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Name : Roll No. : Topic : Mohammed Asif

Ph : 9391326657, 64606657

HEAT TRANSFER 1. Heat can be transmitted from one place to another by three different methods. a) Conduction b) Convection c) Radiation I. > CONDUCTION: - Due to temp. difference between different parts of a body, the transfer of heat takes place from the particle at higher temperature to that at lowest temperature. Hence heat is transmitted by molecules of an object due to limited motion about the equilibrium position. This is called heat conduction. In conduction, molecules do not leave their position. The molecule transfer energy to the neighboring molecules due to mutual contact between them. The heat Conduction (i.e.) transmission of heat in this mode occurs in solids and Hg. E.g: - If one end of a metal bar is heated, then heat flows from the hot end to the colder one by conduction. a) STEADY – STATE: - In conduction, we define steady state which means that temperature of different sections becomes constant in this state. This does not implies that temperature of all section is same, they are different but are constant. Theory of conduction is valid only at steady state. b) ISOTHERMAL – SURFACE: - If we consider a metallic rod in which heat flows from one end to the other so that after sometime, a steady state is reached. In this state temperature of every transverse section of rod is same on the whole area of section. So this transverse section of the rod behaves as a isothermal surface. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 1

At isothermal surface, the temp of all the molecules is same these surface never intersect each other the θ1 θ2 shape of isothermal surface depends on the shape of conductor and nature of I s o t h e r mθ1 > θa2 l flow of heat. It can be 1D, 2D or 3D s u r f a c e heat flow…. The direction of heat flow is always normal to the isothermal surface. C) TEMPERATURE GRADIENT: - The rate of change of temperature with distance along the direction of flow of heat is called temperature gradient. If we consider a metallic rod having θ and θ − ∆θ as the temperature of two isothermal surface which are being at a distance of ∆x from each other, then Temp . gradient =

∆θ −dθ = ∆x dx

The negative sign indicates that temperature decreases with distance in the direction of heat flow. Unit: - o C per meter (i.e) o C / M

θ

θ− ∆θ

θ2

θ1 x

∆x

θ1>θ 2

→ LAW OF HEAT – TRANSFER THROUGH CONDUCTION: dQ In steady state, the rate of flow of heat through the cross – dt section of the conductor is i) Directly proportional to the area A of Surface dQ αA dt −∆θ ii) Directly proportional to the temperature gradient ∆x dQ −dθ α (i.e) dt dx On comb ruing the above two forms, we get dQ −dθ αA dt dx dQ dθ =−k A K = Coefficient of thermal conductivity of the dt dx material Unit of k – J / M – sec – k Now if dH is the amount of heat transfer in time interval dt, then dQ dH = dt dt dH K A ∆ θ ∴ = = rate of heat flow. dt ∆x If heat enters from one end such that temp decreases with distance in the direction of heat flow, then

K A dθ dH Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. = Ph: dt dx040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 2

→ HEAT – CONDUCTION THROUGH STRAIGHT ROD: Consider a metallic rod in which heat is flowing from one end to the other. We assume two isothermal surface at a distance x and x + dx having temperatures θ and θ − ∆θ. ∆θ Ax dH dθ = k A − → ( 1) Now dt dx If we consider whole length of the rod, then to calculate rate of flow of heat, we integrate the above expression. temp gradient =

θ2

l

θ1

o

1

∫ ( −dθ ) = ∫ K A .

θ

θ−∆θ

θ1

θ2

x x

+

θ1>θ 2

d x

l

dH dx. dt l

1 dH . x No remove – ve sign KA dt ∫o 1 dH θ1 − θ2 not θ2 − θ1 . .l ( θ 1 − θ2 ) = KA dt θ −θ 1 dH ∴ 1 2= . → ( 2) l KA dt Comparing (2) with (1) we get dθ θ1 − θ2 − = dx l This implies that temp. gradient for straight homogeneous rod remains same throughout its length. =

For homogeneous straight rod, temp varies linearly. Thermal Resistance of a straight rod is given by l R Th = KA Temp.difference dH ( θ1 − θ2 ) = = Heat current l dt Thermal Resistan ce KA

(

Q)

)

θ θ1

θ2

l

Calculate the amount of heat flowing through the rod when coefficient of a thermal conductivity is given by k = . x

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Sol: -

dH dθ = kA − dt dx dH a dθ = . A − dt x dx

dH x. dx = aA.dθ dt On integrating the above expression we get θ2 l dH x.dx = a A ∫ dθ dt ∫0 θ1 dH dt dH dt

x2 l 2 o = a A θ1 − θ2 l2 . = a A θ 1 − θ2 2

(

(

θ

θ1

)

(

θ2

x

)

dH 2aA θ1 − θ2 = dt l2

θ − ∆θ

d x

l

)

→ COMBINATION OF A STRAIGHT – RODS: 1) SERIES – COMBINATION: - Suppose we have two rods of same cross sectional area A. Length of the rods are l 1 and l 2 and coefficient of thermal conductivity are k 1 and k 2 respectively. Ends of the composite rod are maintained at A ( θ1 ) B ( θ) C ( θ2 ) temperature θ1 and θ2 . Let θ be the common temp of the junction. In steady state, any heat that goes through l2 l1 first rod also through the section rod. So same heat current passer through the two rods. k 1 A ( θ1 − θ ) → ( 1) Heat current in first rod = l1 Heat current in second rod = In steady state k 1A/ θ1 − θ

(

l1

)

=

(

k2A / θ − θ2

k 2 A ( θ − θ2 ) l2

)

l2

k 1 l 2 ( θ 1 − θ ) = k 2 l 1 ( θ − θ2 ) k 1 l 2 θ 1 − k 1 l 2 θ1 = k 2 l 1 θ − k 2 l 1 θ θ=

k 1 l 2 θ1 + k 2 l 1 θ2 k1 l 2 + k2 l 1 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 4

Put the value of θ in equation k 1 l 2 θ 1 + k 2 l 1 θ2 dH k 1A = θ1 − dt l 1 k 1 l 2 + k 2 l 1 k 1 A k 2 l 1 ( θ1 − θ2 ) = l1 k 1 l 2 + k 2 l 1 dH k 1 A = dt l1

(

dH A θ1 − θ2 = l2 l dt + 1 k2 k1

)

dH θ1 − θ2 = dt R equ

(1), we get

k 2 l 1 ( θ1 − θ2 ) k 1 l 2 + k 2 l 1 =

θ1 − θ2 l2 l1 + Ak 2 Ak 1

→

( 2)

l l1 + 2 → ( 4) R equ = R 1 + R 2 = Ak 1 Ak 2 If k equ is the equivalent coefficient of thermal conductivity of the combination of rods, then thermal resistance of the combination becomes l +l R = 1 2 → ( 5) Ak equ ∴

→ (3)

Comparing equations (4) and (5) we get l2 l +l l + 1 = 1 2 Ak 2 Ak 1 Ak equ l 2 l 1 l 1+l 2 + = k2 k1 k equ k equ

l +l = 1 2 l1 l2 + k1 k2

For 2 slabs of equal length 2l 2 l/ k equ = = 1 l 1 l + l k + k k1 k2 2 1 ∴

k equ =

2k 1 k 2 k1 + k2

→ PARALLEL - COMBINATION: - Suppose we have two rods of same length ‘l’ and area of cross – sections A 1 and A 2 . The coefficient of thermal conductivity be k 1 and k 2 respectively. The ends of the rod are maintained at temperature θ1 and θ2 . The same temp difference is maintained between the ends of each rod. dH 2 dH 1 and Let be the heat conducted in first dt dt

A

1

θ1 A

2

θ2

l

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and second rod. Resultant of these is assumed to be dH dH 1 dH 2 = + dt dt dt =

(

k 1 A 1 θ 1 − θ2

)

+

(

k 2 A 2 θ 1 − θ2

dH dt

)

dt dt k A k A dH 1 1 2 2 = + ( θ1 − θ2 ) → ( 1) dt l l → If R equ is the equivalent thermal resistance of combination, then dH θ1 − θ2 = → ( 2) dt R equ Equating (1) and (2), we get k 1 A1 k2 A2 1 = + R equ l l 1 1 1 = + → (3) R equ R 1 R 2 → If K equ is the equivalent coefficient of thermal conductivity of parallel combination, then l R equ = k equ ( A 1 = A 2 ) R equ Using (3) for 1 1 l + = R 1 R 2 k equ ( A 1 + A 2 ) R1R2 R1+ R2

=

l k equ ( A 1 + A 2 )

∴ k equ =

l l × k A k 2 A2 l ⇒ 1 1 = l l k equ ( A 1 + A 2 ) + k 1A 1 k 2 A 2

k 1 A1 + k 2 A2

l l = k 1A 1 + k 2 A 2 k equ ( A 1 + A 2 )

A1 + A2

For 2 slabs of equal area k equ =

k1 +k2 2

→ 2 – D HEAT – CONDUCTION: (RADIAL – THERMAL – CONDUCTION IN CYLINDER)

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θ2 Consider a uniform cylindrical shell having θ1 radius ‘a’ and outer radius ‘b’ which are b maintained at uniform temperature θ1 and θ2 . a r Let length of the shell be ‘l’ and d coefficient of thermal conductivity of material of shell be k. The cylindrical shell is assumed to be made of coaxial cylindrical shells. If the uniform temp at radius r and r + dr be θ and θ − dθ respectively, then radial rate of flow in steady state is dH dθ = k 2 π r l − Here area of cylinder = 2 π r l dt dr dθ dH 1 − = × dr dt k 2π r l

(

)

(

(

−dθ =

r

)

)

dH 1 × .dr dt k 2π r l

Integrating the above expression, we get dH 1 θ=− × .logr + c dt 2 πk l

(

)

At r = a,

θ = θ1 dH 1 θ1 = − × . log a + c → ( 2) dt 2π k l

At r = b,

θ = θ2 dH 1 θ2 = − × . log b + c → ( 3) dt 2π k l

Solving equations (2) and (3), we get dH 1 b θ 1 − θ2 = × . log dt 2π k l a ∴

dH 2π kl ( θ1 − θ2 ) = dt log b/ a

log b/ a Thermal Resistance of the cylinder, R = 2π k l →

3 – D HEAT – CONDUCTION

Consider a sphere having inner radius a and outer radius b. The temperature of inner is θ1 and that of outer sphere be θ2 θ1 > θ2 .

(

)

Assume a concentric sphere of radius r having thickness dr, so that radial flew of heat is given by dH dθ = K ( 4πr 2 ) − dt dr Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 7

2 Here area of sphere = 4 πr 1 dH − dθ = . × dr 2 k ( 4πr ) dt Integrating the above expression, we get

( θ − dθ)

θ2

θ θ1

a

r d r θ1

∫ dθ =

θ2

b

1

∫ k 4π r a

2

.

b

dH .dr dt b

dH 1 1 θ 1 − θ2 = . − dt 4πk r a dH 1 1 1 θ 1 − θ2 = × − dt 4πk a b

(

dH 4πk θ1 − θ2 = dt 1 − 1 a b

)

Thermal Resistance of the sphere ⇒ R =

b− a 4πkab

→ NOTE: - In numericals we may have the situation where the amount of heat traveling to the other end may be required to do some other work. Eg: - it may be required to mew the given amount of ice in that case, we have to equalld Q = ML. ∴ ML = →

kA ( θ1 − θ2 ) t l

WIEDMANN – FRANZ – LAW: -

At a given temperature T, the ratio of thermal conducting to electrical conductivity (a) is Constant. k = Const ( i.e) aT (i. e) a substance which is a good conductor of heat (Eg: - Silver) is also a good conductor of Electricity. Mica is an expectation to this law → THERMOMETRIC – CONDUCTIVITY (OR) DIFFSIVITY (D): - It is a measure of rate of change of temperature (with time) when the body is not in steady state (i.e. invariable state) It is defined as the ratio of the coefficient of thermal conductivity to the thermal capacity per unit volume of the material. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 8

Thermal capacity per unit volume = D= →

k ρC

ELECTRICAL – ANALOGY FOR THERMAL CONDUCTION: ELECTRICAL CONDUCTION dq a) i = dt v −v b) i = 1 2 R ρl l c) R = = A αA

→

MC = ρC V

THERMAL CONDUCTION dQ a) H = dt θ −θ b) H = 1 2 R l c) R = KA

GROWTH OF ICE ON LAKE: - When the temperature of the air is less 0 than 0 C , the cold air near the surface of the pond takes heat from the water which freezes in the forms of layers.

Consequently, the thickness of the ice layer keeps increasing with time. Let x be the thickness of the ice layer at a certain time. If the thickness is increased by dx in time dt, then the amount of heat flowing through the slab in time dt is given by Q=

kA ]0 − ( − T ) dt

x

I c t d x

w

a t e 0 Cr

0

kA T dt

→ ( 1) x x Where A = area of the layer of ice at −T 0C is the temp of the surrounding air If dm = mass of water frozen into ice, then ρ = density of ice Q = dm X L But dm = Aρ .dx ∴ Q = A ρ dx.L → ( 2 ) Equating A ρ L . dx = dt =

=

−T 0C

(1) and (2) KAT .dt x

ρL

.x.dx KT Integrating, we have t ρ L x2 ∫o dt = KT x∫ x.dx 1 x

ρL x2 2 t= KT 2 x1 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 9

t=

ρL x22 − x12 2KT

PROBLEMS 1)

Few rods of material x and y are connected as shown in figure. The cross – sectional area of all the rods are same. If the end A is maintained at 800C and the end F is maintained at 100C. Calculate the end temperatures of junction B and E in steady state. Given that thermal conductivity of material x is double that of Y. 0 0 Ans: - TB = 60.52 C, TE = 19.74 C

C

y

L

A

x

x xx x x x x x x xx xx x x x x x x x x xx x x x x x x xx x x x x x x x x x x xx xx x xx xx x x x x x x x L x

B

y

E

F

L

x D

2)

The space between two thin concentric metallic spherical shells of radii a and b is filled with a thermal conducting medium of conductivity k. The inner shell is maintained at temperature T1 and outer is maintained at a lower temperature T2 . Calculate the rate of flow of heat in radialy outward direction through the medium. 4π kab( T1 − T2 ) dθ Ans: = dt b− a 3)

A slab of stone of area 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 1000 C. A block of ice at 0 0 C rests on the upper surface of the slab. If in one hour 4.8 kg of ice melted, calculate the thermal conductivity of the stone. −1 −1 −1 −1 −1 Ans: - k = 1.24wm k ( or ) 1.24 Js m oC 4)

Three rods AB, BC, and BD having thermal conductivities in the ratio 1: 2: 3 and lengths in the ratio 2: 1: 1 are joined as shown. The ends A, C and D are at temperatures T1 ,T2 and T3 respectively. Find the

temperature of the junction B. Assume steady state. 1 ( T1 + 4T2 + 6T3 ) Ans: - T = 11 5)

D

T

3

T

B

A

2

C T

1

Twelve indentical rods each of length l and cross – sectional area S made of material having thermal conductivity K are arranged as shown. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 10

The thermal resistance between A and B is____ 4l 2l 3l 2l 1) 2) 3) 4) 5K S KS 4K S 3K S

B

A

6)

The figure shows two different metal rods – one straight and bent twice at right angles (P) and the other semi circular (Q). If heat currents drawn by P and Q be the same, then what is the ratio of their T thermal conductivities? Ans: - 4 : π 7) AB, BC and CD are three metal rods identical in shapes and sizes joined end to end as shown. With thermal conductivities k, 2k, and 3k respectively. If end A and D are maintained at temperatures 2T and T, find the temperatures of B and C. Also determine the thermal current through any of the three rods. (Assume S and l as area and length of each rod.) 6k sT 5 2 Ans: - 1 T ; 1 T ; 11 11 11l Seven rods, each having length l and area A are arranged as shown in fig. Their respective thermal conductivities are shown in the figure itself. The free ends are maintained at temperatures T1 and T2 . Determine the thermal current through the system? 60kA T1 : T2 Ans: 97 l

( P

)

R

T

C

1

( Q

A

2

)

K

B 2 K C

3 K

D

8)

(

)

T

T

k 1

2 k 3 k

2

2 k 4 k

3 k

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SOLUTIONS 4)

If T is required temp of junction B and assuming T1 > T > T2 ,T3 wehave ∆Q = ∆Q + ∆Q ∆t AB ∆t BC ∆t BD

(

kA T1 − T 2L T1 − T 2 T =

5)

)

=

(

2kA T − T2

)

L

(

)

(

+

(

3kA T − T3 L

= 2 T − T2 + 3 T − T3

1 T1 + 4T2 + 6T3 11

(

)

)

)

l Thermal Resistance of each rod r = k s Thermal resistance between C & D r ( 2r ) = r+ +r r + ( 2r ) =

8r 3 ( 1i i1 i2

o

ti

i)

i3

A h

3

i3

i3

1

( 1i -

i1

i3 i2

i1

B c o

l d

3 i)

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r

r

C

r

r r r

r

A

B

r

r

E

D

r

r

r

F

∴ Eff . thermal resistan ce R will be 1 3 1 3 = + + R 8r 2r 8r 5 = 4r 4r 4 l 4l R = = × = 5 5 ks 5ks →

CONVECTION Mode of transfer of heat by means of migration of material particles of medium is called convection. It is of two types a) Natural convection b) Forced convection A)

NATURAL – CONVECTION: - This arise due to difference of densities at two places and is a consequence of gravity because on account of gravity the hot light particles arise up and cold heavy particles try setting down. It is mostly occurs on heating a liquid fluid.

B)

FORCED – CONVECTION: - If a fluid is forced to move to take up heat from a hot body then the convection process is called forced convection

In this case Newton’s law of cooling bolds good. According to which rate of loss of heat from a hot body due to moving fluid is directly proportional to the surface area of body and excess temperature of body over its surroundings (i.e) Q ∝ A T − T0 t Q hA T − T0 t

(

(

)

)

Where h = Constant called convection coefficient T = temp of body

T0 = temp of surrounding. h → depends on properties of fluids such as Density, Viscosity, Specific heat and thermal conductivity. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 13

→ Natural convection takes places from bottom to top while forced convection in any direction.

→

RADIATION : - / INFRA – RED – RADIATION The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. Precisely it is electromagnetic – Energy transfer in the form of E.M.W through any medium. It is possible even in vacuum. Eg: - The heat from the sun reaches the earth through radiation. The wavelength of thermal radiation ranges from 7.8 X 10-7 m to 4 X 10-4 m. They belong to infra – red region of the E.m Spec hum. That is why thermal radiations are also called infra – red – radiations.

→

PREVOST – THEORY: In early ages it was assumed that only hot body emits radiation but later a theory of heat radiation was given by PREVOST that every body which is at temperature above O k temp radiate thermal energy in the form of E.M.W of several wavelengths. Simultaneously the body also absorbs thermal radiation from its surroundings. 1 The speed is equal to that of light (C = 3 X 108 m/s) I ∝ 2 . d Just as light waves, they follow laws of Reflection, Refraction, interference, diffraction and palarisation . When these radiations falls on a surface then exert pressure m that surface which is known as Radiation pressure. Spec hum of these radiations cannot be obtained with help of glass prism because it absorbs heat radiations. It is obtained by QUARTZ (or) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 14

ROCK – SALT PRISM, because these materials do net have free electrons and interatonic vibrational frequency is greater than the radiation frequency, hence they do not absorb heat radiation. These travels just like photons. →

DIATHERMANOUS - MEDIUM: - A medium which allows heat radiations to pass through it without absorbing then is called diathermanous medium. Thus the temp of a diathermanous medium does not increase irrespective of the amount of the amount of the thermal radiations passing through it. Eg: - Dry air, So2 , Rock salt (Nacl)

→

ATHERMANOUS - MEDIUM: - A medium which party absorbs heat rays is called a athermanous medium. As a result temp of an athermanous medium increases when heat radiations pass through it. Eg: - Wood, metal, moist air, simple glass, human flesh…….etc

→

COLOUR OF HEATED OBJECT: - When a body is heated, all radiations having wavelengths from zero to infinity are emitted. Radiations of longer wavelengths are predominant at lower temperature. The wavelength corresponding to maximum emission of radiations shifts from longer wavelength to shorter wavelength as the temperature increases. Due to this the colour of a body appears to be changed. c) A blue flame is at higher temp than a yellow flame. a) b)

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HEAT TRANSFER 1. Heat can be transmitted from one place to another by three different methods. a) Conduction b) Convection c) Radiation I. > CONDUCTION: - Due to temp. difference between different parts of a body, the transfer of heat takes place from the particle at higher temperature to that at lowest temperature. Hence heat is transmitted by molecules of an object due to limited motion about the equilibrium position. This is called heat conduction. In conduction, molecules do not leave their position. The molecule transfer energy to the neighboring molecules due to mutual contact between them. The heat Conduction (i.e.) transmission of heat in this mode occurs in solids and Hg. E.g: - If one end of a metal bar is heated, then heat flows from the hot end to the colder one by conduction. a) STEADY – STATE: - In conduction, we define steady state which means that temperature of different sections becomes constant in this state. This does not implies that temperature of all section is same, they are different but are constant. Theory of conduction is valid only at steady state. b) ISOTHERMAL – SURFACE: - If we consider a metallic rod in which heat flows from one end to the other so that after sometime, a steady state is reached. In this state temperature of every transverse section of rod is same on the whole area of section. So this transverse section of the rod behaves as a isothermal surface. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 1

At isothermal surface, the temp of all the molecules is same these surface never intersect each other the θ1 θ2 shape of isothermal surface depends on the shape of conductor and nature of I s o t h e r mθ1 > θa2 l flow of heat. It can be 1D, 2D or 3D s u r f a c e heat flow…. The direction of heat flow is always normal to the isothermal surface. C) TEMPERATURE GRADIENT: - The rate of change of temperature with distance along the direction of flow of heat is called temperature gradient. If we consider a metallic rod having θ and θ − ∆θ as the temperature of two isothermal surface which are being at a distance of ∆x from each other, then Temp . gradient =

∆θ −dθ = ∆x dx

The negative sign indicates that temperature decreases with distance in the direction of heat flow. Unit: - o C per meter (i.e) o C / M

θ

θ− ∆θ

θ2

θ1 x

∆x

θ1>θ 2

→ LAW OF HEAT – TRANSFER THROUGH CONDUCTION: dQ In steady state, the rate of flow of heat through the cross – dt section of the conductor is i) Directly proportional to the area A of Surface dQ αA dt −∆θ ii) Directly proportional to the temperature gradient ∆x dQ −dθ α (i.e) dt dx On comb ruing the above two forms, we get dQ −dθ αA dt dx dQ dθ =−k A K = Coefficient of thermal conductivity of the dt dx material Unit of k – J / M – sec – k Now if dH is the amount of heat transfer in time interval dt, then dQ dH = dt dt dH K A ∆ θ ∴ = = rate of heat flow. dt ∆x If heat enters from one end such that temp decreases with distance in the direction of heat flow, then

K A dθ dH Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. = Ph: dt dx040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 2

→ HEAT – CONDUCTION THROUGH STRAIGHT ROD: Consider a metallic rod in which heat is flowing from one end to the other. We assume two isothermal surface at a distance x and x + dx having temperatures θ and θ − ∆θ. ∆θ Ax dH dθ = k A − → ( 1) Now dt dx If we consider whole length of the rod, then to calculate rate of flow of heat, we integrate the above expression. temp gradient =

θ2

l

θ1

o

1

∫ ( −dθ ) = ∫ K A .

θ

θ−∆θ

θ1

θ2

x x

+

θ1>θ 2

d x

l

dH dx. dt l

1 dH . x No remove – ve sign KA dt ∫o 1 dH θ1 − θ2 not θ2 − θ1 . .l ( θ 1 − θ2 ) = KA dt θ −θ 1 dH ∴ 1 2= . → ( 2) l KA dt Comparing (2) with (1) we get dθ θ1 − θ2 − = dx l This implies that temp. gradient for straight homogeneous rod remains same throughout its length. =

For homogeneous straight rod, temp varies linearly. Thermal Resistance of a straight rod is given by l R Th = KA Temp.difference dH ( θ1 − θ2 ) = = Heat current l dt Thermal Resistan ce KA

(

Q)

)

θ θ1

θ2

l

Calculate the amount of heat flowing through the rod when coefficient of a thermal conductivity is given by k = . x

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Sol: -

dH dθ = kA − dt dx dH a dθ = . A − dt x dx

dH x. dx = aA.dθ dt On integrating the above expression we get θ2 l dH x.dx = a A ∫ dθ dt ∫0 θ1 dH dt dH dt

x2 l 2 o = a A θ1 − θ2 l2 . = a A θ 1 − θ2 2

(

(

θ

θ1

)

(

θ2

x

)

dH 2aA θ1 − θ2 = dt l2

θ − ∆θ

d x

l

)

→ COMBINATION OF A STRAIGHT – RODS: 1) SERIES – COMBINATION: - Suppose we have two rods of same cross sectional area A. Length of the rods are l 1 and l 2 and coefficient of thermal conductivity are k 1 and k 2 respectively. Ends of the composite rod are maintained at A ( θ1 ) B ( θ) C ( θ2 ) temperature θ1 and θ2 . Let θ be the common temp of the junction. In steady state, any heat that goes through l2 l1 first rod also through the section rod. So same heat current passer through the two rods. k 1 A ( θ1 − θ ) → ( 1) Heat current in first rod = l1 Heat current in second rod = In steady state k 1A/ θ1 − θ

(

l1

)

=

(

k2A / θ − θ2

k 2 A ( θ − θ2 ) l2

)

l2

k 1 l 2 ( θ 1 − θ ) = k 2 l 1 ( θ − θ2 ) k 1 l 2 θ 1 − k 1 l 2 θ1 = k 2 l 1 θ − k 2 l 1 θ θ=

k 1 l 2 θ1 + k 2 l 1 θ2 k1 l 2 + k2 l 1 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 4

Put the value of θ in equation k 1 l 2 θ 1 + k 2 l 1 θ2 dH k 1A = θ1 − dt l 1 k 1 l 2 + k 2 l 1 k 1 A k 2 l 1 ( θ1 − θ2 ) = l1 k 1 l 2 + k 2 l 1 dH k 1 A = dt l1

(

dH A θ1 − θ2 = l2 l dt + 1 k2 k1

)

dH θ1 − θ2 = dt R equ

(1), we get

k 2 l 1 ( θ1 − θ2 ) k 1 l 2 + k 2 l 1 =

θ1 − θ2 l2 l1 + Ak 2 Ak 1

→

( 2)

l l1 + 2 → ( 4) R equ = R 1 + R 2 = Ak 1 Ak 2 If k equ is the equivalent coefficient of thermal conductivity of the combination of rods, then thermal resistance of the combination becomes l +l R = 1 2 → ( 5) Ak equ ∴

→ (3)

Comparing equations (4) and (5) we get l2 l +l l + 1 = 1 2 Ak 2 Ak 1 Ak equ l 2 l 1 l 1+l 2 + = k2 k1 k equ k equ

l +l = 1 2 l1 l2 + k1 k2

For 2 slabs of equal length 2l 2 l/ k equ = = 1 l 1 l + l k + k k1 k2 2 1 ∴

k equ =

2k 1 k 2 k1 + k2

→ PARALLEL - COMBINATION: - Suppose we have two rods of same length ‘l’ and area of cross – sections A 1 and A 2 . The coefficient of thermal conductivity be k 1 and k 2 respectively. The ends of the rod are maintained at temperature θ1 and θ2 . The same temp difference is maintained between the ends of each rod. dH 2 dH 1 and Let be the heat conducted in first dt dt

A

1

θ1 A

2

θ2

l

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and second rod. Resultant of these is assumed to be dH dH 1 dH 2 = + dt dt dt =

(

k 1 A 1 θ 1 − θ2

)

+

(

k 2 A 2 θ 1 − θ2

dH dt

)

dt dt k A k A dH 1 1 2 2 = + ( θ1 − θ2 ) → ( 1) dt l l → If R equ is the equivalent thermal resistance of combination, then dH θ1 − θ2 = → ( 2) dt R equ Equating (1) and (2), we get k 1 A1 k2 A2 1 = + R equ l l 1 1 1 = + → (3) R equ R 1 R 2 → If K equ is the equivalent coefficient of thermal conductivity of parallel combination, then l R equ = k equ ( A 1 = A 2 ) R equ Using (3) for 1 1 l + = R 1 R 2 k equ ( A 1 + A 2 ) R1R2 R1+ R2

=

l k equ ( A 1 + A 2 )

∴ k equ =

l l × k A k 2 A2 l ⇒ 1 1 = l l k equ ( A 1 + A 2 ) + k 1A 1 k 2 A 2

k 1 A1 + k 2 A2

l l = k 1A 1 + k 2 A 2 k equ ( A 1 + A 2 )

A1 + A2

For 2 slabs of equal area k equ =

k1 +k2 2

→ 2 – D HEAT – CONDUCTION: (RADIAL – THERMAL – CONDUCTION IN CYLINDER)

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θ2 Consider a uniform cylindrical shell having θ1 radius ‘a’ and outer radius ‘b’ which are b maintained at uniform temperature θ1 and θ2 . a r Let length of the shell be ‘l’ and d coefficient of thermal conductivity of material of shell be k. The cylindrical shell is assumed to be made of coaxial cylindrical shells. If the uniform temp at radius r and r + dr be θ and θ − dθ respectively, then radial rate of flow in steady state is dH dθ = k 2 π r l − Here area of cylinder = 2 π r l dt dr dθ dH 1 − = × dr dt k 2π r l

(

)

(

(

−dθ =

r

)

)

dH 1 × .dr dt k 2π r l

Integrating the above expression, we get dH 1 θ=− × .logr + c dt 2 πk l

(

)

At r = a,

θ = θ1 dH 1 θ1 = − × . log a + c → ( 2) dt 2π k l

At r = b,

θ = θ2 dH 1 θ2 = − × . log b + c → ( 3) dt 2π k l

Solving equations (2) and (3), we get dH 1 b θ 1 − θ2 = × . log dt 2π k l a ∴

dH 2π kl ( θ1 − θ2 ) = dt log b/ a

log b/ a Thermal Resistance of the cylinder, R = 2π k l →

3 – D HEAT – CONDUCTION

Consider a sphere having inner radius a and outer radius b. The temperature of inner is θ1 and that of outer sphere be θ2 θ1 > θ2 .

(

)

Assume a concentric sphere of radius r having thickness dr, so that radial flew of heat is given by dH dθ = K ( 4πr 2 ) − dt dr Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 7

2 Here area of sphere = 4 πr 1 dH − dθ = . × dr 2 k ( 4πr ) dt Integrating the above expression, we get

( θ − dθ)

θ2

θ θ1

a

r d r θ1

∫ dθ =

θ2

b

1

∫ k 4π r a

2

.

b

dH .dr dt b

dH 1 1 θ 1 − θ2 = . − dt 4πk r a dH 1 1 1 θ 1 − θ2 = × − dt 4πk a b

(

dH 4πk θ1 − θ2 = dt 1 − 1 a b

)

Thermal Resistance of the sphere ⇒ R =

b− a 4πkab

→ NOTE: - In numericals we may have the situation where the amount of heat traveling to the other end may be required to do some other work. Eg: - it may be required to mew the given amount of ice in that case, we have to equalld Q = ML. ∴ ML = →

kA ( θ1 − θ2 ) t l

WIEDMANN – FRANZ – LAW: -

At a given temperature T, the ratio of thermal conducting to electrical conductivity (a) is Constant. k = Const ( i.e) aT (i. e) a substance which is a good conductor of heat (Eg: - Silver) is also a good conductor of Electricity. Mica is an expectation to this law → THERMOMETRIC – CONDUCTIVITY (OR) DIFFSIVITY (D): - It is a measure of rate of change of temperature (with time) when the body is not in steady state (i.e. invariable state) It is defined as the ratio of the coefficient of thermal conductivity to the thermal capacity per unit volume of the material. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 8

Thermal capacity per unit volume = D= →

k ρC

ELECTRICAL – ANALOGY FOR THERMAL CONDUCTION: ELECTRICAL CONDUCTION dq a) i = dt v −v b) i = 1 2 R ρl l c) R = = A αA

→

MC = ρC V

THERMAL CONDUCTION dQ a) H = dt θ −θ b) H = 1 2 R l c) R = KA

GROWTH OF ICE ON LAKE: - When the temperature of the air is less 0 than 0 C , the cold air near the surface of the pond takes heat from the water which freezes in the forms of layers.

Consequently, the thickness of the ice layer keeps increasing with time. Let x be the thickness of the ice layer at a certain time. If the thickness is increased by dx in time dt, then the amount of heat flowing through the slab in time dt is given by Q=

kA ]0 − ( − T ) dt

x

I c t d x

w

a t e 0 Cr

0

kA T dt

→ ( 1) x x Where A = area of the layer of ice at −T 0C is the temp of the surrounding air If dm = mass of water frozen into ice, then ρ = density of ice Q = dm X L But dm = Aρ .dx ∴ Q = A ρ dx.L → ( 2 ) Equating A ρ L . dx = dt =

=

−T 0C

(1) and (2) KAT .dt x

ρL

.x.dx KT Integrating, we have t ρ L x2 ∫o dt = KT x∫ x.dx 1 x

ρL x2 2 t= KT 2 x1 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 9

t=

ρL x22 − x12 2KT

PROBLEMS 1)

Few rods of material x and y are connected as shown in figure. The cross – sectional area of all the rods are same. If the end A is maintained at 800C and the end F is maintained at 100C. Calculate the end temperatures of junction B and E in steady state. Given that thermal conductivity of material x is double that of Y. 0 0 Ans: - TB = 60.52 C, TE = 19.74 C

C

y

L

A

x

x xx x x x x x x xx xx x x x x x x x x xx x x x x x x xx x x x x x x x x x x xx xx x xx xx x x x x x x x L x

B

y

E

F

L

x D

2)

The space between two thin concentric metallic spherical shells of radii a and b is filled with a thermal conducting medium of conductivity k. The inner shell is maintained at temperature T1 and outer is maintained at a lower temperature T2 . Calculate the rate of flow of heat in radialy outward direction through the medium. 4π kab( T1 − T2 ) dθ Ans: = dt b− a 3)

A slab of stone of area 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 1000 C. A block of ice at 0 0 C rests on the upper surface of the slab. If in one hour 4.8 kg of ice melted, calculate the thermal conductivity of the stone. −1 −1 −1 −1 −1 Ans: - k = 1.24wm k ( or ) 1.24 Js m oC 4)

Three rods AB, BC, and BD having thermal conductivities in the ratio 1: 2: 3 and lengths in the ratio 2: 1: 1 are joined as shown. The ends A, C and D are at temperatures T1 ,T2 and T3 respectively. Find the

temperature of the junction B. Assume steady state. 1 ( T1 + 4T2 + 6T3 ) Ans: - T = 11 5)

D

T

3

T

B

A

2

C T

1

Twelve indentical rods each of length l and cross – sectional area S made of material having thermal conductivity K are arranged as shown. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 10

The thermal resistance between A and B is____ 4l 2l 3l 2l 1) 2) 3) 4) 5K S KS 4K S 3K S

B

A

6)

The figure shows two different metal rods – one straight and bent twice at right angles (P) and the other semi circular (Q). If heat currents drawn by P and Q be the same, then what is the ratio of their T thermal conductivities? Ans: - 4 : π 7) AB, BC and CD are three metal rods identical in shapes and sizes joined end to end as shown. With thermal conductivities k, 2k, and 3k respectively. If end A and D are maintained at temperatures 2T and T, find the temperatures of B and C. Also determine the thermal current through any of the three rods. (Assume S and l as area and length of each rod.) 6k sT 5 2 Ans: - 1 T ; 1 T ; 11 11 11l Seven rods, each having length l and area A are arranged as shown in fig. Their respective thermal conductivities are shown in the figure itself. The free ends are maintained at temperatures T1 and T2 . Determine the thermal current through the system? 60kA T1 : T2 Ans: 97 l

( P

)

R

T

C

1

( Q

A

2

)

K

B 2 K C

3 K

D

8)

(

)

T

T

k 1

2 k 3 k

2

2 k 4 k

3 k

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SOLUTIONS 4)

If T is required temp of junction B and assuming T1 > T > T2 ,T3 wehave ∆Q = ∆Q + ∆Q ∆t AB ∆t BC ∆t BD

(

kA T1 − T 2L T1 − T 2 T =

5)

)

=

(

2kA T − T2

)

L

(

)

(

+

(

3kA T − T3 L

= 2 T − T2 + 3 T − T3

1 T1 + 4T2 + 6T3 11

(

)

)

)

l Thermal Resistance of each rod r = k s Thermal resistance between C & D r ( 2r ) = r+ +r r + ( 2r ) =

8r 3 ( 1i i1 i2

o

ti

i)

i3

A h

3

i3

i3

1

( 1i -

i1

i3 i2

i1

B c o

l d

3 i)

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r

r

C

r

r r r

r

A

B

r

r

E

D

r

r

r

F

∴ Eff . thermal resistan ce R will be 1 3 1 3 = + + R 8r 2r 8r 5 = 4r 4r 4 l 4l R = = × = 5 5 ks 5ks →

CONVECTION Mode of transfer of heat by means of migration of material particles of medium is called convection. It is of two types a) Natural convection b) Forced convection A)

NATURAL – CONVECTION: - This arise due to difference of densities at two places and is a consequence of gravity because on account of gravity the hot light particles arise up and cold heavy particles try setting down. It is mostly occurs on heating a liquid fluid.

B)

FORCED – CONVECTION: - If a fluid is forced to move to take up heat from a hot body then the convection process is called forced convection

In this case Newton’s law of cooling bolds good. According to which rate of loss of heat from a hot body due to moving fluid is directly proportional to the surface area of body and excess temperature of body over its surroundings (i.e) Q ∝ A T − T0 t Q hA T − T0 t

(

(

)

)

Where h = Constant called convection coefficient T = temp of body

T0 = temp of surrounding. h → depends on properties of fluids such as Density, Viscosity, Specific heat and thermal conductivity. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 13

→ Natural convection takes places from bottom to top while forced convection in any direction.

→

RADIATION : - / INFRA – RED – RADIATION The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. Precisely it is electromagnetic – Energy transfer in the form of E.M.W through any medium. It is possible even in vacuum. Eg: - The heat from the sun reaches the earth through radiation. The wavelength of thermal radiation ranges from 7.8 X 10-7 m to 4 X 10-4 m. They belong to infra – red region of the E.m Spec hum. That is why thermal radiations are also called infra – red – radiations.

→

PREVOST – THEORY: In early ages it was assumed that only hot body emits radiation but later a theory of heat radiation was given by PREVOST that every body which is at temperature above O k temp radiate thermal energy in the form of E.M.W of several wavelengths. Simultaneously the body also absorbs thermal radiation from its surroundings. 1 The speed is equal to that of light (C = 3 X 108 m/s) I ∝ 2 . d Just as light waves, they follow laws of Reflection, Refraction, interference, diffraction and palarisation . When these radiations falls on a surface then exert pressure m that surface which is known as Radiation pressure. Spec hum of these radiations cannot be obtained with help of glass prism because it absorbs heat radiations. It is obtained by QUARTZ (or) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657. www.asifiitphysics.vriti.com 14

ROCK – SALT PRISM, because these materials do net have free electrons and interatonic vibrational frequency is greater than the radiation frequency, hence they do not absorb heat radiation. These travels just like photons. →

DIATHERMANOUS - MEDIUM: - A medium which allows heat radiations to pass through it without absorbing then is called diathermanous medium. Thus the temp of a diathermanous medium does not increase irrespective of the amount of the amount of the thermal radiations passing through it. Eg: - Dry air, So2 , Rock salt (Nacl)

→

ATHERMANOUS - MEDIUM: - A medium which party absorbs heat rays is called a athermanous medium. As a result temp of an athermanous medium increases when heat radiations pass through it. Eg: - Wood, metal, moist air, simple glass, human flesh…….etc

→

COLOUR OF HEATED OBJECT: - When a body is heated, all radiations having wavelengths from zero to infinity are emitted. Radiations of longer wavelengths are predominant at lower temperature. The wavelength corresponding to maximum emission of radiations shifts from longer wavelength to shorter wavelength as the temperature increases. Due to this the colour of a body appears to be changed. c) A blue flame is at higher temp than a yellow flame. a) b)

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