Heat Transfer Theory E

8/12/2019 Heat Transfer Theory E

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PHYSICS

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HEAT TRANSFER

1 . INTRODUCTIONHeat is energy in transit which flows due to temperature difference; from a body at higher temperature to a

bodyat lower temperature. This transfer of heat from one bodyto the other takes place through three routes.(i) Conduction(ii) Convection(iii) Radiation

2 . CONDUCTIONThe process of transmission of heat energy in which heat is trans-ferred from one particle of the medium to the other, but each particleof the medium stays at its own position is called conduction, for ex-ample if you hold an iron rod with one of its end on a fire for some

time, the handle will get hot. The heat is transferred from the fire tothe handle byconduction along the length of iron rod. Thevibrationalamplitude of atoms and electrons of the iron rod at the hot end takeson relatively higher values due to thehigher temperatureof their envi-ronment.These increased vibrationalamplitude are transferredalongthe rod, from atom to atom during collision between adjacent atoms.In this way a region of rising temperature extends itself along the rodto your hand.

TC

Q 2

Q 1

TH

Ox dx

B A

L

Consider a slab of face area A, Lateral thickness L, whose faces have temperatures T H andTC(TH > TC).

Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Lettemperature of faceA be T and that of face B beT + T. Thenexperiments show that Q, the amount of heatcrossing the area A of the slab at position x in time t is given by

Qt

= KAdTdx ... (2.1)

Here K isa constant depending on the material of the slaband is namedthermalconductivityof the material,

and the quantity

dxdT

is called temperature gradient. The () sign in equation (2.1) shows heat flows from

high to low temperature ( T is a ve quantity)

3 . STEADY STATEIf the temperature of a cross-section at any position x in the above slab remains constant with time (remem-ber, it does vary with position x), the slab is said to be in steady state.Remember steady-state is distinct from thermal equilibrium for which temperature at anyposition (x) in theslab must be same.For a conductor in steady state there is no absorption or emission of heat at any cross-section. (as tempera-ture at each point remains constant with time). The left and right face are maintained at constant tempera-tures TH and TC respectively, and all other faces must be covered with adiabatic walls so that no heatescapes through them and same amount of heat flows through each cross-section in a given Intervalof time.Hence Q1 = Q = Q2. Consequently the temperature gradient is constant throughout the slab.

Hence, dxdT

= LT

= LTT if = L

TT HC


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and tQ

= KA LT

tQ

= KA

LTT CH .... (3.1)

Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval of t.

Example 1 . One face of an aluminium cube of edge 2 metre is maintained at 100C and the other end is main-tained at 0C. All other surfaces are covered by adiabatic walls. Find the amount of heat flowingthrough the cube in 5 seconds. (thermal conductivity of aluminium is 209 W/mC)

Solution : Heat will flow from the end at 100C to the end at 0C. Area of cross-section perpendicular to direction of heat flow,

A = 4m2

then tQ

= KA L)TT( CH

Q = m2sec)5)(C0C100)(m4)(Cm/W209( 2 = 209 KJ Ans.

4 . THERMAL RESISTANCE TO CONDUCTIONIf you are interested in insulatingyour house from cold weather or for thatmatter keeping the mealhot inyour tiffin-box, you aremore interested in poor heat conductors, rather than good conductors. For this reason, theconcept of thermal resistance R has been introduced.For a slab of cross-sectionA, Lateral thickness L and thermal conductivity K,

KAL

R ... (4.1)

In terms of R, the amount of heat flowing though a slab in steady-state (in time t)

R)TT(

tQ CH

If we name tQ

as thermal current iT

then, RTTi CHT (4.2)

This is mathematicallyequivalent to OHMs law, with temperatureplaying the role of electric potential. Henceresults derived from OHMs law are also valid for thermalconduction.More over, for a slab in steady state we have seen earlier that the thermal current iL remains same at eachcross-section. This is analogous to kirchoffs current law in electricity, which can now be very convenientlyapplied to thermal conduction.

Example 2. Three identical rods of length 1m each, having cross-sectionareaof 1cm 2 each andmadeofAluminium, copper andsteelrespectivelyaremaintained at temperatures of 12C, 4Cand50C respectivelyat their separate ends.

Find the temperature of their common junction.

50C

4C

s t e

e l

c o p p e r

Aluminium12C

[ KCu = 400 W/m-K , K Al = 200 W/m-K , Ksteel = 50 W/m-K ]


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Solution : R Al = KAL

= 200101

4 = 200104

Similarly Rsteel = 50104

and Rcopper = 400104

Let temperature of common junction = T

then from Kirchoff;s current laws,i Al + isteel + iCu = 0

AlR12T

+steelR

50T+

CuR4T

= 0

(T 12) 200 + (T 50) 50 + (T 4) 400

4(T 12) + (T 50) + 8 (T 4) = 0

13T = 48 + 50 + 32 = 130 T = 10C Ans.

5 . SLABS IN PARALLEL AND SERIES5.1 Slabs in series in steady state)

Consider a composite slab consisting of two materials having different thicknesses L 1 and L2 differ-ent cross-sectional areasA 1 andA2 and different thermal conductivities K1 and K2. The temperatureat the outer surface of the slabs are maintained at T H andTC, and all lateral surfaces are covered byan adiabatic coating.

Let temperature at the junction be T, since steady state has been achieved thermal current through eachslab will be equal. Then thermal current through the first slab.

i = tQ

=1

HR

TT or TH T = iR1 ... (5.1)

and that through the second slab,

i = tQ

=2

CR

TT or T TC = iR2 ....(5.2)

adding eqn. 5.1 and eqn 5.2

TH TL = (R1 + R2) i or i = 21CH

RRTT

Thus these two slabs are equivalent to a single slab of thermal resistance R 1 + R2.If more than two slabs are joined in series and are allowed to attain steady state, then equivalent thermalresistance is given by

R = R1 + R2 + R3 + ....... ...(5.3)


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Example 3 The figure shows the cross-section of the outer wallof a house built in a hill-resort to keep the houseinsulated from the freezing temperature of outside. The wall consists of teak wood of thickness L1and brick of thickness (L2 = 5L1), sandwitching two layers of an unknown material with identicalthermalconductivities andthickness. The thermal conductivityof teak wood is K1 and thatof brick is(K2 = 5K). Heat conduction through the wall has reached a steady state with the temperature of three surfaces being known. (T 1 =25C,T2 = 20C and T5 = 20C). Find the interface temperatureT4 andT3.

T1 T2 T3 T4 T5

L1 L L L4

Solution : Let interface area beA. then thermal resistance of wood,

R1 = AKL1

1

and that of brick wall

R2 = AKL

2

2= AK5

L5

1

1= R1

Let thermal resistance of the each sand witch layer = R. Then the above wall can be visualised as acircuit

R 1 R R R 1

25C 20C T4 20C

iTiT

T3

thermal current through each wall is same.

Hence1R2025

= RT20 3 = R

TT 43 =1

4R

20T

25 20 = T4 + 20 T4 = 15C Ans.

also, 20 T3 = T3 T4

T3 = 2T20 4 = 2.5C Ans.

Example 4 In example 3, K1 = 0.125 W/mC, K2 = 5K1 = 0.625 W/mC and thermal conductivity of theunknown material is K = 0.25 W/mC. L1 = 4cm, L2 = 5L1 = 20cm. If the house consists of a singleroom of total wall area of 100 m2, then find the power of the electric heater being used in the room.


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Solution : I st method R1 = R2 = )m100)(Cm/w125.0()m104(

2

2

= 32 10 4 C/w

1R20 25

= RT 20 3 L = 5

5.17

1KK

L1 = 28 cm

R = KAL

= 112 10 4 C/W

the equivalent thermal resistance of the entire wall = R1 + R2 + 2R = 288 10 4 C/W

Net heat current, i.e. amount of heat flowing out of the house per second = RTT CH

= w/C10288)C20(C25

4 = 2881045 4

watt

= 1.56 KwattHence the heater must supply 1.56 kW to compensate for the outflow of heat. Ans.

IInd method

i =1

21R

T T= 4 1032

20 25 = 1.56 Kwatt

5.2 Slabs in parallel :L

Q 1

Q 2

Heat reservoir at temperature T C

Heat reservoir at temperature T H adiabatic coating

SLAB 1K1 A1

SLAB 2K2 A2

Consider two slabs held between thesame heat reservoirs, their thermalconductivities K 1 andK2 andcross-sectional areas A 1 and A2

then R1 = 11 AKL , R2 = 22 AK

L

thermalcurrent through slab 1

1

CH1 R

TTi

and that through slab 2

2

CH2 R

TTi

Net heat current from the hot to cold reservoir

i =

21CH21 R

1R1)TT(ii


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Comparing with i =eq

CHR

TT , we get,

eqR1

=21 R

1R1

If more than two rods are joined in parallel, the equivalent thermal resistance is given by

eqR1 = 21 R1R1 + 3R

1 + ..... .... (5.4)

Example 5 Three copper rods and three steel rods each of length = 10 cm and areaof cross-section 1 cm 2 areconnected as shown

C

E (0C)

copper

copper

copper

steel

steel

B

steel

D

A(125C)

If endsAandE aremaintainedat temperatures125Cand0Crespectively, calculate theamount of heat flowing per second from the hot to cold function. [ KCu = 400 W/m-K , Ksteel = 50 W/m-K ]

Solution : Rsteel = KAL

= 241

m10)Cm/W(50m10

= 501000

C/W..

Similarly RCu = 4001000

C/W

Junction C andD are identical in everyrespect and both willhave same temperature. Consequently,the rod CD is in thermal equilibrium and no heat will flow through it. Hence it can be neglected infurther analysis.Now rod BC and CE are in series their equivalent resistance is R1 = RS + RCu similarly rods BD andDE are in series with same equivalent resistance R 1 = RS + RCu these two are in parallel giving anequivalent resistance of

2

R1 =2

RR CuS

This resistance is connected in series with rod AB. Hence the net equivalent of the combination is

R = Rsteel + 2R1 = 2

RR3 Custeel

= W/C400

1503500

Now i = RTT CH =

W/C400

1503500

C125

= 4 watt. Ans.


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PHYSICS


Example 6. Two thin concentric shells made of copper with radius r 1 and r 2 (r 2 > r 1) have a material of thermalconductivity K filledbetween them. The inner andouter spheres are maintained at temperatures T Hand TC respectivelybykeeping a heater of power P at the centre of the two spheres. Find the valueof P.

Solution : Heat flowing per second through each cross-section of the sphere = P = i.Thermal resistance of the spherical shell of radius x and thickness dx,

Pdx

r 1

x

r 2dR = 2x4.K

dx

R = 2

1

r

r 2 K.x4

dx= K4

1

21 r 1

r 1

thermal current

i = P = RTT CH = )r r (

r r )TT(K412

21CH . Ans.

Example 7. Acontainer of negligibleheat capacitycontains 1 kg of water. It is connected bya steel rod of length10 m and area of cross-section 10cm 2 to a large steam chamber which is maintained at 100C. If initial temperatureof water is 0C, find the time after which it becomes 50C. (Neglect heat capacityof steel rod and assume no loss of heat to surroundings) (use table 3.1, take specific heat of water = 4180 J/kg C)

Solution : Let temperature of water at time t be T, then thermal current at time t,

i =

RT100

This increases the temperature of water from T to T + dT

i = dtdH

=ms dt

dT

RT100

= msdtdT

50

0 T100dT

= t

0 Rms

dT

n

21

= Rmst

or t = Rms n2 sec

= KAL

ms n2 sec

= 2n)m1010()Cm/w(46)Ckg/J4180)(kg1()m10(

24

= 46418

(0.69) 105

= 6.27 105 sec= 174.16 hours Ans.


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Can you now see how the following facts can be explained by thermal conduction ?(a) In winter, iron chairs appear to be colder than the wooden chairs.(b) Ice is covered in gunny bags to prevent melting.(c) Woolen clothes are warmer.(d) We feel warmer in a fur coat.(e) Two thin blankets are warmer than a single blanket of double the thickness.(f) Birds often swell their feathers in winter.

(g) A new quilt is warmer than old one.(h) Kettles are provided with wooden handles.(i) Eskimo's make double walled ice houses.(j) Thermos flask is made double walled.

6 . CONVECTION *(not in JEE Syllabus)When heat is transferred from one point to the other through actual movement of heated particles, theprocess of heat transfer is called convection. In liquids and gases, some heat may be transported throughconduction. But most of the transfer of heat in them occurs through the process of convection. Convec-tion occurs through the aid of earths gravity. Normally the portion of fluid at greater temperature is lessdense, while that at lower temperature is denser. Hence hot fluid rises up while colder fluid sink down,accounting for convection. In theabsence of gravity convection would not be possible. Also, the anomalous behaviour of water (its densityincreases with temperature in the range 0-4C) give riseto interesting consequences vis-a-vis the processof convection. One of these interesting consequences isthe presence of aquatic life in temperate and polar waters. The other is the rain cycle.

Can you now see how the following facts can be explained by thermal convection ?(a) Oceans freeze top to down and not bottom to up. (this fact is singularly responsible for presence of

aquatic life is temperate and polar waters.)(b) The temperature in the bottom of deep oceans is invariably 4C, whether it is winter or summer.(c) You cannot illuminate the interior of a lift in free fall or an artificial satellite of earth with a candle.(d) You can Illuminate your room with a candle.

7. RADIATION :The process of the transfer of heat from one place to another place without heating the interveningmedium is called radiation. The term radiation used here is another word for electromagnetic waves.These waves are formed due to the superposition of electric and magnetic fields perpendicular to eachother and carry energy.

Properties of Radiation:

(a) All objects emit radiations simply because their temperature is above absolute zero, and all objectsabsorb some of the radiation that falls on them from other objects.

(b) Maxwell on the basis of his electromagnetic theory proved that all radiations are electromagneticwaves and their sources are vibrations of charged particles in atoms and molecules.

(c) More radiations are emitted at higher temperature of a body and lesser at lower temperature.

(d) The wavelength corresponding to maximum emission of radiations shifts from longer wavelength toshorter wavelength as the temperature increases. Due to this the colour of a body appears to bechanging. Radiations from a body at NTP has predominantly infrared waves.

(e) Thermal radiations travels with the speed of light and move in a straight line.

(f) Radiations are electromagnetic waves and can also travel through vacuum.

(g) Similar to light, thermal radiations can be reflected, refracted, diffracted and polarized.

(h) Radiation from a point source obeys inverse square law (intensity 21r

).


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8 . PREVOST THEORY OF EXCHANGE : According to this theory, all bodies radiate thermal radiation at all temperatures. The amount of thermalradiation radiated per unit time depends on the nature of the emitting surface, its area and its temperature.The rate is faster at higher temperatures. Besides, a body also absorbs part of the thermal radiationemitted by the surrounding bodies when this radiation falls on it. If a body radiates more than what itabsorbs, its temperature falls. If a body radiates less than what it absorbs, its temperature rises. Andif the temperature of a body is equal to the temperature of its surroundings it radiates at the same rateas it absorbs.

9 . PERFECTLY BLACK BODY AND BLACK BODY RADIATION

F ERY S B LAC K B ODY ) A perfectly black body is one which absorbs all the heat radiations of whatever wavelength, incident onit. It neither reflects nor transmits any of the incident radiation and therefore appears black whatever bethe colour of the incident radiation.

OP

In actual practice, no natural object possesses strictly the properties of a perfectly black body. But thelamp-black and platinum black are good approximation of black body. They absorb about 99 % of theincident radiation. The most simple and commonly used black body was designed by Ferry. It consistsof an enclosure with a small opening which is painted black from inside. The opening acts as a perfect

black body. Any radiation that falls on the opening goes inside and has very little chance of escapingthe enclosure before getting absorbed through multiple reflections. The cone opposite to the openingensures that no radiation is reflected back directly.

1 0 . ABSORPTION, REFLECTION AND EMISSION OF RADIATIONSQ = Qr + Qt + Qa

QQ

QQ

QQ1 atr

1 = r + t + a

where r = reflecting power , a = absorptive power

Q r (Reflected)

Q t (Transmitted)

Q(Incident)

Q a(Absorbed)and t = transmission power.(i) r = 0, t = 0, a = 1, perfect black body(ii) r = 1, t = 0, a = 0, perfect reflector (iii) r = 0, t = 1, a = 0, perfect transmitter

1 0. 1 Absorptive power :

In particular absorptive power of a body can be defined as the fraction of incident radiation thatis absorbed by the body.

a =incidentEnergy

absorbedEnergy

As all the radiations incident on a black body are absorbed, a = 1 for a black body.


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1 0.2 Emis sive power:Energy radiated per unit time per unit area along the normal to the area is known as emissivepower.

E = t AQ

(Notice that unlike absorptive power, emissive power is not a dimensionless quantity).

1 0.3 Spectral Emissive power E ) :

Emissive power per unit wavelength range at wavelength is known as spectral emissive power,E . If E is the total emissive power and E is spectral emissive power, they are related asfollows,

0

E E d and dE E

d

1 0. 4 E mi ss iv it y:

e = Tetemperatur sameatbodyblackaof power EmissiveTetemperatur atbodyaof power Emissive

=0E

E.

11 . KIRCHOFF S LAW:The ratio of the emissive power to the absorptive power for the radiation of a given wavelength is samefor all substances at the same temperature and is equal to the emissive power of a perfectly black bodyfor the same wavelength and temperature.

)bodyblack(E)body(a)body(E

Hence we can conclude that good emitters are also good absorbers.

1 2 . NATURE OF THERMAL RADIATIONS : WIEN S DISPLACEMENT LAW)From the energy distribution curve of black body radiation, thefollowing conclusions can be drawn :(a) The higher the temperature of a body, the higher is the

area under the curve i.e. more amount of energy is emittedby the body at higher temperature.

(b) The energy emitted by the body at different temperaturesis not uniform. For both long and short wavelengths, the

energy emitted is very small.(c) For a given temperature, there is a particular wavelength ( m) for which the energy emitted (E) is

maximum.

(d) With an increase in the temperature of the black body, the maxima of the curves shift towardsshorter wavelengths.

From the study of energy distribution of black body radiation discussed as above, it was establishedexperimentally that the wavelength ( m) corresponding to maximum intensity of emission decreasesinversely with increase in the temperature of the black body. i.e.

m T1

or m T = bThis is called Wien's displacement law.Here b = 0.282 cm-K, is the Wiens constant.


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Example 8. Solarradiation is foundtohave an intensitymaximum near thewavelength rangeof 470nm.Assumingthe surface of sun to be perfectly absorbing (a = 1), calculate the temperature of solar surface.

Solution : Since a =1, sun can be assumed to be emitting as a black bodyfrom Wiens law for a black body

m . T = b

T = mb

= )cm10470()Kcm(282.0

7

~ 6000 K. Ans.

13 . STEFAN-BOLTZMANNS LAW : According to this law, the amount of radiation emitted per unit time from an area A of a black body atabsolute temperature T is directly proportional to the fourth power of the temperature.

u = A T4 ..... (13.1)where is Stefan's constant = 5.67 x 10-8 W/m2 K4 A body which in not a black body absorbs and hence emits less radiation then

For such a body, 4T Aeu .....(13.2)where e = emissivity (which is equal to absorptive power) which lies between 0 to 1With the surroundings of temperature T0 , net energy radiated by an area A per unit time..

)TT( Aeuuu 404

0 ....(13.3)

Example 9. A bodyof emissivity (e = 0.75), surface area of 300 cm2 and temperature 227C is kept in a room attemperature 27C. Calculate the initial value of net power emitted by the body.

Solution: Usingequation. (13.3) P= e A(T4 T04)= (0.75) (5.67 10 8 W/m2 K4) (300 10 4 m2) {(500 K)4 (300 K)4}= 69.4 Watt. Ans.

Example 10. A hot black body emits the energy at the rate of 16 J m 2 s 1 and its most intense radiationcorresponds to 20,000 . When the temperature of this body is further increased and its mostintense radiation corresponds to 10,000 , then find the value of energy radiated in Jm 2 s 1 .

Solution : Wein's displacement law is :

m .T = b

i.e. T m

1

Here, m becomes half. Temperature doubles.

Also e = T4

4

2

1

2

1TT

ee

e2 =4

1

2TT

.e 1 = (2)4 . 16

= 16.16 = 256 J m 2 s 1 Ans.


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PHYSICS


k = 1034n

for thenext interval(3020)=(3520)e kt

e kt =3

2

kt = n23

10

t34n

= n23

t = 10

34n

23n

minute = 14.096 min Ans.

Aliter : (byapproximate method)for the interval in which temperature falls from 40 to 35C

< > = 23540

= 37.5C

from equation (14.4) dtd

= k(< > 0)

(min)10)C40C35(

= K(37.5C 20C)

K = )(min351 1

for the interval in which temperature falls from 35C to 30C

< > = 23035

= 32.5C

from equation (14.4)

t)C35C30(

= (32.5C 20C) K

required time, t = min355.125

= 14 min Ans.

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Heat Transfer Theory E

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