Notes of Heat and mass transfer for Gate

1. HEAT TRANSFER
GATE & IES Notes
Knowledge Shared is Knowledge Doubled
SouMyth

2. 1
Contents
Introduction............................................................................................................................................................................................2
Modes of Heat transfer .................................................................................................................................................................2
Fourier’s law of conduction .......................................................................................................................................................3
Generalized equation of conduction in different form in Cartesian form ..........................................................4
Steady state 1-Dimensional conduction without heat generation.........................................................................5
Conduction in cylinders...................................................................................................................................................................7
Conduction in cylinders without heat generation .........................................................................................................7
Alternate way to calculate Heat transfer in cylinders.................................................................................................8
Conduction in Spheres......................................................................................................................................................................9
Spheres without Heat generation...........................................................................................................................................9
Conduction through composite cylinders ..........................................................................................................................9
Critical Thickness of Insulation ............................................................................................................................................10
Steady state 1-D conduction with heat transfer..................................................................................................................11
Plane slabs.........................................................................................................................................................................................11
Cylinders ...........................................................................................................................................................................................13
Spheres...............................................................................................................................................................................................14
Heat transfer from Extended surfaces (Fins) ......................................................................................................................15
Generalized equation in differential form for fins .......................................................................................................15
Infinitely Long fin.........................................................................................................................................................................15
When the tip of the fin is insulated .....................................................................................................................................16
Fin with convection at free end.............................................................................................................................................16
Fin with specified temperature at ends.............................................................................................................................17
Fin performance............................................................................................................................................................................17
Effectiveness of Fin .....................................................................................................................................................................18
Unsteady State (Transient) Conduction.................................................................................................................................19
Lumped system Analysis .........................................................................................................................................................19
Heat exchangers ................................................................................................................................................................................21
LMTD Method................................................................................................................................................................................ 22
LMTD for Counter flow Heat exchanger .......................................................................................................................... 22
Effectiveness NTU Method ..................................................................................................................................................... 23
Radiation............................................................................................................................................................................................... 26
Shape factor .................................................................................................................................................................................... 27
Radiosity (J).................................................................................................................................................................................... 28
Convection ........................................................................................................................................................................................... 30
Velocity boundary layer ........................................................................................................................................................... 30
Convection on a flat plate ........................................................................................................................................................ 32
External Forced convection.................................................................................................................................................... 34
Parallel flow over flat plate..................................................................................................................................................... 35
Internal Forced Convection .................................................................................................................................................... 37
Free Convection............................................................................................................................................................................41
Condensation ................................................................................................................................................................................. 42

3. 2
Introduction
Thermodynamics deals with the amount of heat transfer and makes no reference as how long the
process occur. The rate of heat transfer and the temperature variation with time is dealt in heat
transfer. Heat transfer is an extension of thermodynamics. Thermodynamics doesn’t include time as
variable, whereas heat transfer includes time as a variable.
Applications of Heat transfer
1. Civil Engineering → Buildings, Railway tracks.
2. Electrical Engineering → Transformers, Motors
3. Metallurgical Engineering → Furnaces, heat transfer.
4. Mechanical Engineering → Boilers, I.C Engines.
Modes of Heat transfer
Conduction
It is a microscopic phenomenon. The mechanism of heat transfer due to temperature difference in a
stationary medium is known as conduction. Conduction can occur in solids, liquids and gases.
In solids conduction is due to lattice vibrations and free flow of electrons, whereas in liquids and
gases conduction is die to collision of molecules.
The governing equation for conduction is given by Fourier’s law. According to Fourier’s law of
conduction, heat transfer is directly proportional to area (perpendicular to heat flow) and
temperature gradient.
Q ∝ A · (
dT
dx
)
𝐐 = −𝐤. 𝐀 (
𝐝𝐓
𝐝𝐱
)
The negative sign is taken because to keep the heat transfer
positive in the direction of flow of heat.
Convection
It’s a microscopic phenomenon. In this heat transfer, fluid particles
themselves move and carry heat from higher temperature to lower
temperature.
Fluid flow + Heat transfer = Convection
Convection is of two types.
1. Natural or free convection
2. Forced convection
In natural convection, heat transfer occurs due to density
differences.
In forced convection, the heat transfer occurs with the help of
external agent, fan or pump.
Example: Forced convection: blowing of hot food to cool down
The governing equation for convection is given by Newton’s
law of cooling. According to this law,
𝐐𝐜𝐨𝐧𝐯𝐞𝐜𝐭𝐢𝐨𝐧 = 𝐡 ∙ 𝐀 ∙ 𝚫𝐓
A→Surface Area, h→Heat transfer coefficient
hforced conv. > hfree conv. ⟹ Qforced conv. > Qfree conv

4. 3
Radiation
Every surface above ‘0’ K emits radiation in the form of electromagnetic waves and they travel with
speed of light. Radiation emitted in the range of 0.1 μm to 100 μm is known as thermal radiation,
because radiation in this range, when absorbed gets converted into heat.
The governing equation for radiation is given by Stefan- Boltzmann law.
𝐄 = 𝛔𝐓𝟒 (σ = 5.67 ∗ 10−8, T → Kelvin)
As σ is very small. Radiative heat transfer is less significant or negligible at lower temperatures.
For conduction and convection, material medium is required, for radiation no material medium is
required i.e., heat transfer can take place in vacuum.
Fourier’s law of conduction
Assumptions
→Steady heat transfer. →One-dimensional conduction. →Material is homogenous.
→The bounding surfaces are isothermal.
Fourier’s Law is obtained from experiments conducted by Biot.
According to Fourier’s Law,
𝐐
̇ 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐢𝐨𝐧 = −𝐤.𝐀 (
𝐝𝐓
𝐝𝐱
) (k→Thermal conductivity)
Thermal conductivity (k)
It’s numerically equal to heat transfer through an area of 1 m2
of a slab of 1 m thickness when 2 faces
are maintained at a temperature difference of 1 °C or 1 K.
The unit of k is W/m-K.
Thermal conductivity represents the ability of medium to transfer or conduct heat.
Greater the thermal conductivity, greater is the ability of medium to transfer heat.
→ Specific heat represents the ability of medium to absorb/store heat whereas
thermal conductivity represents the ability of medium to transfer heat.
Iron Water
k (W/m-K) 80.6 0.608
c (KJ/kg-K) 0.45 4.18
From this table, it can be observed that though, iron is good conductor of heat, it
has less ability to store heat. Though water is a good medium to absorb heat, but
it’s a bad medium to transfer heat.
Thermal conductivity of solids
In solids conduction is due to lattice vibrations and free flow of electrons
k = klattice + kelectrons
In case of metallic solids, as the availability of free electrons is more, therefore thermal conductivity
of metallic solids is high. With increase in temperature, lattice vibrations increase and this obstructs
free flow of electrons, and hence the thermal conductivity of metallic solids decrease with increase
in temperature.
In case of non-metallic solids, as there is no free flow of electrons, with increase in temperature,
lattice vibrations increase and thereby thermal conductivity of non-metallic solids increase with
increase in temperature.
𝐤𝐭 = 𝐤𝟎(𝟏 + 𝛃𝐭)
KT → Thermal conductivity at any temperature T. k0 → Thermal conductivity at 0 °C.
β → coefficient which depends on material.
Generally, for solids, β is negative and for non-metallic solids β is positive.
Material Diamond Copper Silver Gold Aluminum Brick Glass wool Air
K (W/m-K) 2300 401 429 317 237 0.72 0.043 0.026
A good electrical conductor is always a good thermal conductor, but a good thermal conductor need
not be a good electrical conductor. Thermal conductivity of diamond is highest of known materials,
this is because diamond has highly organized crystalline structure.

5. 4
Thermal conductivity of liquids
Its observed from experiments that for liquids, 𝐤 ∝ 𝛒𝟒/𝟑
,therefore with rise in temperature, density
decreases and hence thermal conductivity of a liquid decreases with increase in temperature.
Thermal conductivity of gases
With increase in temperature, kinetic energy increases and because of this, collisions will also
increase. Therefore, in case of gases with increase in temperature, thermal conductivity increases.
Gases with lower molecular weight have higher velocities and hence thermal conductivity of the gas
with low molecular weight is high.
𝐤𝐬𝐨𝐥𝐢𝐝𝐬 > 𝐤𝐥𝐢𝐪𝐮𝐢𝐝𝐬 > 𝐤𝐠𝐚𝐬
Thermal diffusivity
It’s the ratio of ability of the medium to transfer heat in comparison to its ability to store heat.
𝛂 =
𝐤
𝛒 ∙ 𝐜𝐩
its unit is m2
/s.
Kinematic viscosity and thermal diffusivity have the same units.
Greater the thermal diffusivity, greater is its ability to transfer heat in comparison to storage,
therefore, the larger the thermal diffusivity, the faster the propagation of heat into the medium.
Generalizedequation of conduction in different
form in Cartesian form
If Material is Homogenous
(Heat generating rate)q̇ gen =
Q̇ g
volume
⟹ Q̇ g = q̇ gen ∙ dx ∙ dy ∙ dz
𝐄̇𝐢𝐧 + 𝐄̇𝐠𝐞𝐧 − 𝐄̇𝐨𝐮𝐭 = 𝐄̇𝐬𝐭𝐨𝐫𝐞𝐝
(𝐐̇ 𝐱 + 𝐐̇ 𝐲 + 𝐐̇ 𝐳) + (𝐪̇𝐠 ∙ 𝐝𝐱 ∙ 𝐝𝐲 ∙ 𝐝𝐳) − [𝐐̇ 𝐱+𝐝𝐱 + 𝐐̇ 𝐲+𝐝𝐲 + 𝐐̇ 𝐳+𝐝𝐳] = 𝐄̇𝐬𝐭𝐨𝐫𝐞𝐝
Q̇ x+dx = Q̇ x +
∂Q̇ x
∂x
∙ dx ; Q̇ y+dy = Q̇ y +
∂Q̇ y
∂y
∙ dy ; Q̇ z+dz = Q̇ z +
∂Q̇ z
∂z
∙ dz
Q̇ x = −kx ∙ dy ∙ dz ∙
∂T
∂x
; Q̇ y = −ky ∙ dx ∙ dz ∙
∂T
∂y
; Q̇ z = −kz ∙ dx ∙ dy ∙
∂T
∂z
Ė stored = m ∙ cp ∙
∂T
∂t
= ρ ∙ dx ∙ dy ∙ dz ∙ cp ∙
∂T
∂t
⟹ (Q̇ x + Q̇ y + Q̇ z) + (q̇ gen ∙ dx ∙ dy ∙ dz) − [(Q̇ x +
∂Q̇ x
∂x
∙ dx) + (Q̇ y +
∂Q̇ y
∂y
∙ dy) + (Q̇ z +
∂Q̇ z
∂z
∙ dz)] = Ė stored
⟹ q̇ gen ∙ dx ∙ dy ∙ dz − [
∂Q̇ x
∂x
∙ dx +
∂Q̇ y
∂y
∙ dy +
∂Q̇ z
∂z
∙ dz] = Ė stored
⟹ q̇ gen ∙ dx ∙ dy ∙ dz −
∂
∂x
(−kx ∙ dy ∙ dz ∙
∂T
∂x
) ∙ dx +
∂
∂y
(−ky ∙ dx ∙ dz ∙
∂T
∂y
) ∙ dy +
∂
∂z
(−kz ∙ dx ∙ dy ∙
∂T
∂z
) ∙ dz
= ρ ∙ dx ∙ dy ∙ dz ∙ cp ∙
∂T
∂t
𝐪̇𝐠 +
𝛛
𝛛𝐱
(𝐤𝐱 ∙
𝛛𝐓
𝛛𝐱
) +
𝛛
𝛛𝐲
(𝐤𝐲 ∙
𝛛𝐓
𝛛𝐲
) +
𝛛
𝛛𝐳
(𝐤𝐳 ∙
𝛛𝐓
𝛛𝐳
) = 𝛒 ∙ 𝐜𝐩 ∙
𝛛𝐓
𝛛𝐭
Let us assume the material to be isotropic
kx = ky = kz = k
q̇ g +
∂
∂x
(k ∙
∂T
∂x
) +
∂
∂y
(k ∙
∂T
∂y
) +
∂
∂z
(k ∙
∂T
∂z
) = ρ ∙ cp ∙
∂T
∂t
𝐪̇𝐠 + 𝐤 (
𝛛𝟐
𝐓
𝛛𝐱𝟐
+
𝛛𝟐
𝐓
𝛛𝐲𝟐
+
𝛛𝟐
𝐓
𝛛𝐳𝟐
) = 𝛒 ∙ 𝐜𝐩 ∙
𝛛𝐓
𝛛𝐭

6. 5
Let us assume that thermal condition of material to be independent of temperature (k ≠ f(T))
∂2
T
∂x2
+
∂2
T
∂y2
+
∂2
T
∂z2
+
q̇ g
k
=
ρ ∙ cp
k
∙
∂T
∂t
𝛛𝟐𝐓
𝛛𝐱𝟐
+
𝛛𝟐𝐓
𝛛𝐲𝟐
+
𝛛𝟐𝐓
𝛛𝐳𝟐
+
𝐪̇ 𝐠
𝐤
=
𝟏
𝛂
∙
𝛛𝐓
𝛛𝐭
Case 1: - Steady state with heat generation (Poisson’s equation)
∂2
T
∂x2
+
∂2
T
∂y2
+
∂2
T
∂z2
+
q̇ g
k
= 0
Case 2: - Without heat generation (diffusion equation)
∂2
T
∂x2
+
∂2
T
∂y2
+
∂2
T
∂z2
=
ρ ∙ cp
k
∙
∂T
∂t
Case 3: - Steady state without heat generation (Laplace equation)
∂2
T
∂x2
+
∂2
T
∂y2
+
∂2
T
∂z2
= 0
Steady state 1-Dimensional conductionwithout heat generation
Assumptions
→Steady conduction → 1-D heat transfer → No heat generation →Homogenous material
→Constant thermal conduction
∂2
T
∂𝑥2
+
∂2
T
∂y2
+
∂2
T
∂z2
+
q̇ g
k
=
ρ ∙ cp
k
∙
∂T
∂t
𝛛𝟐
𝐓
𝛛𝐱𝟐
= 𝟎
T = C1 ∙ 𝑥 + C2
At x = 0; T=T1 → C2 = T1;
At x = L; T2 = C1∙L + T1 → C1 = (T2 – T1)/L
𝐓 = (
𝐓𝟐 − 𝐓𝟏
𝐋
) ∙ 𝒙 + 𝐓𝟏
¥ Temperature variation is linear in plane slab when there is no heat
generation and when thermal conductivity is constant.
Q̇ = −k ∙ A ∙
dT
dx
⟹ 𝐐̇ = −𝐤 ∙ 𝐀 ∙
𝐓𝟐 − 𝐓𝟏
𝐋
Concept of thermal resistance
Q̇ = −k ∙ A ∙
T2 − T1
L
⟹ (T1 − T2) = Q̇ ∙ (
L
k ∙ A
) ⟹ ΔT = Q̇ ∙ 𝐑𝐭𝐡𝐞𝐫𝐦𝐚𝐥
𝐑𝐭𝐡𝐞𝐫𝐦𝐚𝐥 =
𝐋
𝐤 ∙ 𝐀
Unit of Rthermal is K/m.
Plane slab with convection at Boundaries
Q̇ = ha ∙ A ∙ (Ta − T1) = k ∙ A ∙
T1 − T2
L
= hb ∙ A ∙ (T2 − Tb )
(Ta − T1) = Q̇ ∙ (
1
ha ∙ A
) (T2 − Tb) = Q̇ ∙ (
L
k ∙ A
) (T2 − Tb ) = Q̇ ∙ (
1
hb ∙ A
)
Rconv res.1 = Q̇ ∙ (
1
ha ∙ A
) Rcond res. = Q̇ ∙ (
L
k ∙ A
) Rconv res.2 = Q̇ ∙ (
1
hb ∙ A
)

7. 6
Conduction in Composite slabs
Assumptions
→ Steady state →1-D heat transfer →Constant thermal conductivity →No heat generation
→Contact between 2 surfaces is perfect and there is no temperature drop at contacting surfaces.
𝐐̇ = ha ∙ A ∙ (Ta − T1) = k ∙ A ∙
T1 − T2
L
= k ∙ A ∙
T2 − T3
L
= k ∙ A ∙
T3 − T4
L
= hb ∙ A ∙ (T2 − Tb)
(𝐓𝐚 − 𝐓𝐛) = 𝐐̇ ∙ (
𝟏
𝐡𝐚 ∙ 𝐀
+
𝐋𝟏
𝐤𝟏 ∙ 𝐀
+
𝐋𝟐
𝐤𝟐 ∙ 𝐀
+
𝐋𝟑
𝐤𝟑 ∙ 𝐀
+
𝟏
𝐡𝐛 ∙ 𝐀
)
Overall Heat transfer
𝐐̇ = 𝐔 ∙ 𝐀 ∙ (𝐓𝐚 − 𝐓𝐛 )
Thermal contact Resistance
When 2 metals surfaces are brought in contact, the contact may
not be perfect due to surface roughness and hence the gap is
filled with air or gas which offers high thermal resistance, due
to lower thermal conductivity. This is known as thermal
contact resistance, because of this resistance, the temperature
profile is not continuous and hence there is a significant
temperature drop at contacting surface.
Steady state 1-D conduction through Variable Area
Assumptions
→Steady state → 1-D Heat transfer →No heat generation
→Homogenous material →Constant thermal conductivity
Q̇ = −k ∙ A ∙
dT
d𝑥
⟹ Q̇ ∙
d𝑥
A
= −k ∙ dT ⟹ Q̇ ∙ ∫
1
A
∙ dx
L
0
= −k ∙ ∫ dT
T2
T1
tan α =
y
x
→ y = x ∙ tan α → R = R1 + y
tan α =
R2 − R1
L
= c
R = R1 + x ∙ c A = π ∙ R2
= π ∙ (R1 + x ∙ c)2
Q̇ ∙ ∫
1
π ∙ (R1 + x ∙ c)2
∙ dx
L
0
= −k ∙ ∫ dT
T2
T1
⟹ Q̇ ∙
1
π
∙
(R1 + x ∙ c)−1
−1 ∙ c
|
0
L
= −k ∙ (T2 − T1)
Q̇
π ∙ c
[
1
R1 + L ∙ (
R2 − R1
L
)
−
1
R1
] = k ∙ (T2 − T1) ⟹
Q̇
π ∙ c
[
1
R2
−
1
R1
] = k ∙ (T2 − T1)
Q̇
π
∙
L
(R2 − R1)
∙ [
R1 − R2
R1 ∙ R2
] = k ∙ (T2 − T1) ⟹ −
Q̇
π
∙
L
R1 ∙ R2
= k ∙ (T2 − T1)
𝐐
̇ =
𝐤 ∙ (𝐓𝟏 − 𝐓𝟐) ∙ 𝛑 ∙ 𝐑𝟏 ∙ 𝐑𝟐
𝐋

8. 7
Conduction in cylinders
Assumptions
→Steady state →1-D heat flow
→constant thermal conductivity →Homogenous material
Q̇ r + Q̇ gen = Q̇ r+dr = (Q̇ r +
∂Q̇ r
∂r
∙ dr) ⟹ Q̇ gen =
∂Q̇ r
∂r
∙ dr
Q̇ gen = q̇ gen ∙ V = q̇ gen ∙ (2πr ∙ dr ∙ L)
q̇ gen ∙ (2πr ∙ dr ∙ L) =
∂
∂r
(−k ∙ (2πrL) ∙
dT
dr
) ∙ dr
𝟏
𝐫
∙
𝛛
𝛛𝐫
(𝐫 ∙
𝐝𝐓
𝐝𝐫
) +
𝐪̇𝐠𝐞𝐧
𝐤
= 𝟎
Conduction in cylinders without heat
generation
Assumptions
→Steady state →1-D heat flow → No heat generation
→constant thermal conductivity →Homogenous material
1
r
∙
∂
∂r
(r ∙
dT
dr
) = 0
r ∙
dT
dr
= c1 → T = c1 · lnr + c2
At r=r1 → T=T1 & r=r2 → T=T2
T1 = c1 ln r1 + c2 & T2 = c1 ln r2 + c2 ⟹ c1 =
T1 − T2
ln
r1
r2
& c2 = T1 − (
T1 − T2
ln
r1
r2
) ∙ ln r1
T = (
T1 − T2
ln
r1
r2
) ∙ ln r + T1 − (
T1 − T2
ln
r1
r2
) ∙ ln r1
𝐓 − 𝐓𝟏
𝐓𝟐 − 𝐓𝟏
=
𝐥𝐧(𝐫/𝐫𝟏)
𝐥𝐧(𝐫𝟐/𝐫𝟏)
Q̇ = −k ∙ A ∙
∂T
∂r
Q̇ = −k ∙ (2πrL) ∙ [
1
r
∙ (
T1 − T2
ln
r1
r2
)] =
2πkL ∙ (T1 − T2)
ln
r2
r1
=
T1 − T2
[
(lnr2 r1
⁄ )
2πkL
⁄ ]
Rthermal =
(ln r2 r1
⁄ )
2πkL
⁄
Concept of logarithmic mean area
Q̇ cylinder =
T1 − T2
[
(lnr2 r1
⁄ )
2πkL
⁄ ]
⟺ Q̇ slab =
k ∙ A ∙ (T1 − T2)
L
T1 − T2
[
(lnr2 r1
⁄ )
2πkL
⁄ ]
=
k ∙ Am ∙ (T1 − T2)
L
Am =
2πr2L − 2πr1L
ln (
2πr2L
2πr1L
)
⟹ 𝐀𝐦 =
𝐀𝟐 − 𝐀𝟏
𝐥𝐧 (
𝐀𝟐
𝐀𝟏
)

9. 8
Heat transfer in composite cylinders
Rconv. a=
1
ha∙Aa
=
1
2πr1L
Rcond 1=
ln (r2
r1
⁄ )
2πk1L
Rcond 2=
ln (r3
r2
⁄ )
2πk2L
Rconv 2=
1
hb∙Ab
=
1
2πr3L
Overall Heat transfer
Q̇ = Ui∙Ai∙(Ta-Tb) = Uo∙Ao∙(Ta-Tb)
Ui∙Ai = Uo∙Ao ⟹ Ui∙(2πriL) = Uo∙(2πroL)
Ui∙ri = Uo∙ro
Alternateway to calculate Heat transferin cylinders
Q̇ = −k ∙ A ∙
dT
dr
⇒ Q̇ = −k ∙ (2πrL) ∙
dT
dr
∫
Q̇
r
∙ dr
r2
r1
= − ∫ k ∙ (2πrL)
T2
T1
∙ dT ⇒ Q̇ ∙ ln
r2
r1
= 2πkL(T2 − T1)
𝐐̇ =
𝟐𝛑𝐤𝐋(𝐓𝟐 − 𝐓𝟏)
𝐥𝐧(𝐫𝟐 𝐫𝟏
⁄ )

10. 9
Conduction in Spheres
Assumptions
→Steady state →1-D heat flow → No heat generation →constant thermal conductivity
→Homogenous material
𝟏
𝐫𝟐
∙
𝛛
𝛛𝐫
(𝐫𝟐
∙
𝐝𝐓
𝐝𝐫
) +
𝐪𝐠𝐞𝐧
𝐤
= 𝟎
Spheres without Heat generation
Q̇ = −k ∙ A ∙
dT
dr
= −k ∙ (4πr2) ∙
dT
dr
∫
Q̇
r2
∙ dr
r2
r1
= −∫ k ∙ 4π
T2
T1
∙ dT ⟹ −Q̇ (
1
r2
−
1
r1
) = −4πk ∙ (T2 − T1)
𝐐̇ =
𝟒𝛑𝐤 ∙ 𝐫𝟏 ∙ 𝐫𝟐 ∙ (𝐓𝟏 − 𝐓𝟐)
𝐫𝟐 − 𝐫𝟏
𝐑𝐭𝐡𝐞𝐫𝐦𝐚𝐥 =
𝐫𝟐 − 𝐫𝟏
𝟒𝛑𝐤 ∙ 𝐫𝟏𝐫𝟐
Geometric mean Area (Am)
Q̇ sphere =
4πk ∙ r1 ∙ r2 ∙ (T1 − T2)
r2 − r1
Q̇ slab =
k ∙ Am ∙ (T1 − T2)
r2 − r1
Q̇ sphere = Q̇ slab ⟹
4πk ∙ r1 ∙ r2 ∙ (T1 − T2)
r2 − r1
=
k ∙ Am ∙ (T1 − T2)
r2 − r1
Am = 4πr1r2 = √4πr1
2
∙ √4πr2
2
⟹ 𝐀𝐦 = √𝐀𝟏 ∙ √𝐀𝟐
Conduction through compositecylinders
Rconv 1 =
1
ha ∙ 4πr1
2 Rcond 1 =
r2 − r1
4πr1r2k
Rcond 2 =
r3 − r2
4πr3 r2k
Rconv 2 =
1
hb ∙ 4πr2
2
(Ta − Tb) = Q̇ ∙ (R1 + R2 + R3 + R4)
Q̇ = U ∙ A ∙ ΔT
Q̇ = Ui ∙ Ai ∙ (Ta − Tb) = Uo ∙ Ao ∙ (Ta − Tb)
Ui ∙ Ai = Uo ∙ Ao ⟹ Ui ∙ (4πri
2) = Uo ∙ (4πr2
2
)
𝐔𝐢𝐫𝐢
𝟐
= 𝐔𝐨𝐫𝐨
𝟐

11. 10
Critical Thickness of Insulation
Plane slabs
In case of plane slabs, there is no change in surface area of Heat transfer with increase in thickness,
so adding insulation would result in increase in conductive resistance, but the convective resistance
remains constant and hence with addition of insulation the total resistance increases and hence
heat transfer increases.
Therefore, addition of insulation in plane slabs always decreases heat transfer. There is no concept
of critical thickness of insulation in plane slabs.
Cylinders
Rcond =
(ln r r1
⁄ )
2πkL
⁄
Rconv =
1
hb ∙ 2πrL
Rtotal = Rcond + Rconv =
(ln r r1
⁄ )
2πkL
+
1
hb ∙ 2πrL
For Minimum Rtotal →
dRtotal
dr
= 0
dRtotal
dr
= (
1
2π ∙ kinsulation ∙ L
×
r1
r
×
1
r1
) + (
1
h ∙ 2πL
× −
1
r2
) = 0
rcritical thickness of insulation ⟹ rc =
kinsulation
h

12. 11
Steady state 1-D conductionwith heat transfer
Examples
Chemical reactions, Electric wire carrying current, Nuclear fuel rods, cutting of a concrete slab,
ripening of fruits etc...
Plane slabs
Uniform Heat generation with both surfaces maintained at same temperature
Assumptions
→ Steady state → 1-D conduction → Homogenous material
→ Constant thermal conductivity
∂2
T
∂x2
+
∂2
T
∂y2
+
∂2
T
∂z2
+
q̇ g
k
=
1
𝛼
∙
∂T
∂t
d2
T
dx2
+
q̇ gen
k
= 0 ⟹
d2
T
dx2
= −
q̇ gen
k
dT
dx
= −
q̇ gen
k
⋅ x + C1
(
dT
dx
= 0) → (x = 0) ⟹ 𝐂𝟏 = 𝟎
dT
dx
= −
q̇ gen
k
T = −
q̇ gen · x2
2k
+ C2
At (x =
L
2
) , T = Ts , 𝐂𝟐 = 𝐓𝐬 +
𝐪̇𝐠𝐞𝐧𝐋𝟐
𝟖𝐤
𝐓 = 𝐓𝐬 + (
𝐪̇𝐠𝐞𝐧
𝟐𝐤
⋅ (
𝐋𝟐
𝟒
− 𝐱𝟐))
At Centre x=0, T=Tmax
𝐓𝐦𝐚𝐱 = 𝐓𝐬 +
𝐪̇𝐠𝐞𝐧𝐋𝟐
𝟖𝐤
Energy Balance
As there is uniform heat generation and both surfaces are maintained at same temperature, half the
generated heat must leave from the right face and remaining heat must leave from the left face.
Q̇ = −k ⋅ A ⋅
dT
dx
|
x=
L
2
= −k ⋅ A ⋅ (−
q̇ genL
2k
) =
q̇ gen ⋅ A ⋅ L
2
Heat generated = Heat conducted (x = L/2) = Heat convected
q̇ gen ⋅ A ⋅ L
2
= −k ⋅ A ⋅
dT
dx
|
x=
L
2
= hbA ⋅ (Ts − Tb)
𝐓𝐬 = 𝐓𝐛 +
𝐪̇𝐠𝐞𝐧𝐋
𝟐𝐡𝐛
𝐓 = 𝐓𝐛 +
𝐪̇𝐠𝐞𝐧 𝐋
𝟐𝐡𝐛
+ (
𝐪̇𝐠𝐞𝐧
𝟐𝐤
⋅ (
𝐋𝟐
𝟒
− 𝐱𝟐))

13. 12
Plane slab with heat generation when both surfaces are
maintained at different temperatures
Let us Assume, T1 >T2
d2
T
dx2
+
q̇ g
k
= 0 →
d2
T
dx2
= −
q̇ g
k
T = −
q̇ gx2
2k
+ C1x + C2
At x = 0, T = T1 → 𝐂𝟐 = 𝐓𝟏
x = L, T = T2 → 𝐂𝟏 =
𝐓𝟐 − 𝐓𝟏
𝐋
+
𝐪̇𝐠𝐋
𝟐𝐤
𝐓 = −
𝐪̇𝐠
𝐱𝟐
𝟐𝐤
+ (
𝐓𝟐 − 𝐓𝟏
𝐋
+
𝐪̇𝐠
𝐋
𝟐𝐤
) ⋅ 𝐱 + 𝐓𝟏
Formax Temperature,
dT
dx
= 0 ⟹ x =
k
q̇g
(
T2 − T1
L
) +
L
2
𝐱𝐦𝐚𝐱,𝐓𝐞𝐦𝐩 =
𝐋
𝟐
−
𝐤
𝐪̇𝐠
(
𝐓𝟏 − 𝐓𝟐
𝐋
)
Energy Balance
Left face:
q̇ g ⋅ A ⋅ xmax = −k ⋅ A ⋅
dT
dx
|
x=0
= ha ⋅ A ⋅ (T1 − Ta)
Right face:
q̇ g ⋅ A ⋅ (L − xmax) = −k ⋅ A ⋅
dT
dx
|
x=L
= hb ⋅ A ⋅ (T2 − Tb)
Plane slab with heat generation when 1 face is insulated
d2
T
dx2
= −
q̇ g
k
T = −
q̇ gx2
2k
+ C1x + C2
At x = 0,
dT
dx
= 0 (Insulated), 𝐂𝟏 = 𝟎
At x = L, T = Ts, 𝐂𝟐 = 𝐓𝐬 +
𝐪̇𝐠𝐋𝟐
𝟐𝐤
𝐓 = 𝐓𝐬 +
𝐪̇𝐠
𝟐𝐤
⋅ (𝐋𝟐
− 𝐱𝟐)
Energy Balance
Heat Generated = Heat conducted (x=L) = Heat convected
q̇ g ⋅ A ⋅ L = −k ⋅ A ⋅
dT
dx
|
x=L
= ha ⋅ A ⋅ (Ts − Ta)

14. 13
Cylinders
Assumptions
→ Steady state → 1-D conduction → Homogenous material → Constant thermal conductivity
Solid cylinder with uniform heat generation
1
r
∙
∂
∂r
(r ∙
dT
dr
) +
q̇ gen
k
= 0 ⟹
∂
∂r
(r ∙
dT
dr
) = −
q̇ gen
k
⋅ r
r ⋅
dT
dr
= −
q̇ gen
r2
2k
+ C1 ⟹
dT
dr
= −
q̇ gen
r
2k
+
C1
r
𝐓 = −
𝐪̇𝐠𝐫𝟐
𝟒𝐤
+ 𝐂𝟏 ⋅ 𝐥𝐧 𝐫 + 𝐂𝟐
At r = 0,
dT
dr
= 0 → 𝐂𝟏 = 𝟎
At r = R, T = Ts → 𝐂𝟐 = 𝐓𝐒 +
𝐪̇𝐠𝐑𝟐
𝟒𝐤
Alternative Method
Qgen = Qconduction
q̇ g ⋅ V = −k ⋅ A ⋅
∂T
dr
→ q̇ g ⋅ πr2
⋅ L = −k ⋅ 2πrL ⋅
∂T
dr
−∫
q̇ gr
2k
⋅ dr = ∫dT → T = −
q̇ gR2
4k
+ C
At r = R → T = Ts → 𝐂 = 𝐓𝐬 +
𝐪̇ 𝐠𝐑𝟐
𝟒𝐤
𝐓 = 𝐓𝐬 +
𝐪̇𝐠
𝟒𝐤
(𝐑𝟐 − 𝐫𝟐)
At the center, r=0 → T=Tmax
𝐓𝐦𝐚𝐱 = 𝐓𝐬 +
𝐪𝐠𝐑
𝟐
𝟒𝐤
Energy Balance
Heat Generated = Heat conducted = Heat convected
q̇ g ⋅ (πR2
L) = −k ⋅ (2πRL) ⋅
dT
dr
= ha ⋅ (2πRL) ⋅ (Ts − T∞)
𝐓𝐬 = 𝐓∞ +
𝐪̇𝐠 · 𝐑
𝟐𝐡𝐚
Hollow cylinder with heat generation when inside
surface is insulated
q̇ gπ(r2
− ri
2)L = −k ⋅ (2πrL) ⋅
dT
dr
∫ dT = − ∫
q̇ g
2k
⋅ (r −
ri
2
r
) ⋅ dr
𝐓 = −
𝐪̇𝐠
𝟐𝐤
⋅ (
𝐫𝟐
𝟐
− 𝐫𝐢
𝟐
⋅ 𝐥𝐧 𝐫) + 𝐂
At r = ri, T = Tmax; r = ro, T = To.
Energy Balance
q̇ g ⋅ (π(ro
2
− ri
2)L) = −k ⋅ (2πroL) ⋅
dT
dr
= ha ⋅ (2πroL) ⋅ (To − T∞ )
𝐓𝐨 = 𝐓∞ +
𝐪̇𝐠
𝟐𝐡𝐚
(𝐫𝐨 −
𝐫𝐢
𝟐
𝐫𝐨
)

15. 14
Hollow Cylinder when outside surface is insulated
Q̇ g ⋅ (π ⋅ (ro
2
− r2) ⋅ L) = +k ⋅ (2πrL) ⋅
dT
dr
As dT/dr is positive, i.e., Maximum temperature is at the outer
surface of the cylinder and temperature decreases radially inwards.
Positive sign is taken.
𝐝𝐓 =
𝐪̇𝐠
𝟐𝐤
⋅ (
𝐫𝐨
𝟐
𝐫
− 𝐫) ⋅ 𝐝𝐫
Energy balance
q̇ g ⋅ (π(ro
2
− ri
2)L) = −k ⋅ (2πriL) ⋅
dT
dr
= ha ⋅ (2πriL) ⋅ (Ti − T∞ )
Hollow cylinder with heat generation when inside and outside surfaces are
maintained at To & Ti
∂
∂r
(r ∙
dT
dr
) = −
q̇ gen
k
⋅ r → 𝐓 = −
𝐪̇𝐠
𝐫𝟐
𝟒𝐤
+ 𝐂𝟏 ⋅ 𝐥𝐧 𝐫 + 𝐂𝟐
At r = ri → T = Ti , At r = ro → T = To , At Tmax,
dT
dr
|
r=rmax
= 0
Q̇ inwards = q̇ gen × (π ⋅ (rmax
2
− ri
2 ) ⋅ L)
Q̇ outwards = q̇ gen × (π ⋅ (ro
2
− rmax
2 ) ⋅ L)
Fraction of total heat moving radially inwards =
rmax
2
− ri
2
ro
2 − ri
2
Fraction of total heat moving radially outwards =
ro
2
− rmax
2
ro
2 − ri
2
Spheres
Assumptions
→ Steady state → 1-D conduction → Homogenous material → Constant thermal conductivity
q̇ g ⋅ (
4
3
πr3) = −k ⋅ (4πr2) ⋅
dT
dr
∫ dT = −∫
q̇ g · r
3k
⋅ dr
T = −
q̇ gr2
6k
+ C1
At r = R → T = Ts → 𝐂𝟏 = 𝐓𝐬 +
𝐪̇𝐠𝐑𝟐
𝟔𝐤
𝐓 = 𝐓𝐬 +
𝐪̇ 𝐠
𝟔𝐤
⋅ (𝐑𝟐
− 𝐫𝟐)
𝐓𝐦𝐚𝐱 = 𝐓𝐬 +
𝐪̇𝐠𝐑𝟐
𝟔𝐤
Energy balance
Heat Generated = Heat conducted = Heat convected
q̇ g ⋅ (
4
3
πr3) = −k ⋅ (4πr2) ⋅
dT
dr
= h ⋅ (4πR2
) ⋅ (Ts − T∞)
𝐓𝐬 = 𝐓∞ +
𝐪̇𝐠𝐑
𝐡
𝐓 = 𝐓∞ +
𝐪̇𝐠𝐑
𝐡
+
𝐪̇𝐠
𝟔𝐤
⋅ (𝐑𝟐
− 𝐫𝟐)

16. 15
Heat transfer from Extendedsurfaces (Fins)
We know that Q
̇ =h⋅A⋅ΔT, to increase heat transfer ΔT can be increased, but in most of the cases ΔT is
fixed and hence it’s not under our control. Further Q
̇ can be increased by increasing h (by forced
convection), we need fan, pump etc., it not only increases initial cost but also increases running and
maintenance cost. Therefore, best option available for increasing heat transfer is to increase surface
area by attaching extended surfaces (fins) to the base area either by welding or by extruding.
If the heat transfer coefficient is high, heat transfer will be more ad hence there is no need of fins.
Therefore, fins are more effective when h is less under free convection environment.
Generalizedequation in differential form for fins
Assumptions
→ Steady state → 1-D conduction → Homogenous material → Constant thermal conductivity
→ 1-D conduction → Constant heat transfer coefficient → No bonding resistance
Q̇ x = Q̇ convection + Q̇ x+dx
Q̇ x = Q̇ conv + (Q̇ x +
∂Q̇ x
∂x
⋅ dx)
Q̇ conv = −
∂Q̇ x
∂x
⋅ dx
[h ⋅ (P ⋅ dx) ⋅ (T − T∞ )] = −[
∂
∂x
(−k ⋅ Ac ⋅
dT
dx
) ⋅ dx]
h ⋅ P ⋅ (T − T∞ ) = k ⋅ Ac ⋅
d2
T
dx2
d2
T
dx2
= [
h ⋅ p
k ⋅ Ac
] ⋅ (T − T∞)
{m = √
h ⋅ p
k ⋅ Ac
}
{θ = T − T∞ ;
𝐝𝟐
𝛉
𝐝𝐱𝟐
=
𝐝𝟐
𝐓
𝐝𝐱𝟐
}
d2
θ
dx2
= m2
⋅ θ
(𝐃𝟐
− 𝐦𝟐 ) ⋅ 𝛉 = 𝟎
The solution for above differential equation is
𝛉 = 𝐂𝟏 ⋅ 𝐞𝐦𝐱
+ 𝐂𝟐 ⋅ 𝐞−𝐦𝐱
𝛉 = 𝐀 ⋅ 𝐜𝐨𝐬𝐡 𝐦𝐱 + 𝐁 ⋅ 𝐬𝐢𝐧𝐡 𝐦𝐱
Infinitely Long fin
Boundary Conditions
At x = 0 → T = To ; θ → 𝛉𝟎 = 𝐓𝐨 − 𝐓∞
θ = C1 ⋅ emx
+ C2 ⋅ e−mx
⟹ θo = C1 ⋅ em⋅0
+ C2 ⋅ e−m⋅0
⟹ 𝛉𝐨 = 𝐂𝟏 + 𝐂𝟐
At x = L → T = T∞ ; 𝐋 = ∞; 𝛉 = 𝟎
θ = C1 ⋅ emx
+ C2 ⋅ e−mx
⟹ 0 = C1 ⋅ em⋅L
+ C2 ⋅ e−m⋅L
→ C1 ⋅ em⋅∞
+ C2 ⋅
1
𝑒𝑚⋅∞
= 0
𝐂𝟏 = 𝟎 ; 𝐂𝟐 = 𝛉𝐨
𝛉 = 𝛉𝟎 ⋅ 𝐞−𝐦𝐱
This equation shows that temperature decreases
exponentially along with the length of the fin.
T − T∞
T0 − T∞
=
θ
θ0
= e−mx
Qconv = h ⋅ Aconv ⋅ (T − T∞)
Aconv = 2 ⋅ (w ⋅ dx + t ⋅ dx) = 2 ⋅ (w + t) ⋅ dx
Aconv = P ⋅ dx
m1> m2> m3

17. 16
Heat transfer
Whatever the heat is entering from fin cross-section area, that should be transferred from fin.
Q̇ = −k ⋅ Ac ⋅
dT
dx
|
x=0
= −k ⋅ Ac ⋅
dθ
dx
|
x=0
dθ
dx
= θo ⋅ e−mx
⋅ (−m) →
dθ
dx
|
x=0
= θo ⋅ e−m⋅0
⋅ (−m) = −m ⋅ θ0
Q̇ = −k ⋅ Ac ⋅ (−m ⋅ θo) = k ⋅ Ac ⋅ √
h ⋅ p
k ⋅ Ac
⋅ θ0 → 𝐐̇ = √𝐡 ⋅ 𝐏 ⋅ 𝐤 ⋅ 𝐀𝐜 ⋅ 𝛉𝐨
When the tip of the fin is insulated
Fins are never insulated because the purpose of the fin is to increase the heat transfer. Here
insulated tip is taken as the heat transfer from the fin tip is negligible compared to large width
surface area, this condition is used to obtain boundary conditions which can be used for
mathematical ease.
As the heat transfer from the tip is zero
Q̇ tip = −k ⋅ Ac ⋅
dT
dx
|
x=L
= 0 ⟹
𝐝𝐓
𝐝𝐱
|
𝐱=𝐋
= 𝟎 →
𝐝𝛉
𝐝𝐱
|
𝐱=𝐋
= 𝟎
θ = A ⋅ coshmx + B ⋅ sinh mx
At x = 0, θ = θ0 ⟹ θ0 = A ⋅ cosh m ⋅ 0 + B ⋅ sinh m ⋅ 0 ⟹ 𝐀 = 𝛉𝟎
dθ
dx
|
x=L
= 0 ⟹
dθ
dx
= A ⋅ cosh mx ⋅ (m) + B ⋅ sinh mx ⋅ m
→
dθ
dx
|
x=L
= m ⋅ (A ⋅ coshm ⋅ L + B ⋅ sinh m ⋅ L) = 0
B = −
A ⋅ sinh mL
cosh mL
𝐁 = −
𝛉𝟎 ⋅ 𝐬𝐢𝐧𝐡 𝐦𝐋
𝐜𝐨𝐬𝐡 𝐦𝐋
θ = 𝛉𝟎 ⋅ cosh mx + (−
𝛉𝟎 ⋅ 𝐬𝐢𝐧𝐡 𝐦𝐋
𝐜𝐨𝐬𝐡 𝐦𝐋
) ⋅ sinh mx θ = θo ⋅ (
cosh mx ⋅ coshmL − sinh mL ⋅ sinh mx
cosh mL
)
𝛉 = 𝛉𝐨 ⋅
𝐜𝐨𝐬𝐡𝐦(𝐋 − 𝐱)
𝐜𝐨𝐬𝐡 𝐦𝐋
Heat transfer
dθ
dx
= θo ⋅
(−sinh m ⋅ (L − x)) ⋅ m
coshmL
𝐝𝛉
𝐝𝐱
|
𝐱=𝟎
= −θo ⋅
(sinh m ⋅ L) ⋅ m
cosh mL
= −𝛉𝟎 ⋅ 𝐦 ⋅ 𝐭𝐚𝐧𝐡 𝐦𝐋
Q̇ = −k ⋅ Ac ⋅
dθ
dx
|
x=0
= −k ⋅ Ac ⋅ −θ0 ⋅ m ⋅ tanh mL = k ⋅ Ac ⋅ √
h ⋅ p
k ⋅ Ac
⋅ θ0 ⋅ tanh mL
𝐐̇ = √𝐤 ⋅ 𝐀𝐜 ⋅ 𝐡 ⋅ 𝐏 ⋅ 𝛉𝟎 ⋅ 𝐭𝐚𝐧𝐡 𝐦𝐋
A fin can be treated as infinitely long fin if tanh mL is 1, i.e., mL>5.
Fin with convection at free end
It is the most practical case, and is mostly used.

18. 17
Assume tip is insulated and the convection which happens from the tip, happens from the width by
adding an extra surface area which is equal to tip surface area.
At x = 0; θ = T − T∞ , θ0 = To − T∞
−k ⋅ Ac ⋅
dT
dx
|
x=L
= hAc ⋅ (T − T∞)
For Rectangular fins → 2 ⋅ (w + t) ⋅ (Lc − L) = t ⋅ w ⟹ 𝐋𝐜 = 𝐋 +
𝐭
𝟐
For Circular fins → 2πr ⋅ (Lc − L) = πr2
⟹ 𝐋𝐜 = 𝐋 +
𝐫
𝟐
𝐐̇ = √𝐤 ⋅ 𝐀𝐜 ⋅ 𝐡 ⋅ 𝐏 ⋅ 𝛉𝟎 ⋅ 𝐭𝐚𝐧𝐡 𝐦𝐋𝐜
Fin with specifiedtemperatureat ends
Boundary conditions
At x = 0 ⇒ T = T1 → θ = θ1
At x = L ⇒ T = T2 → θ = θ2
If T1 & T2 are same, then temperature will be minimum at the
center. i.e., at x=L/2.
As temperature is minimum at the center →
dT
dx
|
x=
L
2
=0
Fin performance
We know that along the length of the fin, temperature decreases and hence the temperature
difference between fin and surrounding decreases along the length and hence heat transfer
decreases along the length of the fin.
Heat transfer from fin will be maximum, when the entire fin is maintained at the base temperature.
The efficiency of a fin is defined as the ratio of actual heat transfer to the maximum heat transfer,
when the entire fin assumed at the base temperature.
η =
Q̇ actual
Q̇ maximum
Q̇ max = h ⋅ Ac ⋅ (To − T∞) = h ⋅ (P ⋅ L) ⋅ θ0
Infinitely long fin
→ η =
Q̇ actual
Q̇ maximum
⟹ η =
√h ⋅ P ⋅ k ⋅ Ac ⋅ θo
h ⋅ (P ⋅ L) ⋅ θ0
= √
kAc
hP
⋅
1
L
𝛈 =
𝟏
𝐦 ⋅ 𝐋
Insulated tip
𝛈 =
𝐭𝐚𝐧𝐡 𝐦𝐋
𝐦𝐋

19. 18
Effectivenessof Fin
When fins are attached to the base area, conductive resistance increases and convective resistance
decreases and hence the total resistance may decrease or increase. If the total resistance increases,
heat transfer decreases. Similarly, if the total resistance decreases, heat transfer increases and
therefore effectiveness of a fin is defined as the ratio of heat transfer with fin by heat transfer
without fin.
𝛜 =
𝐐̇ 𝐰𝐢𝐭𝐡 𝐟𝐢𝐧
𝐐̇ 𝐰𝐢𝐭𝐡𝐨𝐮𝐭 𝐟𝐢𝐧
Effectiveness of infinitely long fin
Q̇ fin = √hPkAc ⋅ θo Q̇ w.o fin = h ⋅ Ac ⋅ (To − T∞) = h ⋅ Ac ⋅ θo
𝛜 =
Q̇with fin
Q̇ without fin
=
√hPkAc ⋅ θo
h ⋅ Ac ⋅ θo
⟹ 𝛜 = √
𝐤 ⋅ 𝐏
𝐡 ⋅ 𝐀𝐜
For better performance, effectiveness must be high. Therefore, thermal conductivity of the fin
material must be high. If thermal conductivity is high, the temperature drop along the length will be
less and hence the temperature drop between fin and the surroundings will be more.
Therefore, the heat transfer will be more.
For better effectiveness, heat transfer coefficient must be less i.e., fins are more effective under free
convection environment (h is less in free convection).
For better effectiveness, P/Ac must be high i.e., effectiveness will be high if Ac is small. Therefore,
thin fins are preferred and fins must be closely spaced (they shouldn’t be placed too closely because
if they are too close, it obstructs the flow of air).
Generally, fins are effective if 𝛜 > 𝟐
√
k ⋅ P
h ⋅ Ac
> 2 ⟹
k ⋅ P
h ⋅ Ac
> 2

20. 19
Unsteady State (Transient) Conduction
Lumped systemAnalysis
It’s such an analysis in which the temperature varies only with time. But not with space.
Example: - A hot copper ball taken out from oven.
Temperature is a function of time → Lumped system Analysis.
Note: - For lumped system analysis, the conductivity must be high and the object must be small.
Time required for a substance to reach a particular temperature
Ti → initial temperature T∞ → Surrounding temperature h → heat transfer coefficient
As → Surface area of the object ρ → Density of the object V → Volume of object
At any instant, let T be the temperature of the temperature of the object
Heat transfer by convection = Energy stored
h ⋅ As ⋅ (T∞ − T) = m ⋅ cp ⋅
dT
dt
h ⋅ As ⋅ (T∞ − T) = ρ ⋅ V ⋅ cp ⋅
dT
dt
h ⋅ As
ρ ⋅ V ⋅ cp
⋅ dt = −
dT
d(T∞ − T)
− ∫
h ⋅ As
ρ ⋅ V ⋅ cp
⋅ dt
t
0
= ∫
d(T − T∞)
(T − T∞)
T
Ti
{dT = d(T − T∞)}
−
h ⋅ As
ρ ⋅ V ⋅ cp
⋅ t = ln (
T − T∞
Ti − T∞
)
𝐓 − 𝐓∞
𝐓𝐢 − 𝐓∞
= 𝐞
−
𝐡⋅𝐀𝐬
𝛒⋅𝐕⋅𝐜𝐩
⋅𝐭
Temperature varies exponentially.
Initially the object takes lesser time, later on for the same temperature drop the time
increases because with the increase in time, the temperature drop between objects and
surroundings decreases. Therefore, it takes more time.
Biot Number
It is the ratio of conductive resistance to convective resistance or it’s also defined as ratio of
temperature drop due to conduction to temperature drop due to convection.
Mathematically,
𝐁𝐢 =
𝐡𝐋
𝐤𝐬𝐨𝐥𝐢𝐝
Q =
ksolid ⋅ A ⋅ (T1 − T2)
L
= h ⋅ A ⋅ (T2 − T∞)
Bi =
Conductive resistance
convective resistance
=
L/(ksolid ⋅ A)
1/(h ⋅ A)
=
h ⋅ L
ksolid
Bi =
Temperature drop due to conduction
Temperature drop due to convection
=
T1 − T2
T2 − T∞
=
h ⋅ L
ksolid

21. 20
For applying lumped system analysis, temperature should not vary with
space and hence to achieve this thermal conductivity (k) should be high i.e.,
conductive resistance should be small. This conductive resistance is also
known as Internal conductive resistance.
Bi =
Rconduction
Rconvection
=
L
ksolid ⋅ A
⁄
1
h ⋅ A
⁄
=
Internal Conductive Resistance
External Conductive Resistance
It is found from experiments that the lumped system analysis is valid, if
𝐁𝐢 < 𝟎.𝟏
L in Biot number is Characteristic dimension (Lc).
Characteristic dimension for various shapes
Plane slab→ Lc=
V
As
=
A⋅L
2⋅A
⟹Lc =
L
2
Cylinder →Lc=
V
As
=
πR2
L
2πRL
⟹Lc=
R
2
=
D
4
Sphere→ Lc=
V
As
=
4
3
πR3
4πR2
⟹Lc=
R
3
→
T − T∞
Ti − T∞
= e
−
h⋅As
ρ⋅V⋅cp
⋅t
= 𝐞−𝐁𝐢⋅𝐅𝐨
h ⋅ As
ρ ⋅ V ⋅ cp
⋅ t =
h
ρ ⋅ cp ⋅ Lc
⋅ t = (
hLc
k
) ⋅ (
k
ρcp
) ⋅ (
t
Lc
2
) = Bi ⋅ (α ⋅
t
Lc
2
) = Bi ⋅ Fo ⟹
𝐡 ⋅ 𝐀𝐬 ⋅ 𝐭
𝛒 ⋅ 𝐕 ⋅ 𝐜𝐩
= 𝐁𝐢 ⋅ 𝐅𝐨
Fourier Number (Fo) represents the ability of heat to penetrate.
Time response of a thermocouple
→
T − T∞
Ti − T∞
= e
−
h⋅As
ρ⋅V⋅cp
⋅t
For reaching equilibrium, T = T∞
0 = e
−
h⋅As
ρ⋅V⋅cp
⋅t
For this, (
h ⋅ As
ρ ⋅ V ⋅ cp
) should be large.
For sphere, For quicker response→ (
h ⋅ 3
ρ ⋅ R ⋅ cp
) should be large and it will be large if
1. Heat transfer coefficient should be large.
2. Diameter / Radius of thermocouple should be small.
3. Density of thermocouple must be small.
4. Specific heat of thermocouple must be small
ρ ⋅cp ⋅ V
h ⋅As
is known as time constant(τ) of a thermocouple.
𝛒 ⋅ 𝐜𝐩 ⋅ 𝐕
𝐡 ⋅ 𝐀𝐬
= 𝛕
T − T∞
Ti − T∞
= e−
t
τ
If t = τ, →
T − T∞
Ti − T∞
= e−1 = 0.368
T − T∞
Ti − T∞
= 0.368 ⟹ 1 −
T − T∞
Ti − T∞
= 1 − 0.368 ⟹
Ti − T
Ti − T∞
= 0.632
𝐓𝐢 − 𝐓 = 𝟎.𝟔𝟑𝟐 ⋅ (𝐓𝐢 − 𝐓∞)
The time taken by the thermocouple to reach 63.2 % of initial temperature difference is known as
sensitivity.
For quicker response, sensitivity must be high.

22. 21
Heat exchangers
Heat exchanger is a device which is used for transferring or exchanging heat from one fluid to other
fluid.
Classification of Heat exchangers
Based on Contact → 1. Direct Contact 2. Indirect contact
Based on Direction of flow → 1. Parallel flow 2. Counter flow 3. Cross flow
Based on Constructional details → 1. Compaq 2. Bulky
Overall Heat transfer coefficient
R =
1
hi ⋅ Ai
+
ln(ro/ri)
2πkL
+
1
ho ⋅ Ao
For better heat transfer, resistance should be low.
Consider, ro = ri, (small thickness of pipe) ⟹less resistance due to conduction.
For, ro = ri ⟹ Ao = Ai ⩵ A
R =
1
hi ⋅ Ai
+
ln(ro/ri)
2πkL
+
1
ho ⋅ Ao
⟹ R =
1
A
(
1
hi
+
1
ho
)
ΔT = Q ⋅ R; → Q = U ⋅ A ⋅ ΔT ⟹ ΔT = Q ⋅ (
1
U ⋅ A
) = Q ⋅ R
R =
1
U ⋅ A
=
1
A
(
1
hi
+
1
ho
) ⟹
𝟏
𝐔
=
𝟏
𝐡𝐢
+
𝟏
𝐡𝐨
Let us assume that hi
>>>ho
,
1
hi
will be negligible ⟹ 𝐔 = 𝐡𝐨
→ 𝐐 = 𝐡𝐨 ⋅ 𝐀 ⋅ 𝚫𝐓
If heat transfer coefficient of inner fluid is very high compared to inner fluid. Heat transfer depends
on the outer fluid.
Fouling factor
Overtime use of pipes, causes deposition of unwanted substances, they
increase the resistance to heat transfer.
R =
1
hi ⋅ Ai
+
1
ho ⋅ Ao
⟹
1
U ⋅ A
=
1
A
(
1
hi
+
1
ho
) ⟹
1
Uclean
=
1
hi
+
1
ho
1
Udirty
=
1
hi
+ Rfoul in + Rfoul out +
1
ho
=
1
Uclean
+ (Rfoul in + Rfoul out)
𝟏
𝐔𝐝𝐢𝐫𝐭𝐲
=
𝟏
𝐔𝐜𝐥𝐞𝐚𝐧
+ 𝐅𝐨𝐮𝐥𝐢𝐧𝐠 𝐑𝐞𝐬𝐢𝐬𝐭𝐚𝐧𝐜𝐞 (𝐅. 𝐑)
Analysis of Heat exchanger
Assumptions
1. Steady flow
2. Kinetic and potential energy change are neglected
3. Specific heats of hotter and colder fluids don’t change with temperature
4. Heat transfer coefficient is assumed to be constant
5. The system/ heat exchanger is completely insulated from the surroundings. Therefore, the
heat lost by hot body is equal to the heat gained by cold body.
The design of a hat exchanger is based on two methods
1. LMTD (Logarithmic Mean temperature difference)
2. Effectiveness NTU method
LMTD is used when the inlet and outlet temperatures are known, and with the help of LMTD, we can
calculate the surface area of the Heat exchanger. It’s used to design a Heat exchanger based on our
needs.
Effectiveness-NTU Method is used to calculate the outlet temperatures of cold and hot fluid streams
or a particular heat exchanger.

23. 22
LMTD Method
𝐐 = U ⋅ A ⋅ ΔTm
đQ = U ⋅ dA ⋅ (Th − Tc)
đQ = −mh ⋅ ch ⋅ dTh ⟹ dTh = −
đQ
mh ⋅ ch
đQ = −mc ⋅ cc ⋅ dTc ⟹ dTc = −
đQ
mc ⋅ cc
mh → mass flow rate of hotter fluid
mc → mass flow rate of colder fluid
ch → Specific heat of hotter fluid
cc → Specific heat of colder fluid C → Heat capacity
d(Th − Tc) = dTh − dTc = −
đQ
mh ⋅ ch
−
đQ
mc ⋅ cc
= −đQ (
1
mh ⋅ ch
+
1
mc ⋅ cc
)
d(Th − Tc) = −U ⋅ dA ⋅ (Th − Tc) ⋅ (
1
mh ⋅ ch
+
1
mc ⋅ cc
) ⟹
d(Th − Tc)
(Th − Tc)
= −U ⋅ dA ⋅ (
1
mh ⋅ ch
+
1
mc ⋅ cc
)
∫
d(Th − Tc)
(Th − Tc)
= − ∫ U ⋅ (
1
Ch
+
1
Cc
) ⋅ dA (Heat capacity)Ch = mh ⋅ ch ; Cc = mc ⋅ cc
∫
d(ΔT)
ΔT
ΔT2
ΔT1
= −∫ U ⋅ (
1
Ch
+
1
Cc
) ⋅ dA ⟹ 𝐥𝐧
𝚫𝐓𝟐
𝚫𝐓𝟏
= −𝐔 ⋅ 𝐀 ⋅ (
𝟏
𝐂𝐡
+
𝟏
𝐂𝐜
)
𝐐 = mh ⋅ ch ⋅ (Th1 − Th2) = Ch ⋅ (Th1 − Th2) 𝐐 = mc ⋅ cc ⋅ (Tc2 − Tc1) = Cc ⋅ (Tc2 − Tc1)
ln
ΔT2
ΔT1
= −U ⋅ A ⋅ (
1
Ch
+
1
Cc
) = −U ⋅ A ⋅ [
Th1 − Th2
Q
−
Tc1 − Tc2
Q
] = −
U ⋅ A
Q
⋅ [(Th1 − Tc1) − (Th2 − Tc2)]
ln
ΔT2
ΔT1
= −
U ⋅ A
Q
⋅ (ΔT1 − ΔT2) ⟹ 𝐐 = 𝐔 ⋅ 𝐀 ⋅ (
𝚫𝐓𝟏 − 𝚫𝐓𝟐
𝐥𝐧
𝚫𝐓𝟏
𝚫𝐓𝟐
) = 𝐔 ⋅ 𝐀 ⋅ 𝚫𝐓𝐦
𝚫𝐓𝐦 =
𝚫𝐓𝟏 − 𝚫𝐓𝟐
𝐥𝐧
𝚫𝐓𝟏
𝚫𝐓𝟐
Generally, in parallel flow heat exchanger, ΔT1≠ΔT2, because the hotter fluid temperature decreases
and colder fluid temperature increases, therefore, ΔT decreases along the length.
LMTD for Counter flow Heat exchanger
𝐐 = U ⋅ A ⋅ ΔTm
In counter flow Heat exchanger, if ΔT1=ΔT2, for that 𝚫𝐓𝟏 = 𝚫𝐓𝟐 = 𝚫𝐓𝐦
As the variation in temperature is large for parallel flow. Heat exchanger and the temperature
difference between hot and cold fluid decreases along the length the length of the heat exchanger.
Therefore, the heat transfer decreases and hence for a given heat transfer more area is required for
parallel flow. Therefore, counter flow heat exchangers are preferred because ΔTm is more and this
results in smaller area, therefore counter flow Heat exchangers are more compact.

24. 23
LMTD for Multi-Tubular heat exchangers
Q = U ⋅ A ⋅ ΔTm ⋅ 𝐅
The L.M.T.D equations derived are for single shell, single
tube pass heat exchanger, for multi-tubular Heat exchanger
and crossflow heat exchanger same equation can be used by
introducing a correction factor F, where ΔTm is LMTD for
counter flow heat exchanger.
Special cases
When Heat capacities of both fluids is same for Counter flow Heat exchanger
Example: - Regenerator of a gas turbine.
Heat capacities are same for both hot and cold fluids ⟹ Ch = Cc
Ch ⋅ (Th1 − Th2 ) = Cc ⋅ (Tc2 − Tc1) ⟹ (Th1 − Th2 ) = (Tc2 − Tc1) ⟹ Th1 − Tc2 = Th2 − Tc1
𝚫𝐓𝟏 = 𝚫𝐓𝟐 = 𝚫𝐓𝐦
Phase change in Heat exchangers
Examples: - Boilers. Evaporators, condensers etc...
Tc1 = Tc2 ⟹ Th1 − Tc2 = Th1 − Tc1
ΔT1 𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥 = ΔT1 𝐜𝐨𝐮𝐧𝐭𝐞𝐫 → ΔT2 𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥 = ΔT2 𝐜𝐨𝐮𝐧𝐭𝐞𝐫
In case of phase change, heat exchangers, ΔTm is same for both parallel and counter flow heat
exchangers.
Parallel and Counter heat exchangers will become same when there is a phase change i.e., in case of
phase change heat exchangers, it is immaterial whether it’s counter flow or parallel flow heat
exchanger.
EffectivenessNTU Method
Effectiveness is defined as the ratio of actual heat transfer to the maximum possible heat transfer.
∈ =
Qactual
Qmaximum
Cc is minimum
Colder fluid will undergo maximum temperature difference.
Qmax = Cc ⋅ (Tc2 − Tc1) = Cmin ⋅ (Th1 − Tc1)
Qactual = Ch ⋅ (Th1 − Th2)
∈ =
Qactual
Qmaximum
=
Ch ⋅ (Th1 − Th2)
Cmin ⋅ (Th1 − Tc1)
𝐓𝐡𝟐 = 𝐓𝐡𝟏 − (
∈⋅ 𝐂𝐦𝐢𝐧 ⋅ (𝐓𝐡𝟏 − 𝐓𝐜𝟏)
𝐂𝐡
)
Ch is minimum
Hotter fluid undergoes maximum temperature difference.
Qmax = Ch ⋅ (Th1 − Th2) = Cmin ⋅ (Th1 − Tc1)

25. 24
Qactual = Cc ⋅ (Tc2 − Tc1)
∈ =
Qactual
Qmaximum
=
Cc ⋅ (Tc2 − Tc1)
Cmin ⋅ (Th1 − Tc1)
𝐓𝐜𝟐 = 𝐓𝐜𝟏 + (
∈⋅ 𝐂𝐦𝐢𝐧 ⋅ (𝐓𝐡𝟏 − 𝐓𝐜𝟏 )
𝐂𝐜
)
The equation for maximum Heat transfer is same, irrespective of
whether Cc is minimum or Ch is minimum.
Effectiveness of parallel flow heat exchanger
ln
ΔT2
ΔT1
= −U ⋅ A ⋅ (
1
Ch
+
1
Cc
)
If Ch is minimum, ln (
Th2 − Tc2
Th1 − Tc1
) = −
U ⋅ A
Ch
⋅ (1 +
Ch
Cc
) = −
U ⋅ A
Cmin
⋅ (1 +
Cmin
Cmax
)
→
Th2 − Tc2
Th1 − Tc1
= 𝑒
−
U⋅A
Cmin
⋅(1+
Cmin
Cmax
)
Let
𝐔 ⋅ 𝐀
𝐂𝐦𝐢𝐧
= 𝐍𝐓𝐔 (Number of Transfer Units),
𝐂𝐦𝐢𝐧
𝐂𝐦𝐚𝐱
= 𝐂
→
𝐓𝐡𝟐 − 𝐓𝐜𝟐
𝐓𝐡𝟏 − 𝐓𝐜𝟏
= 𝒆−𝐍𝐓𝐔⋅(𝟏+𝐂)
Substituting Th2 and Tc2,
⟶
1
Th1 − Tc1
× ((Th1 −
∈⋅ Cmin ⋅ (Th1 − Tc1)
Ch
)− (Tc1 +
∈⋅ Cmin ⋅ (Th1 − Tc1)
Cc
)) = 𝑒−NTU⋅(1+C)
⟶
1
Th1 − Tc1
× ((Th1 − Tc1) − (∈⋅ Cmin ⋅ (Th1 − Tc1) ⋅ (
1
Ch
+
1
Cc
))) = [1 − (∈⋅ Cmin ⋅ (
1
Ch
+
1
Cc
))]
If Ch is minimum → [1 − (∈⋅ Cmin ⋅ (
1
Ch
+
1
Cc
))] ⟹ [1 − (∈⋅
Cmin
Cmin
⋅ (1 +
Cmin
Cmax
))] = [1 − (∈⋅ (1 + C))]
[𝟏 − (∈⋅ (𝟏 + 𝐂))] = 𝒆−𝐍𝐓𝐔⋅(𝟏+𝐂)
∈=
𝟏 − 𝒆−𝐍𝐓𝐔⋅(𝟏+𝐂)
𝟏 + 𝐂
⟶ Parallel flow Heat exchanger
Special cases in parallel flow heat exchanger
Phase change during Heat Exchange
As ΔT = 0, during heat exchange, C for condensing/boiling liquid will be infinity (∞).
𝐂𝐦𝐚𝐱 = ∞ ⟹ C =
Cmin
Cmax
=
Cmin
∞
= 0 ⟹ 𝐂 = 𝟎
∈=
1 − 𝑒−NTU⋅(1+C)
1 + C
⟹ ∈= 𝟏 − 𝒆−𝐍𝐓𝐔
Heat capacities of both fluids are same
Can be found in regenerator of a gas turbine.
𝐂𝐦𝐢𝐧 = 𝐂𝐦𝐚𝐱 ⟶ 𝐂 =
𝐂𝐦𝐢𝐧
𝐂𝐦𝐚𝐱
⟹ 𝐂 = 𝟏
∈=
1 − 𝑒−NTU⋅(1+C)
1 + C
⟹ ∈=
𝟏 − 𝒆−𝟐⋅𝐍𝐓𝐔
𝟐
Effectiveness of counter flow Heat exchanger
Proceeding in previous manner, we get
∈=
𝟏 − 𝒆−𝐍𝐓𝐔⋅(𝟏−𝐂)
𝟏 + 𝐂 ⋅ 𝒆−𝐍𝐓𝐔⋅(𝟏−𝐂)

26. 25
Special cases for counter flow Heat exchangers
Phase change during Heat Exchange
As ΔT = 0, during heat exchange, C for condensing/boiling liquid will be infinity (∞).
𝐂𝐦𝐚𝐱 = ∞ ⟹ C =
Cmin
Cmax
=
Cmin
∞
= 0 ⟹ 𝐂 = 𝟎
∈=
1 − 𝑒−NTU⋅(1−C)
1 + C ⋅ 𝑒−NTU⋅(1−C) ⟹ ∈= 𝟏 − 𝒆−𝐍𝐓𝐔
For phase change, heat exchanger effectiveness will be same for both parallel and counter flow heat
exchanger.
Heat capacities of both fluids are same
Can be found in regenerator of a gas turbine.
𝐂𝐦𝐢𝐧 = 𝐂𝐦𝐚𝐱 ⟶ 𝐂 =
𝐂𝐦𝐢𝐧
𝐂𝐦𝐚𝐱
⟹ 𝐂 = 𝟏
∈=
1 − 𝑒−NTU⋅(1−C)
1 + C ⋅ 𝑒−NTU⋅(1−C) =
0
0
form
By using L′
Hospitals rule,we get ⟶ ∈=
𝐍𝐓𝐔
𝟏 + 𝐍𝐓𝐔
Significance of NTU
For a given overall heat transfer coefficient and for a given fluid, NTU depends on surface area A,
greater the NTU, larger is the surface area i.e., larger is the size of the Heat exchanger. Therefore,
NTU is the measure of size of the heat exchanger.
Note
Baffles are used in the heat exchanger to
1. Support tubes
2. Alter the flow direction
3. Increase turbulence and thereby increasing the heat transfer.

27. 26
Radiation
According to Provost law, every substance above ‘0’ K, emits radiation in the form of electro-
magnetic waves and they travel with the speed of light. Radiation is the fastest mode of transfer.
Thermal radiation is in the range of 0.1 µm to 100 µm. thermal radiation consists of part U.V rays,
Infrared rays and visible rays. Only in this wavelength range, radiation when gets absorbed gets
converted into heat and hence known as thermal radiation.
Properties of radiation
Q = Qreflected + Qabsorbed + Qtransmitted
1 =
Qr
Q
+
Qa
Q
+
Qt
Q
Qr
Q
= ρ (Reflectivity);
Qa
Q
= α (Absorptivity);
Qt
Q
= τ (Transmissivity)
𝛒 + 𝛂 + 𝛕 = 𝟏
Note
1. For solids and liquids τ=0, (α + ρ = 0). Such bodies are known as opaque bodies.
2. For gases, ρ = 0, (α + τ =1).
3. For a blackbody, α = 1, (ρ = τ = 0).
4. For a white body, ρ =1, (α = τ = 0).
5. Color of the object has nothing to do with a black body or white body. It’s the surface
characteristics that tells a body is black or white body.
6. Snow or ice which looks white in color is actually a black body, because snow absorbs
maximum radiation falling on it and is close to a black body.
Greenhouse effect
In cold condition, plants are grown using the concept of greenhouse effect. Glass allows the visible
range of radiation and this is absorbed by substances in green house and subsequently emit
radiation in infrared region, for which the glass is opaque and thereby heat is trapped inside only.
This is known as greenhouse effect.
Important points w.r.t black body
1. It absorbs all the radiation falling on it.
2. It neither transmits nor reflects radiation.
3. It’s observed that at a particular temperature, blackbody emits maximum radiation compared
to other bodies. Therefore, black body is not only a perfect absorber but also a perfect emitter.
Example: - A sphere with a small pin hole.
Laws of radiation
Planck’s law
It gives monochromatic emissive power of a black body at a particular temperature.
Here monochromatic means at a particular wavelength.
𝐞𝐛𝛌 =
𝐂𝟏 ⋅ 𝛌−𝟓
𝐞
𝐂𝟐
𝛌𝐓 − 𝟏
Stefan-Boltzmann law
It gives the total emissive power of a blackbody at a particular
temperature. Here total means summed over all wave lengths.
eb = ∫ ebλ ⋅ dλ
∞
0
= ∫
C1 ⋅ λ−5
e
C2
λT − 1
⋅ dλ
∞
0
𝐞𝐛 = 𝛔 ⋅ 𝐓𝟒 (σ = 5.67 × 10−8
W
m2K
)
Wein’s Displacement Law
According to Wien’s displacement law, the product of wave length at
which maximum emissive power occurs and the corresponding
temperature is constant.
𝐞𝐛𝛌 =
𝐂𝟏 ⋅ 𝛌−𝟓
𝐞
𝐂𝟐
𝛌𝐓 − 𝟏

28. 27
For maximum emissivity,
d(ebλ)
dλ
= 0 ⟹ 𝛌𝐦𝐚𝐱 ⋅ 𝐓 = 𝛋
The emissive power of a black body is continuous overall wave length.
The wavelength at which the maximum emissive power occurs decreases with increase in
temperature.
Irradiation (G)
The total radiation incident on the surface is known as irradiation.
Emissivity (∈)
It is defined as the ratio of emissive power of a body to emissive power of a black body at the same
temperature.
∈=
e
eb
⟹ e =∈⋅ eb ⟹ 𝐞 =∈⋅ 𝛔𝐓𝟒
For a black body, ∈ = 1, α = 1, ρ = 0, τ = 0.
Types of reflection
1. Specular ⟶ Incident angle is equal to Reflected angle.
2. Directional ⟶Reflected angle depends on the incident angle.
3. Diffuse ⟶Reflected angle is independent of incident angle.
A surface is said to be diffuse surface, when the radiation properties are independent if direction.
Grey surface
A surface is said to be gray surface if radiation properties are independent of wavelength.
Comparison of the emissivity (a) and emissive power (b) of a real surface with those of a gray surface and a blackbody at
the same temperature
Kirchhoff’s Law
Under thermal equilibrium, ratio of emissive power to the absorptivity is constant.
e
α
= κ ⟹
e1
α1
=
e2
α2
=
e3
α3
For black body,
e1
α1
=
eb
αb
⟹
e
α
=
eb
1
⟹ α =
e
eb
=∈ ⟶ ∈= 𝛂
Under thermal equilibrium, emissivity and absorptivity are same.
Shape factor
Shape factor F12 is the fraction of radiation emitted by the surface 1 and falling on 2, the shape factor
depends only on geometry and orientation. It has nothing to do with surface characteristics and
temperature of surface.
Reciprocity theorem ⟶ 𝐀𝟏 ⋅ 𝐅𝟏𝟐 = 𝐀𝟐 ⋅ 𝐅𝟐𝟏
Summation rule ⟶ 𝐅𝟏𝟏 + 𝐅𝟏𝟐 + 𝐅𝟏𝟑 + 𝐅𝟏𝟒 + ⋯ = 𝟏
Self-view factor for flat surfaces and convex surfaces is zero.
If the surface us bounded by similar surfaces, then the shape factor will be
same.

29. 28
Net radiative heat transfer between 2 black bodies
Radiation emitted by 1 is 𝛔 ⋅ 𝐀𝟏 ⋅ 𝐓𝟏
𝟒
, the fraction if radiation emitted by 1 falling on 2 is 𝐅𝟏𝟐 ⋅ 𝛔 ⋅ 𝐀𝟏 ⋅ 𝐓𝟏
𝟒
.
Similarly, radiation emitted by 2, falling in 1 and absorbed by 1 is 𝐅𝟐𝟏 ⋅ 𝛔 ⋅ 𝐀𝟏 ⋅ 𝐓𝟏
𝟒
.
Net radiative heat transfer between 1 and 2 is,
Q12 = σ ⋅ A1 ⋅ T1
4
⋅ F12 − σ ⋅ A2 ⋅ T2
4
⋅ F21
𝐐𝟏𝟐 = 𝛔 ⋅ 𝐀𝟏 ⋅ 𝐅𝟏𝟐 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒) = 𝛔 ⋅ 𝐀𝟐 ⋅ 𝐅𝟐𝟏 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒)
Radiosity (J)
The total radiation leaving the surface is called radiosity.
J = ρ ⋅ G + ∈⋅ eb ⟹ G =
J−∈⋅ eb
ρ
α + ρ + τ = 1 → α + ρ + 0 = 1 (τ = 0 → Opaque surface) → α + ρ = 1 → ρ = 1 − α
𝛒 = 𝟏−∈ (α = ρ → Kirchoff′
s law at thermal equilibrium)
𝐆 =
𝐉−∈⋅ 𝐞𝐛
𝟏−∈
Net radiation leaving the surface = J − G ⟹ J −
J−∈⋅ eb
1−∈
=
J − J ⋅∈ −J+∈⋅ eb
1−∈
=
∈⋅ (eb − J)
1−∈
Total radiation leaving the surface(Q) =
∈⋅ (eb − J)
1−∈
⋅ A
(𝐞𝐛 − 𝐉) =
𝐐
(𝟏−∈
∈⋅ 𝐀
⁄ )
(1‒∈ / ∈⋅A) is known as surface resistance. For a black body ∈=1, i.e., surface resistance is zero.
Total radiation leaving 1 and falling on 2 is J1⋅A1⋅F12, similarly the total radiation leaving 2 and falling
on 1 is J2⋅A2⋅F21.
Net heat exchange between 1 and 2 is
Q12 = J1 ⋅ A1 ⋅ F12 − J2 ⋅ A2 ⋅ F21 = (J1 − J2) ⋅ A1 ⋅ F12
→ 𝐉𝟏 − 𝐉𝟐 = 𝐐𝟏𝟐 ⋅ (
𝟏
𝐀𝟏 ⋅ 𝐅𝟏𝟐
)
→ eb1 − eb2 = Q12 ⋅ [
1 −∈1
∈1⋅ A1
+
1
A1 ⋅ F12
+
1 −∈2
∈2⋅ A2
] ⟹ 𝐐𝟏𝟐 =
𝐞𝐛𝟏 − 𝐞𝐛𝟐
[
𝟏 −∈𝟏
∈𝟏⋅ 𝐀𝟏
+
𝟏
𝐀𝟏 ⋅ 𝐅𝟏𝟐
+
𝟏 −∈𝟐
∈𝟐⋅ 𝐀𝟐
]
Radiative heat exchange between 2 body enclosure
Here 2 body enclosure means heat exchange is occurring between 2 bodies only.
A small body kept in larger enclosure
A1
A2
≈ 0 → F12 = 1
Q12 =
eb1 − eb2
[
1 −∈1
∈1⋅ A1
+
1
A1 ⋅ F12
+
1 −∈2
∈2⋅ A2
]
=
eb1 − eb2
1
A1
⋅ [
1 −∈1
∈1⋅
+
1
F12
+
A1
A2
⋅
1 −∈2
∈2
]
Q12 =
σ ⋅ A1 ⋅ (T1
4
− T2
4
)
(
1
∈1
− 1) +
1
1
+ 0 ⋅ (
1
∈2
− 1)
𝐐𝟏𝟐 =∈𝟏⋅ 𝛔 ⋅ 𝐀𝟏 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒

30. 29
Radiative heat transfer between 2 large parallel plates
A1 = A2 = A
F11 = 0 → F12 = 1
Q12 =
eb1 − eb2
1
A1
⋅ [
1 −∈1
∈1⋅
+
1
F12
+
A1
A2
⋅
1 −∈2
∈2
]
=
σ ⋅ A1 ⋅ (T1
4
− T2
4
)
(
1
∈1
− 1) +
1
1
+ 1 ⋅ (
1
∈2
− 1)
𝐐𝟏𝟐 =
𝛔 ⋅ 𝐀𝟏 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒
)
𝟏
∈𝟏
+
𝟏
∈𝟐
− 𝟏
Radiative heat exchange between two constant concentric cylinders
A1 = π ⋅ D1 ⋅ L A2 = π ⋅ D2 ⋅ L
F11 = 0 ⟹ F12 = 1
Q12 =
eb1 − eb2
1
A1
⋅ [
1 −∈1
∈1⋅
+
1
F12
+
A1
A2
⋅
1 −∈2
∈2
]
=
σ ⋅ A1 ⋅ (T1
4
− T2
4)
(
1
∈1
− 1) +
1
1
+
D1
D2
⋅ (
1
∈2
− 1)
𝐐𝟏𝟐 =
𝛔 ⋅ 𝐀𝟏 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒)
𝟏
∈𝟏
+
𝐃𝟏
𝐃𝟐
⋅ (
𝟏
∈𝟐
− 𝟏)
Radiative exchange between two concentric spheres
A1 = 4 ⋅ π ⋅ r1
2
A2 = 4 ⋅ π ⋅ r2
2
F11 = 0 ⟹ F12 = 1
Q12 =
eb1 − eb2
1
A1
⋅ [
1 −∈1
∈1⋅
+
1
F12
+
A1
A2
⋅
1 −∈2
∈2
]
=
σ ⋅ A1 ⋅ (T1
4
− T2
4)
(
1
∈1
− 1) +
1
1
+
r1
2
r2
2 ⋅ (
1
∈2
− 1)
𝐐𝟏𝟐 =
𝛔 ⋅ 𝐀𝟏 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒)
𝟏
∈𝟏
+
𝐫𝟏
𝟐
𝐫𝟐
𝟐 ⋅ (
𝟏
∈𝟐
− 𝟏)
Radiation shields
Radiation shields are used for reducing heat transfer. Therefore, radiation shields must have high
reflectivity.
Without shield
A1 = A2 = A
Q12 =
σ ⋅ A ⋅ (T1
4
− T2
4)
1
∈1
+
1
∈2
− 1
(Special case → ∈𝟏=∈𝟐=∈) ⟹ 𝐐𝟏𝟐 =
𝛔 ⋅ 𝐀 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒)
𝟐
∈
− 𝟏
With shield
Q13 = Q32
σ ⋅ A ⋅ (T1
4
− T3
4)
1
∈1
+
1
∈3
− 1
=
σ ⋅ A ⋅ (T3
4
− T2
4)
1
∈3
+
1
∈2
− 1
Special case → ∈𝟏=∈𝟐=∈𝟑=∈
→
σ ⋅ A ⋅ (T1
4
− T3
4)
1
∈
+
1
∈
− 1
=
σ ⋅ A ⋅ (T3
4
− T2
4)
1
∈
+
1
∈
− 1
⟹ T1
4
− T3
4
= T1
4
− T2
4
⟹ 𝐓𝟑
𝟒
=
𝐓𝟏
𝟒
+ 𝐓𝟐
𝟒
𝟐
𝑄𝑤𝑖𝑡ℎ 𝑠ℎ𝑖𝑒𝑙𝑑 =
σ ⋅ A ⋅ (T1
4
− T3
4)
1
∈
+
1
∈
− 1
=
σ ⋅ A ⋅ (T1
4
− (
T1
4
+ T2
4
2
))
2
∈
− 1
=
𝟏
𝟐
×
𝛔 ⋅ 𝐀 ⋅ (𝐓𝟏
𝟒
− 𝐓𝟐
𝟒)
𝟐
∈
− 𝟏
𝐐𝐰𝐢𝐭𝐡 𝐬𝐡𝐢𝐞𝐥𝐝
𝐐𝐰𝐢𝐭𝐡𝐨𝐮𝐭 𝐬𝐡𝐢𝐞𝐥𝐝
=
𝟏
𝟐
If there are 𝐧 similar shields,
𝐐𝐰𝐢𝐭𝐡 𝐬𝐡𝐢𝐞𝐥𝐝
𝐐𝐰𝐢𝐭𝐡𝐨𝐮𝐭 𝐬𝐡𝐢𝐞𝐥𝐝
=
𝟏
𝐧 + 𝟏

31. 30
Convection
It is the mechanism of heat transfer through fluid in the presence of bulk fluid motion,
There are 2 types of convection. Natural convection and Forced convection.
Heat transfer through fluid occurs by convection in the presence of bulk fluid motion and occurs by
conduction in the absence of it.
Rate of heat transfer through a fluid is much higher for convection than it’s by conduction.
Temperature difference is more constant in convection.
Q̇ convection = h ⋅ As ⋅ (Ts − T∞)
Heat transfer coefficient (h) can be defined as the rate of heat transfer between a solid surface and a
fluid per unit surface area per unit temperature difference.
Fluid in motion comes to a complete stop at the surface and assumes a zero-velocity relative to the
surface, due to the viscous effects. This is known as No-Slip condition.
Heat transfer from solid surface to fluid layer adjacent to the surface is by pure conduction.
q̇ convection = q̇ convection = −kf ⋅
∂T
∂y
|
y=0
(
W
m2
)
h ⋅ As ⋅ (Ts − T∞) = −kf ⋅ As ⋅
∂T
∂y
|
y=0
⟹ h =
−kf ⋅
∂T
∂y
|
y=0
Ts − T∞
=
1
L
⋅ ∫ hx ⋅ dx
L
0
Nusselt number
Nu =
h ⋅ Lc
k
Non-dimensionalization the heat transfer coefficient h with the Nusselt
number.
Nusselt number represents the enhancement of heat transfer through a fluid
layer as a result of convection relative to conduction across the same fluid layer.
q̇ convection = h ⋅ ΔT q̇ convection = −kf ⋅
ΔT
L
𝐪𝐜𝐨𝐧𝐯
𝐪𝐜𝐨𝐧𝐝
=
𝐡 ⋅ 𝚫𝐓
𝐤 ⋅
𝚫𝐓
𝐋
=
𝐡 ⋅ 𝐋
𝐤
Velocity boundary layer
Consider the parallel flow of a fluid over a flat plate.
The development of the boundary layer for flow over a flat plate, and the different flow regimes .
Due to no-slip condition, velocity of fluid layer adjacent to plate becomes zero.
This layer (V=0), slows down the particles of neighbouring fluid layer due to friction between these
two layers, this continues till velocity reaches free stream velocity.
The region of flow above the plate bounded by δ in which the effects of viscous shearing forces
caused by fluid viscosity are felt is called as the velocity boundary layer.
Boundary layer thickness δ, is usually defined as the distance y from surface at which u = 0.99× V.
Hypothetical line of u = 0.99× V, divides flow over the plate into 2 regions,
1. Boundary layer region (viscous effects)
2. Irrotational flow region (V= κ)

32. 31
Wall shear stress
τw = μ ⋅
du
dy
|
y=0
(
N
m2
)
From experiments, it'
s found that τ ∝
du
dy
,
τ = Cf ⋅
ρV2
2
(
N
m2
) (Cf → Skin friction coefficient)
Friction force (𝐅𝐟) = Cf ⋅ As ⋅
ρ ⋅ V2
2
Thermal boundary layer
Thermal boundary layer develops when a fluid at a specified temperature
flows over a surface that is at a different temperature.
Fluid layer adjacent to wall, assumes wall temperature (thermal
equilibrium) fluid particles exchange heat with adjacent layers, till T∞. Flow
region over the surface in which temperature variation in the direction
normal to surface is significant is thermal boundary layer. The thickness of
thermal boundary layer δt, at any location along the surface is defined as the
distance from the surface at which the temperature difference T-Ts equals
0.99⋅ (T∞ ‒ Ts).
Thermal boundary layer thickness increases as the length of the plate increases. Shape of the
temperature profile in thermal boundary layer dictates the convection heat transfer between a solid
surface and fluid flowing over it.
Prandtl Number
Relative thickness of velocity and thermal boundary layer is best described by dimensionless
parameter, Prandtl number
Pr=
Molecular diffusivity of momentum
molecular diffusivity of heat
=
ν
α
=
μ ρ
⁄
k ρ⋅cp
⁄
=
μ⋅cp
k
High Prandtl number (ex- Oils) has better momentum diffusivity than thermal diffusion. More
friction in fluid layers than heat exchange between layers of fluid.
Low Prandtl number (ex- Liquid metals) has better thermal diffusivity than momentum diffusivity,
thermal boundary layer is developed faster than velocity boundary layer. As the thermal
conductivity will be higher for liquid metals, more heat transfer from surface to liquid metal.
Reynolds number
Osborn Reynolds discovered that the flow regime depends mainly in the ratio of inertia forces to
viscous forces in the fluid.
Re=
Inertia force
viscous force
=
V⋅Lc
ν
=
ρ⋅V⋅Lc
μ
V⟶ upstream velocity, μ → dynamic viscosity, Lc→ characteristic length of geometry.
➢ Laminar ⟶ Viscous forces dominate.
➢ Transition ⟶ Viscous and inertia forces go up and down
➢ Turbulent ⟶ Inertia forces dominate, rapid mixing of fluid particles from adjacent layers.
The intense mixing of fluid in turbulent flow as a result of rapid fluctuations increases heat and
momentum transfer between fluid particles, which increases the friction force on the surface and
the convection heat transfer rate. The enhancement in heat transfer in turbulent flow comes with
penalty, we need to use larger pump to overcome the large friction forces accompanying the higher
heat transfer rate.
Convection equations
∂u
∂x
+
∂v
∂y
= 0
ρ ⋅ (u ⋅
∂u
∂x
+ v ⋅
∂v
∂y
) = μ ⋅
∂2
u
∂y2
−
∂P
∂x
ρ ⋅ cp ⋅ (u ⋅
∂T
∂x
+ v ⋅
∂T
∂y
) = k ⋅ (
∂2
T
∂x2
+
∂2
T
∂y2
) + μϕ

33. 32
Convectionon a flat plate
Consider laminar flow over a flat plate. The x-coordinate is measured along the plate surface from
the leading edge of the plate in the direction of the flow, and y is measured from the surface in the
normal direction. The fluid approaches the plate in the x-direction with a uniform upstream velocity,
which is equivalent to the free stream velocity V.
Viscous dissipation is negligible, flow is steady and incompressible.
𝐜𝐨𝐧𝐭𝐢𝐧𝐮𝐢𝐭𝐲 →
∂u
∂x
+
∂v
∂y
= 0 ; 𝐱‒𝐦𝐨𝐦𝐞𝐧𝐭𝐮𝐦 → u ⋅
∂u
∂x
+ v ⋅
∂v
∂y
= ν ⋅
∂2
u
∂y2
; 𝐄𝐧𝐞𝐫𝐠𝐲 → u ⋅
∂T
∂x
+ v ⋅
∂T
∂y
= α ⋅
∂2
T
∂y2
Boundary conditions
At x = 0, u(0, y) = V, T(0, y) = T∞
At y = 0, u(x,0) = 0, v(x,0) = 0 T(x,0) = Tw
At y → ∞, u(x,∞) = V, T(x,∞) = T∞
When fluid properties are assumed to be constant and independent of
temperature, continuity and x-momentum equations are solved separately
for velocity components u & v, (Velocity distribution will be available) and Cf & δ can be calculated
from u & v.
Continuity and momentum equations are solved by Blasius. By transforming 2 partial differential
equations into single ordinary differential equation, by using a new independent variable, called the
similarity variable.
Noticing that the general shape of velocity profile remains same along the plate, Blasius reasoned
that non-dimensional velocity profile u/v remains unchanged when plotted against non-
dimensional y/δ, as general shape of velocity profile remains same along the plate. He defined
(Dimensionless Similarity variable) 𝛈 = 𝐲 ⋅ √
𝐕
𝛎 ⋅ 𝐱
(
y
δ
= η, From stokes exp → δ ∝ √
ν ⋅ x
V
)
And thus u/v = function (η),
Stream function ψ(x , y)→ u =
∂ψ
∂y
; v = −
∂ψ
∂x
, Continuity eq.n is satisfied.
Blasius defined, 𝐟(𝛈) =
𝛙
𝐕 · √𝛎 · 𝐱
𝐕
⁄
𝐮 =
𝛛𝛙
𝛛𝐲
= (
∂ψ
∂η
) · (
∂η
∂y
) = (V · √
νx
V
·
df
dη
) · (√
V
νx
) = 𝐕 ·
𝐝𝐟
𝐝𝛈
; 𝐯 = −
𝛛𝛙
𝛛𝐱
=
𝟏
𝟐
· √
𝐕
𝛎𝐱
· (𝛈 ·
𝐝𝐟
𝐝𝛈
− 𝐟)
𝛛𝐮
𝛛𝐱
= −
𝐕
𝟐𝐱
· 𝛈 ·
𝐝𝟐
𝐟
𝐝𝛈𝟐
;
𝛛𝐮
𝛛𝐲
= 𝐕 · √
𝐕
𝛎𝐱
·
𝐝𝟐
𝐟
𝐝𝛈𝟐
;
𝛛𝟐
𝐮
𝛛𝐲𝟐
=
𝐕𝟐
𝛎𝐱
·
𝐝𝟑
𝐟
𝐝𝛈𝟑
Substituting the 𝐮, 𝐯,
𝛛𝐮
𝛛𝐱
,
𝛛𝐮
𝛛𝐲
,
𝛛𝟐
𝐮
𝛛𝐲𝟐
in momentum equation and solving,
𝟐
𝐝𝟑𝐟
𝐝𝛈𝟑
+ 𝐟
𝐝𝟐𝐟
𝐝𝛈𝟐
= 𝟎
We obtain a third-order nonlinear differential equation. Therefore, the system of two partial
differential equations is transformed into a single ordinary differential equation by the use of a
similarity variable.
The boundary conditions obtained in terms of similarity variables can be expressed as
f(0) = 0
df
dη
|
η=0
= 0
df
dη
|
η=∞
= 1
Transformed equation can’t be solved analytically. Blasius used power series expansion approach to
solve and later its solved by using Numerical approaches. Results obtained are
η = 4.91, y = δ when
df
dη
=
u
v
= 0.99 ⟹ (u = 0.99 · V at y = δ)

34. 33
→ η = y ⋅ √
V
ν ⋅ x
(η = 4.91,y = δ) ⟹ 𝛅 =
𝟒.𝟗𝟏
√𝐕
𝛎𝐱
⁄
=
𝟒. 𝟗𝟏𝐱
√𝐑𝐞𝐱
τw = μ ·
∂u
∂x
|
y=0
= μ · V · √
V
νx
·
d2
f
dη2
|
η=0
⟹ 𝛕𝐰 = 𝟎. 𝟑𝟑𝟐 · 𝐕 · √
𝐕
𝛎𝐱
𝐂𝐟,𝐱 =
𝛕𝐰
𝛒𝐕𝟐/𝟐
= 𝟎. 𝟔𝟔𝟒 · 𝐑𝐞𝐱
−𝟏/𝟐
The Energy Equation
Knowing the velocity profile, we are ready to solve the energy equation
for temperature distribution for the case of constant wall temperature
Ts, first we introduce the dimensionless temperature θ as
Dimensionless temperature ⟹ 𝛉(𝐱, 𝐲) =
T(x,y) − Ts
T∞ − Ts
Energy equation is transformed to ⟹ u ⋅
∂θ
∂x
+ v ⋅
∂θ
∂y
= α ⋅
∂2
θ
∂y2
Temperature profiles for flow over an isothermal flat plate are similar, just like the velocity profiles,
and thus we expect a similarity solution for temperature to exist.
→ δt ∝ √
ν ⋅ x
V
→ η = y ⋅ √
V
ν ⋅ x
→ θ = θ(η)
Using the chain rule and substituting the u and v expressions from previous results into the energy
equation gives
𝐮 ⋅
𝛛𝛉
𝛛𝐱
+ 𝐯 ⋅
𝛛𝛉
𝛛𝐲
= 𝛂 ⋅
𝛛𝟐
𝛉
𝛛𝐲𝟐
(V ·
df
dη
) · (
dθ
dη
·
∂η
∂x
) + [
1
2
· √
V
νx
· (η ·
df
dη
− f)] · (
dθ
dη
·
∂η
∂y
) = α · [
d2
θ
dη2
(
∂η
∂y
)
2
]
Simplifying and noting that Pr = ν/α gives,
𝟐
𝐝𝟐
𝛉
𝐝𝛈𝟐
+ 𝐏𝐫 · 𝐟 ·
𝐝𝛉
𝐝𝛈
= 𝟎
For Pr=1, and when θ is replaced
df
dη
(
𝐮
𝐯
) , this equation reduces to velocity boundary layer eq.n
Velocity and thermal boundary layers coincide and non-dimensional velocity and temperature
profiles (u/v and θ) are identical for steady, incompressible, laminar flow of a fluid with constant
properties and Pr = 1 over an isothermal plate.
Boundary conditions are θ (0) = 0, θ (∞) = 1
→
dθ
dη
|
𝑦=0
η=0
=
d2
θ
dη2
|
𝑦=0
η=0
= 0.332
Solving the simplified energy equation for numerous values of Prandtl numbers, it’s found
→
𝐝𝛉
𝐝𝛈
∝ 𝐏𝐫
𝟏
𝟑
→
𝐝𝛉
𝐝𝛈
|
𝐲=𝟎
𝛈=𝟎
= 𝟎. 𝟑𝟑𝟐 × 𝐏𝐫
𝟏
𝟑 (𝐏𝐫 > 𝟎. 𝟔)
Temperature gradient at the surface,
𝐝𝐓
𝐝𝐲
|
𝐲=𝟎
= (T∞ − Ts) ·
∂θ
∂y
|
y=0
= (T∞ − Ts) ·
∂θ
∂η
|
η=0
·
∂η
∂y
|
y=0
= 𝟎. 𝟑𝟑𝟐 · 𝐏𝐫𝟏/𝟑
· (𝐓𝐬 − 𝐓∞) · √
𝐕
𝛎𝐱

The local convection coefficient and the Nusselt number becomes,

35. 34
→ hx =
qs
̇
Ts − T∞
=
−k · (
∂T
∂y
|
y=0
)
Ts − T∞
⟹ 𝐡𝐱 = 𝟎.𝟑𝟑𝟐 · 𝐏𝐫𝟏/𝟑
· 𝐤 · √
𝐕
𝛎𝐱

→ Nux =
hx · x
k
⟹ 𝐍𝐮𝐱 = 𝟎. 𝟑𝟑𝟐 · 𝐏𝐫𝟏/𝟑
· 𝐑𝐞𝒙
𝟏/𝟐
Solving simplified energy equation numerically for the temperature profile for different Prandtl
numbers, and using the definition of the thermal boundary layer, it is determined that
𝛅
𝛅𝐭
≅ 𝐏𝐫𝟏/𝟑
Then the thermal boundary layer thickness becomes
𝛅𝐭 =
𝟒. 𝟗𝟏 · 𝐱
𝐏𝐫𝟏/𝟑 · 𝐑𝐞𝐱
𝟏/𝟐
External Forced convection
Developing a good understanding of external flow and external forced convection is important in the
mechanical and thermal design of many engineering systems.
We need to rely on experimental data for solving most external flow problems as flow fields and
geometries are too complicated. High speed computers are used to solve problems numerically in a
very less time.
Velocity if the fluid relative to an immersed solid body sufficiently far from body (outside boundary
layer) is called free stream velocity. It’s actually assumed to be uniform and steady.
Friction and pressure drag
The force a flowing fluid body exerts on a body is called drag.
Stationary fluid exerts normal forces. Moving fluid exerts Normal and tangential shear forces.
Drag force is equal to pressure forces and wall shear force in flow direction.
Lift forces is equal to pressure force and wall shear force in Normal direction.
Drag coefficient (𝐂𝐝) =
𝐅𝐝
𝟏
𝟐
𝛒𝐕𝟐𝐀
Drag coefficient mainly depends in shape of the body, nut may depend on Reynolds number and
surface roughness.
𝐂𝐃 = 𝐂𝐃, 𝐟𝐫𝐢𝐜𝐭𝐢𝐨𝐧 + 𝐂𝐃, 𝐩𝐫𝐞𝐬𝐬𝐮𝐫𝐞
For parallel flow over a flat plate → CD = CD, Friction + Cf
Friction drag is a strong function of viscosity. Friction drag is also proportional to surface area. In
laminar flow, FD is independent of surface roughness. In turbulent flow, FD depends on surface
roughness very much.
Heat transfer
𝐍𝐮 = 𝐂 · 𝐑𝐞𝐋
𝐦
· 𝐏𝐫𝐧
m & n are exponential constants and value of C depends on geometry and flow.
In order to account for variation of properties with temperature, fluid properties are usually
evaluated at so called film temperature.
Tf =
Ts + T∞
2
We are usually interested in drag force and heat transfer for the entire surface, so
Cf =
1
L
∫ Cf,x
L
0
· dx
h =
1
L
∫ hx
L
0
· dx
Q̇ = h · As · (Ts − T∞)

36. 35
Parallel flow over flat plate
Reynolds number at a distance x from the leading edge if a flat plate is
expressed as
𝐑𝐞𝐱 =
𝛒𝐕𝐱
𝛍
=
𝐕𝐱
𝛎
The value of critical Reynolds number for a flat plate may vary from 105
to
3 × 106
, depending upon the surface roughness, turbulence level and
variation of pressure along the surface.
Laminar → 𝛅 =
𝟒.𝟗𝟏𝐱
𝐑𝐞𝐱
𝟏/𝟐 𝐂𝐟,𝐱 =
𝟎. 𝟔𝟔𝟒
𝐑𝐞𝐱
𝟏/𝟐 Rex < 5 × 105
Turbulent → 𝛅 =
𝟎. 𝟑𝟖𝐱
𝐑𝐞𝐱
𝟏/𝟓 𝐂𝐟,𝐱 =
𝟎. 𝟎𝟓𝟗
𝐑𝐞𝐱
𝟏/𝟓 5 × 105
< Rex < 107
Cf, x is higher in turbulent than in laminar. Cf, x reaches its highest value when flow becomes fully
turbulent and then decreases by a factor of x-1/5
in flow direction.
Average friction coefficient over entire plate ①(𝐂𝐟) =
𝟏. 𝟑𝟑
𝐑𝐞𝐋
𝟏/𝟐 𝐋𝐚𝐦𝐢𝐧𝐚𝐫 (ReL < 5 × 105)
②𝐂𝐟 =
𝟎. 𝟎𝟕𝟒
𝐑𝐞𝐋
𝟏/𝟓 𝐓𝐮𝐫𝐛𝐮𝐥𝐞𝐧𝐭 (5 × 105
< ReL < 107)
Note that the average friction coefficient over the entire plate in case of laminar flow is twice the
value of local friction coefficient at the end of the plate,
Laminar flow → 𝐂𝐟
̅ = 𝟐 · 𝐂𝐟,𝐱=𝐋
Laminar eqn.① for surface roughness is used when total flow is laminar over entire plate or if
turbulence length is negligible. Turbulence equation ② equation is used, when the flow is turbulent
over the entire plate, or when the laminar flow region of the plate is too small relative to the
turbulent flow region (that is, xcr ≪ L).
In some cases, a flat plate is sufficiently long for the flow to become turbulent, but not long enough to
disregard the laminar flow region. In such cases, the average friction coefficient over the entire plate
is determined by,
𝐂𝐟 =
𝟏
𝐋
· (∫ 𝐂𝐟,𝐱 𝐥𝐚𝐦𝐢𝐧𝐚𝐫 · 𝐝𝐱
𝐱𝐜𝐫
𝟎
+ ∫ 𝐂𝐟,𝐱 𝐭𝐮𝐫𝐛𝐮𝐥𝐞𝐧𝐭 · 𝐝𝐱
𝐋
𝐱𝐜𝐫
)
Assuming Recr = 5 × 105
, 𝐂𝐟 =
𝟎. 𝟎𝟕𝟒
𝐑𝐞𝐋
𝟏/𝟓 −
𝟏𝟕𝟒𝟐
𝐑𝐞𝐋
Smooth surface (5 × 105
< ReL < 107)
For laminar flow, Surface roughness has no effect on friction coefficient and laminar flow depends
only on Reynolds number.
For turbulent flow, Surface roughness causes the friction coefficient to increase several folds, till the
point it depends only on surface roughness.
⟹ 𝐂𝐟 = (𝟏. 𝟖𝟗 − 𝟏.𝟔𝟐 × 𝐥𝐨𝐠
𝛆
𝐋
)
−𝟐.𝟓
Rough surface (Re > 106),(
Surface roughness (ε)
Length of plate (L)
> 104)
Heat transfer coefficient
Laminar → Nux =
hxx
k
= 0.332 · Rex
0.5
· Pr
1
3 (0.6 < Pr, Rex < 5 × 105)
Turbulent → Nux = 0.0296 · Rex
0.8
· Pr
1
3 (0.6 < Pr < 60, 5 × 105
< Rex < 107)
Laminar → Nu
̅̅̅̅ =
hL
k
= 0.664 · ReL
0.5
· Pr
1
3 (0.6 < Pr, Rex < 5 × 105)
Turbulent → Nu
̅̅̅̅ =
hL
k
= 0.037 · ReL
0.8
· Pr
1
3 (0.6 < Pr < 60, 5 × 105
< Rex < 107)
Hx reaches its highest value when flow becomes fully turbulent and then decreases by a factor of x-0.2
in the flow direction.
Laminar flow → Nu
̅̅̅̅ = 2 · Nux=L h
̅ = 2 · hx=L

37. 36
For considerable laminar and turbulent flow lengths,
𝐡 =
𝟏
𝐋
· (∫ 𝐡𝐱,𝐥𝐚𝐦𝐢𝐧𝐚𝐫 · 𝐝𝐱
𝐱𝐜𝐫
𝟎
+ ∫ 𝐡𝐱 𝐭𝐮𝐫𝐛𝐮𝐥𝐞𝐧𝐭 · 𝐝𝐱
𝐋
𝐱𝐜𝐫
)
Assuming Recr
= 5 × 10
5
, 𝐍𝐮 =
𝐡𝐋
𝐤
= (𝟎.𝟎𝟑𝟕 · 𝐑𝐞𝐋
𝟎.𝟖
− 𝟖𝟕𝟏) · 𝐏𝐫
𝟏
𝟑 (0.6 < Pr < 60, 5×10
5
<ReL < 10
7
)
For liquid metals → 𝐍𝐮𝐱 = 𝟎. 𝟓𝟔𝟓 · (𝐑𝐞𝐱 · 𝐏𝐫)𝟏/𝟐
= 𝟎. 𝟓𝟔𝟓 · 𝐏𝐞𝐱
𝟏/𝟐
(Pr ≤ 0.05, (Peclet number) Pex≥100)
𝐍𝐮𝐱 =
𝐡𝐱𝐱
𝐤
=
𝟎. 𝟑𝟑𝟖𝟕 · 𝐏𝐫
𝟏
𝟑 · 𝐑𝐞𝐱
𝟏
𝟐
[𝟏 + (𝟎.𝟎𝟒𝟔𝟖
𝐏𝐫
⁄ )
𝟐
𝟑
]
𝟏
𝟒
[Rex · Pr ≥ 100]
Applicable for all Prandtl Numbers
.
Flat plate with unheated starting length
Many practical situations involve surfaces with an unheated starting
length section of length ξ, thus there is no heat transfer for 0 < x < ξ. In
such case, velocity boundary layer starts to develop at x = 0, but
thermal boundary layer starts at x = ξ.
Laminar: 𝐍𝐮𝐱 =
Nux|ξ=0
[1 − (ξ
x
⁄ )
3 4
⁄
]
1/3 =
𝟎. 𝟑𝟑𝟖𝟕 · 𝐏𝐫
𝟏
𝟑 · 𝐑𝐞𝐱
𝟏
𝟐
[𝟏 − (𝛏
𝐱
⁄ )
𝟑 𝟒
⁄
]
𝟏/𝟑
Turbulent: 𝐍𝐮𝐱 =
Nux|ξ=0
[1 − (ξ
x
⁄ )
9 10
⁄
]
1/9 =
𝟎. 𝟎𝟐𝟗𝟔 · 𝐏𝐫
𝟏
𝟑 · 𝐑𝐞𝐱
𝟎.𝟖
[𝟏 − (𝛏
𝐱
⁄ )
𝟗 𝟏𝟎
⁄
]
𝟏/𝟗
Terms in denominator serve as correction factor plates with unheated starting lengths.
Laminar: 𝐍𝐮𝐱 =
𝟐 · [𝟏 − (𝛏
𝐱
⁄ )
𝟑 𝟒
⁄
]
𝟏 −
𝛏
𝐋
⁄
· 𝐡𝐱=𝐋
Turbulent: 𝐍𝐮𝐱 =
𝟓 · [𝟏 − (
𝛏
𝐱
⁄ )
𝟗 𝟏𝟎
⁄
]
𝟒 · [𝟏 −
𝛏
𝐋
⁄ ]
· 𝐡𝐱=𝐋
The variation of the local friction and heat transfer
coefficients for flow over a flat plate
Graphical representation of the average heat transfer
coefficient for a flat plate with combined laminar and
turbulent flow.

38. 37
Uniform Heat flux
When a flat plate is subjected to uniform heat flux instead of uniform temperature,
Laminar: 𝐍𝐮𝐱 = 𝟎. 𝟒𝟓𝟑 · 𝐏𝐫𝟏 𝟑
⁄
· 𝐑𝐞𝐱
𝟏 𝟐
⁄
(Pr > 0.6, Rex < 5 × 105
)
Turbulent: 𝐍𝐮𝐱 = 𝟎.𝟎𝟑𝟎𝟖 · 𝐏𝐫𝟏 𝟑
⁄
· 𝐑𝐞𝐱
𝟎.𝟖 (0.6 < Pr < 60,5 × 105
< Rex < 107)
→ Q̇ = q̇ s · As → q̇ s = hx · (Ts(x) − T∞ ) ⟹ Ts(x) = T∞ +
q̇ s
hx
Internal Forced Convection
Small diameter pipes are usually referred to as tubes. Pipes with circular cross-section can
withstand large pressure differences between inside and outside without undergoing significant
distortion. For a fixed surface area, circular tubes give the most heat transfer for the least pressure
drop. We must rely on experimental results and empirical relations for most fluid flow problems
rather than closed analytical solutions. So, solutions obtained from these empirical relations are not
exact and have some error.
Because of no-slip condition, velocity gradient is developed. So, it’s convenient to work with average
velocity (Vavg) which remains constant in incompressible flow when the cross-sectional area of pipe
is constant. Because of friction in the fluid flow, there will be pressure drop.
Average velocity and temperature
In the absence of V∞ (free stream velocity), it’s convenient to work with Vavg.
⟶ m
̇ = ρ · Vavg · Ac = ∫ ρ · u(r) · dAc
Ac
⟶ Vavg =
∫ ρ · u(r) · dAc
Ac
ρ · Ac
=
∫ u(r) · 2πr · dr
Ac
πR2
𝐕𝐚𝐯𝐠 =
𝟐
𝐑𝟐
· ∫ 𝐮(𝐫) · 𝐫 · 𝐝𝐫
𝐑
𝟎
When the fluid is heated is cooled, temperature of fluid at any cross-section changes from Ts at the
surface of the wall to some maximum/minimum at the tube center, it’s convenient to work with
average or mean temperature (Tmean) which remains constant at any cross-section.
Rate at which energy is transferred is same in both cases.
⟶ Ė fluid = m
̇ · cp · Tmean = ∫ cp · T(r) · δm
̇
m
̇
= ∫ ρ · cp · T(r) · u(r) · dAc
Ac
Tmean =
∫ ρ · cp · T(r) · u(r) · dAc
Ac
m
̇ · cp
=
∫ ρ · cp · T(r) · u(r) · 2πr · dr
Ac
(ρ · πR2 · Vavg) · cp
⟹ 𝐓𝐦 =
𝟐
𝐕𝐚𝐯𝐠 · 𝐑𝟐
· ∫ 𝐓(𝐫) · 𝐮(𝐫) · 𝐫 · 𝐝𝐫
𝐀𝐜
For flow through circular pipes, Reynolds number is
→ Re =
Vavg · D
ν
=
ρD
ν
· Vavg =
ρD
ν
·
m
̇
ρ · πD2/4
⟹ 𝐑𝐞 =
𝟒 · 𝐦̇
𝛒𝛑𝐃𝟐
=
𝐦̇
𝛒𝛑𝐑𝟐
For flow through Non-circular pipes, f & Re & Nu are based on Hydraulic diameter.
(Hydraulic diameter) 𝐃𝐡 =
𝟒𝐀𝐜
𝐏𝐰
Ac → Cross − sectional Area
Pw → Wetted perimeter
Entrance Region
Due to no-slip condition, fluid at surface will remain stationary, friction slows down the adjacent
layers. But to keep up the constant mass flow rate, velocity at center of pipe increases and velocity
gradient is developed. The region of flow in which the effects if viscous shearing forces caused by
fluid viscosity are felt is called Velocity Boundary layer. Thickness id boundary layer increases in
the flow direction until it reaches pipe center and thus fills the entire pipe. The region from pipe inlet
to point at which the velocity profile is fully developed is called Hydrodynamic entrance region.

39. 38
Length of this region is called Hydrodynamic entry length. The region beyond the entrance region in
which the velocity profile is fully developed and remains unchanged is called Hydrodynamically
fully developed Region.
When fluid enters a pipe, whose wall temperature is at Ts, fluid layer adjacent to wall assumes wall
temperature and due to heat transfer between fluid layers there is a temperature gradient developed
in fluid and thermal boundary layer is developed along the tube. Its thickness increases until it
reaches center and fills whole tube.
The region of flow over which the thermal boundary layer develops and reaches the tube center is
called thermal entrance region and the length is called Thermal entry length Lt.
The region beyond the thermal entrance region, in which the dimensionless temperature profile
remains unchanged is called Thermally fully developed region.
→Dimensionless temperature profile →
𝐓𝐬 − 𝐓
𝐓𝐒 − 𝐓𝐦𝐞𝐚𝐧
The region in which the flow is both hydrodynamically and thermally fully developed and both
velocity and dimensionless temperature profiles remain constant is called Fully developed flow.
⟶
∂u(r, x)
∂x
= 0 → u = u(r) ⟶
∂
∂x
[
Ts(x) − T(r, x)
Ts(x) − Tm (x)
] = 0
As velocity gradient (∂u/∂y)|y=0 at the surface remains constant in
hydrodynamically developed region. Wall shear stress remains constant in
Hydrodynamically developed region.
∂
∂r
[
Ts − T
Ts − Tm
]|
r=R
=
−(
∂T
∂r
)|
r=R
Ts − Tm
≠ f(x)
𝐪̇𝐬 = 𝐡𝐱 · (𝐓𝐬 − 𝐓𝐦 ) = 𝐤 ·
𝛛𝐓
𝛛𝐫
|
𝐫=𝐑
→ 𝐡𝐱 =
𝐤 ·
𝛛𝐓
𝛛𝐫
|
𝐫=𝐑
𝐓𝐬 − 𝐓𝐦
≠ 𝐟(𝐱)
Both fx (local friction factor) and hx (local convection coefficient), remain
constant in hydrodynamic and thermally fully developed regions. (Pr>1)
Temperature profile in thermally fully developed region may vary with x in
the flow direction. Although dimensionless temperature profile remains
unchanged in thermally fully developed region.

40. 39
Entry Lengths
Hydrodynamic entry length is usually taken to be the distance from tube entrance, where the wall
shear stress (thus friction factor) reaches within about 2% of fully developed value.
Lh,laminar ≈ 0.05 · Re · D
Lt,turbulent ≈ 0.05 · Re · Pr · D ≈ Pr · Lh, laminar
General thermal Analysis
In the absence of any work interactions, Q̇ = m
̇ · cp · (Te − Ti)
Ti & Te are mean fluid temperatures at inlet and exit, approximated to be constant surface
temperature (Ts = constant) or constant heat flux (q̇s=constant)
Ts = κ, ⟹ during condensation/Boiling
q̇s= κ, ⟹ Radiation or electrical resistance heating
𝐪̇𝐬 = 𝐡𝐱 · (𝐓𝐬 − 𝐓𝐦) W/m2
hx → local heat transfer coefficient. Ts → surface temperature. Tm → Mean temperature
Tm varies when pipe is heated or cooled.
If Ts = constant, ⟹ q̇s varies along the pipe
If q̇ s = constant, ⟹ Ts varies along the pipe
Constant Surface heat flux
Q̇ = q̇ s · As = m
̇ · Cp · (Te − Ti)
Te = Ti +
q̇ s · As
m
̇ · cp
⟹ 𝐓𝐞 = 𝐓𝐢 +
𝐪̇𝐬 · (𝟐𝛑𝐑𝐋)
𝐦̇ · 𝐜𝐩
Mean temperature increases linearly in the flow direction in the case of
constant heat flux.
q̇ s = h · (Ts − Tm) ⟹ 𝐓𝐬 = 𝐓𝐦 +
𝐪̇𝐬
𝐡
In a fully developed flow, h is constant, therefore, Ts increases with Tm.
⟹ (Ts – Tm) is constant.
⟶ m
̇ · cp · dTm = q̇ s · (P · dx) ⟹
𝐝𝐓𝐦
𝐝𝐱
=
𝐪̇𝐬 · 𝐏
𝐦̇ · 𝐜𝐩
⟶ Ts − Tm = κ →
𝐝𝐓𝐒
𝐝𝐱
=
𝐝𝐓𝐦
𝐝𝐱
For temperature profile to remain constant,
∂
∂x
(
Ts − T
Ts − Tm
) = 0 ⟹
𝛛𝐓𝐬
𝛛𝐱
=
𝛛𝐓
𝛛𝐱
𝛛𝐓
𝛛𝐱
=
𝐝𝐓𝐒
𝐝𝐱
=
𝐝𝐓𝐦
𝐝𝐱
=
𝐪̇𝐬 · 𝐏
𝐦̇ · 𝐜𝐩
= 𝛋 (h = constant)
In fully developed flow subjected to constant heat flux, temperature gradient
is independent of x and thus shape of temperature profile doesn’t change
along the tube.
𝐓𝐦 = 𝐓𝐢 + (
𝐪̇𝐬 · 𝐏
𝐦̇ · 𝐜𝐩
) · 𝐱
Constant surface temperature
Q̇ = h · As · ΔTavg = h · As · (TS − Tm)avg
[Tm varies along the length, for heating/cooling]
[Tsremains constant for boiling/condensation]
In the constant surface temperature (Ts = constant) case, ΔTavg can be expressed approximately by
the arithmetic mean temperature difference ΔTam as
⟶ 𝚫𝐓𝐚𝐯𝐠 ≈ 𝚫𝐓𝐚𝐦 =
ΔTi + ΔTe
2
=
(Ts − Tm ) + (Ts − Te)
2
= Ts − (
𝑇𝑖 + 𝑇𝑒
2
) = 𝐓𝐬 − (𝑻𝒃)
This approximation often gives acceptable results, but not reasonable results.

41. 40
From previous analysis (Heating fluid, Ts = κ),
→ m
̇ · cp · dTm = h · (Ts − Tm) · dAs = h · (Ts − Tm) · (P · dx)
We can write, dTm = −d(Ts − Tm)(Ts = κ)
⟶
d(Ts − Tm)
Ts − Tm
= −
h · P
m
̇ · cp
· dx
⟶ ∫
d(Ts − Tm )
Ts − Tm
= − ∫
h · P
m
̇ · cp
· dx
L
0
⟹ ln (
Ts − Te
Ts − Ti
) = −
h · As
m
̇ · cp
Te
Ti
→ 𝐓𝐞 = 𝐓𝐬 − [(𝐓𝐬 − 𝐓𝐢) · 𝐞
−(
𝐡·𝐀𝐬
𝐦
̇ ·𝐜𝐩
)
]
⟶ m
̇ · cp = −
h · As
ln [
(Ts − Te)
(Ts − Ti)
⁄ ]
We know, Q̇ = m
̇ · cp · (Te − Ti) = q̇ s · As = h · As · ΔTm
(Logarthmic Mean Temperature) 𝚫𝐓𝐥.𝐦 =
𝐓𝐢 − 𝐓𝐦
𝐥𝐧 [
(𝐓𝐬 − 𝐓𝐞)
(𝐓𝐬 − 𝐓𝐢)
⁄ ]
=
𝚫𝐓𝐞 − 𝚫𝐓𝐢
𝐥𝐧 (
𝚫𝐓𝐞
𝚫𝐓𝐢
)

42. 41
Free Convection
P = ρ · R · T →
P
ρ
= R · T = κ
Free convection occurs due to density difference.
Coefficient of Volume expansivity (β)
m = κ ⟹ ρ · V = κ ⟹ ρ · dV + V · ρ = 0 ⟹
dV
V
= −
dρ
ρ
β =
1
V
· (
∂V
∂T
)|
P
= −
dρ
ρ
·
1
dT
⟹ 𝐝𝛒 = −𝛃 · 𝛒 · 𝐝𝐓
In heat transfer studies, primary variable is tempeature and it’s desirable to express net buoyancy
force (ρf·Vd·g) in terms of temperature difference. Volume expansivity coefficient (β) represents the
variation of density of fluid with temperature at constant pressure.
If density difference is more, convection will be faster. Therefore, β must be large for effective free
convection.
β for ideal gas
PV = mRT → V =
mR
P
· T ⟹ (
∂V
∂T
)
P
=
mR
P
=
V
T
β =
1
V
· (
∂V
∂T
)
P
⟹ β =
1
V
· (
V
T
) ⟹ 𝛃 =
𝟏
𝐓
Grashoff Number (Gr)
It’s the ratio of buoyancy force to the viscous force.
Grashof Number (Gr)=
Buoyancy force
Viscous force
→ 𝐆𝐫 =
𝐠 · 𝛃 · 𝚫𝐓 · 𝛅𝟑
𝛎𝟐
From experiments, equation for Nusselt number is developed and is given by
→ 𝐍𝐮 = 𝐂 · (𝐆𝐫 · 𝐏𝐫)𝐧
= 𝐂 · 𝐑𝐚𝐧 (Gr. Pr = Ra(Rayleigh number))
In free convection, Grashof number and Prandtl number are important. In forced convection, Prandtl
number and Reynolds number are important.
Evaporation
Evaporation occurs at the liquid–vapor interface, when the vapor pressure is less than the saturation
pressure of liquid at given temperature.
Boiling
Boiling occurs at solid–liquid interface when a liquid is brought into contact with a surface
maintained at Ts sufficiently above saturation temperature Tsat of liquid.
Boiling heat flux
qb = h · (Ts − Tsat )
Initially when temperature is increases, boiling heat flux qb increases due to natural convection i.e.,
up to A, this region is known as natural convection region.
If the surface temperature is sufficiently high, vapor packets are formed at surface ad these vapor
packets try to move un upward direction. But they can’t reach top surface. Because when vapor
moves in upwards direction, it’s surrounded by low temperature liquid and vapor gets condensed,
they act as energy carriers. Therefore, heat flux increases from A to B.
If the surface temperature increases further, vapor packets reach to the top and hence heat flux will
reach maximum (C).

43. 42
The region from AC is called Nucleate boiling. If the temperature further increases, heat flux
decreases, because vapor forms a blanket and offers thermal resistance, thereby decreasing flux. The
point at which heat flux is minimum is called Leiden frost point.
If the temperature is further increased, radiation becomes significant and heat flux increases.
Condensation
Condensation occurs if the surface temperature of plate is less than saturation temperature.
Condensation is of two types. 1. Film wise 2. Drop wise.
In case of film wise condensation, a thick film is formed on surface and thickness of film increases
with depth and thereby offers large thermal resistance and hence along the depth δ increases and
heat transfer coefficient decreases.
In case of drop wise condensation, certain chemicals are coated on surface, so that continuous film
is not formed, but droplets are formed, this results in the sufficient free surface and direct heat
transfer takes place and heat transfer will be effective.