Tutorial 1 – Conduction (Solutions) Problem - 1 The fuel rod can be considered considered as a solid cylinder with a uniformly distributed heat source. qg = 74.52 MW/m h = 5!.7" #W/m 2 $ # = %.% #W/m $ Ta&e = '%o( )*&erage fluid tem+erature while +assing through the ,ac#et-
The oissons equation0 ∇
2
T +
q g k
=0
r e t a W
5ac#et
r e t a W
4uel rod
1n cylindrical cylindrical coordinates coordinates r dT + q g = 0 r dr dr k
1 d
1ntegrating twice with boundary condition )i- r = ro T = To dT )iir = % dr 0 gi&es =
T ( r )
− T o
T o
=
q g r o2 4 kT o
r 2 1 − r o
)'-
Ma3imum tem+erature tem+erature occurs at the center and is gi&en q g r o2 by T ( 0) = T max = 4k + T o )21
To find T max max we need to #now T o Ma#ing an energy balance at the surface of the rod dT = 2 r h( T − T ) 2 r o − k o o w ) dr π
π
rom 6q.)'-0 rom 6q.)-0
−
T o
k
=T w
q g r o
dT dr
=
r r o =
2
−k dT dr r =r o + h
=
= '%'!.4 = '4!.4 °(
2
8ence Tmax max
= T o +
q g r o
4k
MW/m2
0.931
2
= 146.4 +
(74.52)(0.025) 4 × 30
= '4!.497." = 54. °(
2
Problem-2
*l rame T
T ∞
T
o
c
Transistor case
Thermal grease and gas#et
;con ;tc
T ∞
;rad
To
;T Tc
T ∞
Given conditions: :+erating +ower ≤ 4.5 W ;T = (ase to ambient resistance = 29 °(/W Tcma3 = Ma3imum allowable case tem+erature = 75 °( *f = = *rea of the frame )heat sin#;tc = contact resistance between transistor and frame = %.75°(/W
3
or the transistor case tem+erature T c = 75°( the rate of heat transfer directly to the surroundings Tc − T ∞ 75 − 26 qT = = = '.75 W RT 28 1n addition heat is dissi+ated through the heat sin# )frame-0 )rom electrical networ# analogy-0
;con ;tc
T ∞
;rad
qcon
= hA∆T =
∆T 1
=
To
Tc
∆T Rcon
hA
qrad
= εσ AFs −a ( To4 − T∞4 )
*ssume emittance emittance ε = '
>ased on this result the tem+erature tem+erature of the heat sin# )T ois calculated qheat sink To
=
Tc
− T o
Rt ,c
= Tc − qheat sink Rt ,c = 75 ? '.4×%.75 = 7.""°(
@ou can reiterate with with this this tem+e tem+erat rature ure until until assum assumed ed &alue of To is a++r a++ro3 o3im imat atel ely y same same as the the calc calcul ulat ated ed &alue. Asing the abo&e &alue of q heat sin# = '.4 W we can see that the total +ower that can be safely dissi+ated = '.75 '.4 = .%" W B 4.5 W
5
Without the heat sin# the transistor can only o+erate at '.75 W. To o+erate the transistor at the rated +ower of 4.5W a bigger heat sin# is required.
Problem-3 Tb = Ts = '2%o( # = 245 W/m $
Ta = 25o( h = 9% W/m 2 $
in0 d = 2 mm C = '% mm
(a) 1f the heat losses from the ti+ is considered considered negligible )adiabatic ti+-the rate of heat transfer is gi&en by q f θ b
=
hPkAc θ b tanh mL
= T b − T ∞ =
hPkAc
=
'2% ? 25 = "5 o(
80 × π × 0.002 × 24 245 × (π / 4) × 0. 0 .002 2
= %.%'"! W/$ 6
hP
=
2
80 × π × 0.002
=
2 = !5.%! )'/m kAc 245 × (π / 4) × 0.002 m = 25.55 )'/mmC = 25.55 ×%.%' = %.255
m
=
q f
2
hPkAc θ b tanh mL = %.%'"! ×"5×%.24"! = %.4!45 W
(b) The (b) The efficiency efficiency of of the fin η f =
q f qmax
=
q f
Af = = total surface area of the fin
hA f θ b
1n the most general case for which there is heat transfer from the fin ti+ the fin surface area should include the fin ti+ area i.e. * f = = DC *ti+. :nly when there is no heat transfer from the fin ti+ e.g. in the long fin/adiabatic fin ti+ case *f = = DC. A f
= π DL
A f η f
= π (0.002)(0.01) = !.293'%5 m2 =
0.4645 −5
80 × 6.283 × 10
× 95
= "7.E
(c) The (c) The effecti&eness of the fin ε f
=
q f hAc,0θ b
=
0.4645 80 × (π / 4)(0.002)
2
× 95
= 19.45
(d) *rea (d) *rea occu+ied by fin = F * fb where *fb = base area occu+ied by each fin = * co F = no of fins +er m 2 Anfinned area = ' F * fb = *uf Total heat transfer from finned and unfinned area = Nq f + hAuf θ b 7
8eat transfer from ' m 2 of +late = 9%)W/m2 $- 3 ' )m 2- 3 "5 )$- = 7!%% W 8ence 7!%% ×2 = F)%.4!5!- 9% )' F* fb- "5
9
