Department of polymer And Petrochemical Engineering Heat ... · convection. Obviously, an analysis of combined conduction-convection systems is Obviously, an analysis of combined

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




]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




....(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




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.




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




]/208[97.60)35.74)(82.0( hBtuWqact




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




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.




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




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

Documents

Department of polymer And Petrochemical Engineering Heat ... · convection. Obviously, an analysis of combined conduction-convection systems is Obviously, an analysis of combined