Unit –II Thermal Physics Introduction- Modes of Heat Transfer Normally there are three modes of transfer of heat from one place to another viz., conduction, convection and radiation. Conduction : Conduction refers to the heat transfer that occurs across the medium. Medium can be solid or a fluid. Convection: It is the process in which heat is transferred from hotter end to colder end by the actual movement of heated particles. .Radiation : In radiation, in the absence of intervening medium, there is net heat transfer between two surfaces at different temperatures in the form of electromagnetic waves. Specific Heat Capacity The specific heat capacity of a material is the amount of energy (in Joules) needed to increase the temperature of one kilogram of mass of the material by one Kelvin. RECTILINEAR FLOW OF HEAT THROUGH A ROD Consider a long rod AB of uniform cross section heated at one end A as shown in figure. Then there is flow of heat along the length of the bar and heat is also radiated from its surface. B is the cold end. www.studentsfocus.com
Transcript
Unit –II Thermal Physics Introduction- Modes of Heat Transfer
Normally there are three modes of transfer of heat from one place to another viz., conduction, convection and radiation.
Conduction : Conduction refers to the heat transfer that occurs across the medium. Medium can be solid or a fluid.
Convection: It is the process in which heat is transferred from hotter end to colder end by the actual movement of heated particles.
.Radiation : In radiation, in the absence of intervening medium, there is net heat transfer between two surfaces at different temperatures in the form of electromagnetic waves.
Specific Heat Capacity
The specific heat capacity of a material is the amount of energy (in Joules) needed to increase the temperature of one kilogram of mass of the material by one Kelvin.
RECTILINEAR FLOW OF HEAT THROUGH A ROD
Consider a long rod AB of uniform cross section heated at one end A as shown in figure.
Then there is flow of heat along the length of the bar and heat is also radiated from its surface. B is the cold end.
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Consider the flow of heat between the sections P and Q at distance x and x+δx from the hot end.
Excess temperature above the surroundings at section P = 0
Before the steady state is reached, the amount of heat Q is used in two ways.
A part of the heat is used in raising the temperature of the rod and the remaining heat is lost by radiation from the surface.
Heat absorbed per second to raise the temperature of the rod
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= mass x specific heat capacity x dtdT
= (A x δx)ρ x S x …………(5) where A – Area of the cross-section of the rod ρ – Density of the rod S – Specific heat capacity of the rod - Rate of rise in temperature Heat lost per second due to radiation = E p δx T …………..(6) Where E – Emissive power of the surface p - Perimeter of the bar δx – Surface area of the element T - Average excess of temperature of the element over that of the surroundings
Amount of heat (Q) = Amount of heat absorbed + Amount of heat lost Q = (A x δx)ρ x S x + E p δx T …….(7)
On Comparing the eqns (4) and (7)
xdxdKA GT2
2
= (A x δx)ρ x S x + E p δx T………(8)
Dividing both LHS and RHS of the above equation by KA, we have,
xKA
xExKAdtdSxA
xKA
xdxdKA
GTUG
G
TUG
G
GT
��2
2
TTUTKAEp
dtd
KS
dxd
� 2
2
………………………(9)
The above equation is standard differential equation for the flow of heat through the rod. Special cases:-
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Case – 1: when heat lost by radiation is negligible.
If the rod is completely covered by insulating materials, then there is no loss of heat due to radiation.
Hence EpδxT = 0
dtd
hdtd
KS
dxd TTUT 1
2
2
� …………….(10)
where, hS
K
U, thermal diffusivity of the rod.
\ Case – 2: After the steady state is reached.
After the steady state is reached, there is no raise of temperature
Hence, dtdT
= 0
? equation (9) becomes
TTKAEp
dxd
2
2
Substituting,
2P KAEp
, we have
TPT 22
2
dxd
……….(11)
022
2
� TPTdxd
(This represent second order differential equation).
The general solution of this equation is
xx BeAe PPT �� � ……..(12) Where A and B are two unknown constants which can be determined from the
boundary conditions of the problem.
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Suppose the bar is of infinite length, Excess temperature above the surrounding of the rod of the hot end = T0
Excess temperature above the surrounding of the rod at the cold end = 0
Boundary condition
(i) When x = 0, T = T0 (ii) when x = f, T = 0
T0 = A+B 0 = Aef + Be-f
0 = Aef
As we know ef cannot be zero, therefore A should be zero i.e., A=0 then, T0 = B Substituting A and B in equation (12), we have T = T0e-μx
………(13) The above equation represents the excess temperature of a point at a distance x from
the hot end after the steady state is reached and it exponentially falls from hot end.
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HEAT CONDUCTION THROUGH A COMPOUND MEDIA (SERIES AND PARALLEL)
Consider a composite slab of two different materials, A & B of
thermal conductivity K1 & K2 respectively. Let the thickness of these two layers A & B be d1 and d2 respectively
Let the temperature of the end faces be θ1 & θ2 and temperature at the contact surface be θ, which is unknown. Heat will flow from A to B through the surface of contact only if θ1 > θ2. After steady state is reached heat flowing per second (Q) through every layer is same. A is the area of cross section of both layers
Amount of heat flowing per sec through A
Q = 1
11 )(x
AK TT � …………(1)
Amount of heat flowing per sec through B
Q= 1
12 )(x
AK TT � …………(2)
The amount of heat flowing through the materials A and B is equal in steady
conditions
Hence (1) and (2) are equal
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1
11 )(x
AK TT �=
1
12 )(x
AK TT � ……….(3)
Rearranging the (3), we have
K1A(T1-T)x2 = K2A(T-T2)x1
K1T1x2 - K1Tx2 = K2Tx1- K2T2x1
K1T1x2 +K2T2x1 = K2Tx1+ K1Tx2
K1T1x2 +K2T2x1 = T( K2x1+ K1x2)
T = 2112
122211
xKxKxKxK
�� TT ………..(4)
This is the expression for interface temperature of two composite slabs in
series.
Substituting T from equation (4) in equation (1), we get
Let us consider a compound wall of two different materials A and B of thermal
conductivities K1 and K2 and of thickness d1 and d2 respectively. These two material layers are arranged in parallel.
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The temperatures θ1 is maintained at one faces of the material A and B and opposite faces of the material A and B are at temperature θ2 . A1 & A2 be the areas of cross-section of the materials.
Amount of heat flowing through the first material (A) in one second.
Q1 = 1
2111 )(x
AK TT � …………….(1)
Amount of heat flowing through the second material (B) in one second.
Q2 = 2
2122 )(x
AK TT � …………….(2)
The total heat flowing through these materials per second is equal to the sum of Q1 and Q2
Q = Q1+Q2 ……………………….(3)
Substituting equations (1) and (2) in (3), we get,
In general, the net amount of heat flowing per second parallel to the composite slabs is given by
x
KAQ ¦� ¦ )( 21 TT
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RADIAL FLOW OF HEAT
In this method heat flows from the inner side towards the other side along the radius of the cylindrical shell. This method is useful in determining the thermal conductivity of bad conductors taken in the powder form.
CYLINDRICAL SHELL METHOD (or) RUBBER TUBE METHOD
Consider a cylindrical tube of length l, innerradius r1 andouter radius r2. The tube carries steam or some hot liquid. After the steady state is reached, the temperature on the innersurface is θ1 and on the outer surface is θ2 in such a way θ1>θ2. Heat is conducted radially across the wall of the tube. Consider an element of thickness dr and length l at a distance r from the axis.
Working: Steam is allowed to pass through the axis of the cylindrical shell. The heat flows from the inner surface to the other surface radially.After the steady state is reached, the temperature at the inner surface is noted as T1 and on the outer surface is noted as T2.
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Calculation: The cylinder may be considered to consists of a large number of co-axial cylinders of increasing radii. Consider such an elemental cylindrical shell of thr thickness dr at a distance ‘r’ from the axis. Let the temperatures of inner and outer surfaces of the elemental shell be T and T+dT. Then,
The Amount ofheat conducted per second drdKAQ T
�
Here Area of cross section A = 2πrl
�?drdrlKQ TS2�
Rearranging the above equation we have
TS d
QlK
rdr 2�
…………(1)
? The Thermal conductivity of the whole cylinder can be got by, integrating equation (1) within the limits r1 to r2 and T1 to T2,
By knowing the values in RHS, the thermal conductivity of the given material can be found.
DETERMINATION OF THERMAL CONDUCTIVITY OF RUBBER
It is based on the principle of radial flow of heat through a cylindrical shell. Description: It consists of a calorimeter, stirrer with a thermometer. The setup is kept inside the woodenbox. The space between the calorimeter and the box is filled with insulating materials such as cotton, wool, etc. to avoid radiation loss, as shown in fig.
Working:
� The empty calorimeter is weighed, let it be (w1). � It is filled with two third of water and is again weighed, let it be (w2) � A known length of rubber tube is immersed inside the water
contained in the calorimeter. � Steam is passed through one end of the rubber tube and let out
through the other end of the tube. � The heat flows from the inner layer of the rubber tube to the outer
layer and is radiated. � The radiated heat is gained by the water in the calorimeter. � The time taken for the steam flow to raise the temperature of the
water about 10qC is noted, let it be ‘t’ seconds.
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Observation and calculation: Let w1 Weight of calorimeter w2 Weight of calorimeter and water w2 – w1 Weight of the water alone T1 Initial temperature of the water T2 Final temperature of the water T2 -T1 Rise in temperature of the water TS Temperature of the steam l Length of the rubber tube (immersed) r1 Inner radius of the rubber tube r2 Outer radius of the rubber tube s1 Specific heat capacity of the calorimeter s2 Specific heat capacity of the water T3 Average temperature of the rubber tube.
T3 =2
21 TT �
We know from the theory of cylindrical shell method the amount of heat conducted by the rubber tube per second is given by
? The amount of heat gained by the water and calorimeter per second is obtained by (2) +(3)
Q = t
swsww )()()( 121112212 TTTT ����
Q = � �> @t
swwsw 2121112 )( ���TT ..............(4)
Under steady state The amount of heat conducted by The amount of heat gained by the rubber tube per second = the water and the calorimeter per escond Hence, equation (1) = Equation (4)
By substituting the values in RHS, the thermal conductivity of the rubber can be determined.
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Methods to determine thermal conductivity The thermal conductivity of a material is determined by various methods 1. Searle’s method – for good conductors like metallic rods 2. Forbe’s method - for determining the absolute conductivity of metals 3. Lee’s disc method – for bad conductors 4. Radial flow method – for bad conductors
LEE’S DISC METHOD FOR DETERMINATION OF THERMAL CONDUCTIVITY OF BAD CONDUCTOR The thermal conductivity of bad conductor like ebonite or card board is determined by this method. Description:
The given bad conductor (B) is shaped with the diameter as that of the circular slab (or) disc ‘D’. The bad conductor is placed inbetween the steam chamber (S) and the disc (D), provided the bad conductor, steam chamber and the slab should be of same diameter. Holes are provided in the steam chamber (S) and the disc (D) in which
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thermometer are inserted to measure the temperatures. The total arrangement is hanged over the stand as shown in fig.
Working:
Steam is passed through the steam chamber till the steady state is reached . Let the temperature of the steam chamber (hot end) and the disc (cold end) be T1 and T2 respectively. Observation and Calculation:
Let ‘x’ be the thickness of the bad conductor (B), ‘m’ is the mass of the slab, ‘s’ be the specific heat capacity of the slab. ‘r’ is the radius of the slab and ‘h’ be the height of the slab, then
Amount of heat conducted by the
Bad conductor per second = x
KA )( 21 TT � …….(1)
Area of the cross section is = πr2 ……………… (2)
Amount of heat conducted per second = x
rK )( 212 TTS � ……(3)
The amount of heat lost by slab per second = m x s x Rate of cooling = msRc…………..(4)
Under steady state
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The amount of heat conducted by the = Amount of heat lost by the slab Bad conductyor (B) per second (D) Per second
Hence, we can write equation (3) = equation (4)
x
rK )( 212 TTS � = msRc
K = )( 21
2 TTS �rmsxRc …….(5)
To find the rate of cooling Rc Rc in equation (3) represents the rate of cooling of the disc along with the steam chamber. To find the rate of cooling for the disc alone, the bad conductor is removed and the steam chamber is directly placed over the disc and heated. When the temperature of the slab attains 5qC higher than T2, the steam chamber is removed. The slab is allowed to cool, simulataneously a stop watch is switched ON. A graph is plotted taking time along ‘x’ axis and temperature along ‘y’ axis, the
The rate of cooling is directly proportional to the surface area exposed.
Case(i) Steam chamber and bad conductor are placed over slab, in which radiation takes
place from the bottom surface of area (πr2) of the slab and the sides of the of area (2πrh). ? Rc = 2 πr2 + 2πrh Rc = πr(r+2h)………(6)
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Case(ii) The heat is radiated by the slab alone, (i.e) from the bottom of area(πr2), top surface of the slab of area (πr2) and also through the sides of the slab of area 2πrh.
