Heat_mass_Transfer by naveen choudhary.pptx

Heat and Mass Transfer
Course No: FT-236
Course Instructor:
❖ Er. Ashish Dhiman

Course Contents
Heat Conduction
ONE DIMENSIONAL STEADY STATE HEAT CONDUCTION
HEAT TRANSFER THROUGH COMPOSITE WALL AND
INSULATED PIPE
Dimensionless numbers
Introduction of condensation & Boiling Heat Transfer

Heat conduction and
fourier’s law

Heat transfer
• Conduction
Thermal energy is transferred by the
direct contact of molecules ,not by the
movement of material.
• Convection
Thermal energy is transferred by the
mass motion of group of molecule resulting
in transport and mixing of properties
• Radiation
Thermal energy is transferred by
electromagnetic radiation [WAVES]
Heat transfer is the process of transfer of heat from high temperature system to low
temperature system.
Mechanisms of heat transfer:

FOURIER’S LAW OF HEAT CONDUCTION
• It states that “the rate of heat transfer is directly proportional to the
area normal to the area normal to the direction of heat flow and
temperature gradient”
Q A
∝
∝
q=-
where ,k= Thermal conductivity of material and negative sign indicate
the drop in temperature
Q=heat flux density

Derivation of fourier’s law
• Rate of heat conduction [area][temperature difference]/thickness
∝
Suppose T1 and T2 are different temperature through a short distance of an area where
the distance is Δx the area is A and K is thermal conductivity of material.
Therefore is one dimension equation can be represent as
• Q=
=-KA
• q=-k T [Differential form]
∇
• Q=- [One dimension form]
• [Integral form]

Assumptions in fourier law of heat conduction:
• The thermal conductivity of the material is constant throughout the material.
• There is no internal heat generation that occurs in the body .
• The temperature gradient is considered as constant.
• The heat flow is unidirectional and takes place under steady state conditions.
• The surfaces are isothermal.
Essential features of fourier’s law of heat conduction:
• It is a vector expression. Negative sign indicates that heat flow in direction of
decreasing temperature.
• It is applicable to heat conduction in all materials regardless of their state.
• It helps to determine the thermal conductivity ‘k’ of material through which
heat transfer.
• It is not an expression derived from the first place ; instead, it is based on
experimental evidence.

STEADY STATE HEAT TRANSFER VS UNSTEADY STATE
HEAT TRANSFER
• STEADY STATE HEAT TRANSFER:
The heat transfer process that does not affected by the time interval is
know as steady state heat transfer . In this type of heat transfer , the
temperature of the object does not change with respect of time .
• Unsteady state heat transfer:
The heat transfer process that get affected by the time interval is
know as unsteady state heat transfer . In this type of heat transfer , the
temperature of the object increases with respect to time .

STEADY STATE Vs UNSTEADY STATE
Steady state heat transfer unsteady heat transfer
1. The steady state heat transfer is denoted by =0 1. The unsteady state heat transfer is denoted by,
≠0
2 The temperature of the object doesn’t vary with
respect to time.
2. The temperature of the object changes with
respect to time.
3. No heat diffusion occurs in the object. 3. Heat diffusion occurs in the object.
4. For the steady state heat transfer, the energy entering
the object is equal to the energy leaving from
the ,Ein=Eout.
4. For the unsteady state heat transfer, the energy
entering the object is not equal to the energy leaving
from the object, Ein≠Eout.
5. The change in the internal energy of the object is zero. 5. The change in the internal energy of the object is
not equal to zero.
6. Example: heat generation because of electrical
current.
6. Example: heating or cooling of water, heating of
metal in a furnace.

FREE AND FORCED CONVECTION
Free Convection:
• Free convection is also know as natural convection.
• It is the type of convection heat transfer in which the fluid
molecules move due to density and temperature gradient to
transfer the heat
• Natural convection is the convection occurring due to buoyant
forces with the distinction in densities caused by distinction in
temperature . The oceanic wind is an example of natural convection
Forced convection
• Forced convection is the type of convective heat transfer in which
the motion in molecules is generated by the use of an external
sources for the transfer of heat.
• Example ; exhaust fan or a ceiling fan. The exhaust fan helps eject
the heat inside the room by generating motion into air molecules

FREE VS FORCED CONVECTION
FREE CONVECTION FORCED COVECTION
• In this, the molecules move due to density and
temperature variation.
• In this, the fluid molecules are forced to move by
an external source.
• The rate of heat transfer is lower.
• It has less overall heat transfer coefficient.
• The rate of heat transfer is higher.
• The overall heat transfer coefficient is higher.
• No external equipment is required. • External equipment is necessary for convective
heat transfer.
• The motion of molecules is comparatively slower. • Molecules of fluid are forced to move faster.
• Equipment based on natural convection is larger in
size.
• Equipment based on forced convection is compact
in size.
• The flow of molecules cannot be controlled.
• E.g. movement of water molecules while boiling.
• The flow of molecules can be controlled by
controlling the fan, pump, blower.
• E.g. movement of molecules due to fan or blower.

Fourier law and heat
conduction
Numerical Practice

Problem 1: A 10 cm thick block of ice with a temperature of 0 °C lies on the upper surface of 2400 cm2
slab
of stone. The slab is steam-exposed on the lower surface at a temperature of 100 °C. Find the heat
conductivity of stone if 4000 g of ice is melted in one hour given that the latent heat of fusion of ice is 80
cal ⁄ gm.
Solution:
Given:
Area of slab, A = 2400 cm2
Thickness of ice, d = 10 cm
Temperature difference, Th – Tc = 100 °C – 0 °C 100 °C
Time of heat transfer, t = 1 hr = 3600 s
Amount of heat transfer, Q = m L = 4000 × 80 320000 cal
Heat transfer rate, q = Q ⁄ t = 320000 cal ⁄ 3600 s  89 cal ⁄ s
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of K.
K = q d ⁄ A (Th – Tc)
 (89 × 10) ⁄ (2400 × 100) cal ⁄ cm s °C
3.7 × 10-3
cal ⁄ cm s °C
Hence, the thermal conductivity of stone is 3.7 × 10-3
cal ⁄ cm s °C.

Problem 2: A metal rod 0.4 m long & 0.04 m in diameter has one end at 373 K & another end at 273
K. Calculate the total amount of heat conducted in 1 minute. (Given K = 385 J ⁄ m s °C)
Solution:
Given:
Thermal conductivity, K = 385 J ⁄ m s °C
Length of rod, d = 0.4 m
Diameter of rod, D = 0.04 m
Area of slab, A = π D2 ⁄ 4 = 0.001256 m2 note: cross-sectional inside pipe area= pi*(d/2)2
Temperature difference,
Th – Tc = 373 K – 273 K = 100 K
Time of heat transfer, t = 1 min = 60 s
Q ⁄ t = K A (Th – Tc) ⁄ d
Q = K A t (Th – Tc) ⁄ d
= (385 × 0.001256 × 60 × 100) ⁄ 0.4 J
 7.25 × 103
J
Hence, the total amount of heat transfer is 7.25 × 103
J.

Problem 3: An aluminium rod and a copper rod of equal length 2.0 m and cross-sectional area 2 cm2
are
welded together in parallel. One end is kept at a temperature of 10 °C and the other at 30 °C . Calculate the
amount of heat taken out per second from the hot end . (Thermal conductivity of aluminium is 200 W ⁄ m °C
and of copper is 390 W ⁄ m °C).
Solution:
Given:
Thermal conductivity of aluminium, Kal = 200 W ⁄ m °C
Thermal conductivity of copper, Kcu = 390 W ⁄ m °C
Combined thermal conductivity for parallel combination, K = 200 W ⁄ m °C + 390
W ⁄ m °C = 590 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2
= 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C
q = K A (Th – Tc) ⁄ d
= (590 × 2 × 10-4 × 20) ⁄ 2 W
 1.18 W
Hence, the total amount of heat transfer is 1.18 W.

Problem 4: The average rate at which energy is conducted outward through the ground surface at a place is
50.0 mW ⁄ m2
, and the average thermal conductivity of the near-surface rocks is 2.00 W ⁄ m K. Assuming
surface temperature of 20.0 °C, find the temperature at a depth of 25.0 km.
solution
Given:
Average thermal conductivity, K = 2.00 W ⁄ m K
Depth, d = 25.0 km = 2.50 × 104 m
Surface temperature, Tc = 20.0 °C = (20 + 273) K = 293 K
Heat transfer rate per unit area, q ⁄ A = 50.0 mW ⁄ m2
= 50.0 × 10-3
W ⁄ m2
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of Th.
Th = q d ⁄ KA + Tc
((50.0 × 10-3 × 2.00 × 104) ⁄ 2.00) + 293
 (500 + 293) K
 893 – 273 K
520 °C
Hence, the temperature at depth of 25.0 km is 520 °C.

Problem 5: The energy lost from a 10 cm thick slab of steel is 50 W. Assuming the temperature
difference of 10.0 K, find the area of the slab. (Thermal conductivity of steel = 45 W ⁄ m K).
solution
Given:
Thermal conductivity, K = 45 W ⁄ m K
Thickness of slab, d = 10 cm = 0.1 m
Temperature difference, Th – Tc = 10.0 K
Energy lost per sec, q = 50 W
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of A.
A = q d ⁄ K (Th – Tc)
 (50 × 0.1) ⁄ (45 × 10.0) m2
 0.011 m2
Hence, the area of the slab is 0.011 m2.

Problem 6: One face of an aluminium cube of edge 5 meters is maintained at 60 ºC and the other
end is maintained at 0 ºC. All other surfaces are covered by adiabatic walls. Find the amount of
heat flowing through the cube in 2 seconds. (Thermal conductivity of aluminium is 209 W ⁄ m ºC).
solution
Given:
Edge length of cube, d = 5 m
Surface area of cube, A = d2
= (5 m)2 = 25 m2
Temperature difference, Th – Tc = 60 ºC – 0 ºC = 60 ºC
Thermal conductivity, K = 209 W ⁄ m ºC
Heat transfer time, t = 2 sec
q = K A (Th – Tc) ⁄ d
(209 × 25 × 60) ⁄ 5 J
 62700 J
 62.7 KJ
Hence, the amount of heat that flows through the cube is 62.7 KJ.

Problem 7: An aluminium rod and a copper rod of equal length 2.0 m and cross-sectional area 2 cm2
are welded together in series. One end is kept at a temperature of 10 °C and the other at 30 °C .
Calculate the amount of heat taken out per second from the hot end . (Thermal conductivity of
aluminium is 200 W ⁄ m °C and of copper is 390 W ⁄ m °C).
solution
Given:
Thermal conductivity of aluminium, Kal = 200 W ⁄ m °C
Thermal conductivity of aluminium, Kcu = 390 W ⁄ m °C
Combined thermal conductivity for series combination, 1 ⁄ K = 1 ⁄ 200 W ⁄ m °C + 1 ⁄ 390 W ⁄ m °C
K = (200 × 390) ⁄ (200 + 390) W ⁄ m °C
= 132.2 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2 = 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C
The formula for heat transfer rate is given as:p
q = K A (Th – Tc) ⁄ d
 (132.2 × 2 × 10-4 × 20) ⁄ 2 W
0.2644 W
Hence, the total amount of heat transfer is 0.2644 W.

ONE DIMENSIONAL
STEADY STATE HEAT
CONDUCTION

Steady State One Dimensional Conduction
• The formula for steady state one dimensional conduction
describing heat transfer through a material is given by Fourier’s
Law :
• Where :
• Q is the heat transfer rate,
• k is the material’s thermal conductivity,
• A is the cross sectional area through which the heat is conducted,
• is the temperature difference across the material,
• d is the thickness of material.

Relation with Fourier’s Law :
• Steady-state one-dimensional conduction is a concept in
heat transfer that deals with the steady flow of heat through
a material along a single direction, typically in a straight line.
• The term "steady-state" implies a constant and unchanging
condition over time.
• In this context, the temperature distribution within the
material remains constant, and there is a continuous
transfer of heat from one side to the other.
• The governing principle is Fourier's Law, which relates the
heat transfer rate to the material's thermal conductivity,
cross-sectional area, temperature difference, and thickness.
This phenomenon is crucial in understanding how heat
moves through solid objects under stable conditions.

For the following geometries, one-dimensional, steady-state heat conduction is taken into consideration :
SLAB CYLINDER SPHERE

One Dimensional Steady State Heat Conduction through a Slab :
• One-dimensional steady-state heat conduction through a
slab refers to the controlled and constant flow of heat in a
straight path through a material with a uniform cross-
section.
• Key points include :
1. Steady-state : The temperature distribution within the slab
remains constant with time. There is no change in the
temperature profile as time progresses.
2. One-dimensional : Heat transfer occurs primarily in one
direction, usually along the thickness of the slab. This simplifies the
analysis by focusing on a single dimension.
3. Slab : The material through which heat is conducted has a flat and
often rectangular shape. This could represent various objects like
walls, windows, or other planar structures.

Boundary Conditions for Heat Flow through Slab :
• Thickness = ‘L’ , Area = ‘A’ and Conductivity = ‘k’
• In the case of one-dimensional, steady heat conduction through a wall of
uniform conductivity without heat generation :
, and (1)
Equation (1) reduces to : (2)
Integrating equation (2) twice with respect to x :
(3)
where, and are constants of integration
Now using the boundary conditions :
At x = 0,
Equation (3) is written as = (4)
At x = L,

Equation (3) can be written as
OR
(5)
Now substituting the values of and in equation (3)
(6)
Equation (6) represents temperature distribution in the wall.
Rate of heat transfer can be determined by using Fourier’s Law and can be expressed as :
(7)
Integrating equation (6) with respect to x to obtain the expression for temperature gradient

Substituting the value of from above equation in equation (7), we get
(8)
Equation (8) represents the heat transfer rate through the wall.

One Dimensional Steady State Heat Conduction through
a Cylinder :
cylinder involves the controlled and constant flow of heat along
the radial direction of the cylindrical shape.
• Key points include :
1. Steady-state : Similar to the slab case, the temperature
distribution within the cylinder remains constant with time. This
implies that there is no change in the temperature profile as time
progresses.
2. One-dimensional : Heat transfer primarily occurs in the radial
direction, perpendicular to the length of the cylinder. This
simplifies the analysis by focusing on a single dimension.
3. Cylinder : The material through which heat is conducted has a
cylindrical shape. This could represent objects like pipes, rods, or
other cylindrical structures.

Boundary Conditions for Heat Flow through a
Cylinder :
• Inner Radii = a, Outer Radii = b, Length = H,
Inner Surface Temp = , Outer Surface Temp =
T = at r = a
T = at r = b
• Upon integrating, we get
i.e, (1)

Upon integrating further, we get
(2)
By using the boundary conditions at r = a , b gives
(3)
(4)
Subtracting equation (4) from equation (3)
i.e, (5)
Now using equation (5) in equation (1), we get

From the definition of heat flux, and from , we get
= (6)
where,
Equation (6) represents the heat transfer rate through the cylinder.

One Dimensional Steady State Heat Conduction through
a Sphere :
sphere involves the controlled and constant flow of heat in
the radial direction of a spherical object.
Key Points Include :
1. Steady-state : Similar to the slab and cylinder cases, the
temperature distribution within the sphere remains constant with
time. There is no change in the temperature profile as time
progresses.
2. One-dimensional : Heat transfer primarily occurs in the radial
direction, from the center to the outer surface of the sphere. This
simplifies the analysis by focusing on a single dimension.
3. Sphere : The material through which heat is conducted has a
spherical shape. This could represent objects like a solid sphere or a
spherical container.

Boundary Conditions for Heat Flow through a
Sphere :
• Inner Radii = a, Outer Radii = b
T = at r = a
T = at r = b
Upon integrating, we get
i.e, (1)

Upon integrating further, we get
(2)
Using the boundary conditions at r = a, b gives
(3)
(4)
Subtracting equation (3) from equation (4)
- = - = =
i.e, =

From the definition of heat flux, and from equation (1), we get
(5)
Where,
Equation (5) represents the heat transfer rate through the sphere.

HEAT TRANSFER
THROUGH
COMPOSITE WALL
AND INSULATED PIPE

HEAT TRANSFER THROUGH COMPOSITE WALL
• Consider a composite slab made of three different materials
having conductivity k1, k2 and k3, length L1, L2 and L3 as
shown in Figure 3. wall is exposed to a hot fluid having
temperature Tf and on the other side is atmospheric air at
temperature Ta.
• Convective heat transfer coefficient between the hot fluid
and inside surface of wall is hi (inside convective heat transfer
coefficient) and ho is the convective heat transfer coefficient
between atmospheric air and outside surface of the wall
(outside convective heat transfer coefficient).
• Temperatures at inner and outer surfaces of the composite
wall are T1 and T4 whereas at the interface of the constituent
materials of the slab are T2
• One side of the and T3 respectively.

Heat is transferred from hot fluid to atmospheric air and involves following steps:
(A) Heat transfer from hot fluid to inside surface of the composite wall by
convection
(B) Heat transfer from inside surface to first interface by conduction
(C) Heat transfer from first interface to second interface by conduction

(D) Heat transfer from second interface to outer surface of the composite wall by conduction
(E) Heat transfer from outer surface of composite wall to atmospheric air by convection
Adding equations (1), (2), (3) and (5), we get

If composite slab is made of ‘n’ number of materials, then equation (6) reduces to
If inside and outside convective heat transfer coefficients are not to be considered, then equation (6) is
expressed as

HEAT TRANSFRER THROUGH INSULATED PIPE
• Consider a composite cylinder consisting of inner and outer
cylinders of radii r1, r2 and thermal conductivity k1, k2
respectively as shown in Figure 4.
• Length of the composite cylinder is L.
• Hot fluid at temperature Tf is flowing inside the composite
cylinder. Temperature at the inner surface of the composite
cylinder exposed to hot fluid is T1 and outer surface of the
composite cylinder is at temperature T3 and is exposed to
atmospheric air at temperature Ta.
• The interface temperature of the composite cylinder is T2.
• Convective heat transfer coefficient between the hot fluid
and inside surface of composite cylinder is hi (inside
convective heat transfer coefficient) and ho is the convective
heat transfer coefficient between atmospheric air and
outside surface of the composite cylinder (outside
convective heat transfer coefficient).
• Heat is transferred from hot fluid to atmospheric air and
involves following steps:

(A) Heat transfer from hot fluid to inside surface of the composite cylinder by convection
(B) Heat transfer from inside surface to interface by conduction
1
2

(C) Heat transfer from interface to outer surface of the composite cylinder by conduction
(D) Heat transfer from outer surface of composite wall to atmospheric air by convection
3
4

Adding both sides of equations (1), (2),(3) and (4), we get
5

If the composite cylinder consists of ‘n’ cylinders, then equation (5) can be expressed as:
If inside and outside convective heat transfer coefficients are not to be considered, then equation (5) is
expressed as
6
7

THERMAL RESISTANCE ANALOGY
Consider heat flowing through a slab of thickness ‘L’ and area ‘A’ and T1 and T2 are the temperatures on
the two faces of the slab as shown .
Heat transfer from high temperature to low temperature side is expressed as
-1
where (T1-T2) is the thermal potential kA/L is the thermal resistance.
Now consider an electric circuit having resistance ‘R’ and electric potential E1 and
E2 at the ends as shown . Current ‘I’ passing through the circuit can be expressed as
-2

Equations 1 and 2 are found to be symmetrical on comparison. ‘Q’ amount of heat flows
through the slab having thermal resistance when a thermal potential (T1-T2) exits. Similarly,
‘I’ amount of current passes through the circuit having resistance ‘R’ when an electric
potential (E1-E2) exists. Therefore, flow of heat through the slab can be represented by an
electric circuit as shown below.
If a hot gas at temperature Tg is in contact with one side of the slab and air at temperature Ta
at the other side , then heat transfer from the hot gas to air through this slab of thickness can
be represented by an electric circuit as shown.

Heat transfer from hot gas to air can be expressed as
- 3
For a slab made of 3 material having thermal conductivities K1 , K2 and K3 respectively and is exposed
to a hot gas on one side and atmospheric air on the other side as shown.
Heat transfer from the hot gas to atmospheric air is expressed as
- 4

Similarly for the composite slab shown below , an equivalent electric circuit has been shown
here,

Plane and composite walls
Numerical Practice

Problem 1: Determine the loss of heat Q (watts) through a wall laid of lead brick of length 5 m,
height 4 m and thickness 0.25 m, if the temperature of the surfaces of the wall are maintained at
110C & 40C. Assume thermal conductivity of brick = 0.7 watt/m-C.
Solution
λ thermal conductivity = 0.7 watt/m-C
δ = thickness = 0.25 m
∆t, temperature difference of surfaces = (110-40) = 70C
= 196 watt/m2
Q, loss of heat through 20 m2
area of wall = q x A= 196 x 20 = 3920 watt
Q 3920 watt

Problem 2: The inner surface of furnace wall is at 2000C and outer surface at 500C. Calculate the
heat lost per m^2 area of the wall. If thermal conductivity of the brick is 0.5 W/m0C & the wall
thickness is 200mm.
solution
Given T1 = 200°C , T2 = 50°C, L = 200mm = 0.2 , k = 0.5 w/m°c
Heat loss =-ka dt/dx
heat lost / m^2 -k(t2-t1)/L
. - 0.5(50-200)/0.2
.  375 W/m^2

Problem 3: Heat transfer through a composite wall is shown in figure. Both the sections of the wall
have equal thickness (l). The conductivity of one section is k and that of the other is 2k. The left
face of the wall is at 600 K and the right face is at 300 K . The interface temperature Ti (in K) of the
composite wall is __________?
Solution
When heat flows from a solid wall then total heat transfer from one face
to another,
Q =( t1 – t2)/(B1/k1a + B2/k2a)
For interference temperature calculation,
(t1 – t I)/(b1 /k1a) = ( Ti – t2)/(b2/k2a)........eq 1
where,
Ti = The interference temperature
Given:
k1 = k, k2 = 2k, T1 = 600 K, T2 = 300 K, b1 = b2, A1 = A2
From equation (1)
(600 – ti)/ (1/k) =. (ti – 300)/(1/2k)
⇒ Ti = 400 K

Problem 4: Two surfaces of a plane wall of 15cm thickness and 5 m2 area are maintained at
240°C and 90°C respectively. Determine the heat transfer between the surfaces and temperature
gradient across the wall if conductivity of the wall material is18.5 W/(m-K).
Solution
Given
T1= 2 40°C, T2= 90°C, Thickness of wall, x = 15 cms.=0.15 m,
Area, A= 5 m2
Thermal conducivity, k=18.5 W/(m-K)
To determine: i) Temperature gradient, ,
ii) Heat transfer rate, ,
i) The temperature gradient in the direction of heat flow is
(b) Heat flow across the wall is given by Fourier’s heat conduction equation
=92500 W or 92.50 kW

Problem 5: The bond between two plates, 2.5 cm and 15 cm thick, heat is uniformly applied through the
thinner plate by a radiant heat source. The bonding epoxy must be held at 320 K for a short time. When the
heat source is adjusted to have a steady value of 43.5 kW/m2, a thermocouple installed on the side of the
thinner plate next to source indicates a temperature of 345 K. Calculate the temperature gradient for heat
conduction through thinner plate and thermal conductivity of its material.
Solution:
t1 = 345 K ; t2 = 320 K
δ = 2.5 cm = 0.025m
# Temperature gradient,
dt/dx = T2-T1/δ
320 – 345/0.025
 - 1000°C/m

Problem 2: A hollow cylinder 5 cm inner radius and 10 cm outer radius has inner surface
temperature of 200°C and outer surface temperature of 100°C. If the thermal conductivity is 70
W/m K, find heat transfer per unit length.
solution
Given: inner radius, r1= 5 cm = 0.05 m Outer radius, r2= 10 cm = 0.1 m Inner surface
temperature, T1= 200 + 273 = 473 K outer surface temperature, T2=100 + 273=373 K
Thermal conductivity, k = 70 W/m K .
to find:Heat flow per uniT length :
solution : heat transfer through hollow cylinder is given by
q= ∆t overall/r
Where ∆t = t1-t2 and r= (1/2πlk) ln R2/r1
 q= [t1-t2]/[(1/2πlk) Ln [ r2 / r1]]
 q/l = 63453.04 w/m

Problem 1: If the inner and outer walls of a sphere having surface areas of A1 and A2 and inner
and outer radii r1 and r2 are maintained at t1 and t2, then rate of heat flow will be ?
Solution
Heat conduction:
The transfer of heat between two different bodies or two different locations of the same body through molecular vibrations is
known as heat conduction.
The governing law for heat conduction is Fourier’s law of heat conduction.
Fourier’s law of heat conduction:
It states that ‘The rate of heat transfer is directly proportional to the temperature gradient’ and it is given as
qk = -K × Dt/dx
Where x = direction of heat flow, k = thermal conductivity, T = temperature
The heat conduction equation of the spherical wall is given as:
Q= 4πk(t1-t2)R1r2/(R2-r1)
So if we consider the surface areas at radius r1 and r2 then A1 = 4πr21 and A2 = 4πr22
q= K × (√4πR1^2)×(√4πr2^2 ) × T1 – t2/ R2 – R1
∴ The heat conduction in a sphere having surface areas of A1 and A2 and inner and outer radii r1 and r2 are maintained at
temperatures t1 and t2 is given as:
Q = K × √A1 × √a2 × T1-t2 / r2-r1

problem 2: What will be the geometric radius of heat transfer for a hollow sphere of inner and outer radii r1 and r2?
solution
Heat transfer through the composite sphere is given by,
Q = 4πkr1r2 (t1-t2)/ r2-r1
as Q =-K Am T2-t1/R2-r1
Am = 4πr^2M = 4πR1r2
. Therefore ,Rm = √r1r2

Problem 3: A spherical shaped vessel of 1.4 m outer diameter is 90 mm thick. Find the rate of
heat leakage, if the temperature difference between the inner and outer surfaces is 220°C.
Thermal conductivity of the material of the sphere is 0.083 W/m K.
Solution
R2 = 0.7 m
r1 = 0.7 – 0.09 = 0.61 m
dT = 220°C, k = 0.083 W/mK
rate of heat leakage:
q= dt/r
 r = r2-r1/4π(kr1r2)
 0.09/4× 3.14× 0.083× 0.7× 0.61
. 0.20218
 q= 220/2020.  1.088Kw

Problem 1: A plane brick wall, 25 cm thick, is faced with 5 cm thick concrete layer. If the temperature of
the exposed brick face is 70°C and that of the concrete is 25°C, find out the heat lost per hour through a
wall of 15 m x10 m. Also, determine the interface temperature. Thermal conductivity of the brick and
concrete are 0.7W/m.K and 0.95 W/m.
Solution
Heat loss Q = (Ta- Tb) / r
where Ta = 70°c; Tb = 25°c
and R = R(brick)+ R(concrete)
R(brick) = L(brick)/(Ak( brick)
R (concrete)= L(concrete) /(Ak(Concrete)
A = 15 x 10  150 m2
R(brick )= 0.25/(150 x 0.7)  2.381 10^-3 °c/w
r(concrete) = 0.05/(150×0.95)  3.509 × 10^-4 °c/w
R= 2.381 × 10^-3 + 3.509 × 10^-4 2.732 × 10^-3 °c/w
q = ( 70-25) / (2.732 × 10^-3)  16471.4 J/sec  59.3 Mj/hr
heat loss per hour = 59.3 Mj
at steady state , this is the amount of Heat transferred through the Brick wall or concrete per hour .
Q = A k(brick) Ta – ti /l(brick) = 16471.4W
150× 0.7× (70-ti) / 0.25 16471.4
70- Ti 39.2
Ti = 70- 39.2  30.8 °c
interface temperature
ti = 30.8°c

Problem 2: The wall of a boiler is made up of 250mm of the brick, KFB= 1.05 W/m K; 120 mm of
insulation brick KIB = 0.15W/mK, and 200 mm of red brick, KRB= 0.85 W/m K. The inner and outer
surface temperatures of the wall are 8500C and 650C respectively. Calculate the temperatures at
the contact surfaces.
solution :-
Q/a = Q= (t1-t4)
x1 + _X2 + X3_
. K(fb) K(m) K(rb)
Q/a = 1×(850-65)
(0.250 + 0.120 + 0.200)
1.05. 0.15 0.85
 616.5 W/m^2
Q/a = T1-T2 = T2-t3 = t3-t4
. X1. X2. X3
k(fb). K(ib). K(Rb)
 Q/a = T1-T2  616.5  850- t2
X1 0.250
. K(fb) 1.05
t2 = 703°c
Similarly t3 can be determined From the relation:-
q/a = T2-T3  616.5 = 703 – t3
X2. 0.120
K(ib). 0.15
. T3 = 209.8°c

Problem 3: A steam pipe of inner diameter 200 mm is covered with 50mm thick high insulated
material of thermal conductivity k = 0.01 W/m0C. The inner and outer surface temperatures
maintained at 5000C and 1000C respectively. Calculate the total heat loss per meter length of pipe?
Solution:-
Given R1 = 200/2 = 100mm = 0.1m
R2 = 200+100/ 2 = 150mm = 0.15m
K = 0.01W/ M°c
t1 = 500°c , t2 = 100°c
heat transfer is given by :-
q = t1-t2. = 500 – 100
ln[ R2/r1] ln[0.15/0.1]
2πk l 2π × 0.01× 1
 61.98W .

Problem 4: A spherical shaped vessel of 1.2 m diameter is 100 mm thick. Find the rate of heat
leakage, if the temperature difference between the inner and outer surface is 200oC. Thermal
conductivity of the material of sphere is 0.3 kJ/m-h °c.
Solution
Given , outside diameter Of sphere = D2 = 1.2 m
= R2 = 0.6m
Inside diameter of sphere
 1.2 – 2 × 0.100 = 1m
. R1= 0.5m
T1-T2 = 200°c, K = 0.3kJ/m°C
 heat transfer in hollow sphere is given by :-
q= 4πk(r1)(R2)(t1-t2)
R2-R1
=. 4π × 0.3× 0.5×0.6 ×200.
0.100
 2262Kj/hr

Problem 5: A cylinder having its diameter 30 cm and length 60 cm, has hemispherical ends, giving
an overall length of 90 cm. The cylinder which is maintained at a steady temperature of 60oC is
covered to a depth of 5 cm with lagging which has a coefficient of conductivity of 0.14 W/mK.
Calculate the rate of heat loss if the outer surface of lagging is at 30oC.
solution
Rate of heat loss From cylindrical portion
Q= 2πLK(t1-t2)
Ln[r2/r1]
 2×π×0.6×0.14(60-30)  55w
ln(20/15)
Rate of heat loss from spherical portion
q = 4πk(r1)(R2)(t1-t2)
(r2-r1)
 4π×0.14×0.15×0.20(60-30)
0.20-0.15
 31.667w
Hence total heat loss
55 + 31.66  86.667 W

Problem 6: The meat rolls of 25 mm diameter having k=1 W/moC are heated up with the help of
microwave heating for roasting. The centre temperature of the rolls in maintained at 100oC when
the surrounding temperature is 30oC. The heat transfer coefficient on the surface of the meat
roll is 20 W/moC. Find the microwave heating capacity required in W/m3.
Solution
Given , R = 0.0125M , k =1 w/M°C;
T(max) = 100°c , t(a) = 30°c , H= 20 W/m^2°c
Microwave heating capacity ,q(g):
The maximum temperature occurs at the centre and is given by :-
T(max) = t(a) + q(g) × r + q(g) × r^2
. 2h. 4k
100. =. 30 + Q(g) ×0.0125 + q(g) × 0.0125^2
. 2×20 4×1
q(g) = (100-30)_______
0.0003125 + 0.00003906)
. 1.991 × 10^5 w/m^3

Combined convection and radiation

Problem 1 : A surface is at 200°C and is exposed to surroundings at 60°C and convects and radiates
heat to the surroundings. The convection coefficient is 80W/m2K. The radiation factor is one. If the heat is
conducted to the surface through a solid of conductivity 12 W/mK, determine the temperature gradient at the
surface in the solid.
Solution
 Heat convected + heat radiated = heat conducted
considering,1m^2 ,
h(T1 – T2) + sigma(T1^2 – T2^4) = – kdT/dx
Therefore, 80(200 – 60) + 5.67 {[(200 + 273)/100]^4 – [(60 + 273)/100]^4} = – 12 dT/dx
Therefore dT/dx = – (11200 + 2140.9)/12
 – 1111.7°C/m

Problem 2: Heat is conducted through a material with a temperature gradient of – 9000 °C/m. The
conductivity of the material is 25W/mK. If this heat is convected to surroundings at 30°C with
a convection coefficient of 345W/m2K, determine the surface temperature. If the heat is radiated to
the surroundings at 30°C determine the surface temperature.
Solution
: In this case only convection and conduction are involved.
– kAdT/dx = hA(T1 – T2). Considering unit area,
– 25 × 1 × (– 9000) = 345 × 1 (T1 – 30)
Therefore, T1 = 682.17°C
in this case conduction and radiation are involved.
Heat conducted = Heat radiated
– 25 × 1 × (– 9000) = 5.67 [(T1/100)^4 – (303/100)^4]
Therefore, T1 = 1412.14K = 1139°C

Dimensionless numbers
 Reynolds number
 Prandtl number
.Grashof Number
 NussElt number

Problem 1:- Calculate Reynolds number, if a fluid having viscosity of 0.4 Ns/m2 and relative density
of 900 Kg/m3 through a pipe of 20 mm with a velocity of 2.5 m.
Solution
Given that,
viscosity of the fluid (mu)
mu = 0.4 ns
. M^2
Density of fluid (Rho)
Rho = 900 Kg/m^2
diameter of the fluid
l = 20 × 10^-3 M.
So,
 900×2.5× 20× 10^-3
. 0.4
 112.5
From the above answer, we observe that the Reynolds number value is less than 2000. Therefore,
the flow of liquid is laminar.

Problem 2: Determine the flow of fluid having a relative density of 100 kg/m3, the viscosity of 0.5
Ns/m2 with a velocity of 5 m/s through a pipe of 0.2 m.
Solution
Given:
Velocity of fluid, V=5 m/s
Diameter of pipe, D= 0.2 m
Relative density of fluid, p=100 kg/m^3
Viscosity of fluid=0.5 Ns/m^2
The formula of Reynolds number is given as:
Re = pVD/u
Substitute all the values in the formula to calculate the Reynolds number.
Re (100 kg/m3)(5 m/s)(0.2 m)/(0.5 Ns/m2)
 200
Since, the Reynolds number is less than 2000, the flow of liquid is laminar.

Problem 3:Calculate the Reynolds number, Re, for oil flow in a circular pipe. The diameter of the
pipe is 60 mm, the density of the oil is 910 kg/m3, the volumetric oil flow rate is 60 L/min, and
the dynamic viscosity of the oil is 50 m Pa s.
Given:
Diameter of the pipe, D = 60 mm = 0.06 m.
Density of the oil, p = 910 kg/m3
Volume flow rate, Q = 60 L/min = 0.01 m3/s.
Dynamic viscosity, u=50 m Pa s =0.05 Pa s
The area of the pipe is given as:
A = π(D/2)2
= π(0.06/2)2
= 0.0283 m2
The formula for volume flow rate is given as:
Q = Av
0.01 = 0.0283 × V
V = 0.353 m/s
So,
Re = pVD/u
Re = (910 kg/m3)(0.353 m/s)(0.06 m)/(0.05 Pa s)
 386
Thus, the Reynolds number of the flow is 386.

Problem 4: The Reynold’s number for the flow of a fluid in a horizontal circular tube of constant
diameter is 1200. If the diameter of the tube and the kinematic viscosity of the fluid are doubled and
that discharged at the pipe exit is unchanged, then the new Reynold’s number for the flow in the tube
will be ?
Given:
Initially Reynold’s number, Re1 = 1200, D1 = D, D2 = 2 × D, (kinematic viscosity)2 = 2 × (kinematic
viscosity)1 v2 = 2×v1
⇒
Also, discharge at the pipe exit is unchanged.
Q1=q2
So, A1×v1 = A2×v2
D1^2 × v1 = d2^2×v2
putting the value of D1 and D2 in the above equation, we get
V2 = v1/4
Re = Vd/v
re1/re2 = d1×v1 ×. V2 = D×v1 ×. 2×v1
. V1. D2×v2. V1. 2×d×v1
. 4
Re(2) = Re(1)/4  300
So, New Reynold’s number for the flow in the tube will be 300.

Problem 5: In a circular tube of diameter 100 mm and length 13 m with laminar flow, the friction
factor is estimated to be 0.05. Calculate the Reynolds number?
Solution
Laminar flow Friction factor ⇒ F = 64/re
Where Re is the Reynolds number.
For laminar flow friction factor depends only upon the Reynolds number of the flow .
Calculation:-
Given, f = 0.05
0.05 = 64  Re = 64  1280
. Re. 0.05
.

Prandtl number
In the case of boundary layer flow, the Prandtl number is given by,
Pr =

Problem 1: The water is flowing over the heated plate. The Prandtl number of water is 6. Find the
relation between velocity boundary layer thickness and thermal boundary layer thickness.
Solution
Given: Pr = 6
The Prandtl number is given by,
Therefore the velocity boundary layer
thickness is 1.817 times the thermal
boundary layer thickness.

Grashof number
Problem 1: Check the flow is laminar or turbulent. The properties of the fluid are given below,
Characteristics length of body, L=6 m.
Surface temperature = 650 K
Fluid temperature= 450 K
Kinematic viscosity= 0.2 * 10^-3
Solution
ẞ = 2 = _. 2_____
Ts + t∞ 650 + 450
ẞ =. 1.81 × 10^-3
The Grashof number is given by,
gr = GL^3ẞ(ts - t∞)
V2
gr = 9.81 × 63.
× (1.81 × 10^-3) (650-450)
(0.2 × 10^-3)^2
Gr=. 1.917 × 10^10

Problem 2: The sphere at 600 K with a characteristic length of 5.5 m is cooled by the gas at 300 K
by free convection. If the cooling gas is considered as ideal and if the kinematic viscosity is 0.25 ×
10^-3, find the type of flow over the surface.
Solution
l = 5.5 m, ts = 600k, t∞ = 300k, v= 0.25 × 10^-3M^2/s
ß = 2. =. 2.____
ts + T∞. 600 + 300
ß = 2.22 × 10^-3
For free convection, the Grashof number is given by,
Gr = Gl^3ẞ(ts – t∞)
V^2
Gr = 9.81 × 5.5^3 × ( 2.22× 10^-3) (600 – 300)
(0.25 × 10^-3)^2
gr = 1.73 × 10^10

Nusselt number
Problem 1: at 1 atmospheric Pressure And 27°c blows accross a 12mm Diameter sphere At a small
heater inside The sphere maintains The surface temperature at 77°c. With k = 0.026 w/mk And with
nu = 31.4 , the heat Loss by the sphere would be ?
Solution
Given , nu = 31.4 ,K =0.026 w/mk , D = 12Mm  r= 6 × 10^-3 M , Ts = 27°c , t = 77°c
nu = Hlc/k
H = Nu × k/ Lc
h = 31.4 × 0.026. . 68.03 w/m^2k
. 0.012
surface area of sphere a = 4×π ×r^2 = 4×3.14 × (6×10^-3)^2
q(loss) = 68.03× 4× 3.14× (6×10^-3)^2 × (77-27)
 1.54 j/s

Problem 2:For a hydrodynamically and thermally fully developed laminar flow through a circular
pipe of constant cross-section, the Nusselt number at constant wall heat flux (Nuq) and that at
constant wall temperature (NuT) are related as
Solution:-
Part 1) For constant surface heat flux (qs = constant)
hydronamically and Thermally Fully developed laminar flow Through a circular pipe Of constant
cross section
nuq = 4.36  eq (1)

Part 2) For constant wall temperature (tw= constant)
for this case
nut = 3.66  eq (2)
Comparing (1) and (2)
Nuq > nut

Introduction of condensation & Boiling Heat Transfer
Introduction
Boiling:
 Liquid – to – vapors phase change
 Like Evaporation
Condensation:
 Gas phase into liquid phase
 Water vapors to liquid water when in contact with a liquid or solid surface
 Reverse of evaporation

BOILING
• Boiling occurs at the solid–liquid interface when a liquid is brought into contact with a surface
maintained at a temperature sufficiently above the saturation temperature of the liquid.
• Evaporation occurs at the liquid–vapor interface when the vapor pressure is less than the
saturation pressure of the liquid at a given temperature.
Types-
Nucleate
Critical heat flux
Transition
Film

Classification
• Pool Boiling:-
Motion of bubbles under the influence of buoyancy.
Boiling is called pool boiling in the absence of bulk fluid flow.
• Flow Boiling:-
 Continuous flow of bulk fluid.
 Boiling is called flow boiling in
the presence of bulk fluid flow.

Classification
Subcooled Boiling:-
 When the temperature of the main body of the liquid is below the saturation temperature.
Saturated Boiling:-
 When the temperature of the liquid is equal to the saturation temperature

POOL BOILING
In pool boiling, the fluid is not forced to flow by a mover such as a pump. Any
motion of the fluid is due to natural convection currents and the motion of the
bubbles under the influence of buoyancy.
Boiling Regimes and the Boiling Curve
Boiling takes different forms, depending on the
Dt excess = Ts - Tsat

Flow Boiling
In flow boiling, the fluid is forced to move by an external source such as a pump as it
undergoes a phase-change process.
External flow boiling, over a plate or cylinder is similar to pool boiling, but the added
motion increased both the nucleate boiling heat flux and the maximum heat flux considerably.
Internal flow boiling, commonly referred to as two-phase flow, is much more complicated
in natural because there is no free surface for the vapour to escape, and thus both the liquid and
the vapour are forced to flow together.

Film Boiling (beyond Point D)
• Beyond point D the heater surface is completely covered
by a continuous stable vapor film.
• Point D, where the heat flux reaches a minimum is called
the Leidenfrost point .
• The presence of a vapor film between the heater surface
and the liquid is responsible for the low heat transfer
rates in the film boiling region.
• The heat transfer rate increases with increasing excess
temperature due to radiation to the liquid.

Condensation
• Condensation of vapor is the reverse phenomenon of the evaporation of liquid.
• Condensation (Vapor to Liquid): Heat energy is released Boiling (Liquid to Vapor): Heat
energy is absorbed
• When a saturated vapor is brought in contact with a surface at a lower temperature, heat is
received by the surface from the vapor and condensation occurs. The heat exchange is
equivalent to the latent heat.

Condensation
When does condensation occurs on a surface ?
• Consider a vertical flat plate which is exposed to a condensable vapour.
• If the temperature of the plate is below the saturation temperature of the vapour, condensate
will form on the surface and flows down the plate due to gravity.
• It is to be noted that a liquid at its boiling point is a saturated liquid and the vapour in equilibrium
with the saturated liquid is saturated vapour.
• A liquid or vapour above the saturation temperature is called superheated.
• If the non-condensable gases will present in the vapour the rate of condensation of the vapour
will reduce significantly.

Types of condensation:
• Film-wise Condensation
• Drop wise Condensation

Film wise vs Drop-wise condensation
Film wise condensation Drop-wise condensation
The surface over which the steam condenses is wet-
able and hence, as the steam condenses, a film of
condensate is formed.
The surface over which condensation takes place is
non wet-able. In this mode, when steam condenses,
the droplets are formed.
Low heat transfer rates as the film of condensate
impedes the heat transfer.
High heat transfer rates are achieved and hence, many
times, chemicals are used to ensure that condensation
takes place drop wise.
The thickness of the film formed depends on many
parameters including orientation of the surface,
viscosity, rate of condensation etc.
When the drops become bigger, they simply fall under
the gravity.

WHAT IS RADIATION
• Heat is transferred between two bodies even though they are
not in direct physical contact and vacuum exists between these
bodies. Such a heat transfer is termed as radiation.
• It involves the transfer of heat energy via electromagnetic
waves. The electromagnetic radiation that can be detected as
heat is termed as thermal radiation.
• Thermal radiation emitted by a body or substance is directly
proportional to its temperature.
• Heat transfer by radiation occurs through open space and is
most significant when temperature differences are substantial.
• For example: In food processing, heat is transferred by:
• Radiation: From the oven walls to the product surface.

Absorptivity, Transmissivity and Emissivity
• Absorptivity(α): refers to the ability of a material to absorb
electromagnetic radiations.
• It is defined as the ratio of the absorbed radiant energy to
the incident radiant energy.
• Materials with high absorptivity are efficient at absorbing
radiation across a range of wavelengths.
• Transmissivity(T): it is the measure of how much
electromagnetic radiation can pass through a material.
• It is defined as the ratio of transmitted radiant energy to
incident radiant energy.
• Transparent materials have high transmissivity whereas
opaque materials have low transmissivity.

• Emissivity(ε): describes the
ability of a material to emit
electromagnetic radiation.
• It is defined as the ratio of the
radiant emittance of a
particular surface to that of a
particular surface to that of a
blackbody at the same
temperature.
• Materials with high emissivity
radiate thermal energy more
effectively than those with low
emissivity
TYPE ABSORPTIVITY TRANSMISSIVITY EMISSIVITY
Black Body 1 0 1
Transparent
body
0 1 LESS THAN 1
Opaque
body
LESS THAN 1 0 BETWEEN 0-1

BLACK BODY
• A perfect black body is one that absorbs all the thermal
radiation, irrespective of wavelength, received by it.
• It does not reflect or transmit incident thermal radiation;
therefore, absorptivity of such a body is 100%.
• At a given temperature and wavelength, a black body emits
more energy as compared to any other body.
• A black body absorbs all incident radiation irrespective of their
wavelength and direction.
• The radiation emitted by a black body depends upon
wavelength and temperature, but it is independent of
direction.

Net radiation heat exchange between black bodies
• Total emissive power for black body 1 per unit area:
• = W/m2
K4
Total emissive power for black body 1:
ε1= A1 T1
4
For black body 2:
ε2 A2T2
4
• Total radiation leaving from surface 1 and falling on surface 2:
Q12 = A1 T1
4
F12 α
From surface 2-1:
Q21 = A2 T2
4
F21 α
As we are dealing with black bodies so whatever radiations coming on surface will absorb
completely:
(α=1)
[ Q12-Q21 ]: NET RADIATION HEAT EXCHANGE
Q net = A1 T1
4
F12 - A2 T2
4
F21
Q net = A1 F12 [T1
4
- T2
4
]

GRAY BODY
• A Gray body is defined as a surface whose emissivity is constant at all temperatures and
throughout the entire range of wavelength.
• For a Gray body emissivity and absorptivity are independent of wavelength.
• Therefore, a Gray body, like black body, is an ideal body and values of its absorptivity and
emissivity are less than unity.
NET RADIATION HEAT EXCHANGE BETWEEN GREY
BODIES-
• Radiation emitted by S1 in the form of electromagnetic waves.
• Radiation emitted by grey surface having emissivity ε1.
• = ε
= ε1 = ε × εb (εb = A1 T1
4
)
= ε1 = ε × A1 T1
4
Fraction of radiation emitted by surface 1 and absorbed factor α
E1 = ε1 A1 T1
4
F12 α2
Similarly for surface 2, fraction of radiation is
E2 = ε2 A2T2
4
F21 α1

• Net radiation heat exchange between two surfaces:
• Q net = E1 - E2
ε1 A1 T1
4
F12 α2 - ε2 A2T2
4
F21 α1
• Q net = A1F12 [ ε1T1
4
α2 = ε2T2
4
α1 ]
• According to kirchoff’s law of thermal equilibrium-
• α = ε
• Considering α1 = ε1
And α2 = ε2
Q net = A1F12 [ ε1ε2T1
4
= ε2ε1T2
4
]
Q net = A1F12ε1ε2 { T1
4
- T2
4
}

BASIC LAWS OF THERMAL RADIATION
• STEFAN-BOLTZMAN LAW:
• It states that total energy emitted by black body per unit area and per unit time is directly
proportional to the fourth power of its absolute temperature and is expressed as:
b=4
• Eb- Energy emitted from a black body per unit area per unit time, W/m2
• b- Stefan- Boltzman constant
• =5.6710-8
W/(m2
-K4
)
• T= Absolute temperature of the emitting surface, K.
• Total emissive power ‘E’ of a body is defined as total energy radiated by the body in all
directions over entire range of wavelength per unit surface area per unit time.
• The total emissive power ‘E’ of a black body can be determined easily by using Stefan-
Boltzman law if absolute temperature of the black body is known.

PLANCK’S LAW:
• In 1900, German physicist Max Planck proposed Planck’s law.
• Planck’s constant h=6.625×10-34
Joule-Sec.
• It describes the spectral density of electromagnetic radiation emitted by a black
body in thermal equilibrium at a given temperature T, when there is no net flow
of matter or energy between the body and its environment.
• The higher the temperature of a body, the more radiation it emits at every
wavelength.
• It relates the energy of electromagnetic radiation to its frequency.
• In the context of food processing, Planck’s law can be applied to various
aspects, such as thermal processing, heating and cooling, which are essential in
preserving food quality and safety.

Planck’s law of distribution
equation:
• Where
• C1=2c2
h=37.4041017
J-m2
/K
• C2=ch/k=1.438710⁻2
m-K
• An increase in temperature results in
decrease in value of wavelength for
which emissive power is maximum.
• In this figure, area under each curve
represents energy emitted by the black
body at a particular temperature for the
range of wavelength considered. With
increase in temperature area under the
curve increases as energy emitted
increases.

Kirchoff’s Law:
• According to kirchoff’s law ratio of total emissive power to absorptivity is constant for all
bodies which are in thermal equilibrium with their surroundings.
• For three bodies which are in thermal equilibrium with each other:
= = equation(1)
• Assuming third body to be black body, equation(1) can be written as:
= = []→ equation(2)
= = → equation(3)
However, according to definition of emissivity,
= =→ equation(4)
• Comparing equation(3) and equation(4), we can write:
= , = → equation(5)
• Therefore, kirchoff’s law states that for a body in thermal equilibrium with its
surroundings, its absorptivity is equal to its emissivity.

WEIN’S DISPLACEMENT LAW:
• It states that product of absolute temperature and wavelength at which emissive
power is maximum, is constant.
• It has been established that monochromatic emissive power of a black body
depends upon its temperature and wavelength of emitted radiations.
• For a given temperature, emissive power initially increases with increase in
wavelength, attains a maximum value corresponding to particular wavelength of
emitted radiations.
• With increase in temperature, maximum emissive power occurs at smaller
wavelengths.
• Wein’s displacement law gives the value of wavelength at which emissive power of a
body is maximum for a given temperature and is expressed as:
• = wavelength at which monochromatic power is maximum corresponding to temperature “T”.

SHAPE FACTOR
• Shape factor, also known as form factor is a dimensionless quantity that
characterizes the shape and size of a heat source, and it is used to determine the
heat radiation emitted from the source.
• Heat radiation is the transfer of heat energy from one object to another through
electromagnetic waves, which are mainly in the form of infrared radiation. The
amount of heat radiation emitted from a surface is determined by its temperature,
emissivity, and view factor.
• The view factor is a dimensionless quantity that accounts for the geometric
configuration of surfaces and their orientations, while emissivity is a measure of a
surface’s ability to emit radiant energy.
• The shape factor is related to the view factor, and it is used to determine the heat
radiation from non-uniform heat sources.
• The shape factor, denoted as σ, is defined as the ratio of the heat transfer rate
through a surface to the heat transfer rate through a reference surface of unit heat
transfer coefficient and the same surface area.
• In other words, the shape factor is a measure of how efficiently a heat source
transfers heat to its surroundings.

DETERMINATION OF SHAPE FACTOR
• There are three rules for the determination of the shape factor:
1.Summation
2.Reciprocity
3.Superposition
The shape factor is purely a function of geometrical parameters.
When two bodies radiating energy with each other only, the shape factor relation is expressed as :
A1 F12 = A2 F21
F12= Heat emitted from Body 1 reached Body 2.
The shape factor of convex surface or flat surface with the other surface enclosing the first is always
unity. This is because all the radiation coming out from the convex surface is intercepted by the
enclosing surface but not vice versa. (F11= 0)
A concave surface has a shape factor with itself because the radiation energy coming out from one part
of the surface is intercepted by another part of the same surface. The shape factor of a surface with
respect to itself is denoted by F11.
If a surface of area A1 is completely enclosed by a second surface of area A2 and if A1 does not see
itself(F11=0)
Then F12 = 1.

Shape factor for spherical surface
F21 = 1 (inner to outer)
F11 + F21 = 1
Reciprocity theorem:
A1F12 = A2F21
F12 = × F21
F11 + × F21 = 1
× F21 = 1 - F11
F11 = 1 - × F21 (F21 = 1)
F11 = 1-

NUMERICALS
QUESTION 1. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is
290 K. the plate attains a temperature of 295 K when one of its face receives radiant energy from a heat
source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m2
-
deg, workout the reflectivity of the plates.
• Solution: Heat lost by convection from both sides of the plate
= 2hA t
The factors 2 accounts for two sides of the plate
=2 87.5 (295 – 290) = 1.1 W
For most of solids, the transmissivity is zero.
Energy lost be reflection = 2.0 – 1.1 = 0.9 W
Reflectivity = = 0.45

QUESTION 2. A black body of total area 0.045 m2
is completely enclosed in a space bounded by 5 cm
thick walls. The walls have a surface area 0.5 m2
and thermal conductivity 1.07 W/m-deg. If the inner
surface of the enveloping wall is to be maintained at 215°C and the outer wall surface is at 30°C, calculate
the temperature of the black body. Neglect the difference between inner and outer surfaces areas of
enveloping material.
Solution: Net heat radiated by the black body to the enclosing wall,
Qr = b A(Tb
4
– Tw
4
)
= 5.67-8
Where is the temperature of the black body is kelvin
Heat conducted through the wall,
Qc = = = 1979.5 W
Under steady state conditions, the heat conducted through the wall must equal the
net radiation loss from the black body. Thus
= 5.67
Tb4 = + 4884
= 8349.47
temperature of the black body, Tb = 955.9 K

QUESTION 3. A furnace radiation at 2000K. Treating it as a black body radiation, calculate the
(i) Monochromatic radiant flux density at 1 μm wavelength.
(ii) Wavelength at which emission is maximum and the corresponding radiant flux density
(iii) Total emissive power
Solution: (a) From Planck’s law of distribution,
(E
=
= = 2.81
Wavelength
(b) From Wein’s displacement law:
= 1.449
Maximum radiant flux density,
1.285×10⁻ T5 = 1.285×10⁻ ×(2000)5
𝟓 𝟓
=4.11×1011 / 2
𝑊 𝑚
(c) From Stefan- Boltzman law,
E= 4
𝝈𝒃 𝑻
= 5.67×10⁻⁸×(2000)⁴=907200 / 2
𝑊 𝑚

• Maximum radiant flux density,
1.28510 T5 = 1.2855
(c) From Stefan- Boltzman law,
E= 4
= 5.672

QUESTION 4. A gray surface has an emissivity at a temperature of 550 K source. If the surface is opaque, calculate its reflectivity for a
black body radiation coming from a 550 K source.
(b) A small 25 mm square hole is made in the thin-walled door of a furnace whose inside walls are at 920 K. if the emissivity of the walls
is 0.72, calculate the rate at which radiant energy escapes from the furnace through the hole to the room.
Solution: =1
Here,
(i)
(ii)
This is in accordance with Kirchoff’s law which states that absorptivity equals emissivity under the same temperature
conditions.
Reflectivity
Thus the surface reflects 65 percent of incident energy coming from a source at 550K.
(b) The small hole acts as a black body and accordingly the rate at which radiant energy leaves the hole is
E=4
=
= 25.38 watts

Heat_mass_Transfer by naveen choudhary.pptx

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