MEC 3609 H T Conduction

MEC 3609: HEAT TRANSFER

Conduction

Convection

Radiation

Dr M N A Hawlader

Office: E1-05 8.1

e-mail: [email protected]

Tel. 6196 6518 (O)

0102796097 (M)

What is Heat Transfer?

• Energy in Transit is called Heat.

• Heat Transfer is a process by which energy transfer takes place.

• A knowledge of heat transfer is necessary in order to evaluate cost, the feasibility, and the size of the equipment to transfer a specified amount of heat in a given time.

• For example, power generation, heating and cooling, and many industrial and domestic applications.

Conduction, Convection and

Radiation

Heat Transfer Conduction Heat Transfer:

Fourier's law of conduction, one dimensional heat conduction through composite wall, tubes and spheres.

Derivation of general transient conduction equation with a heat source. Steady state 1D conduction with and without energy generation; overall heat transfer coefficient, critical and economic thickness of insulation.

Extended Surfaces: derivation of fin equation for simpler cases, fin efficiency and effectiveness.

Unsteady heat conduction, lumped -system analysis, numerical method.

Convection Heat Transfer

Forced and free convection;

Laminar and turbulent flow;

Mass, Momentum and Energy Equation;

External and internal flows;

Heat Exchangers;

Boiling and condensation.

Radiation Heat Transfer

Laws of blackbody and gray body radiation; semi-

transparent and opaque material.

Intensity, emissive power, emittance, absorptance,

reflectance, transmittance; Shape factor.

Radiation exchange between blackbody and gray

surfaces; radiation shield.

Thermal Power Plant

Steam Turbine

Heat Exchangers

Basic Modes of Heat Transfer

• Three distinct modes of heat transfer: conduction,

convection and radiation.

• Heat transfer in a solid or fluid at rest takes place by

conduction.

• Heat transfer in a fluid in motion takes place by

convection.

• Both conduction and convection require a medium,

whereas, radiation heat transfer requires no medium.

Conduction Heat Transfer

• Conduction heat transfer

Qx A dT/dx

• Heat transfer in x direction is given by

Qx = -kA dT/dx

(Fourier Law of Conduction)

k, constant of proportionality, called thermal

conductivity, W/m.K

T

x

Conduction Heat Transfer

T

x

T

x

dT/dx >0

Q<0 dT/dx <0

Q>0

Sign Convention for the direction of heat flow

Thermal Conductivities

Materials k at 300K (W/m.K)

• Copper 399.0

• Aluminium 237.0

• Carbon steel, 1%C 43.0

• Glass 0.81

• Plastics 0.2 - 0.3

• Water 0.6

• Air 0.026

• Styrofoam 0.030

Convective heat transfer between a surface

and a fluid can be calculated by

Qc = hc A T, Newton Law of Cooling

Qc = rate of heat transfer, W

hc = average convective heat

transfer coefficient, W/m2.K

A = heat transfer area, m2

T = Temp diff between surface

(Ts) and the fluid (Tf), K

Convective Heat Transfer

Fluid W/m2K

• Air, free convection 6 - 30

• Superheated steam or

air, forced convection 30 - 300

• Oil, forced convection 60 - 1,800

• Water, forced convection 300 - 6,000

• Water, boiling 3,000 - 60,000

• Steam, condensing 6,000 - 120,000

Convective heat transfer

coefficient

Radiation Heat Transfer • The amount of radiation leaving a surface

depends on the absolute temperature (K)

and the nature of the surface.

• A perfect radiator (blackbody) emits

radiation from the surface at the rate

Qr = A1T14, W

A1 = surface area, m2

T1= surface temperature, K

= Stefan-Boltzmann constant

5.67x10-8,W/m2K4

Basic Equations of Heat

Transfer

• Conduction: Fourier Law of Conduction

Qx = -kA dT/dx, W

• Convection: Newton Law of Cooling

Qc = hc A T, W

• Radiation: Stefan-Boltzmann Equation

Qr = A1T14, W

Example 1: In a manufacturing plant, the walls and ceiling of an oven are made of 200 mm thick fire-clay brick having a thermal conductivity of 1.5 W/m.K. During steady-state operation, measurements reveal an inner surface temperature of 1200oC and an outer surface temperature of 200oC. The internal dimensions of the oven are as follows: Length = 4m, Width=3m and the Height =3 m. What is the rate of heat input required to maintain steady-state temperature inside the oven?

Solution of Example 1:

4 m

3 m 3 m

1200oC 200oC

200 mm

Tm = 0.5(1200+200)

= 700oC

km = 1.5 2(4x3+3x3)+4x3 = Internal HT area = 54 m2

Four walls Ceiling

Heat Flux, Q/A = -k dT/dx = - k (To - Ti)/t

= 1.5(W/m.K)[(1200 - 200)(K)]/0.20( m)

= 7500 W/m2

Rate of heat input required = 7500 x 54 = 405 kW

Example 2: An insulated pipe supplying steam from a boiler runs through a room where the air and walls are at 30oC. The outer diameter of the pipe is 100 mm and its surface temperature is 250oC. The natural convection heat transfer coefficient from the surface to the air is 20 W/m2K. Find the rate of heat transfer from the surface due to convection and radiation per unit length of pipe. For radiation heat loss, the outer surface of the pipe may be treated as blackbody surface.

Solution for Example 2:

Tœ=30oC

250oC Do = 100 mm

h=20 W/m2K

Heat loss due to convection, Q

= hA (To - Tœ),(W/m2K)(m2)(K);

= h( DoL) (To - Tœ)

Q/L = 1382.3 W/m = Qcon

Heat loss due to radiation,Q

= A (To 4 - Tœ

4 ), Temp, T in K

= ( DoL) (To 4 - Tœ

4 )

Q/L = (5.67x10-8,W/m2K4) (0.1m)( 5234 -3034)(K4)

= 1182.58, W/m = Qrad

Qtotal = Qcon + Qrad =1382.3 + 1182.58 = 2564.88, W/m

Example 3 • The temperature in a house located at latitude 40o N is

maintained at 23oC with a temperature controller. The temperature of the inner surfaces of walls, floors and the ceiling of the house are found to be at an average temperature of 12oC in winter and 27oC in summer. A person with an external body surface area of 1.2 m2 and temperature of 32oC remains in standing position for fifteen minutes inside the room, where the temperature is 23oC. Find the rate of radiation exchange between the person and surrounding surfaces.

Solution Example 3

The emittance of person (external surfaces of the skin) =0.95

The rate of radiation heat exchange is given by

)(44

surrss TTAQ

For summer,

))(300305)(2.1)(./1067.5)(95.0( 4442428 KmKmWxQ

= 35.78 W

For winter,

))(285305)(2.1)(./1067.5)(95.0( 4442428 KmKmWxQ

=132.9W

Note: T is the Absolute Temperature in K

Although the thermostat setting is same, one feels chilly in winter

and warm in summer.


Transfer


Qx = -kA dT/dx, W

•Convection: Newton Law of Cooling

Qc = hc A T, W

•Radiation: Stefan-Boltzmann Equation

Qr = A1T14, W

Exercise-1A

• 1. In a cold climate, a house is heated either using electricity or gas or coal to maintain

the desired temperature. The roof of such a house is 5 m long and 7 m wide, and 0.20 m thick, and is made of concrete having a thermal conductivity of 0.8 W/m.K. On a winter night , the temperatures of the inner and outer surfaces of the roof are measured as 16oC and 2oC, respectively, for a period of 8 hours. Determine (i) the rate of heat loss through the roof and (ii) the cost of heat loss to the home owner if the cost of electricity is $0.17 per kWh.

• 2. An electrical heater, which consists of a rod 300 mm long and 10 mm in diameter, is

placed in room at 12oC in steady state operation. Heat is generated in the rod as a result

of resistance heating and the surface temperature is 140oC under steady state operation. The voltage drop and the current through the rod are measured and found to be 50 V and 2 A, respectively. Considering negligible heat losses by radiation, estimate the convective heat transfer coefficient between the outer surface of the rod and the air in the room

• 3. A blackbody at 25oC is exposed to solar radiation and the temperature increased to 95o C. Estimate the increase in radiation heat transfer.

Exercise-1A

• 1. In a cold climate, a house is heated either using electricity or gas or coal to maintain the desired temperature. The roof of such a house is 5 m long and 7 m wide, and 0.20 m thick, and is made of concrete having a thermal conductivity of 0.8 W/m.K. On a winter night , the temperatures of the inner and outer surfaces of the roof are measured as 16oC and 2oC, respectively, for a period of 8 hours. Determine (i) the rate of heat loss through the roof and (ii) the cost of heat loss to the home owner if the cost of electricity is $0.17 per kWh.

Exercise-1A

Brief outline of solution

• Question 1

• Heat loss from the roof

• q = kA[(T1 –T2)/t], where t = thickness of

in kW the roof

Amount of heat lost during 8 hours

Q = q x no of hours, kWh

Cost = (amount of energy in kWh) x(unit cost of

energy)

Exercise-1A

• Question 2

• An electrical heater, which consists of a rod 300 mm long and 10 mm in diameter, is placed in room at 12oC in steady state operation. Heat is generated in the rod as a result of resistance heating and the surface temperature is 140oC under steady state operation. The voltage drop and the current through the rod are measured and found to be 50 V and 2 A, respectively. Considering negligible heat losses by radiation, estimate the convective heat transfer coefficient between the outer surface of the rod and the air in the room.

•

Exercise-1A


• Question 2

• Neglect radiation heat loss

• Under steady state operation, heat loss from the surface by convection equals energy generated within the rod due to resistance heating.

• q = energy generated = VI = (voltage drop, V) x

• (current, A), W

• = heat lost =hAs(Ts – Ta), W

• h = q /[As(Ts – Ta)] = ,W/m2 K

Exercise-1A

• Question 3

• A blackbody at 25oC is exposed to solar

radiation and the temperature increased to

95o C. Estimate the increase in radiation

heat transfer.

Exercise-1A


• Question 3

• Calculate emissive power at both temperatures. Increase in radiant heat transfer is equal to the difference in emissive power.

• E1 = σ T14 , W/m2 E2 = σ T2

4 , W/m2

• Increase in radiant heat transfer = E2 –E1

SECTION 2 CONDUCTION EQUATIONS

Conduction equation is a mathematical tool.

Describe energy distribution within the body

of the material.

Derived by performing an energy

balance on an elemental volume.

Steady state problem - the temperature

of the body of the material is independent

of time.

Transient or unsteady problem - the

temperature of the material is a function

of time.

CONDUCTION EQUATIONS (contd)

• T = f(x,t) ...one dimensional,

transient problem

T = f(x) One dimensional,

steady state

• T = f(x,y,z,t) three dimensional,

transient problem

• T = f(r,,t) Cylindrical co-

ordinate, 2D, transient

• T = f(r,) 2D, steady

x

θ

r

1-D CONDUCTION EQUATIONS

y

x

z

dx

dy dz

x x

q q+dq

Figure 1.1: Control volume in rectangular co-ordinates

T(x,t)

Qx >>Qy, Qz

CONDUCTION EQUATIONS

Plane Wall

Considering temperature variation in the control

volume in x-direction only, T = f(x)

Rate of energy conducted into control volume

= Rate of energy conducted out of control volume

x

y

At x, heat conducted into the element

= Qx

=-kA dT/dx

At x+dx, heat conducted out of the element

= Qx+dx = -kA dT/dx

+[d(-kAdT/dx)/dx]dx

x = 0

T=To

x=L

T=TL

L

For steady state conduction,

Qx – Qx+dx =0

x x+dx

21

2

2

tan

0

0

CxCT

solution

tconsk

dx

Tdk

dx

dTk

dx

d

becomes

d2T/dx2=0

Integrate

twice

STEADY STATE CONDUCTIO0N

1-D system without heat generation

Rectangular Coordinates

Boundary Conditions:

1. T(0) = T0

2. T(L) = TL

Fluid Flow

at T , h

x

x=0 x=L

T0 TL

T

Heat Flux

W/m2K

T = C1x + C2

x = 0, T=To, C2 = To;

x = L, T=TL,

TL = C1 L + To C1 =( TL –To)/L

Hence, T = x ( TL –To)/L + To

(T – To)/(TL – To) = x/L



Rectangular Coordinates

Solution:

T(x) – T0

TL - T0

T0 TL

Fluid Flow

at T , h

x

x=0 x=L

T0 TL

T

Heat Flux

W/m2K

= x/L (2.3)

Q = -kA dT/dx

=kA(T0 - TL)/L

= (T0 - TL)/(L/ kA)

Rt.cond = L/kA (2.6)

Rt,cond

When convection at x=0 and x=L is

taken into account

1/hiA L/kA 1/hoA

Ti To TL T

Qc=hiA(Ti-T0)

= (Ti-T0)/(1/hiA)



Rectangular Coordinates-Composite walls

T1 T2 T3 T4 T

Fluid Flow

at T , h

x

T

Heat Flux

W/m2K

1 2 3 4

L1

L2 L3

q = Ttotal/ Rt

Rt = R1 + R2 + R3 + R4

R1 = L1/k1A1

R2= L2/k2A2

R3= L3/k3A3

R4= 1/hA

Q = hA(T4 - T)

R4= 1/hA

R1 = L1/k1A1 R2= L2/k2A2 R3= L3/k3A3 R4= 1/hA

q

Taking convection into account on

the left surface

• Qx=

Ti-T1 =Qx(1/hiA)

T1-T2= Qx(L1/k1A)

T2-T3 = Qx(L2/k2A)

T3-T4=Qx(L3/k3A)

T4-To =Qx(1/hoA)

Ti –To = Qx(1/hiA+L1/k1A+L2/k2A+L3/k3A+1/hoA) =∆T

Qx = UA ∆T where UA= 1/ Rtotal (2.13)

1 2 3

4

Ti To

Example 2.1B

• A double glazed window (height:1m and

width:1.5m consists of two 4mm-thick layers of glass (k=0.78 W/mK) separated by a 10mm thick stagnant air space (k=0.026 W/mK). Determine the steady rate of heat transfer through this double-glazed window. The temperature inside the room is maintained at 22oC while the ambient (outdoor) is 32oC. The convective heat transfer coefficient of the inner and outer surfaces of the window are hi=12 W/m2K and h0=48 W/m2K, respectively, which include the effect of radiation.

Solution Example 2.1B

Solution:

W

WK

WK2

mK2

mW

WK

WK2

mmKW

m

WK2

mmKW

m

WK2

mK2

mW

15.303317.0

2232

R

TQ

3317.0RR

0139.05.148

1

Ah

1R

0034.0RR

256.05.1026.0

001.0

AkR

0034.05.178.0

004.0

AkR

055.05.112

1

Ah

1R

totoal

total

o

0

13

2

22

1

11

i

i

l

l

l1 l3 l2

22oC 32

oC

Ri R1 R2 R3 Ro

Exercise 2A • 1. In an aluminum pan placed on a heater, heat is transferred

steadily to the boiling water. Find the outer surface temperature of the bottom of the pan and the boiling heat transfer coefficient under the following conditions:

• The inner surface temperature of the bottom of the pan: 108oC.

• Rate of heat transfer to the bottom of the pan : 600 J/s

• Thermal conductivity of the aluminum pan material : 237 W/m.K

• The diameter of the pan : 250 mm

• Temperature of water inside the pan : 95 oC

• The thickness of the pan material : 5 mm

•

• Ans: 942 W/m2K;108 oC

Exercise 2A-contd

• 2. The temperature inside a house is maintained at 20 oC with a gas heater, where the outside temperature is 10 oC on a winter night. Heat is lost through roof, 300 m2 in area and thickness 150 mm, to the surroundings air at 10 oC. The roof may be considered black with an emittance value of 1. The thermal conductivity of the roof material is 2 W/m.K. The roof exchange radiation with the sky at 260K. The convective heat transfer coefficients for the inner and outer surfaces of the roof are 5 and 12 W/m2K, respectively.

• Determine – the rate of heat loss from the roof;

– for a period of 14 hours, find the amount of money lost, if the energy cost is 2 cents/MJ.

• Ans:37.44 kW; $38.00

Steady state conduction

Cylindrical coordinate

Fig 2.4

Heat Trasnsfer from a

hollow cylinder

Steady state heat conduction

Cylindrical coordinate system

• For steady heat transfer in radial direction

• qr - qr+dr = 0

• qr – [qr+dr] = 0 (2.16)

• qr – [qr+(dqr/dr)dr]=0

• Fourier law of conduction gives

• qr = - kA (2.17)

•

•

•

dx

dT

where A = 2 r.L

L = length of cylinder

Substitution of equation (2.17) into equation (2.16) results in the following equation:

-(2 rL)kdr

dT + (2 rL)k

dr

dT + (2 kL)

d dTr dr

dr dr

=0

(2 kL)d dT

r drdr dr

=0

d

drrdT

dr

0

d

drrdT

dr

0 (2.18)

Integrating equation (2.18) twice gives

T= C1Ln r + C2


(1) T(ri) =Ti

(2) T(ro ) =To 21

1

1

0

CLnrCT

drr

CdT

Cdr

dTr

Integrate

dr

dTr

dr

d

21 CLnrCT

Boundary Conditions (BC)

1 ii TrT

2 oTrT

BC 1 gives 21 CLnrCT ii (1)

BC 2 gives 21 CLnrCT oo (2)

Subtract (2) from (1)

o

i

oioir

rLnCLnrLnrCTT 11 )(

Hence, )/(1

oi

oi

rrLn

TTC

Insert C1 in equation 1, giving

2

2

)/(

)/(

CLnrrrLn

TTT

CLnrrrLn

TTT

i

oi

oii

i

oi

oii

Hence, the solution for the local temperature, T(r) is

T(r T T T

Ln r ri

Ln r ri i o

i o

)

where,

CT T

Ln r r

i o

i o

1

and

C T T TLn r

Ln r ri i o

i

i o

2

Hence, heat flow through the wall

Q k A rd T

d rk rL

d T

d r

kLT T

L n r r

T T

L n r r

k L

T T

RR

L n r r

k L

r

i o

o i

i o

o i

i o

tt

o i

( ) ( )

;

2

2

2

2

Rr h L1

1 1

1

2

R

Ln r r

k L2

2 1

12

R

Ln r r

k L3

3 2

22

Lk

rrR

3

34

42

)/ln(

LhrR

o4

52

1

Rt = R1 + R2 + R3 + R4 + R5

Example 2.2B Steam at a temperature of 300oC flows through a

cast iron pipe (k=75 W/mK) whose inner and

outer diameters are 50mm and 55mm

respectively. The pipe is covered with 25mm-

thick glass wool insulation (k=0.05 W/mK).

Heat is lost to the surroundings at 30oC by

convection and radiation, with a combined heat

transfer coefficient of 25 W/m2K. The heat

transfer coefficient at the inner wall of the pipe

is 65 W/m2K. Find the rate of heat loss from the

steam per unit length of the pipe. Find the

temperature drop across the wall of the pipe and

the insulation. Also find (UA) for the pipe.

Example 2.2B: Solution

Ti = Fluid temperature inside the tube T1 = Temperature of the inner wall of the tube

T2 = Temperature of the outer wall of the tube

T3 = Temperature of the outer surface of

T3

To

T2 r2

r1

Ti T1 T2 T3 T0

R1

=1/h1A1

R2 R3 R4

T3

Ti

T1

T2

To

insulation

To = Surrounding fluid temperature

1i

1Ah

1R ;

Lk2

rrLnR

1

12

2

;

Lk2

rrLnR

2

23

3

; oo

4Ah

1R

Considering L=1m , A1 = 2r1 =

2(0.025) = 0.1571 m2

A0 = A3 = 2r3 = 2(0.0525) = 0.3299 m

2

WK

2mmKW

098.01571.065

1R1

WKmKW

0002.0752

255.27Ln

R 2

WKmKW

058.205.02

5.275.52Ln

R 3

WK1212.03299.0x25

1R 4

WK

RRtotal

2774.2

1212.0058.2002.0098.0

0

Q = steady state heat loss

= (300 – 30) / 2.2774 = 118.55W

Temperature drop across pipe wall (pw)

2pw QRT

= (118.55W)(0.0002 K/W)

= 0.02 oC

3insulation QRT

= (118.5W)(2.058 K/W)

= 243.97 oC

T1 T2

Q

R2

total

oi

R

TTTUAQ

, where oi TTT

Hence, totalR

1UA

mKW439.02774.2

1


Transfer


Qx = -kA dT/dx, W

• Convection: Newton Law of Cooling

Qc = hc A T, W

• Radiation: Stefan-Boltzmann Equation

Qr = A1T14, W


Spherical coordinate


Spherical coordinate

dr

dTkAqr

drdr

dTkA

dr

d

dr

dTkAq drr

For steady state, qr – q r+dr = 0

For A=4πr2

04 2

dr

dr

dTkr

dr

d

For constant thermal conductivity

02

dr

dTr

dr

d


Spherical Coordinates

ro

ri


Spherical Coordinates

21)( C

r

CrT


1) T(ri) = Ti; 2

i

1

rC

CT i

2) T(ro) = To To = - 2

o

1

rC

C

Hence,

)/1()/1(

1

io

oi

rr

TTC

and

)/1()/1()/1(2

io

oi

iirr

TTrTC

]1[r

r

rr

r

TT

TT i

oi

o

oi

i



Summary of Results

Confuguration Heat flow rate q Thermal Resistance

Plane wall q kA

L T

L

kA

Hollow Cylinder

q

kL

Ln r r T

o i

2

Ln r r

kL

o i

2

Hollow sphere q kr r

r r T

i o

i o

4

r r

kr r

o i

i o

4

Class Test: upto this point

(Just before this slide)

Date of Class test:

Thursday 6 Sept 2007 at 5pm

Class Test: 20%

Lab: 20%

Final Exam:60%


Overall Heat Transfer Coefficient

q = UA (T)total (21)

Like Newton's law of cooling

q = hA (T)

both U and h have the same dimensions, W/m2K.



Overall Heat Transfer Coefficient For composite walls:

)/()/()/(

1

,

)/()/()/(

1

332211

321

333222111

kLkLkLU

AAAFor

AkLAkLAkLUA



Overall Heat Transfer Coefficient For composite cylinders, Figure 2.5

q UA T)T T

R t

(1

;where UAR t

1

(2.23)

and

LrhLk

rrLn

Lk

rrLn

Lk

rrLn

LrhR

oii

t

33

34

2

23

1

12

2

1

2

)/(

222

1

Ti

Steady state heat conduction

critical thickness of insulation

LrhkL

rrLn

TTq

o

io

i

2

1

2

q q LT T

Ln r r

k h r

i

o i

o

2

1

2

(2.24)

1/2hπro Ln(ro/ri)/2πk

Critical thickness of insulation

Resistance

Condition for optimum heat flow

The condition of optimum heat flow can be obtained by

differentiating equation (2.24) with respect to ro and setting

the resultant equation to zero.

This condition is h ro/k = 1.0, where h ro/k is known as

Biot number, Bi

Hence the optimum heat transfer occurs from the cylinder

when the Biot Number is 1 and the critical radius,

rcrit = k/h (2.25)

At Bi = 1, heat transfer rate reaches a maximum;

resistance to heat transfer is minimum.

At Bi < 1, addition of insulation increases heat

transfer rate. Electric cables are designed

for maximum heat dissipation, hence the

insulation value should be around the

critical value.

At Bi > 1, addition of insulation decreases heat

transfer rate. To reduce heat losses, the

insulation thickness should be much

greater than the critical thickness of

insulation.

Example 2.2C An electric wire, diameter d=3mm and length L=5m, is

tightly wrapped with a 2mm-thick plastic cover (thermal

conductivity k=0.15 W/mK). Measurements indicate that

a current of 10A passes through the wire causing a

voltage drop of 8V. The wire is exposed to an

environment at 32oC with a convective heat transfer

coefficient h=12 W/m2K. Determine the temperature at

the interface of the wire and the plastic cover in steady

operation. Also, evaluate whether doubling the thickness

of the plastic cover will increase or decrease this

interface temperature.

Example 2.2C - solution

Under steady operating condition, rate of heat transfer is equal to the heat generated within the wire.

Q = VI = (8V)(10A)

= 80 W

A2 = (2r2)L = 2(0.0035m)(5m)

= 0.1099 m2

WK2

mmKW758.0

)1099.0)(12(

1

hA

1R

2

con

mmKW 515.02

1535Ln

kL2

rrLnR 12

plastic

= 0.1798 WK

WKplasticcontotal RRRR 9378.01798.0758.0

r1

k

T1

T2

r2

T1 T2 Ta

Rplastic Rcon

Q

totalo1

total

o1 QRTTR

TTQ

WKW 9378.08032

= 107.02 oC

The critical thickness of insulation:

K

2mW

mKW

12

15.0

h

krcr

= 0.0125 m

= 12.5 mm

1-D steady state heat conduction

with energy generation

Considering an element in a wall with a uniformly

distributed heat source

in x-direction dxdx

dTkA

dx

d

dx

dTkA

dx

dTkAq

xnet )(

energy generated within the control volume = qg A dx

Energy balance gives, k 02

2

gqdx

Td

STEADY STATE CONDUCTIO0N 1-D system with heat generation

]

2 0Tq

k

g

Poisson’s Equation T1

qg

X= L

X= 0

x

insulated

Boundary Conditions

d2T/dx2 +qg/k = 0

1. T(0) = T1

2. q x=L= - k dT/dx at x=L

= 0 T(x) = - [qg/2k]x2 + C1x +C2

2

2

1

2

1 2

0

2

g

g

g

qd T

dx k

qdTx C

dx k

qT x C x C

k

1

1 2 1

2

1

(1) (0)

(2)( ) 0

;

2

x L

g

g g

T T

dT

dx

q LC C T

k

q q LT x T

k k


1-D system with heat generation

L

x

kT

xLq

T

TxT g

21

)(

11

1

The temperature distribution is parabolic

with x

and the maximum value occurs at the

insulated

surface, x = L.

k

LqTTLT

g

2)(

2

1max

Conduction in cylinders with

energy generation

1-D steady state conduction equation in cylindrical coordinate is given by

0 gdrrr QQQ

02

])(2)(2[)(2

grLdrq

drdr

dTr

dr

dkL

dr

dTrkL

dr

dTrkL

After rearrangements

0)(1

k

q

dr

dTr

dr

d

r

g


Cylindrical Coordinates

Current flowing through a wire

qg = I2R/V

d (rdT/dr)/rdr + qg/k = 0

T(r) = C1 Ln r - qgr2/4k +C2

ro

Boundary conditions:

1. T(ro) = To

2. dT/drr=o = 0



2

22

14

)(

oo

og

o

o

r

r

kT

rq

T

TrT

Maximum temperature occurs at the centre

T(o) = Tmax = To + qgro2/4k



Example 5:

A 2-kW resistance water heater is used to boil

water in a kettle. The cylindrical heating element

has a diameter of 5 mm and length 0.6 m, where

thermal conductivity, k= 15 W/m.K. If the outer

surface temperature is 110oC, estimate the

temperature at the centre of the element.



Solution:

qg = Qgen/Vele =2000(W)/π (0.0025)2(m2) 0.6 (m)

= 0.1697x109 W/m3

T(r=0) = To + qgro2/4k

=110+0. 1697x109 (W/m3) (0.0025)2(m2)

/4x15 (W/m.K)

= 127.7oC

STEADY STATE CONDUCTIO0N 1-D system without heat generation

Heat Transfer from Fins

• Heat transfer from a surface to air or gas is

rather slow due to lower heat transfer

coefficient.

• One of the ways of increasing heat transfer

is to provide additional surface area,

perhaps in the form of fins, which may be

of different designs and geometries.



Fins in Electronic Cooling



x x

-kA dT/dx -kA dT/dx

+d(-kAdT/dx) x/dx

hAs[T(x ])-T

qx qc qx+ x

Energy balance across

the element:

,)(

TxThAdxdx

dTkA

dx

d

dx

dTkA

dx

dTkA

qqq

s

xx

cxxx

or

d

dxkA

dT

dxh P T x T

( ) = 0 (2.34)

where A = Pdx P= Perimeter

As



02

2

2

mdx

dEquation representing

temperature distribution in

a straight fin of constant

x-section Where = T - T

m2 = h P/ kA

Solution, = C1 e-mx + C2 e mx

= C1 cosh mx + C2 sinh mx



Two boundary conditions are required:

1. x=0, = To - T

2. Other boundary condition depends

on the physical situation

Case 1: Very long fin, T x= T

Case 2: Negligible heat losses from the tip of the fin,

dT/dx] x=L= 0

Case 3: Fin of finite length, heat losses from the tip by

convection, -k dT/dx] x=L= h [ TL - T]

x

Different Boundary Conditions



Case 1. Long Fin

x , = To - T 0

x=0, = o

Temperature distribution,

= C1 e-mx + C2 e mx

C1 = o and C2 = 0

/ o = (T- T)/( To - T) = e -mx

Heat Transfer

dxhPQLx

x

0 = -kA d/dx| x=0

x

Q = kA om = o[hpkA] 1/2



Case 2. Fins with insulated tip

x

= C1 e-mx + C2 e mx

C1= o /(1+e -2mL)

C2= o /(1+e 2mL)

/ o = [e-mx /(1+e -2mL)] + [e mx /(1+e 2mL)]

Q = -kA d/dx| x=0

= kA om[1/(1+e 2mL) - 1/(1+e -2mL)]

= o [hpkA] 1/2 tanh (mL)

dT/dx] x=L= 0

x=0, T = To

= o



Case 3. Fins with convection at the tip

x

-kdT/dx]x=L

= hL(T x=L - T)

Temp Distribution:

/ o = (T- T)/( To - T) = A/B

where,

A = cosh m(L-x) +(h/mk) sinh m(L-x)

B = cosh mL + (h/mk) sinh mL

Heat Transfer:

Q = (hpkA)1/2 ( To - T) C/D

where,

C = sinh mL + (h/mk) cosh mL

D = cosh mL + (h/mk) sinh mL



Fin Efficiency:

A ratio of actual heat transfer to heat which would

be transferred if the entire fin were at the base

temperature

f = o (hpkA)1/2/(hpL o )

=(1/L)(kA/hp)1/2

=1/mL, for Long Fins



Effectiveness of Fins:

= ( Q with fin)/ (Q without fin)

= f Af h o /(hAb o )

= f Af h /(hAb)

= f Af/Ab = [kP/hA]0.5

where, Af = PL

Ab = A



Temperature Profile and Heat Transfer Equation

Straight Fins with constant area

Infinite fin (L) = 0

θ/ θb = e-mx θ = T - T

θb= Tb - T

qf = M, where M= (hpkAc)1/2 θb

Finite fin – no heat loss from the tip,

dθ/dx = 0 at x=L

θ/ θb = A/B, A=cosh m(L-x); B=cosh mL

qf = M tanh mL m2= hp/kAc

Temperature Profile and Heat Transfer Equation

Straight Fins with constant area

Finite fin – convective heat loss from the tip of

the fin

h(L) = -k d/dx x=0

θ/ θb = C/D, C= coshm(L-x)+(h/mk)sinh m(L-x)

D= coshmL+(h/mk)sinmL

qf = M /, = sinhmL+(h/mk)coshmL

= coshmL+(h/mk)sinmL



Example 2.3A:

A copper pin fin 2.5 mm in diameter protrudes from a

wall at 100oC into an air at 28oC. The heat transfer is

mainly by natural convection with a heat transfer

coefficient of 12 W/m2.K.

Calculate heat losses from the fin assuming-

i) the fin is infinitely long;

ii) the fin is 30 mm long and the tip of the

fin is insulated;

iii) the fin 30 mm long having convective

losses, with h = 12 W/m2.K

Solution 2.3 A

The fin temperature is a function of length and varies

between 100oC and 28

oC. We evaluate thermal

conductivity at an average temperature of 642

28100

oC,

k=396.0 W/mK.

PhkA = (0.0025m) x (12 W/mK)

x (396 W/mK) x (/4)(0.0025m)2

= 1.834x10-4

(W2/K

2)

phkA = 0.0135 W/K

i) Considering the fin infinitely long, L

PhkAQ 0 = To-T = 72

= 72 x 0.0135 W

= 0.9745 W

2.3A- (ii)

•

i) The tip of the fin is insulated, 0dx

d

Lx

mLtanhPhkAQ 0

= 0.9745 x 0.2058

= 0.2005 W

kA

hPm

963.6

)0025.0(4

x396

)0025.0(x12

2

mL = 6.963 x 0.03

= 0.2088

2.3A-(iii)

i) Convective heat transfer from the tip,

mLsinh

mkhmLcosh

mLcoshmk

hmLsinhPhkAQ 0

W205.0

02.1

2144.0x9745.0

21.0x00435.002.1

02.1x00435.021.0x9745.0

sinh mL = 0.21

cosh mL = 1.02

396x963.6

12

mk

h

= 0.00435

Example 2.3B

An array of 10 aluminum alloy fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 67oC and the ambient is at 27oC, how much power do they dissipate if the combined convection and radiation heat transfer coefficient is estimated to be 8 W/m2K? The alloy has a conductivity of 175 W/m.K. The heat transfer from the tip of the fin is negligible. Also, find the efficiency and effectiveness of the fin.

Solution Outline For one fin:

A= (0.003)(0.0004)=1.2x10-6m2

P=2(0.003+0.0004)=6.8x10-3m

m2= hP/kA =(8.0W/m2.K)(6.8x10-3m) /(175W/m.K)(1.2x10-6m2)

=259 m -2

m = 16.1 m -1; mL= (16.1 m-1)(0.040 m)

= 0.644

PL=(6.8x10-3m)(0.040 m)= 2.72x10-4m2

1. Finite fin – negligible tip losses:

Qmax = h(PL)(To – Tα)

= (8 W/m2K)(2.72x10-4m2)(340-300)K

= 0.087 W for 10 fins, QT=0.87 W

η = (1/mL)tanh mL

= (1/0.644)tanh (0.644)

= 0.881

Q = ηQmax= 0.0764W per fin

QT= 10x Q=0.764 W

TRANSIENT HAET CONDUCTIO0N

General Heat Transfer Equation:

2T + qg/k = (1/)T/ t

Newton’s Law,

q/As =h (Ts - T) = T/(1/h)

Fourier Law,

q/A = -k dT/dx = T/(Lc/k)

Bi = Biot Number

= h Lc/k = (Lc/k)/(1/h) < 0.1


Lumped Capacitance Method

q = hAs (T - T) = Vc dT/dt

dT/(T - T) = -(hAs/ Vc) dt

Integration between, t=0, T=Ti

and t=t, T =T

(T - T)/(Ti - T) =e - t

where, = (hAs/ Vc)

T

=density

c=sp ht

V=volume

T



T

t

T

Heat transfer over a period of time t’

Qc = dtTThA

t

s )(

'

0

Qc = Vc (Ti - T)

[1 - exp(-hAs/ Vc)t’]

Qmax = Vc (Ti - T)

Qc/Qmax =1- exp(-hAs/Vc)t’

• Steel ball bearings are required to be subjected to heat treatment to obtain the desired surface characteristics. The balls are heated to a temperature of 650oC and then quenched in a pool of oil that has a temperature 55oC. The ball bearings have a diameter of 40 mm. The convective heat transfer coefficient between the ball bearings and oil is 300 W/m2.K. Determine

• a) the length of time that the bearings must remain in the oil before their temperature drops to 200oC,

• b) total amount of heat removed from each bearing during this time interval, and

• c) instantaneous heat transfer rate from the bearings when they are first placed in the oil and when they reach 200oC.

• The properties of steel ball bearings are as follows:

• k = 50W/m.K; α = k/ρcp =1.3x10-5 m2/s

• Outline of Solution:

• hAst/ρVcp =(hLc/k) (αt/Lc2) = (Bi) (Fo)

• Where, (Bi) = (hLc/k = h(ro/3)/k =0.04

• (Fo) = (αt/Lc2) = αt/(ro/3)2 = 0.293t

• a) (200-55)/(650-55) = e –(0.04)(0.293t)

• t = 120.5 s

• Find t from above eqn

• b) Q =hAs(Ti - T)[1- e -(Bi) (Fo) ] t/ (Bi) (Fo)

• = 5.79x104 J

• t = 0, Q = 897 W

• t = 120.5 s, Q = 218 W



Example:

The hot plate of a cooker has a surface area of 0.05 m2 and

is made of steel (density:7,820 kg/m3) having a total weight

of 1.4 kg. The convection heat transfer coefficient is 17

W/m2K between the plate and its surroundings at 27oC.

How long, after being switched on, would the plate take to

attain a temperature of 117oC? The plate heater is rated at

500Watts and initially at the temperature of the

surroundings. The thermal conductivity of the plate is 17.3

W/m.K.

Solution

sg hAVq

dt

dVc

where = T - T

badt

d

where c

qa

g

Vc

hAb s

Separating the variable and integrating

t

oodtb

ba

bd

bta

baLn

sec1261

a

baLn

bt

where,

775.0c

qa

g

310318.1 xVc

hAb s


Example 2

• The temperature of a stream of natural gas flowing through a pipe at 100oC is to be measured by a thermocouple whose junction can be approximated as a 1-mm diameter sphere. The properties of the junction are as follows: k=35 W/mK, ρ=8500 kg/m3 and Cp = 320 J/kg.K. The convection heat transfer coefficient between the junction and the gas, h=210 W/m2K. The thermocouple is initially at 28oC. Determine the time constant of the thermocouple. Also, find the time taken to read 99% of the initial gas temperature difference.

Example 2- solution outline

3

4

2

1

/ 61.67 10

0.1

0.01

0.462

10

c

s

c

i

s

bt

i

V DL x m

A D

hLBi check

k

T T

T T

hAs b

Vc

T Te t s

T T

Documents

MEC 3609 H T Conduction