Fourier Law and Thermal Conductivity MCQ Quiz in मल्याळम - Objective Question with Answer for Fourier Law and Thermal Conductivity - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Fourier’s law of heat conduction states that;

\(\dot Q = kA\frac{{\left( {{T_1} - {T_2}} \right)}}{L} = \frac{{{\rm{\Delta }}T}}{{{R_{cond}}}}\) …1)

DIAGRAM

\({R_{cond}} = \frac{L}{{kA}} = Conduction\;resistance\;of\;plane\;wall\)    …2)

Calculation:

Given configuration

Heat transfer in radial direction is negligible, contact resistance is also negligible. We can express given composite bar by equivalent thermal circuit as →

Heat transfer rate can be given by,

\({\dot Q_{cond}} = \frac{{{T_1} - {T_2}}}{{\frac{L}{{{k_1}A}} + \frac{L}{{{k_2}A}}}} = \frac{{{\rm{\Delta }}T}}{{\frac{L}{A}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}} \right]}} = \left( {\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)\left( {\frac{A}{L}} \right)\left( {{\rm{\Delta }}T} \right)\) …3)

Now, we have to replace the composite bar by a bar of length 2L, area A and thermal conductivity k_eq.

Figure: Equivalent bar having thermal conductivity k_eq

\({\dot Q_{cond}} = {k_{eq}}\frac{{A\left( {{T_1} - {T_2}} \right)}}{{2L}}\) …4)

Heat transfer through both the bars must be same.

Comparing 3) and 4)

\(\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}.\frac{A}{L}{\rm{\Delta }}T = {k_{eq}}A\frac{{\left( {{\rm{\Delta }}T} \right)}}{{2L}}\) 

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

Key Points:

In such questions, confusion arises due to cylindrical rod.

We generally confused whether should we use

\(\dot Q = \frac{{\left( {2\pi kL} \right){\rm{\Delta }}T}}{{{\rm{ln}}\left( {{r_2}/{r_1}} \right)}}\;or\;\dot Q = \frac{{kA\left( {{\rm{\Delta }}T} \right)}}{L}\) 

If geometry is cylindrical and heat transfer is in radial direction, then use \(\dot Q = \frac{{\left( {2\pi kL} \right)\left( {{\rm{\Delta }}T} \right)}}{{{\rm{ln}}\left( {{r_2}/{r_1}} \right)}}\)

If geometry is cylindrical and heat transfer is in axial direction than use \(\dot Q = kA\left( {\frac{{{\rm{\Delta }}T}}{L}} \right)\)

Concept:

1) Convection heat flow through the surface is given by:

Q = hA (T_s1 – T_∞)

Where,

A = surface area through heat flow, h = convective coefficient of heat transfer

T_s1 = surface temperature, T_∞ = surrounding temperature

∴ A = 2πr₁L

And Q = (2πr₁L) h (T_s1 – T_∞)

Or

\(\left( {\frac{Q}{L}} \right) = \left( {2\pi {r_1}} \right)h\left( {{T_{s1}} - {T_\infty }} \right){\rm{\;\;\;}} \ldots {\rm{i}})\)

2) Radial conduction, convection heat transfer through a composite cylinder with uniform heat generation is given by

Thermal circuit:

\(\therefore \left( {\frac{Q}{L}} \right) = \frac{{{T_{s1}} - {T_\infty }}}{{\frac{{\ln \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}{{2\pi K}} + \frac{1}{{h\left( {2\pi {r_2}} \right)}}}}\)

Where,

h = Convective coefficient of wire, k = Thermal conductivity of the material,

T_s2 = Surface temperature of the wire

Calculation:

Given:

Wire radius (r) = 1 mm, Heat generation (Q / L) = 5 W/m, Wire surface temperature (T_s1) = 75°C

Surrounding temperature (T_∞) = 25°C

After applying insulation:

P_VC insulation thickness, t = 1 mm

New wire surface temperature, T_s2 = 55°C

Now, using concept 1) & Equation i)

\(\left( {\frac{Q}{L}} \right) = \left( {2\pi {r_1}} \right)h\left( {{T_{s1}} - {T_\infty }} \right)\)

S = h(2π × 10^-3) (75 - 25)

h = 15.9 W/m²

Now,

Using concept 2) & equation ii)

\(\left( {\frac{Q}{L}} \right) = \frac{{{T_{s2}} - {T_\infty }}}{{\frac{{\ln \left( {{r_2}/{r_1}} \right)}}{{2\pi K}} + \frac{1}{{h\left( {2\pi {r_2}} \right)}}}}\)

Now,

After applying insulation,

r₂ = r₁ + t = 2 mm

\(S = \frac{{55 - 25}}{{\frac{{\ln \left( {\frac{2}{1}} \right)}}{{2\pi K}} + \frac{1}{{15.9 \times \left( {2\pi \times 2 \times {{10}^{ - 3}}} \right)}}}}\)

\(\frac{{\ln 2}}{{2\pi K}} + \frac{{250}}{{15.9\pi }} = 6\)

\(k = \frac{{\ln 2}}{{2\pi }} = 0.11\;W/mK\)

Explanation:

Given k₁ : k₂ : k₃ = 2k : 4K : 6k

\(Q = - kA \frac{dT}{dx} \rightarrow \frac{{kA\Delta T}}{l}\)

Where

K = thermal conductivity of the material

A = cross-section area wall

ΔT = temperature difference between wall

l = thickness of the plate

Since, A, Q, dx are constant so,

ΔT α \(\frac{1}{k}\)

Thus, ΔT₁ : ΔT₂ : ΔT₃ = 1/k₁ : 1/k₂ : 1/k₃ = ½ : ¼ : 1/6

Concept:

Both the cylinders are in parallel combination as given that heat is flowing along the length of the cylinder. In parallel, the equivalent thermal conductivity of ystem is given by

\(K_{eq} = \frac{K_1A_1 + K_2A_2}{A_1 + A_2}\)

Calculation:

Given:

Radius of Cylinder = R/2, Thermal Conductivity of Inner Cylinder = K

Inner and Outer Radius of Cylindrical Shell = R/2 & R respectively, Thermal Conductivity of Shell = 2K

Area of Inner Cylinder, \(A_1 = \frac{\pi R^2}{4}\), K₁ = K

Area of Cylindrical Shell, \(A_2 = \pi R^2 - \frac{\pi R^2}{4} = \frac{3\pi R^2}{4}\), K₂ = 2K

\(\Rightarrow K_{eq} = \frac{(K)(\frac{\pi R^2}{4}) + (2K)(\frac{3\pi R^2}{4})}{(\frac{\pi R^2}{4}) + (\frac{3\pi R^2}{4})}\)

\(\Rightarrow K_{eq} = \frac{K + 6K}{4}\)

\(\Rightarrow K_{eq} = \frac{7K}{4}\)

Explanation:

The relationship of thermal conductivity with temperature for one dimension is given by

\(k=k_{o}(1+β \ T)\),

where k_o is the thermal conductivity at zero temperature 

β is the slope of temperature variation line

and T is the temperature in K.

for metal and liquid β is -ve, hence as T↑, k↓.

and for gas and non-metal amorphous solid β is +ve, hence as T↑, k↑

or as T↓, k↓.

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:

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

Calculation:

Given:

Surface area (A_s) = 2 m², Thickness (t) = 10 mm = 10 × 10^-3 m, Thermal conductivity (k) =200 W/m-K

we know that, for solid plate 

\({{\bf{R}}_{{\bf{thermal}}}} = \frac{{\bf{L}}}{{{\bf{kAs}}}} = \frac{{10\; \times \;{{10}^{ - 3}}\;\;}}{{200\; \times 2}} = 2.5 \times {10^{ - 5}}\;K/W\)

Explanation:

The variation in thermal conductivity of a material with temperature in the temperature range of interest is given by:

k(T) = k₀ (1 + αT) where α 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:

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:

Given:

k1 = 2k2

As nothing is mentioned about the other variables like Area, Temperature change across the wall so consider this as a fixed parameter.

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

\(Q \propto k\)

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

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

Explanation:

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.

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

As we know that heat flows from high temperature to low temperature hence dt/dx is negative along x-direction

Important Points

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