Fourier Law and Thermal Conductivity MCQ Quiz - Objective Question with Answer for Fourier Law and Thermal Conductivity - Download Free PDF

Correct option is: (3) 5 / 3

In series, R_eq = R₁ + R₂ + R₃

= 1 / (2KA) + 1 / (KA) + 1 / (2KA)

= 4 / (2KA)

R_eq = 2 / (KA)

In series rate of heat flow is same

(3T − T₁) / R₁ = (3T − T) / R_eq

((3T − T₁) KA) / 1 = (2T) KA / 2

⇒ 6T − 2T₁ = T

⇒ T₁ = 5T / 2         ...(1)

Now, equate heat flow rate in 3^rd section and total section

(T₂ − T) / R₃ = (3T − T) / R_eq

((T₂ − T)(2KA)) / 1 = (2T KA) / 2

⇒ 2T₂ − 2T = T

⇒ T₂ = 3T / 2         ...(2)

By equation (1) and (2)

T₁ / T₂ = (5T / 2) / (3T / 2) = 5 / 3

Explanation:

Fourier's Law of Heat Conduction

• Fourier's law of heat conduction is a fundamental principle that describes the flow of heat through a material.
• It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat flows. Mathematically, it is expressed as:

q = -k × ΔT

where:
q is the heat flux (amount of heat per unit area per unit time), k is the thermal conductivity of the material, ΔT is the temperature gradient (difference in temperature divided by the distance).

• The negative sign in the equation is crucial as it indicates the direction of heat flow.
• Heat naturally flows from regions of higher temperature to regions of lower temperature.
• The negative sign ensures that this directionality is correctly represented in the mathematical formulation.
• Essentially, it indicates that the heat flux is in the direction opposite to the temperature gradient.

Explanation:

Thermal Conductivity of Materials

• Thermal conductivity is a measure of a material's ability to conduct heat.
• It is typically represented by the symbol 'k' and is measured in watts per meter-kelvin (W/m·K).
• Materials with high thermal conductivity can transfer heat quickly and efficiently, while those with low thermal conductivity are good insulators.
• Copper is known for its excellent thermal conductivity, which is one of the highest among common materials.
• The thermal conductivity of copper is approximately 401 W/m·K.
• This high value makes copper an ideal choice for applications that require efficient heat transfer, such as in heat exchangers, radiators, and electronic components.
• Copper's ability to conduct heat effectively is due to its free electrons, which can move easily within the metal and transfer thermal energy quickly.

Importance of Thermal Conductivity in Copper:

1. Heat Exchangers: Copper is widely used in heat exchangers due to its high thermal conductivity. It allows for efficient transfer of heat between fluids or gases, making the process of heating or cooling more effective.

2. Electronic Components: In electronics, copper is used in printed circuit boards (PCBs) and heat sinks to dissipate heat generated by components. This helps in maintaining the performance and longevity of electronic devices.

3. Plumbing: Copper pipes are commonly used in plumbing systems due to their ability to conduct heat efficiently, which is beneficial for hot water supply systems.

4. Industrial Applications: Copper's thermal conductivity makes it suitable for various industrial applications, including the manufacturing of machinery and equipment where heat transfer is crucial.

Additional Information

Rubber

• Rubber has a very low thermal conductivity, typically around 0.1 to 0.2 W/m·K. This makes it an excellent insulator rather than a conductor of heat. Rubber is often used in applications where heat insulation is required, such as in gaskets, seals, and insulating mats.

Glass

• Glass has a moderate thermal conductivity, generally around 1 W/m·K. It is not as efficient as metals like copper but is better than materials like rubber and wood. Glass is used in applications where moderate heat transfer is acceptable, such as in windows and some types of cookware.

Wood

• Wood has a low thermal conductivity, typically ranging from 0.1 to 0.2 W/m·K, similar to rubber. This property makes wood a good insulator. It is used in construction for its insulating properties and in various applications where heat retention is important.

Explanation:

Let radii be r, 2r, and 3r

Case 1:

\({R_1} = \frac{{\ln \left( {\frac{{2r}}{r}} \right)}}{{2\pi kL}}\)

\(∴ {R_1} = \frac{{\ln \left( 2 \right)}}{{2\pi kL}}\)

\({R_2} = \frac{{\ln \left( {\frac{{3r}}{{2r}}} \right)}}{{2\pi kL}}\)

\(∴ {R_2} = \frac{{\ln \left( {1.5} \right)}}{{2\pi \left( {3k} \right)L}}\)

\(∴ {Q_1} = \frac{{{\rm{\Delta }}T\left( {2\pi kL} \right)}}{{\ln \left( 2 \right) + \frac{{{\rm{ln}}\left( {1.5} \right)}}{6}}}\)

Case 2:

\({R_3} = \frac{{\ln \left( {\frac{{2r}}{r}} \right)}}{{2\pi kL}}\)

\({R_3} = \frac{{\ln \left( 2 \right)}}{{2\pi \left( {3k} \right)L}}\)

\({R_4} = \frac{{\ln \left( {\frac{{3r}}{{2r}}} \right)}}{{2\pi kL}}\)

\({R_4} = \frac{{\ln \left( {1.5k} \right)}}{{2\pi kL}}\)

\(∴ {Q_2} = \frac{{{\rm{\Delta }}T\left( {2\pi kL} \right)}}{{\ln \left( {1.5} \right) + \frac{{{\rm{ln}}\left( 2 \right)}}{6}}}\)

Now when you compare the values of Q1 and Q2

We can see that Q1 < Q­2

∴ We can say that when a material with lower thermal conductivity is used for the inner layer and one with higher thermal conductivity for the outer layer then the heat transfer is less compared to the reverse case.

Concept:

The efficiency of a fin (\( \eta_f \)) is defined as the ratio of the actual heat transfer from the fin to the heat transfer that would occur if the entire fin were at the base temperature.

Calculation:

For a long straight fin with a uniform cross-sectional area, the efficiency can be expressed as:

\( \eta_f = \frac{\tanh(mL)}{mL} \)

where:

 \( m = \sqrt{\frac{hP}{kA_c}} \)

 \( h \) = Convective heat transfer coefficient

 \( P \) = Perimeter of the fin cross-section

 \( k \) = Thermal conductivity of the fin material

 \( A_c \) = Cross-sectional area of the fin

 \( L \) = Length of the fin

The impact of increasing thermal conductivity \( k \):

From the above relation, it is clear that \( m \) depends on \( k \):

\( m = \sqrt{\frac{hP}{kA_c}} \)

As the thermal conductivity \( k \) increases, the value of \( m \) decreases because \( m \) is inversely proportional to the square root of \( k \).

When \( m \) decreases, the argument \( mL \) in the tanh function also decreases.

For large values of \( mL \):

\( \tanh(mL) \approx 1 \)

Therefore, \( \eta_f \approx \frac{1}{mL} \)

Thus, as \( m \) (and hence \( mL \)) decreases, \( \eta_f \) increases because the argument of the tanh function decreases, making the tanh function closer to the argument itself for small values.

The impact of increased thermal conductivity on fin efficiency:

When the thermal conductivity of the fin material increases, \( m \) decreases, leading to an increase in fin efficiency \( \eta_f \).

 

Hence, the correct answer is 4) The efficiency increases.

Explanation:

Gases transfer heat by the collision of molecules.

As the temperature increases, the kinetic energy of molecules of gases also increases and eventually collision between molecules also increases which increases the thermal conductivity of gases.

∴ As temperature increases the thermal conductivity of gases increases.

For liquid and solids, generally as the temperature increases, the thermal conductivity decreases.

Explanation:

• The variation in thermal conductivity of a material with temperature in the temperature range of interest is given by:
• k(T) = k0 (1 + βT) where β is called the temperature coefficient of thermal conductivity.
• The variation of temperature in a plane wall during steady one - dimensional heat conduction for the cases of constant and variable thermal conductivity is

Concept:

• Thermal Conductivity: When one end of a metal rod is heated, heat flows by conduction from the hot end to the cold end. In this process, each cross-section of the rod receives some heat from the adjacent cross-section towards the hot end.

It is found that the amount of heat Q that flows from hot to cold face during steady-state:

Or, \(Q = \frac{{KA\left( {{T_1} - {T_2}} \right)t}}{x}\)

where K = Coefficient of thermal conductivity of the material.

Rate of conduction of heat energy is given by:

\(\frac{{dQ}}{t} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{x} = KA\frac{{{\bf{\Delta }}T}}{x}\)

Calculation:

Given:

Rate of conduction of heat energy is given by:

\(\frac{{dQ}}{t} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{x} = KA\frac{{{\bf{\Delta }}T}}{x}\)

Coefficient of thermal conductivity of material will be,

\(K = \frac{{dQ \times x}}{{t \times A \times {\rm{\Delta }}T}}\)

SI unit of Q = J/s = W, A = m2, x = m and ΔT = K

\(\therefore K = \frac{{dQ \times x}}{{t \times A \times {\rm{\Delta }}T}} = \frac{{J \cdot m}}{{sec \cdot {m^2} \cdot K}} = W{m^{ - 1}}{K^{ - 1}}\)

Therefore, the SI unit of the thermal conductivity is Wm-1K-1.

Explanation:

\(Q = \frac{{{T_1} - {T_2}}}{{\frac{L}{{kA}}}} = kA\frac{{{T_1} - {T_2}}}{L}\)

\(Q = kA\frac{{dT}}{{dx}}\)

For same heat transfer:

\(k \propto \frac{1}{{\frac{{dT}}{{dx}}}}\)

\({\rm{k}} \propto \frac{1}{{{\rm{Temperature\;gradient}}}}\)

\({\left[ {\frac{{dT}}{{dx}}} \right]_1} > {\left[ {\frac{{dT}}{{dx}}} \right]_2} \Rightarrow {k_1} < {k_2}\)

 

Heat conduction through the plane wall	\(Q = \frac{{{T_1} - {T_2}}}{{\frac{L}{{kA}}}}\)
Heat conduction through a hollow cylinder	\(Q = \frac{{{T_1} - {T_2}}}{{\frac{{\ln \left( {\frac{{{r_o}}}{{{r_i}}}} \right)}}{{2\pi kL}}}}\)
Heat conduction through the hollow sphere	\(Q = \frac{{{T_1} - {T_2}}}{{\frac{{{r_o} - {r_i}}}{{4\pi k{r_o}{r_i}}}}}\)

Concept:

Thermal conductivity: 

• Thermal conductivity is the property of a particular substance and shows the ease by which the process takes place
• Higher the thermal conductivity more easily will be the heat conduction through the substance
• The thermal conductivity of a medium strongly depends on the Atomic arrangement of medium and Operating temperature.
• It is denoted by K and the SI unit is watt per (meter.kelvin) i.e. (W/m-K).
• Some values of the thermal conductivity of common materials are.

1. Diamond -  2200 W/m-K
2. Silver - 430 W/m-K 
3. Copper – 385 W/m-K
4. Aluminium – 209 W/m-K
5. Brass – 109 W/m-K
6. Ice - 202 W/m-K
7. Air – 0.0238 W/m-K

Explanation:

Heat flow has an analogy in the flow of electricity.

Ohm’s law states that the current ‘I’ flowing through a wire is equal to the voltage potential (E1 – E2) divided by the electrical resistance Re.

\(I = \frac{{{E_1} - {E_2}}}{{{R_e}}}\)

Since the temperature difference and heat flux in conduction are similar to the potential difference and electrical current respectively, the rate of heat conduction through the wall can be written as

\(Q = \frac{{{T_1} - {T_2}}}{{L/kA}} = \frac{{{T_1} - {T_2}}}{{{R_{th}}}}\)

Where Rth = \(\frac{{L}}{{kA}}\) is the conductive thermal resistance to heat flow offered by the wall.

Fourier's law of heat conduction is analogous to Ohm's law for electrical circuits. In the analogy:

• the heat flow (Q) corresponds to the electrical current (I)
• the thermal resistance to the electrical resistance
• temperature (T) to the electrical voltage (V)
• thermal conductivity to the electrical conductivity
• heat capacity to the capacitance

Concept:

According to Fourier’s law, the rate of heat flow, Q through a homogeneous solid is directly proportional to the area A, of the section at the right angles to the direction of the heat flow, and to the temperature difference dT along the path of heat flow.

\(Q = - kA\frac{{dT}}{{dx}}\;\)

Calculation:

\(\begin{array}{l} \frac{{{k_1}}}{{{k_2}}} = \frac{1}{2}\\ \frac{{{\rm{Δ }}{T_1}}}{{{\rm{Δ }}{T_2}}} = 1\\ Q = \frac{{kA{\rm{Δ }}T}}{t} \end{array}\)

As A1 = A2 and t1 = t2 hand ΔT₁ = ΔT₂  

\(\therefore \frac{{{Q_1}}}{{{Q_2}}} = \frac{{{k_1}}}{{{k_2}}} = \frac{1}{2} \)

Q1 : Q2 = 1 : 2

A liquid's density is a measure of how heavy it is for the amount measured. When equivalent amounts or volumes of two distinct liquids are weighed, the liquid that weighs more is more dense. The transfer of vibrational forces between molecules is the mechanism of conduction in fluid flow. As the temperature of the liquid rises, the kinetic energy of the molecules rises as well. This resulted in an increase in vibrational forces.Due this density difference heat transfer in fluid flow occurs.

Concept:

Fourier law of heat conduction:- 

\(Q = \frac{{{\bf{\Delta }}T}}{{\frac{b}{{kA}}}}\)

Calculation:

Given:

b = 0.6 m, k = 0.6 W/mK, A = 1 m², T_i = 1840°C, T_o = 340°C 

\(Q = \frac{{{T_i} - {T_o}}}{{\frac{b}{{kA}}}}\)

\(Q = \;\frac{{\left( {1840 - 340} \right)}}{{\frac{{0.6}}{{0.6 \times 1}}}}\)

Q = 1500 W

Heat conduction throw a hollow cylinder

\(Q = \frac{{{\rm{\Delta }}T}}{{\frac{{\left( {\ln \frac{{{r_2}}}{{{r_1}}}} \right)}}{{2\pi kl}}}}\)

Heat conduction throw a hollow sphere

\(Q = \frac{{{\rm{\Delta }}T}}{{\frac{{{r_2} - {r_1}}}{{4\pi k{r_1}{r_2}}}}}\)

 

 

Concept:

Thermal resistance:

• Thermal resistance is defined as the ratio of the temperature difference between the two faces of a material to the rate of flow per unit area.

          \({\bf{i}}.{\bf{e}}.\;{{\bf{R}}_{{\bf{thermal}}}} = \frac{{{\bf{\Delta T}}}}{{\bf{Q}}}\)

• Its unit is K/W 
• The concept of thermal resistance is used to solve composite layer problems.

​For a solid plate, thermal resistance is given by:

\({{\bf{R}}_{{\bf{thermal}}}} = \frac{{\bf{L}}}{{{\bf{kA}}}} \)

Calculation:

Given:

k₁ = k, k₂ = 2k

Since the slabs are in contact with each other side by side. Therefore the thermal resistances will be in series.

The equivalent thermal resistance to the flow of heat is given by,

\({R_{eq}} = {R_1} + {R_2}\)

\(\frac{{{L_{eq}}}}{{{k_{eq}}A}} = \frac{{{L_1}}}{{{k_1}A}} + \frac{{{L_2}}}{{{k_2}A}}\)

Since \({L_1} = {L_2} = L\)

\(\frac{{2L}}{{{k_{eq}}}} = \frac{L}{{{k_1}}} + \frac{L}{{{k_2}}}\)

\(\frac{{{k_{eq}}}}{2} = \frac{{{k}\times{2k}}}{{{k} + {2k}}}\\\)

\({k_{eq}} = \frac{{4k}}{{3}}\)