Heat Conduction Through Plane and Composite Walls MCQ Quiz - Objective Question with Answer for Heat Conduction Through Plane and Composite Walls - Download Free PDF

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:

k = A + Bx

where A and B are positive real number

So with an increase in x, k increases

Energy equation;
\(\frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) + \dot{q} = 0\)

\(\frac{\partial}{\partial x}\left((A+B x) \cdot \frac{\partial T}{\partial x}\right)=0\)

On integration

\((A+B x) \cdot \frac{\partial T}{\partial x}=C_{1}\)

\(\frac{\partial T}{\partial x}=\frac{C_{1}}{A+B x}\)

On integration,

T = logarithmic function of x

Hence, variation of T with x is given as;

Explanation:

In the case of radial conduction through a cylinder with steady state conditions and constant thermal conductivity, the temperature distribution follows a logarithmic relationship. This is derived from the heat conduction equation in cylindrical coordinates, which simplifies to a logarithmic form when considering steady state and constant thermal conductivity.

Explanation:

One-Dimensional Steady-State Heat Conduction Through a Plane Wall

Definition: One-dimensional steady-state heat conduction through a plane wall refers to the process where heat is transferred through the thickness of a wall in a direction perpendicular to the wall surfaces. In this scenario, the temperature distribution does not change with time (steady-state), and the heat transfer occurs in only one dimension (perpendicular to the wall surfaces).

Assumptions: The analysis of this heat conduction problem is based on several key assumptions:

• No heat generation within the wall: The wall does not have any internal heat sources.
• Constant thermal conductivity: The thermal conductivity of the wall material remains constant throughout the wall.
• Steady-state conditions: The temperature distribution does not change with time.
• One-dimensional heat transfer: Heat conduction occurs only in the direction perpendicular to the wall surfaces.

Heat Conduction Equation:

Under the given assumptions, the heat conduction through a plane wall can be described by Fourier's law of heat conduction:

q = \(kA\frac{dT}{dx}\)

Where:

• q is the heat flux (amount of heat transfer per unit area per unit time).
• k is the thermal conductivity of the wall material.
• A is the cross-sectional area through which heat is conducted.
• dT/dx is the temperature gradient in the direction of heat transfer.

For one-dimensional heat conduction through a plane wall with thickness L, the temperature distribution can be determined by integrating Fourier's law. The resulting temperature distribution, T(x), is given by the linear equation:

\(\frac{{T - {T_1}}}{{{T_2} - {T_1}}} = \frac{x}{L}\)

Where:

• T₁ is the temperature at one surface of the wall (at x = 0).
• T₂ is the temperature at the other surface of the wall (at x = L).

This equation shows that the temperature distribution across the wall is linear, meaning that the temperature varies linearly from T₁ at one surface to T₂ at the other surface.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: The temperature remains constant across the wall.

This option is incorrect because it implies that there is no temperature gradient across the wall, which contradicts the condition of heat transfer. For heat conduction to occur, there must be a temperature difference between the two surfaces of the wall.

Option 3: The temperature increases exponentially across the wall.

This option is incorrect because it suggests an exponential temperature distribution, which is not characteristic of one-dimensional steady-state heat conduction through a plane wall with constant thermal conductivity. Instead, the temperature distribution is linear.

Option 4: The temperature distribution is exponential.

Similar to option 3, this option is incorrect because it describes an exponential temperature distribution. As derived, the temperature variation across the wall is linear, not exponential.

Explanation

The conductivity of the wall, k = k₀ + bT

And T₂ > T₁

Since conductivity is a function of temperature, As we move from left to right temperature increases. Thus thermal conductivity of the wall also increases.

Using Fourier's law of heat conduction.

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

As thermal conductivity, k is increasing so to keep the heat transfer constant the temperature gradient \(\left( {\frac{{dT}}{{dx}}} \right)\)  must decrease.

Concept:

• Heat is transferred from one side of a wall to the other side of the wall. Therefore,

Total heat transfer = Heat transfer by conduction only in this case.

∴ Heat transfer = \(\frac{{Total\;temperature\;difference}}{{Total\;resistance}}\)  = \(\frac{{{T_1} - \;{T_2}}}{{\frac{t}{{kA}}}}\)

where T₁ and T₂ are temperatures on the sides of the wall, K = Thermal conductivity, A = Area of the wall.

Calculation:

Given:

Length, L = 5 m, Width, b = 10 m, Thickness, t = 0.25 m, Thermal conductivity, K = 1 W/mK, T₁ = 25° C, T₂ = 15° C, Time = 10 hours, Cost of electricity = Rs 10/KWh.

Heat loss, Q = \(\frac{{25 - 15}}{{\frac{{0.25}}{{1\; × 5\; × 10}}}}\) = 2000 W = 2 KW

Cost of electricity = Heat loss × time × Cost of electricity = 2 × 10 × 10 = 200

So, the cost of electricity in rupees is 200.

Concept:

The general equation for unsteady heat conduction in a slab with no heat generation in one dimension is given by

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{α }\times \frac{{\partial T}}{{\partial \tau }}\)

Calculation:

Given:

T = 4 -10x + 20x2 + 10x3, α = 2 × 10-3 m2/hr

\( \frac{{\partial T}}{{\partial x }}= -10 +40x+30x^2\)

\(\frac {\partial^2T}{\partial x^2} = 40 + 60x\)

Now at the other face of the wall x = 1,

 \(\frac {\partial^2T}{\partial x^2} = 100\)

we have,

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{α }\times \frac{{\partial T}}{{\partial \tau }}\)

\( \frac{{\partial T}}{{\partial \tau }}=α\times( \frac{{{\partial ^2}T}}{{\partial {x^2}}}) =100 × 2 × 10^{-3} = 0.2 °C/hr \)

The rate of change of temperature at the other face of the wall is 0.2°C/hr.

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

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

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

where, x = Thickness of the wall, k = Thermal conductivity, A = Area of the wall.

 Important Points

Concept:

Heat flux through the wall is given by

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

Where k is the thermal conductivity, A is the area perpendicular to the direction of heat flow, dT is the temperature difference and dx is the thickness of the wall.
Heat flux flowing through wall remains same.

∴ Heat flux through Brick wall = Heat flux through insulation

Calculation:

Given,

For brick wall, k₁ = 2.5 W/m.k, dT₁ = 700 – 1100 = -400 K

For Insulation, dT­₂ = 200 – 700 = -500 K and L₂ = L₁/4

Since the heat flux through the wall remains same,

\({k_1}\frac{{d{T_1}}}{{d{x_1}}} = {k_2}\frac{{d{T_2}}}{{d{x_2}}}\;\)

\( - 2.5 \times \frac{{\left( { - 400} \right)}}{{{L_1}}} = - \frac{{{k_2}\left( { - 500} \right)}}{{{L_1}/4}}\)

Explanation:

The temperature distribution for various types of geometry with constant thermal conductivity and no internal heat generation is given in the table below:

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:

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

For number of pkates connected in series, 

RTotal = R1 + R2 + R3

\( {R_{Total}} = \frac{{{l_1}}}{{{A_1}{k_1}}} + \frac{{{l_2}}}{{{A_2}{k_2}}} + \frac{{{l_3}}}{{{A_3}{k_3}}}\)

Calculation:

Given:

l₁ = 200 mm = 0.2 m,  l₂ = 10 mm = 0.01 m,  l₃ = 150 mm = 0.15 m

k₁ = 4.5 kcal/m-hr-°C,  k₂ = 15 kcal/m-hr-°C,  k₃ = 1.75 kcal/m-hr-°C

A₁ = A₂ = A₃ = 1 m² 

R_Total = R₁ + R₂ + R₃

\( {R_{Total}} = \frac{{{l_1}}}{{{A_1}{k_1}}} + \frac{{{l_2}}}{{{A_2}{k_2}}} + \frac{{{l_3}}}{{{A_3}{k_3}}}\)

\( {R_{Total}} = \frac{{0.2}}{{4.5}} + \frac{{0.01}}{{15}} + \frac{{0.15}}{{1.75}}\)

R_Total = 0.13 °C-hr/kcal

Concept:

According to Fourier’s Law of heat conduction, 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.

\(Q = \frac{{kAdT}}{{dx}}\)

where k is thermal conductivity which depends on the material of the body.

Assumptions of Fourier equation:

• Steady-state heat conduction
• One directional heat flow
• Bounding surfaces are isothermal in character that is constant and uniform temperatures are maintained at the two faces
• Isotropic and homogeneous material and thermal conductivity ‘k’ is constant
• Constant temperature gradient and linear temperature profile
• No internal heat generation

Concept:

COP in terms of Temperature of a Refrigerator \(= \frac{{{T_L}}}{{{T_H} - {T_L}}}\)

Amount of heat conducted \(\left( Q \right) = - kA\frac{{dT}}{{dx}}\)

Where T_L is the temperature of refrigerated space and T_H is the room temperature, k is the thermal conductivity of the wall, A is the area perpendicular to heat flow and dx is the wall thickness.

Calculation:

Given, COP = 5, A = 100 × 100 cm² = 1 m², dx = 10 cm = 0.1 m,

\(K = 1\frac{W}{{cmK}} = 100\frac{W}{{mK}}\)

First, let us try to find out TL with the help of COP

\(5 = \frac{{{T_L}}}{{300 - {T_L}}}\)

T_L = 250 K

Now for heat intake

\(Q = - k\;A \cdot \frac{{dT}}{{dx}}\)

\(Q = - 100 \times 1 \times \frac{{\left( {250 - 300} \right)}}{{0.1}} = 50000\;W\)

Concept:

Heat balance and resistance networks;

\(\dot{Q}=kA\frac{\text{ }\!\!\Delta\!\!\text{ }T}{L}=\frac{\text{ }\!\!\Delta\!\!\text{ }T}{{{R}_{th}}}\)

\(⇒ {{R}_{th}}=\frac{L}{kA}\) → conduction resistance

Newton’s law of cooling; \(\dot{Q}=hA\left( {{T}_{s}}-{{T}_{\infty }} \right)=\frac{{{T}_{s}}-{{T}_{\infty }}}{\frac{1}{hA}}=\frac{{{T}_{s}}-{{T}_{\infty }}}{{{R}_{conve}}}\)

\(⇒ {{R}_{conve}}=\frac{1}{hA}\) → convection resistance

Calculation:

Heat generated in central slab dissipates through left and right faces.

Q̇_gen = Q̇_L + Q̇_R     …(1)

Be careful while using the dimension.

The center slab experiences a non-uniform internal heat generation with an average value equal to 10000 Wm−3

Q̇gen  = 10000  Wm−3 × V = 0000  Wm−3 × AL

Q̇gen  =  10000 × A × 1 = (10000A) W

⇒ qgen  =  10000 W/m2

Where, A is in m² and L is the thickness of the central slab.

Q̇_L = hA (T₁ – T_∞)

\(\overset{}{\mathop{\frac{{\dot{Q}_{L}}}{A}}}\,={{\dot{q}}_{L}}=h\left( {{T}_{1}}-{{T}_{\infty }} \right)\)

\(⇒ {{\dot{q}}_{L}}=\left( 100 \right)\left( 100-30 \right)=7000~W/{{m}^{2}}\)

For a unit area, (1) can be written as;

\({{\dot{q}}_{gen}}={{\dot{q}}_{L}}+{{\dot{q}}_{R}}\)

\({{\dot{q}}_{gen}}=10000=7000+{{\dot{q}}_{R}}\)

\(⇒ {{\dot{q}}_{R}}=3000=\left( 100 \right)\left( {{T}_{2}}-30 \right)\)