# Conduction

 Question 1
Consider steady state, one-dimensional heat conduction in an infinite slab of thickness $2L (L = 1 m)$ as shown in the figure. The conductivity $(k)$ of the material varies with temperature as $k = CT$, where $T$ is the temperature in $K$, and $C$ is a constant equal to $2 W\dot m^{-1} \dot K^{-2}$. There is a uniform heat generation of $1280 kW/m^3$ in the slab. If both faces of the slab are maintained at 600 K, then the temperature at $x = 0$ is ________ K (in integer). A 825 B 1126 C 1000 D 1256
GATE ME 2022 SET-2   Heat Transfer
Question 1 Explanation:
General Heat conduction equation in Cartesian coordinates is
$\frac{\partial }{\partial x}\left ( K_x\frac{\partial T}{\partial x} \right )+\frac{\partial }{\partial y}\left ( K_y\frac{\partial T}{\partial y} \right )\frac{\partial }{\partial z}\left ( K_z\frac{\partial T}{\partial z} \right )+q_g=\rho c\frac{\partial T}{\partial t}$
For one dimensional steady state heat generation with variable thermal conductivity, the above equation can be written as
\begin{aligned} \frac{\partial }{\partial x}\left ( CT\frac{\partial T}{\partial x} \right )+q_g&=0\\ C\frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+q_g&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )+\frac{q_g}{C}&=0\\ \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\frac{q_g}{C}\\ \int \frac{\partial }{\partial x}\left ( T\frac{\partial T}{\partial x} \right )&=-\int \frac{q_g}{C}\\ \text{For 1-D heat flow}\\ \frac{\partial T}{\partial x}&=\frac{dT }{\partial x}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}+C_1\\ x=0\;\;\frac{dT}{dx}&=0\;\;C_1=0 \text{ (At centerline)}\\ T\frac{dT}{dx}&=-\frac{q_gx}{C}\\ \int TdT&=\int -\frac{q_gx}{C}dx\\ \frac{T^2}{2}&=\frac{-q_gx^2}{2C}+C_2 . . . (1)\\ T=T_s&=\text{Outside surface temperature}\\ \frac{T_S^2}{2}&=\frac{-q_gL^2}{2C}+C_2\\ C_2&=\frac{T-s^2}{2}+\frac{q_gL^2}{2C}\\ \text{Put }&c_2 \text{ in equation (1)}\\ \frac{T^2}{2}&=\frac{-q_ax^2}{2C}+\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \text{At }&x=0,T=T_{max}\\ \frac{T_{max}^2}{2}&=\frac{T_s^2}{2}+\frac{q_gL^2}{2C}\\ \frac{T_{max}^2}{2}&=\frac{600^2}{2}+\frac{128010^3 \times 1^2}{2 \times 2}\\ T_{max}&=1000K \end{aligned}
 Question 2
Consider a solid slab (thermal conductivity, $k =10 W \dot m^{-1}\dot K^{-1}$) with thickness 0.2 m and of infinite extent in the other two directions as shown in the figure. Surface 2, at 300 K, is exposed to a fluid flow at a free stream temperature ($T_\infty$) of 293 K, with a convective heat transfer coefficient (h) of $100 W\dot m^{-2} \dot K^{-1}$. Surface 2 is opaque, diffuse and gray with an emissivity ($\varepsilon$) of 0.5 and exchanges heat by radiation with very large surroundings at 0 K. Radiative heat transfer inside the solid slab is neglected. The Stefan-Boltzmann constant is $5.67 \times 10^{-8} W \dot m^{-2}\dot K^{-4}$. The temperature $T_1$ of Surface 1 of the slab, under steady-state conditions, is _________ K (round off to the nearest integer). A 254 B 632 C 985 D 319
GATE ME 2022 SET-1   Heat Transfer
Question 2 Explanation:
$T_\infty =293K,T_{surrounding}=0K, h=100W/m^2K, K_{slab}=10W/mK$ \begin{aligned} Q_{stored}&=0\\ Q_{in}&=Q_{out}\\ Q_{cond}&=Q_{conv}+Q_{rad}\\ K\left [ \frac{T_1-T_2}{L} \right ]&=h(T_2-T_\infty )+\varepsilon \sigma (T_2^4-T_{surr}^4)\\ 10\left [ \frac{T_1-300}{0.2} \right ]&=100(300-293 )+0.5 \times 5.67 \times 10^{-8} (300^4-0^4)\\ T_1-300&=\frac{929.635}{50}\\ T_1&=18.59+300\\ T_1&=318.59\\ T_1&=318.6K=319K \end{aligned}
 Question 3
Consider a one-dimensional steady heat conduction process through a solid slab of thickness $0.1 m$. The higher temperature side A has a surface temperature of $80^{\circ}C$, and the heat transfer rate per unit area to low temperature side B is $4.5 \; kW/m^2$. The thermal conductivity of the slab is $15 \; W/m.K$. The rate of entropy generation per unit area during the heat transfer process is ________ $W/m^2.K$(round off to 2 decimal places).
 A 2.25 B 3.36 C 1.18 D 1.85
GATE ME 2022 SET-1   Heat Transfer
Question 3 Explanation:
1D Steady heat conduction \begin{aligned} \frac{Q}{A} &=4.5\frac{kW}{m^2}=4500 \frac{W}{m^2}\\ k&=15W/mk \\ S_{gen}&=? \\ \frac{Q}{A} &= k\left [ \frac{T_1-T_2}{dx} \right ]\\ 4500&= 15\left [ \frac{80-T_2}{0.1} \right ]\\ T_2 &=50^{\circ}C=323K \\ T_1 &=80^{\circ}C=353K \\ \therefore \;\; \frac{S_{gen}}{Area}&=\frac{-Q}{AT_1}+\frac{Q}{AT_2} \\ &=\frac{-4500}{353}+\frac{4500}{323} \\ S_{gen} &= 1.18\frac{W}{m^2K} \end{aligned}
 Question 4
A tiny temperature probe is fully immersed in a flowing fluid and is moving with zero relative velocity with respect to the fluid. The velocity field in the fluid is $\vec{V}=2x\hat{i}+(y+3t)\hat{j}$,and the temperature field in the fluid is $T=2x^2+xy+4t$ where $x$ and $y$ are the spatial coordinates, and $t$ is the time. The time rate of change of temperature recorded by the probe at ($x=,y=1,t=1$) is _______.
 A 4 B 0 C 18 D 14
GATE ME 2022 SET-1   Heat Transfer
Question 4 Explanation:
\begin{aligned} \frac{dT}{dt}&=? \text{ at }x=1,y=1,t=1\\ \frac{dT}{dt}&=\frac{\partial T}{\partial t}\frac{dt}{dt}+\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{\partial T}{\partial y}\frac{dy}{dt}\\ &=\frac{\partial T}{\partial t}+u\frac{\partial T}{\partial x}+v\frac{\partial T}{\partial y}\\ &=4+(2x)(4x+y)+(y+3t)(x)\\ &=4+8x^2+3xy+3xt\\ &\text{at }x=1,y=1,t=1\\ \frac{dT}{dt}&=4+8(1)^2+3(1)+3(1)\\ &=18 \end{aligned}
 Question 5
In a furnace, the inner and outer sides of the brick wall ($k_1 = 2.5 W/mK$) are maintained at $1100^{\circ}C$ and $700^{\circ}C$ respectively as shown in figure. The brick wall is covered by an insulating material of thermal conductivity $k_2$. The thickness of the insulation is $1/4^{th}$ of the thickness of the brick wall. The outer surface of the insulation is at $200^{\circ}C$. The heat flux through the composite walls is 2500 $W/m^2$.

The value of $k_2$ is ________ W/mK (round off to 2 decimal places).
 A 0.25 B 0.5 C 0.75 D 1
GATE ME 2020 SET-2   Heat Transfer
Question 5 Explanation:
Given, $L_{2}=\frac{L_{1}}{4}$
Assuming steady state, one-dimensional conduction heat transfer through composite slab,
Thermal circuit:
$\begin{array}{l} \Rightarrow \quad q=\frac{1100-700}{\frac{L_{1}}{k_{1} A}}=\frac{700-200}{\frac{L_{2}}{k_{2} A}} \\ \Rightarrow \quad \frac{400}{\frac{L_{2}}{2.5}}=\frac{500}{\frac{L_{2}}{k_{2}}}\\ \Rightarrow \quad k_{2}=0.5 \mathrm{W}-\mathrm{mK} \end{array}$
 Question 6
One-dimensional steady state heat conduction takes place through a solid whose cross-sectional area varies linearly in the direction of heat transfer. Assume there is no heat generation in the solid and the thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is
 A Linear B Quadratic C Logarithmic D Exponential
GATE ME 2019 SET-2   Heat Transfer
Question 6 Explanation:
$\begin{array}{c} \text { Area }(\mathrm{A}) \propto \mathrm{x} \\ \mathrm{A}=\mathrm{c} \mathrm{x} \end{array}$
According to Fourier's law of heat conduction
$\mathrm{Q}=-\mathrm{k} \mathrm{A} \frac{\mathrm{d} \mathrm{T}}{\mathrm{d} \mathrm{x}}$
$\begin{array}{l} \mathrm{Q}=-\mathrm{k} \cdot \mathrm{c} \mathrm{x} \frac{\mathrm{d} \mathrm{T}}{\mathrm{d} \mathrm{x}} \\ \mathrm{Q} \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}=-\mathrm{k} \mathrm{c} \mathrm{d} \mathrm{T} \\ \int \mathrm{d} \mathrm{T}=-\frac{\mathrm{Q}}{\mathrm{k} \mathrm{c}} \int \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}} \\ \mathrm{T}=-\frac{\mathrm{Q}}{\mathrm{k}_{\mathrm{r}}} \ln \mathrm{x}+\mathrm{c}_{1} \end{array}$
Temperature distribution is logarithmic.
(or)
In hollow cylinder, area is linearly proportional to radius.
$A = 2\pi rL$
Temperature profile is logarithmic in case of hollow cylinder with no heat generation.
 Question 7
A slender rod of length L, diameter d ($L \gt \gt d$) and thermal conductivity $k_1$ is joined with another rod of identical dimensions, but of thermal conductivity $k_2$, to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is
 A $k_1+k_2$ B $\sqrt{k_1k_2}$ C $\frac{k_1k_2}{k_1+k_2}$ D $\frac{2k_1k_2}{k_1+k_2}$
GATE ME 2019 SET-1   Heat Transfer
Question 7 Explanation: Area is constant in the direction of heat flow. (similar like slab)
Thermal circuit : $\begin{array}{l} \text { Heat transfer rate, }(Q)=\frac{T_{1}-T_{3}}{\frac{L}{k_{1} A}+\frac{L}{k_{2} A}}=\frac{T_{1}-T_{3}}{\frac{2 L}{k_{e q} A}} \\ \frac{L}{k_{1} A}+\frac{L}{k_{2} A}=\frac{2 L}{k_{e q} A} \\ \frac{2}{k_{e q}}=\frac{1}{k_{1}}+\frac{1}{k_{2}} \\ k_{e q}=\frac{2 k_{1} k_{2}}{k_{1}+k_{2}} \end{array}$
 Question 8
A 0.2 m thick infinite black plate having a thermal conductivity of 3.96 W/m-K is exposed to two infinite black surfaces at 300 K and 400 K as shown in the figure. At steady state, the surface temperature of the plate facing the cold side is 350 K. The value of Stefan- Boltzmann constant, $\sigma$ , is $5.67 \times 10^{-8} \: W/m^{2}\: K^{4}$ Assuming 1-D heat conduction, the magnitude of heat flux through the plate (in W/$m^{2}$) is ________ (correct to two decimal places). A 258.78 B 963.25 C 391.61 D 147.93
GATE ME 2018 SET-2   Heat Transfer
Question 8 Explanation:
Under steady state condition, all rate of heat transfer i.e. from surface at 400 K to black plate (via radiation), inside black plate (via conduction) and from black plate to surface at 300 K (via radiation) are equal.
So, heat flux through wall= flux from wall to surface at 300 K
$=\frac{\sigma\left(350^{4}-300^{4}\right)}{\frac{1}{1}+\frac{1}{1}-1}=391.612 \text { watt/m }^{2}$
 Question 9
Heat is generated uniformly in a long solid cylindrical rod (diameter = 10mm) at the rate of $4\,\times\,10^{7}$ W/$m^{3}$. The thermal conductivity of the rod material is 25W/m.K. Under steady state conditions, the temperature difference between the centre and the surface of the rod is _________ $^{\circ}C$.
 A 7 B 8 C 9 D 10
GATE ME 2017 SET-1   Heat Transfer
Question 9 Explanation: \begin{aligned} T_{0}&= \text{ Temp. of cylinder at the axis}\\ T_{S} &=\text { Surface Temp. of Rod (cylindel) } \\ T_{O}-T_{S} &=\frac{\dot{q}}{4 k} R^{2} \end{aligned}
$\dot{q}$ uniform heat generation rate per unit volumne
$\left(w / m^{3}\right)$
$T_{0}-T_{S}=\frac{4 \times 10^{7}}{4 \times 25} \times\left(\frac{5}{1000}\right)^{2}=10^{\circ} \mathrm{C}$
 Question 10
Steady one-dimensional heat conduction takes place across the faces 1 and 3 of a composite slab consisting of slabs A and B in perfect contact as shown in the figure, where $k_{A},k_{B}$ denote the respective thermal conductivities. Using the data as given in the figure, the interface temperature T2 (in $^{\circ}C$) is __________ A 67.5 C B 68.5 C C 70.6 C D 54.6 C
GATE ME 2016 SET-3   Heat Transfer
Question 10 Explanation: \begin{aligned} Q_{A} &=Q_{B} \\ \frac{k_{A}\left[T_{1}-T_{2}\right] A}{L_{1}} &=\frac{k_{B}\left[T_{2}-T_{3}\right] A}{L_{2}} \\ \frac{20 \times\left[130-T_{2}\right]}{0.1} &=\frac{100 \times\left[T_{2}-30\right]}{0.3} \\ 3\left[130-T_{2}\right] &=5 T_{2}-150 \\ 390-3 T_{2} &=5 T_{2}-150 \\ 540 &=8 T_{2} \\ T_{2} &=67.5^{\circ} \mathrm{C} \end{aligned}
There are 10 questions to complete.

### 4 thoughts on “Conduction”

1. 9 number answer A hoga

• As per Official answer key answer is C.

2. • 