Conduction
Fouriers Law of Heat Conduction
Learning Objectives:

Understand and are able to recognize conduction, convection and radiation as modes of heat transfer.

Fourier’s law of conduction.

Able to solve onedimensional heat conduction problems using the energy equation and Fourier’s law.
Conduction expressed by Fourier’s law of heat conduction as written below:
\[\dot{Q}_{\mathrm{conduction}} =  k A \frac{dT}{dx}\]
Where, \(\displaystyle \dfrac{dT}{dx}\) is the temperature gradient. k is the thermal conductivity A is the area which is normal to the direction of heat transfer.
Conduction expressed by Fourier’s law of heat conduction as written below:
Where, \(\displaystyle \dfrac{dT}{dx}\) is the temperature gradient. k is the thermal conductivity A is the area which is normal to the direction of heat transfer.
Solved Example: 80101
Fourier's law of heat conduction is valid for:
A. One dimensional cases only
B. Two dimensional cases only
C. Three dimensional cases only
D. Regular surfaces having nonuniform temperature gradients
Among the assumptions for Fourier's law of heat conduction, the main ones are: Steady state heat conduction, one directional heat flow and no internal heat generation. For two dimensional steady state heat conduction, Laplace equation is used.
Correct Answer: A
Solved Example: 80102
Unit of thermal conductivity in S.I. units is: (ISRO URSC Tech Asst Mech Nov 2016)
A. J/m$^2$ sec
B. J/m $^\circ$K sec
C. W/m $^\circ$K
D. (B) and (C) above.
In the International System of Units (SI), thermal conductivity is measured in Watts per meterKelvin (W/(m$\cdot$K)). Note that Watt is Joule/second.
In Imperial units, thermal conductivity is measured in BTU/(h $\cdot$ ft $\cdot$ $^\circ$F).
Correct Answer: D
Solved Example: 80103
Thermal conductivity of nonmetallic amorphous solids with decrease in temperature:
A. Increases
B. Decreases
C. Remains constant
D. May increase or decrease depending on temperature
Correct Answer: B
Solved Example: 80104
The insulation ability of an insulator with the presence of moisture would: (BPSC AE ME 2012Part V)
A. Increases
B. Decrease
C. Remain unaffected
D. May increase/decrease depending on temperature and thickness of insulation
Correct Answer: B
Solved Example: 80105
Heat is closely related with:
A. Liquids
B. Energy
C. Temperature
D. Entropy
Correct Answer: C
Solved Example: 80106
Pick up the wrong case. Heat flowing from one side to other depends directly on:
A. Face area
B. Time
C. Thickness
D. Temperature difference
Correct Answer: C
Solved Example: 80107
Metals are good conductors of heat because: (RSMSSB JE Mech Dec 2020)
A. Their atoms collide frequently
B. Their atoms are relatively far apart
C. They contain free electrons
D. They have high density
Correct Answer: A
Solved Example: 80108
Which of the following is a case of steady state heat transfer: (SSC JE ME Paper 6 Mar 2017)
A. I.C. engine
B. Air preheaters
C. Heating of building in winter
D. None of the above.
Correct Answer: D
Solved Example: 80109
Heat flows from one body to other when they have: (ISRO VSSC Tech Asst Mech Feb 2015)
A. Different heat contents
B. Different specific heat
C. Different atomic structure
D. Different temperatures
Heat flow takes places when the temperatures of two bodies are different. This is true even when the body with lower temperature has higher heat content. Heat flow is analogous to fluid flow, where fluid flows from higher level to lower level, even though the lower level reservoir has more fluid content. The 'level' in fluid flow is similar to 'temperature' in heat flow.
Correct Answer: D
Solved Example: 80110
In heat transfer, conductance equals conductivity (kcal/hr/sqm/$^\circ$C/cm) divided by:
A. hr (time)
B. sq m (area)
C. $^\circ$C (temperature)
D. cm (thickness)
Correct Answer: D
Solved Example: 80111
The amount of heat flow through a body by conduction is:
A. Directly proportional to the surface area of the body
B. Directly proportional to the temperature difference on the two faces of the body
C. Inversely proportional to the thickness of the body
D. All of the above.
Correct Answer: D
Solved Example: 80112
Which of the following has least value of conductivity? (UPPSC AE Mech 2013 Paper II)
A. Water
B. Plastic
C. Rubber
D. Air
The thermal conductivity of a material is a measure of its ability to conduct heat. A material with high resistance to thermal conduction will have least value of conductivity. Water has thermal conductivity = 0.6089 W/m K compared to that of air = 0.026 W/m K, fiber reinforced plastic has 0.23 W/m K, whereas CR rubber has 0.16 W/m K. (Values referred from Wikipedia.)
Correct Answer: D
Solved Example: 80113
Which of the following is expected to have highest thermal conductivity? (SSC JE ME Mar 2017 Paper 6)
A. Steam
B. Solid ice
C. Melting ice
D. Water
Correct Answer: B
Solved Example: 80114
Thermal conductivity of a material may be defined as the:
A. Quantity of heat flowing in one second through one cm cube of material when opposite faces are maintained at a temperature difference of 1$^\circ$C
B. Quantity of heat flowing in one second through a slab of the material of area one cm square, thickness 1 cm when its faces differ in temperature by 1$^\circ$C
C. Heat conducted in unit time across unit area through unit thickness when a temperature difference of unity is maintained between opposite faces
D. All of the above.
Correct Answer: D
Solved Example: 80115
Which of the following has maximum value of thermal conductivity?
A. Aluminum
B. Steel
C. Brass
D. Copper
Correct Answer: A
Solved Example: 80116
Moisture would find its way into insulation by vapour pressure unless it is prevented by:
A. High thickness of insulation
B. High vapour pressure
C. Less thermal conductivity insulator
D. A vapour seal
Correct Answer: D
Solved Example: 80117
Thermal diffusivity is: (TANGEDCO AE ME 2015)
A. A dimensionless parameter
B. Function of temperature
C. Used as mathematical model
D. A physical property of the material
Correct Answer: D
Solved Example: 80118
Thermal diffusivity of a substance is:
A. Proportional of thermal conductivity
B. Inversely proportional to k
C. Proportional to (ln k)
D. Inversely proportional to k$^2$
Correct Answer: A
Solved Example: 80119
Unit of thermal diffusivity is:
A. m$^2$/hr
B. m$^2$/hr$^\circ$C
C. kcal/m$^2$ hr
D. kcal/m.hr$^\circ$C
Correct Answer: A
Solved Example: 80120
Thermal conductivity of wood depends on:
A. Moisture
B. Density
C. Temperature
D. All of the above.
Correct Answer: D
Solved Example: 80121
Heat conducted through unit area and unit thick face per unit time when temperature difference between opposite faces is unity,is called:
A. Thermal resistance
B. Thermal coefficient
C. Temperature gradient
D. Thermal conductivity
Correct Answer: D
Solved Example: 80122
Which of the following property of air does not increase with rise in temperature:
A. Thermal conductivity
B. Thermal diffusivity
C. Density
D. Dynamic viscosity
Correct Answer: C
Solved Example: 80123
The thermal diffusivities for gases are generally:
A. More than those for liquids
B. Less than those for liquids
C. More than those for solids
D. Dependent on the viscosity
Correct Answer: A
Solved Example: 80124
The thermal diffusivities for solids are generally:
A. Less than those for gases
B. Less than those for liquids
C. More than those for liquids and gases
D. More or less same as for liquids and gases
Correct Answer: C
Solved Example: 80125
Thermal diffusivity of a substance is:
A. Directly proportional to thermal conductivity
B. Inversely proportional to density of substance
C. Inversely proportional to specific heat
D. All of the above.
Correct Answer: D
Solved Example: 80126
Which one of the following states the fundamental modes of heat transfer?
A. Friction, Convection, Nuclear reactions.
B. Conduction, Convection, Radiation
C. Ionization, Evaporation, Radiation
D. Chemical reaction, Electrical resistance, Nuclear reactions.
Correct Answer: B
Conduction through a Plane Wall
Learning Objectives:

To find rate of heat transfer by steady heat conduction through multilayer plane walls.
\[\dot{Q} =  k A \frac{T_2T_1}{L}\]
where, \(T_1\) and \(T_2\) are the temperature of surfaces of the wall
Solved Example: 80201
A furnace is made of a red brick wall of thickness 0.5 m and conductivity 0.7 W/mK. For the same heat loss and temperature drop, this can be replaced by a layer of diatomite earth of conductivity 0.14 W/mK and thickness: (ISRO RAC 2017)
A. 0.5 m
B. 0.1 m
C. 0.2 m
D. 0.25 m
Here to achieve the same heat transfer and temperature drop, the conductive resistance should be kept same.
\[R= \dfrac{t}{k}\]
where t= thickness of wall and k= conductivity of material.
So,
\[R = \dfrac{0.5}{0.7}= \dfrac{t}{0.14}, \ \ t = 0.1\ m\]
This is thickness for wall of new material.
Correct Answer: B
Solved Example: 80202
A square silicon chip (k=150 W/m.k) is of width W=5mm on a side and of thickness t=1mm. the chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 4W are being dissipated in circuits mounted to the back surface of the chip, what is the steadystate temperature difference between back and front surfaces?
A. 2.1$^\circ C$
B. 1.1$^\circ C$
C. 1.85$^\circ C$
D. 1.33$^\circ C$
\begin{align*} \Delta t &= \dfrac {0.001\times 4}{150\times \left( 0.005\right)} =1.1^\circ C \end{align*}
Correct Answer: B
Conduction through a Cylindrical Wall
Learning Objectives:

To find rate of heat transfer by steady heat conduction through multilayer cylindrical walls.
\[\dot{Q} = \dfrac{2 \pi k L (T_1  T_2)}{ln\left(\dfrac{r_2}{r_1} \right )}\]
\[R_{cyl} = \dfrac{ln\left(\dfrac{r_2}{r_1}\right)}{2 \pi k L }\] where, \(R_{cyl}\) is the conduction resistance of the cylinder layer.
Solved Example: 80301
In a hollow cylindrical tube of length L, conductivity k, inner radius r and outer radius $r_o$, if $T_i$ is greater than $T_o$, the thermal resistance is: (GATE ME 2016 Shift II)
A. $\dfrac{1}{2\pi kL} ln \left ( \dfrac{r_i}{r_o} \right )$
B. $\dfrac{1}{2\pi kL} ln \left ( \dfrac{r_o}{r_i} \right )$
C. $\dfrac{1}{2\pi kL} ln \left ( r_i \times r_o \right )$
D. $2\pi kL ln \left ( \dfrac{r_i}{r_o} \right )$
Correct Answer: B
Solved Example: 80302
A 160 mm diameter pipe carrying saturated steam is covered by layer of lagging of thickness of 40 mm (k=0.8 W/m $^{\circ}\mathrm{C}$) later, an extra layer of lagging 10 mm thick (k=1.2 W/m $^{\circ}\mathrm{C}$) is added. If the surrounding temperature remains constant and heat transfer coefficient for both the lagging materials is 10 W/sq m $^{\circ}\mathrm{C}$, determine the percentage change in the rate of heat loss due to extra lagging layer.
A. 0.193%
B. 0.113%
C. 0.223%
D. 0.563%
Correct Answer: C
Conduction through a Sphere
Learning Objectives:

To be able to solve onedimensional problems in radial systems.
\[R_{\mathrm{sphere}} = \dfrac{r_2  r_1}{4 \pi r_1 r_2 k}\] The convection resistance remains the same in both cylindrical and spherical coordinates, \(R_{conv}\) = \(\dfrac{1}{hA}\). However, note that the surface area A = 2 \(\pi\)rL (cylindrical) and A = 4\(\pi\)r\(^2\) (spherical) are functions of radius.
Solved Example: 80401
Consider onedimensional steady state heat conduction along xaxis (0 < x < L), through a plane wall with the boundary surfaces (x=0 and x=L) maintained at temperatures of 0$^{\circ}\mathrm{C}$ and 100$^{\circ}\mathrm{C}$. Heat is generated uniformly throughout the wall. Choose the CORRECT statement.
A. The temperature distribution is symmetric about the midplane of the wall.
B. The direction of heat transfer will be from the surface at 100$^{\circ}\mathrm{C}$ to the surface at 0$^{\circ}\mathrm{C}$.
C. The temperature distribution is linear within the wall.
D. The maximum temperature inside the wall must be greater than 100$^{\circ}\mathrm{C}$.
Since heat is generated uniformly throughout the wall, option B is not correct. Also, this heat generation will make some portion of the wall at higher temp. than 100 deg. Celsius.
Correct Answer: D
Solved Example: 80402
A freezer compartment consists of a cubical cavity that is 2 m on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of Styrofoam insulation (k=0.030 W/m.K) which must be applied to the top and side walls to ensure a heat load less than 500 W, when the inner and outer surfaces are 10 $^\circ$ C and 35$^\circ$C?
A. 27mm
B. 36mm
C. 13.5mm
D. 54mm
\begin{align*} q &=kA\dfrac {\Delta t}{\Delta x}\\ \Delta x &=\dfrac {5\times 0.03\times 45\times 4}{500}\\ \Delta x &=0.054\ m =54\ mm \end{align*}
Correct Answer: D
Solved Example: 80403
If the inner and outer walls of a sphere having surface areas of A1 and A2 and inner and outer radii r1 and r2 are maintained at t1 and t2, then rate of heat flow will be: (ISRO Refrigeration and Air Conditioning 2013)
A. $\dfrac{k}{\sqrt{A_1A_2}}\dfrac{t_1\ \ t_2}{r_2\ \ r_1}$
B. $k\sqrt{A_1A_2}\dfrac{t_1\ \ t_2}{r_2\ \ r_1}$
C. $4\pi k\dfrac{t_1\ \ t_2}{\sqrt{A_1A_2}}$
D. $4\pi kr_1r_2\dfrac{t_1\ \ t_2}{\sqrt{A_1A_2}}$
Correct Answer: B
Thermal Resistance of Various Objects
Learning Objectives:

Explain the concept of thermal conductivity and thermal resistance.

Explain the similarities and relationship of thermal and electrical conductivity.

Compute thermal resistance for individual items such as plane wall, pipes and spheres with or without insulation.

Describe the concept of thermal resistance of air layers.

Identify series and parallel combinations from reallife thermal insulation scenarios for application of thermal resistance concepts.

Evaluate thermal resistance of simple shapes such as plane wall, cylindrical wall.

Evaluate thermal resistance of thin air film layer.
There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. Just as an electrical resistance is associated with the conduction of electricity, a thermal resistance may be associated with the conduction of heat.
The rate of heat transfer through a plane wall corresponds to the electric current, the thermal resistance corresponds to electrical resistance and the temperature difference corresponds to voltage difference across the plane wall.
Equivalent thermal circuits may be used for more complex systems, such as composite walls. Such walls may involve any number of series and parallel thermal resistances due to layers of different materials.
The thermal resistance is defined as the ratio of the temperature difference, dT, to the heat transfer Q. This is analogous to Ohm’s law, in which the electrical resistance is defined as the ratio of the voltage drop across a resistor to the current flow across the resistor. Thermal resistance is a heat property and a measurement of a temperature difference by which an object or material resists a heat flow (heat per time unit or thermal resistance). Thermal resistance is the reciprocal of thermal conductance.
Defining resistance as the ratio of a driving potential to the corresponding transfer rate, it follows that the thermal resistance for conduction is: \[R = \dfrac{T_1  T_2}{q_x} = \dfrac{L}{kA}\]
Plane Wall:
\(R = \dfrac{L}{kA}\)
Cylindrical Wall:
\[R = \dfrac{ln\left(\dfrac{r_2}{r_1}\right)}{2\pi k L}\]
Convection Resistance:
\[R = \dfrac{1}{hA}\]
Solved Example: 80501
Hot air at a temp. of 60 $^{\circ}\mathrm{C}$ is flowing through a steel pipe of 100 mm diameter. The pipe is covered with two layers of different insulating layers of thickness 50 and 30 mm and their thermal conductivities are 0.23 and 0.37 W/m $^{\circ}\mathrm{C}$. The inside and outside heat transfer coefficients are 58 and 12 W/sq m$^{\circ}\mathrm{C}$.the atm temperature is at 25$^{\circ}\mathrm{C}$. Find the rate of heat loss from a 50 m length of pipe.
A. 1.971 kW
B. 2.334 kW
C. 3.018 kW
D. 1.441 kW
Inside heat transfer resistance \[\begin{split}
\dfrac{1}{h_i A} & = \dfrac{1}{58 \times 2\pi \times 50 \times
10^{3} \times 50} = 1.097 \times 10^{3} \dfrac{K}{W}
\end{split}\] Outside heat transfer resistance \[\begin{split}
\dfrac{1}{h_o A} & = \dfrac{1}{12 \times 2\pi \times 130 \times
10^{3} \times 50} = 2.04 \times 10^{3} \dfrac{K}{W}
\end{split}\] Inside insulation heat transfer resistance \[\begin{split}
\dfrac{ln \left(\dfrac{r_2}{r_1}\right)}{2 \pi K L} & = \dfrac{ln
\left(\dfrac{100}{50}\right)}{2 \pi \times 0.23 \times 50} = 9.59 \times
10^{3} \dfrac{K}{W}
\end{split}\] Outside insulation heat transfer resistance \[\begin{split}
\dfrac{ln \left(\dfrac{r_3}{r_2}\right)}{2 \pi K L} & = \dfrac{ln
\left(\dfrac{130}{100}\right)}{2 \pi \times 0.37 \times 50} = 2.257
\times 10^{3} \dfrac{K}{W}
\end{split}\] Total heat transfer resistance = 14.984 \(\times 10^{3} \dfrac{K}{W}\)
Rate of Heat loss = \(\dfrac{T_2 
T_1}{R_{Total}}\) = \(\dfrac{6025}{14.984 \times 10^{3} }\)=
2.334 kW
Correct Answer: B
Composite Plane Wall
Learning Objectives:

Draw the thermal circuit for three layer composite wall.
Steps for calculating heat transfer rate for a composite wall having multiple layers of insulation. For more details, refer solved problem for the same chapter.

Calculate convection resistance for the inside air film using \(R_i = \dfrac{1}{h_iA}\)

Calculate convection resistance for the outside air film using \(R_o = \dfrac{1}{h_oA}\)
If these h values are not specified in the problem, then it is assumed that they do not contribute much towards the overall heat transfer resistance and hence the above two terms can be neglected. 
Calculate conduction resistance for first insulation using \(R = \dfrac{L}{kA}\) for plane wall or \(R = \dfrac{ln\left(\dfrac{r_2}{r_1}\right)}{2\pi k L}\) for cylindrical wall.

repeat this process for all insulation.

Add all these resistances in case of series combination. If any two or more resistances are parallel then use parallel resistance formula for that combination only.

Calculate the overall heat transfer.

if junction (boundary) temperatures are required, then consider resistances only upto that point.
Solved Example: 80601
A composite wall is made of two layers of 0.3 m and 0.15 m thickness with surfaces held at 600$^{\circ}\mathrm{C}$ and 20$^{\circ}\mathrm{C}$ resp. If the conductivities are 20 and 50 W/mK, determine the heat conducted.
A. 2.85W/mK
B. 0.23W/mK
C. 1.53W/mK
D. 1.23W/mK
Correct Answer: C
Solved Example: 80602
A flat wall of a furnace is made up of fire brick, insulating brick and building brick of thicknesses 25 cm, 12.5 cm and 25 cm resp. The inside wall is at a temperature of 600 $^{\circ}\mathrm{C}$ and atm. Temperature is 20 $^{\circ}\mathrm{C}$. If the heat transfer coefficient for the outer surface is 10 W/sq m deg C, calculate the loss per sq meter of wall area. Take: K of wall brick=1.4 W/m $^{\circ}\mathrm{C}$, K of insulating brick= 0.2 W/m $^{\circ}\mathrm{C}$, K of fire brick= 0.85 W/m $^{\circ}\mathrm{C}$.
A. 0.48 kW per sq m
B. 1.16 kW per sq m
C. 0.86 kW per sq m
D. 2.94 kW per sq m
Total heat transfer resistance \begin{equation*} \begin{split} & = \dfrac{25 \times 10^{2}}{0.85 \times 1} + \dfrac{12.5 \times 10^{2}}{0.2 \times 1}\\ & + \dfrac{25 \times 10^{2}}{1.4 \times 1} + \dfrac{1}{10 \times 1} = 1.198 \dfrac{K}{W} \end{split} \end{equation*} Rate of heat transfer = $\dfrac{600  20}{1.198}$ = 484 W per sq m = 0.48 kW per sq m
Correct Answer: A