# Conduction

 Question 1
Consider a laterally insulated rod of length L and constant thermal conductivity. Assuming one-dimensional heat conduction in the rod, which of the following steady-state temperature profile(s) can occur without internal heat generation?

 A A B B C C D D
GATE ME 2023   Heat Transfer
Question 1 Explanation:

This is simply heat transfer through plane surface along it's length.
So, Governing equation for it is
$\frac{d^{2} T}{d x^{2}}=0$
or $\frac{\mathrm{d} T}{\mathrm{dx}}=4$
or $\mathrm{T}=\mathrm{C}_{1}+\mathrm{C}_{2}$
a straight line equation} So, both option (A) and (B) are correct.
 Question 2
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 2 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 3
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 3 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 4
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 4 Explanation:

\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 5
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 5 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}

There are 5 questions to complete.

### 4 thoughts on “Conduction”

1. 9 number answer A hoga