Conduction MCQ Quiz - Objective Question with Answer for Conduction - 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:

Thermal Conductivity of Materials

• Thermal conductivity is a physical property of materials that measures their ability to conduct heat. It is defined as the amount of heat (in watts) that passes through a material with a given area and thickness when a temperature difference exists across the material. The unit of thermal conductivity is watts per meter per degree Celsius (W/m·°C).

Working Principle:

• The working principle of thermal conductivity involves the transfer of heat energy from the hotter part of the material to the cooler part. This transfer occurs through the vibration of atoms and molecules or through the movement of electrons in the case of metals. Materials with high thermal conductivity can transfer heat quickly and efficiently, while materials with low thermal conductivity are better insulators.

Aluminium:

• Aluminium is known for its high thermal conductivity. It is a metal with a thermal conductivity value of approximately 237 W/m·°C, which is significantly higher compared to other materials listed in the options. This high thermal conductivity makes aluminium an excellent material for applications requiring efficient heat transfer, such as heat exchangers, cooling systems, and electronic components.
• Aluminium's ability to conduct heat quickly is due to the presence of free electrons that can move easily through the material, transferring thermal energy from one part to another. This property is particularly useful in industries where rapid heat dissipation is critical to prevent overheating and maintain optimal performance.

Additional Information Option 1: Rubber

• Rubber is an insulating material with a very low thermal conductivity. Its thermal conductivity value is typically around 0.1-0.2 W/m·°C, which means it does not conduct heat well. This property makes rubber suitable for applications requiring thermal insulation, such as in the manufacture of heat-resistant gloves, mats, and seals.

Option 2: Air

• Air also has low thermal conductivity, approximately 0.024 W/m·°C. This makes it a poor conductor of heat and an excellent insulator. Air is commonly used in insulating applications, such as in double-glazed windows, where the trapped air layer helps reduce heat transfer between the interior and exterior of buildings.

Option 4: Wood

• Wood has a moderate thermal conductivity, typically around 0.12-0.15 W/m·°C. While not as insulating as rubber or air, wood does not conduct heat as efficiently as metals like aluminium. Wood’s thermal conductivity depends on its density and moisture content, but it is generally used in applications where moderate insulation is required, such as in building construction and furniture making.

Concept:

The rate of heat transfer per unit area through a solid plate is calculated using Fourier's law:

\( q = \frac{k \cdot \Delta T}{L} \)

Where,

• \( q \) = heat transfer rate per unit area (W/m²)
• \( k \) = thermal conductivity (W/m°C)
• \( \Delta T \) = temperature difference across the plate (°C)
• \( L \) = thickness of the plate (m)

Given:

\( k = 370~\text{W/m}^\circ\text{C}, \quad \Delta T = 350 - 50 = 300^\circ\text{C}, \quad L = 45~\text{mm} = 0.045~\text{m} \)

Calculation:

\( q = \frac{370 \cdot 300}{0.045} = \frac{111000}{0.045} = 2466666.67~\text{W/m}^2 \)

\( q \approx 2.466 \times 10^6~\text{W/m}^2 \)

Concept:

The heat transfer through a composite wall is governed by thermal resistance. The total thermal resistance is the sum of the resistances of individual layers and the thermal contact resistance at the interface.

The total thermal resistance is given by:

\( R_{\text{total}} = R_A + R_B + R_{\text{contact}} \)

The heat transfer rate is calculated using Fourier’s Law:

\( Q = \frac{T_1 - T_2}{R_{\text{total}}} \)

Given:

• Thickness of material A: \(L_A = 100 mm = 0.1 m\)
• Thermal conductivity of material A: \(k_A = 50 W/m·K\)
• Thickness of material B: L_B = 10 mm = 0.01 m
• Thermal conductivity of material B: \(k_B = 2 W/m·K\)
• Thermal contact resistance: \(R_{\text{contact}} = 0.003 ~m²K/W\)
• Temperature difference: \(T_1 = 300^\circ C ,~ T_2 = 50^\circ C\)
• Cross-sectional area: A = 1 m²

Calculation:

Step 1: Compute the thermal resistance of each material

Thermal resistance of material A:

\( R_A = \frac{L_A}{k_A A} = \frac{0.1}{50 \times 1} = 0.002~K/W\)

Thermal resistance of material B:

\( R_B = \frac{L_B}{k_B A} = \frac{0.01}{2 \times 1} = 0.005~K/W\)

Step 2: Compute total thermal resistance

\( R_{\text{total}} = R_A + R_B + R_{\text{contact}} \)

\( R_{\text{total}} = 0.002 + 0.005 + 0.003 = 0.01~K/W\)

Step 3: Compute heat transfer rate

\( Q = \frac{T_1 - T_2}{R_{\text{total}}} \)

\( Q = \frac{300 - 50}{0.01} = \frac{250}{0.01} = 25,000~W\)

\( Q = 25~kW\)

Explanation:

Thermal Diffusivity

• Thermal diffusivity is a material-specific property that measures the rate at which heat moves through a material. It is a key parameter in heat transfer analysis and is defined as the ratio of thermal conductivity to the product of density and specific heat capacity at constant pressure. Mathematically, it is expressed as:

\(α = \frac{k }{ (ρ \times c_p)}\)

where:

• α is the thermal diffusivity (m²/s)
• k is the thermal conductivity (W/m·K)
• ρ is the density of the material (kg/m³)
• c_p is the specific heat capacity at constant pressure (J/kg·K)

Factors Affecting Thermal Diffusivity:

From the definition above, it is clear that thermal diffusivity depends on three main factors:

1. Thermal conductivity of the material
2. Density of the material
3. Specific heat capacity of the material

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.

Thermal conductivity of different metals in liquid state is given below

Sodium (Na) – 140 W/m-K

Potassium (K) – 100 W/m-K

Lithium (Li) – 85 W/m-K

Tin (Sn) – 64 W/m-K

Lead (Pb) – 36 W/m-K

Mercury ( Hg) – 8 W/m-K

So out of given options Sodium has highest thermal conductivity.

Explanation:

Insulation

• It is defined as a process of preventing the flow of heat from the body by applying insulator materials to the surface which controls the rate of heat transfer.
• The insulating ability of an insulator depends on various factors:
  • thickness of insulator
  • material of insulator
  • surrounding conditions
  • temperature difference  
• Generally, air packets are present in porous insulating materials.
• Since water which is a more conductive material is replacing air which is a less conductive material, so the overall insulating ability of the insulator will decrease. Most insulators are porous in nature.
• If it has been about Non-porous insulators then the insulating ability will remain unaffected.

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:

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

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

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

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

Concept:

For maximum heat dissipation, the thickness of insulation should be critical thickness.

The critical radius for wire is given as, \(r_c=\frac{k}{h}\)

Calculation:

Given:

k = 0.1 W/mK, h = 100 W/m2K

\({r_c} = \frac{k}{h} = \frac{{0.1}}{{100}} = 0.001~m = 1~mm\)

Thickness of insulation = rc - r

⇒ 1 – 0.5 = 0.5 mm

Concept:

Heat transfer rate through a cylinder is given by:

\(\dot Q = \frac{{2\pi kl\left( {{T_1} - {T_2}} \right)}}{{ln\left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}\)

where Q̇ = Heat transfer rate, T₁, T₂ = Temperatures of the inner and outer cylindrical surface, r₁, r₂ = Inner and outer radius, k = Thermal conductivity of the material, l = Length of pipe

Calculation:

Given:

r₁ = 2 cm, r₂ = 4 cm , Q₂ = \(\frac{1}{3}\)× Q₁

Heat loss with existing insulation,

\({Q_1} = \frac{{2\pi kl\left( {{T_1} - {T_2}} \right)}}{{ln\left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}\)

Heat loss with additional insulation,

\({Q_2} = \frac{{2\pi kl\left( {{T_1} - {T_2}} \right)}}{{ln\left( {\frac{{{r_2} + x}}{{{r_1}}}} \right)}}\)

Where x is the additional thickness of insulation.

According to the given condition,

Q2 = \(\frac{1}{3}\)× Q1

\(\frac{{2\pi kl\left( {{T_1} - {T_2}} \right)}}{{ln\left( {\frac{{{r_2} + x}}{{{r_1}}}} \right)}} = \frac{1}{3} \times \frac{{2\pi kl\left( {{T_1} - {T_2}} \right)}}{{ln\left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}\)

\(\frac{{{r_2} + x}}{{{r_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^3} = {\left( {\frac{4}{2}} \right)^3} = 8\)

r₂ + x = 8r₁

x = 8 × 2 – 4 = 12 cm

∴ The additional thickness of insulation required will be 12 cm.

Concept:

Thermal conductivity of any material is dependent on two things:

i. The motion of free electrons

ii. Lattice vibrations

The thermal conductivity of gases increases with temperature.

Thermal conductivity of liquids decreases with increasing temperature as the liquid expands and the molecules move apart.

In the case of solids, because of lattice distortions, higher temperatures make it difficult for electrons to flow, hence the thermal conductivity of metals decreases.

In non-metals, electronic conductivity is practically non-existent and due to an increase in phonon conduction, the thermal conductivity increases.