i R5
q6 R6 q6
q7 R7 q7
Combined Modes of Heat Transfer
• Combined Convection and Radiation
qnet qconv qrad
q rad hr A (Ts T f )
q rad (Ts4 Te4 )
hr Fse
A (Ts T f ) (Ts T f )
(Ts4 T f4 )
hr Fse
(Ts T f )
(Ts2 T f2 ) (Ts T f ) (Ts T f )
hr Fse
(Ts T f )
Combined Modes of Heat Transfer
• Combined Convection and Radiation
hr Fse (Ts2 T f2 ) (Ts T f )
hr Fse ((Ts T f ) 2 2TsT f ) (Ts T f )
Now if we define the arithmetic mean temperature as:
Ts T f
Tm
If further Ts-Te<<Ts then 2
Tm Ts T f
So we may define the radiation heat transfer coefficient as:
hr 4 Fse Tm3
And finally;
htot hconv hrad
Where q net htot A (Ts T f )
Combined Modes of Heat Transfer
• Combined Convection and Conduction
• This combination is likely to occur with the use of
extended surfaces where the primary surface exchanges
heat by convection to the adjacent fluid flow and by
conduction through the extended surfaces. This case
may be considered in a similar manner as the above, but
here the problem doesn't need extra work as the
conduction thermal resistance is predefined.