Find in this article, everything you need to know about Heat Transfer.

**Prepared by: Er. Ashok Subedi, Mechanical Engineer (Institute of Engineering)**

It deals with heat energy, which occurs due to the temperature difference involving the relation between the matter, time, energy and space. Transfer of heat takes place in three different modes which are conduction, convection and radiation. These all modes of heat transfers are influenced by thermophysical characteristics, geometrical constraints and temperature-related with the source and sink of the heat employed to drive the heat transfer.

**Conduction Heat Transfer:**

The process of transferring the heat from the one to another body or within a body because of the vibration of the molecules at their mean positions. The bodies must be in contact with each other through which the heat is required to be transferred. During the transfer of the heat, the matter does not move from one position to other.

Mainly the conduction takes place in solids bodies where the molecules are tightly packed by intermolecular force of attraction among themselves that’s why the vibration of the body takes place about their mean position when the heart energy is received and thus passing the heat to the surrounding molecules by the vibrations.

The conduction heat transfer is explained by the Fourier Law which is given as;

Q _x=-kA ∂T/∂x.......... (1)

Where A= differential element’s cross sectional area

**1D steady-state heat conduction through the plane wall:**

Let us consider a plane wall having thickness L, cross-sectional area A and thermal conductivity k whose either faces are exposed surrounding having the uniform temperatures say T1 and T2 respectively. Since it is considered as 1D, so the temperature gradient exists only in x-direction which is given by the Fourier law. Rearranging the equation (1) we get;

Q _x dx=-kAdT ..... .... .... (2)

Integrating the above equation for the boundary conditions we get;

Assuming the k as the constant and A for the plane wall we get;

Q _x L=-kA(T_2 -T_1 )=kA(T_1 -T_2 )

∴Q _x=kA/L (T_1 -T_2 ) ........... (4)

Equation (4) gives the magnitude of the conduction heat transfer via the plane wall.

**Radial Steady-state heat conduction via a hollow cylinder:**

Let us consider the hollow cylinder with length L, r1 and r2 as the inner and outer radii respectively having k as the thermal conductivity which is exposed to the surrounding having the uniform temperature T1 and T2 as inner and outer temperatures respectively.

Here the gradient of temperature exists radially so from **Fourier Law** can be applied as;

Q =-kA dT/dr .............. (5)

During this case, the area is given by the area of the cylinder so we can write, then equation (5) becomes;

Q =-k(2πrL)dT/dr ........... (6)

Rearranging the equation (6) we get;

Q (dr/r)=-k(2πL)dT ............ (7)

Integrating the above equation for the associated boundary conditions we get;

Therefore the magnitude of the radial conduction heat transfer via the hollow cylinder is given as;

Q =2πkL/ln(r_2/r_1 ) (T_1 -T_2 )………(10)

**Heat conduction via composite structure:**

Let us consider a plane wall having three layers of various materials where the thickness and thermal conductivity of each layer are in the pair of L1, L2, L3 and k1, k2, k3 respectively whose two faces are exposed the to the temperature having T1 and T4 and the T2 and T3 being the interface temperature respectively.

For the steady-state heat transfer, heat flowing via every layer should be the same and given as;

Q =(k_1 A)/L_1 (T_1 -T_2 )

Q =(k_2 A)/L_2 (T_2 -T_3 ) .......... (11)

Q =(k_3 A)/L_3 (T_3 -T_4 )

Rearranging the above equations in terms of temperature difference we get,

Adding the above equations we get;

Equation (14) is used for determining the conduction heat transfer through the composite plane wall.

**Heat transfer through composite hollow cylinder:**

Let us consider a composite hollow cylinder having three cylindrical layers of different materials. The inner and outer radii of the first cylinder are supposed as r1 and r2 respectively with k1 as the thermal conductivity.

Similarly, for the second and third cylinders, the parameters are r2, r3, k2 and r3, r4, and k3 as the inner radius, outer radius, and thermal conductivities respectively. T1 and T4 are the temperatures of the inner and outer temperature in which the inner and outer surface is exposed and T2 and T3 as the interface temperatures respectively.

For the steady-state heat transfer, heat flowing via every layer should be the same so;

Rearranging the above equations in terms of temperature we get;

Adding the above equations we get;

Again rearranging the above equation, we get;

The above equation (18) can be used to determine the conduction heat transfer for the composite hollow cylinders.

**Heat transfer from the finned surface:**

Extended fins surface are added to improve the heat transfer rate of the body. For creating the equation of the heat transfer of the fin several assumptions are made as below;

- Steady-state
- Properties of the materials are constant
- No internal generation of heat
- 1D conduction
- Uniform cross-section area and uniform convection across the surface area

The temperature difference between the surrounding and the fin is given as;

θ = T (x)- T∞ .......... (19)

Which permits the 1D fin equation as;

(d^2 θ)/dx^2 -m^2 θ=0 .......... (20)

Where the fin parameters m is given as;

And the boundary conditions are given as;

The solution to the differential equation for is;

θ(x)=θ_b cosh[m(L-x)]/cosh(mL)............ (22)

The heat transfer flowing via the fin’s base can be determined as;

**Efficiency and effectiveness of fin:**

The efficiency of the fin as defined as the dimensionless parameters which compare the actual to the ideal heat transfer from the fin and is given as;

When the fin having constant cross section then the efficiency of the fin is given as;

η=tanh(mL)/mL ………(27)

Similarly, the effectiveness of the fin is given as;

The efficiency of the fin is one of the factors which needs to be considered while designing the heat sink as the efficiency of the fin increases when the aspect ratio of the fin is decreased, and materials with high thermal conductivity are used.

**Convection Heat transfer:**

Convection heat transfer takes place in liquid and gases with the actual movement of the matter from one place to others within the body. A common example of convection heat transfer is the boiling of water. Using Newton’s Law of Cooling is used for determining the equation for the convection heat transfer which is given as;

**Types of Convection:**

**Forced convection:**The convection in which the flow is induced using the external source like compressor, fans, pumps and so on**Natural convection:**The convection in which the flow is induced by natural means, i.e. without using the external sources. It is instigated by the variation in the fluid density, which occurred as the result of the heating.**Mixed flow:**The convection which is the combination of the natural and forced convection

The following table shows the process and the coefficient of heat transfer in each process;

Process | Value of coefficient of heat transfer(W/m^{2}.K) |

Forced convection | Gas= 30 to 300 Oils = 60 to 1800 Water= 100 to 1500 |

Boiling | Water = 3000 to 100000 |

Natural convection | Gas= 3 to 20 Water= 60 to 900 |

Condensation | Steam= 3000 to 100000 |

In forced convention different dimensionless groups are used to determine the types of flow which are explained as below;

**Prandtl number:**It is the ratio of diffusion of momentum to the diffusion of heat and given as;

Pr=v/α

where, Pr=v/α

**2. Reynolds number:** It is the measure of the balance between the inertial and viscous forces and given as;

Re=ρUL/v

The flow of the liquid is determined according to the Reynolds number.

**3. Peclet number: **It is the product of the Prandtl and Reynolds number.

**4. Nusselt Number:** It is considered as the dimensionless heat transfer coefficient and is given as;

Nu=hL/k_f

Like above dimensionless numbers, Grashof number is used for the natural convection which is given as;

Further, the coefficient of the volumetric expansion is used for expressing the change in the fluid’s density w.r.t temperature which is given as;

β=-[1/ρ] (∂ρ/∂T)_P ............. (31)

**Natural convection from plate-fin heat sinks:**

It is widely used in the natural convection for increasing the surface areas of heat transfer which in turns reduce the resistance of the boundary layers which is given as;

R↓=1/(hA↑)

For the given base plate area, *W*L* , two factors must be considered during the selection of the fins numbers, and they are;

- More fins result in the addition of the surface area and reduction in the resistance of the boundary layer

R↓=1/(hA↑)

- More fins result in a decrease in space between the fin S which in turn decrease the coefficient of the heat transfer

R↑=1/(h↓A)

So the optimization of the space between the fins can be determined as;

Q =hA(T_w-T_∞ ) .......... (32)

Where the fins are considered to be isothermal, and the surface area of the fins is* 2nHL*, with the fin edge area neglected.

For isothermal fins with* t<S*

The corresponding value of the coefficient of the heat transfer is given as;

All the properties of the fluids are evaluated at the temperature of the film.

The following table shows the distribution of temperature and rate of heat transfer for the fin of uniform cross-sectional areas;

**Radiation heat transfer:**

Radiation heat transfer takes place without any medium and used for transferring the heat in the vacuum. It employed the electromagnetic waves for transferring the heat from one place to another. Solar radiation is the example of radiation where the heat and light reach the earth from the sun.

Blackbody radiation is the ideal radiator that absorbs all the radiation that is incident. No surface emits energy more than a blackbody at a particular wavelength and temperature, and the blackbody is independent of the direction as it is the diffuse emitter.

The radiation heat transfer rate for the large parallel plates is given as;

The radiation heat transfer rate for long concentric cylinders is;

The radiation heat transfer rate for concentric spheres is;

**Properties of radiation surface:**

Surface possess different properties, i.e. they can emit, absorbs, reflect, or transmit radiant energy. For any surface the radiation received must be reflected, absorbed, or transmitted via the materials and mathematically it is given as;

Dividing both sides *E_incident *by we get;

i.e.α+ρ+τ=1

α=absorptivity

Absorptivity is defined as the ratio of absorbed energy to the total incident energy

ρ=reflectivity

Reflectivity is defined as the ratio of reflected energy to total incident energy

τ=transmissivity

Transmissivity is defined as the ratio of transmitted energy to total incident energy

Therefore for any surface sum of absorptivity, reflectivity and transmissivity must be equal to one.

For grey body or surface the emissivity, is the ratio of grey surface emission to the black surface-emission must be less than one which is given as;

ε_S=E_S/E_B <1

**Electric Analogy for Heat Transfer and Thermal Resistance:**

Heat transfer problem which is complex and complicated can be computed using the equivalent electric circuit and applying the proper network theorem. For comparing the analogous parameters we can use Ohm’s Law as;

I=ΔV/R ……… (41)

Which indicates that the current flows because of the potential difference and the property of the substance by virtue of which it opposes the flow of the current via it, called electric resistance.

Like above equation (41) we can expresses the equation of heat transfer as;

Q =∆T/R_th ………. (42)

Which indicates that the heat flow occurs because of the temperature difference and the property of the substance by virtue of which it opposes the heat flow via it, called **thermal resistance.**

**Thermal Resistance of plane wall:**

Heat transfer via the plane wall is;

Q =(T_1-T_2)/(L/Ak) ……… (43)

Comparing the equation (42) and (43) we get;

Similarly the thermal resistance of Hollow cylinder ca be expressed as;

Thermal resistance of convective layer:

Convection heat transfer via the fluid layer is given as;

Q =(T_s-T_∞)/(1/Ah) ………. (45)

Comparing this equation with equation (42) we get;

Heat transfer via composite plane wall using electric analogy approach:

The electric equivalent circuit for a composite plane wall is;

During the composite plane wall the same quantity of heal flows via all layers so the thermal resistance are arranged in series. When the heat flow from the resistance then drop in temperature occurs and resistance for every layer is given as;

Since the thermal resistance are connected in series so the equivalent thermal resistance is equal to the sum of the individual resistance and is given as;

Then overall heat transfer for the composite plane wall becomes as;

This equation (48) is identical to the equation (14)

Heat transfer via composite cylinder using electric analogy approach:

The electric equivalent circuit for a composite cylinder is;

During the composite cylinder the same quantity of heal flows via all layers so the thermal resistance are arranged in series. When the heat flow from the resistance then drop in temperature occurs and resistance for every layer is given as;

Since the thermal resistance are connected in series so the equivalent thermal resistance is equal to the sum of the individual resistance and is given as;

Then the overall heat transfer for the composite cylinder is give as;

This equation (50) is identical to equation (18).

**Combined heat transfer and overall coefficient of heat transfer:**

**Plane wall subjected to convective medium on both sides:**

Let us consider a plane wall having L and k as the thickness and thermal conductivity respectively where one face is subjected to the hot fluid with temperature T_{A} and h_{A} as the coefficient of convective heat transfer whereas the next face is subjected to a cold fluid having T_{B} and h_{B} as the temperature and coefficient of heat transfer respectively as shown in the figure below;

For steady state heat transfer, heat flow through every layers should be same so;

Rearranging the above equations in terms of temperature difference;

Adding above equations we get;

Rearranging the above equation to get the equation for the overall heat transfer we get;

The above equation (54) can be expressed in simpler form as;

Is called the overall coefficient of heat transfer.

**Electrical Analogy Approach:**

Equivalent thermal resistance for the circuit is given as;

The overall heat transfer for composite plane wall is given as;

The above equation (58) is identical to equation (54).

**Hollow cylinder subjected to convective medium on both sides:**

Let us consider a hollow cylinder having r1 and r2 as the inner and outer radii with L and k as length and thermal conductivity respectively. Let T_{A} and h_{A} be the temperature and convective coefficient of heat transfer of the hot fluid where the inner surface is exposed.

Similarly, T_{B} and h_{B} be the temperature and convective coefficient of heat transfer of the cold fluid at which the outer surface is exposed as shown in the figure:

For steady state heat transfer, heat flow via the every layer should same so;

Rearranging the above equations in terms of temperature difference we get;

Adding above equations we get;

Rearranging the above equation to get the overall heat transfer we get;

The above equation (62) can be expressed in two different ways as below;

Further the equation (63) and (64) ca ne expressed as;

Where,

The above equations (67) and (68) are inner and outer overall coefficient of heat transfer respectively.

## Electric Analogy Approach:

Equivalent thermal resistance is given as;

Overall heat transfer is given as;

Which is identical to the equation (62).