department of polymer and petrochemical engineering heat ... · convection. obviously, an analysis...
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Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
CONVECTION SYSTEMS-CONDUCTION The heat that is conducted through a body must frequently be removed (or delivered)
by some convection process. For example, the heat lost by conduction through a
furnace wall must be dissipated to the surroundings through convection. In heat-
exchanger applications a finned-tube arrangement might be used to remove heat from
a hot liquid. The heat transfer from the liquid to the finned tube is by convection. The
heat is conducted through the material and finally dissipated to the surroundings by
convection. Obviously, an analysis of combined conduction-convection systems is
very important from a practical standpoint.
Fin equation: Consider the one-dimensional fin exposed to a surrounding fluid at a temperature T∞
as shown in figure . The temperature of the base of the fin is Tb. We approach the
problem by making an energy balance on an element of the fin of thickness dx as
shown in the figure. The defining equation for the convection heat-transfer coefficient
is recalled as )ThA(Tq w where the area in this equation is the surface area for
convection. Let the cross-sectional area of the fin be (Ac) and the perimeter be (P).
Then the energy quantities are Energy in left face = dx
dTkAq cx ,
Energy out right face = dx)dx
Td
dx
dT(kA]
dx
dTkAq
2
2
cdxxcdxx
Energy lost by convection = )ThPdx(T
Energy in left face= Energy out right face+ Energy lost by convection
)T(TkA
hp
dx
Td
*
1])dxThp(Tdx
dx
TdkA
2
2
2
2
dxkA
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
]equation governor of [eCeCθ
Ceθ
mD0)θmD(θmθDDdx
d
equation)(governorθmdx
θd
dx
θd
dx
Td
dx
dθ
dx
)Td(θ
dx
dT
]kA
hpm[]TT[θlet
mx
2
mx
1
Dx
2222
2
2
2
2
2
2
2
2
solutiongeneral
and
Can be applied to three types of fins, depending on physical situation
CASE 1 The fin is very long, and the temperature at the end of the fin is
essentially that of the surrounding fluid.
CASE 2 The end of the fin is insulated
CASE 3 The fin is of finite length and loses heat by convection from its end.
For case1 :- The boundary conditions are
at at [x = 0] TTθθ(0) bb
x =L=∞
Tfin tip=T∞ i.e (TL =T∞) then,
0TTθ(L) L
Appling this boundary conditions in general solution equation, yields
b21
m(0)
2
-m(0)
1(0) θCCeCeCθ
at x =L=∞ 0C2)(C(0)CeCeC0θ 21
)m(
2
)m(
1(L)
mx
b(x)
b1
m(0)
1b
eθθ
θCeCθ
And the solution becomes ckA
hpx
mx
bb
(x)ee
TT
TT
θ
θ
For case2:- negligible heat loss from the fin tip (insulated fin tip ,qfin tip=0
)the boundary conditions are
at x = 0 bθθ(0)
x=0 θ=θb
x=L
cond=conv
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
....(a) θCCeCeCθ b21
m(0)
2
-m(0)
1(0)
Lx
0x
θ 0qfin(tip)
Lxat
.......(b)]emCeC0 m(L)
2
-mL
1 mm
Solving equations (a and b) for C1 and C2 the solution is obtained as
mLmLmL
where
ee
ee
ee
e
ee
e
θ
θ
thene1
θC2
e1
θC1
e1
e
e1
e
θ
θ
mL
L)-m(xL)-m(x
mL
L)-m(x
mL
L)-m(x
b
(x)
2mL
b
2mL
b
2mL
mx
2mL
mx
b
(x)
Or mL
ee
ee
θ
θmL
x)--m(Lx)-m(L
b
(x)
coshmL
x)]cosh[m(L
TT
TT
θ
θ
b
(x)
b
(x)
Case 3 the boundary conditions are
at x = 0 b(0) θθ
qconduction(x=L)= qconvection(x=L)
LxLLx
LxLxc
)Th(Tx
θk
θ hAc
x
θkA
c
c
b
(x)
coshmL
x)coshm(L
TT
TT
All of the heat lost by the fin must be conducted into the base at x = 0.
Using the equations for the temperature distribution, we can compute the
heat loss from 0x]dx
dTkAq
An alternative method of integrating the convection heat loss could be
used:
L L
dxhPdxTThPq0 0
)( In most cases, however, the first equation is
easier to apply.
qfin
p
Ac
Lc
a-Actual fin with convection
at the tip
qfin
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
The heat flow for case 1 is bc
m(0)
b θhPkA)emθkA(q
For case 2, tanhmLθhPkA)e1
1
e1
1m(kAθq bc2mL2mLb
For case 3 cbc tanhmLθhPkAq
-Fin efficiency:
temp)baseatwereareafinentired(iftransferrebewouldwhichHeat
dtransferreheatActualηf
finmax
finf
q
qη the fin efficiency for case1 infinite long fin
mL
1
kA
hp
L
1
)T(ThA
)T(ThPkAη
cbfin
bc
fin)f(verylong
mL
tanhmL
hPLθ
tanhmLθhPkA
q
qη
b
b
maxfin,
findtipfin)f(insulate
where (z)is the depth of the fin and (t) is the thickness. Now, if the fin is
sufficiently deep, the term (2z) will be large compared with (2t ), and
Lkt
2hL
ktz
2hzmL
Multiplying numerator and denominator by (L1/2 ) gives 3/2L
kLt
2hmL
(Lt) is the profile area of the fin, which we define as [Am = Lt] So that
3/2
m
LkA
2hmL
A corrected length Lc is then used in all the equations which apply for the case
of the fin with an insulated tip.
2z
t*zLLtzift)2(zpt*zA
p
ALL cc
cc
2
tLLc ,The error which results from this approximation will be less than 8
percent when 2
1)
2k
ht( 1/2 , d/4L
πd
/4πdLL
2
c
where Af in is the total surface area of the fin. Since[Afin =pL] for fins with
constant cross section.
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
Example (1) An aluminum fin [k = 200 W/m·℃] 3.0 mm thick and 7.5 cm long
protrudes from a wall. The base is maintained at 300℃, and the ambient
temperature is 50℃ with h = 10 W/m2·℃. Calculate the heat loss from the fin
per unit depth of material.
Solution:- We may use the approximate method of solution by extending the
fin a fictitious length t/2 and then computing the heat transfer from a fin with
insulated tip.
7.65cm0.157.5t/2LLc , 774.5ktz
2t)h(2z
kA
hPm
bc θhPkA)(tanhmLq , ]65.4[103)103)(1( 2233 inmA
And q = (5.774)(200)(3*10-3
)(300-50)tanh[(5.774)(0.0765)] = 359 W/m
[373.5 Btu/h·ft]
Circumferential fins of rectangular profile
L= r2 − r1 Lc = L +t/2 r2c= r1 + Lc Am= tLc
Example (2) Aluminum fins 1.5 cm wide and 1.0 mm thick are placed on a 2.5-
cm-diameter tube to dissipate the heat. The tube surface temperature is 170℃,
and the ambient-fluid temperature is 25℃.Calculate the heat loss per fin for h =
130 W/m2·℃ .assume k=200W/m.oC for aluminum.
Solution. For this example we can compute the heat transfer by using the fin-
efficiency curves . The parameters needed are
Lc = L + t / 2 = 1.5 + 0.05 = 1.55 cm , r1 = 2.5/2 = 1.25 ,r2c = r1 + Lc = 1.25 +
1.55=2.80 cm
r2c/ r1 = 2.80 / 1.25 = 2.24, Am = t(r2c - r1 ) = (0.001)(2.8 – 1.25)(10-2) = 1.55*10-5
m2
396.0)1055.1)(200(
130)0155.0()(
5
2/32/12/3
m
ckA
hL
From figηf = 82 percent.
qf=hAf(Tb-T∞) ,Af=surface area Af=2π(rc22-r1)
.7Btu/h]74.35W[25325))(130)(170)(101.25(2.82)T)h(Tr(r 2q 422
0
2
1
2
2cmax
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
]/208[97.60)35.74)(82.0( hBtuWqact
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
FIN EFFECTIVENESS :-
The performance of the fins is judged on the basic of the enhancement in heat
transfer relative to the no fin casebb
bfinfin
withoutfin
withfinfin
θhA
hθAη
q
qε For the insulated-tip
fin. Where Afin=pL is the total surface area of the fin and Ab=A is the base area.
ɛfin=1 indicated that the addition of fins to the surface does not affect heat transfer.
ɛfin<1 indicated that the fin actually acts as insulation slowing down the heat transfer
from the surface. This situation can occur when fins made of low thermal conductivity
materials are used .
ɛfin>1 indicated that the fins are enhancing heat transfer from the surface.
THERMAL CONTRCT RESISTANCE
B
BB
C
BA
A
AA
x
TTAk
Ah
TT
x
TTAkq
322221
/1or
AkxAhAkx
TTq
BBcAA //1/
31
1/hc A is called the thermal contact
resistance and hc is called the contact
coefficient.
Example :Two 3.0cm diameter 304
stainless-steel bars, 10 cm long, have ground surfaces and are exposed to air
with a surface roughness of about 1 μm. If the surfaces are pressed together
with a pressure of 50 atm and the two-bar combination is exposed to an overall
temperature difference of 100℃, calculate the axial heat flow and temperature
drop across the contact surface.
Solution:- The overall heat flow is subject to three thermal resistances, one
conduction resistance for each bar, and the contact resistance. For the bars,
679.8)103()3.16(
)4)(1.0(22
kA
xRth
℃/W
XX AA
q q
A B
1 2 3
1T
T
T2A
2BT
T3
X
(a)
(b)
qfin
Tb
Ab
Tb
Ab
qfin
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
From Table the contact resistance is 747.0)103(
)4)(1028.5(122
4
AhR
c
c ℃/W
The total thermal resistance is thereforeΣRth = (2)(8.679) + 0.747 = 18.105
and the overall heat flow is ]/83.18[52.5105.18
100hBtuW
R
Tq
th
The temperature drop across the contact is found by taking the ratio of the
contact resistance to the total thermal resistance:
13.4105.18
)100)(747.0(
T
R
RT
th
cc ℃ [39.43℉]
In this problem the contact resistance represents about 4 percent of the total
resistance.
Problems
Q1)Water and air are separated by a mild-steel plane wall. It is proposed to
increase the heat-transfer rate between these fluids by adding straight
rectangular fins of 1.27-mm thickness and 2.5-cm length spaced 1.27 cm apart.
The air-side and water-side heat-transfer coefficients may be assumed constant
with values of 11.4 and 256 W/m^2.K respectively. Determine the percent
change in total heat transfer when fins are placed on (a) the water side, (b) the
air side, and (c) both sides
Q2)Straight rectangular fin ,made of aluminum(k=208 W/m.0c)4mm
thickness(t)and 100mm width(w).The environment condition is h=60 W/m2. 0
C
and T2=3000C .Its found that the efficiency of the fin is(0.85)and the base
temperature Tb=12000C? 1-calculate the heat loss from the base of the fin.2-
Detrmine the performance of the fin. Note Neglect the convection at the tip
of the fin
Q3)To increase the heat dissipation from a 2.5 cm_O.D tube ,circumferential
fins made of aluminum (k=200 W/m.K) are soldered to the outer surface .The
fins are 0.1 cm thick and have an outer diameter of 5.5 cm as shown in fig .If
the tube temperature is 100 0C , the environmental temperature is 25 0C,and the
heat transfer coefficient between the fin and , the environmental is 25
W/m2.K,calculate the rate of heat loss from the fin.
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
Q4)An experimental device that produces excess heat is passively cooled. The
additional of pin fins to the casing of this device is being considered to
augment the rate of cooling .Consider a copper pin fin 0.25cm in a diameter
that protrudes from a wall at 95 0C into ambient air temperature as shown in
figure .The heat transfer is mainly by natural convection with a coefficient
equal to 10W/m2.k .Calculate the heat loss, assuming (a)the fin is "infinitely
long"(b)the fin is 2.5 cm long and the coefficient at the end is the same as
around circumferences .Finally(c)how long would the fin have to be for the
infinitely long solution to be correct within 5percent?
Q5)
Q6)
Q7)The figure below shows part of a set of radial aluminum
fins(k=180W/m.K)that are to be fitted to a small air compressor .The device
dissipates 1 Kw by convection to surrounding air which is at 20 0C .each fin is
100mm long ,30mm high and 5mm thick .The tip of each fin may be assumed
to be adiabatic and heat transfer coefficient of h=15 W/m2.K acts over the
Department of polymer And Petrochemical Engineering
Heat And Mass Transfer
Assistant lecture:Qusai A.Mahdi
remaining surfaces. Estimate the number of fins required to ensure the base
temperature drop not exceed 120 0C?
Q8)A hot plate is to be cooled by attaching aluminum fins of square cross
section (0.15mx0.2m)on one side ,k=237 w/m.c ,h=20 w/m2.c .Assuming heat
transfer from the fin tips is negligible .Determined the number of fins needed to
triple the rate of heat transfer?
Q9) Steam in a heating system flows through tubes whose outer diameter is 3
cm and whose walls are maintained at a temperature of 120°C. Circular
aluminum fins (k =180 W/m · °C) of outer diameter 6 cm and constant
thickness of 2 mm are attached to the tube. The space between the fins is 3 mm,
and thus there are 200 fins per meter length of the tube. Heat is transferred to
the surrounding air at 25°C, with a combined heat transfer coefficient of h = 60
W/m2· °C. Determine the increase in heat transfer from the tube per meter of its
length as a result of adding fins. (Fin efficiency = 95%).
Q10)A hot surface at 100°C is to be cooled by attaching 3-cm-long, 0.25-cm-
diameter aluminum pin fins (k =237 W/m · °C) to it, with a center-to-center
distance of 0.6 cm. The temperature of the surrounding medium is 30°C, and
the heat transfer coefficient on the surfaces is 35 W/m2°C. Determine the rate
of heat transfer from the surface for a 1-m 1-m section of the plate. Also
determine the overall effectiveness of the fins. .(Fin efficiency = 95.9%).
Tb=850C T∞=250C
2mmX2mm
4cm