# Fins and Unsteady Heat Transfer

 Question 1
Consider a rod of uniform thermal conductivity whose one end $(x = 0)$ is insulated and the other end $(x = L)$ is exposed to flow of air at temperature $T_\infty$ with convective heat transfer coefficient $h$. The cylindrical surface of the rod is insulated so that the heat transfer is strictly along the axis of the rod. The rate of internal heat generation per unit volume inside the rod is given as
$\dot{q}=\cos \frac{2 \pi x}{L}$
The steady state temperature at the mid-location of the rod is given as $T_A$. What will be the temperature at the same location, if the convective heat transfer coefficient increases to $2h$?
 A $T_A+\frac{\dot{q}L}{2h}$ B $2T_A$ C $T_A$ D $T_A \left (1-\frac{\dot{q}L}{4 \pi h} \right )+\frac{\dot{q}L}{4 \pi h}T_\infty$
GATE ME 2022 SET-1   Heat Transfer
Question 1 Explanation: $\dot{q}=\frac{\cos 2 \pi x}{L}$
Total heat generation $=Q_g=\int_{o}^{L}\dot{q}d(\text{Volume})$
\begin{aligned} Q_g&=\int_{o}^{L}\dot{q}Adx\\ &=\int_{o}^{L}\left ( \frac{\cos 2\pi x}{L} \right )Adx\\ &=A\left [ \frac{\sin \left (\frac{2 \pi x}{L} \right ) }{\frac{ 2 \pi }{L}} \right ]_{0}^{L}\\ &=\left ( \frac{L}{2 \pi} \right )A\left [ \sin \left (\frac{2 \pi x}{L} \right )- \sin (0) \right ] \end{aligned}
$Q_g=0$
At steady state $Q_{stored}=0$
$Q_{in}+Q_{gen}-Q_{out}=Q_{stored}$
$Q_{gen}=Q_{out}$
$0=hA_S(T_S-T_\infty$
$h \text{ and } A_S$ can't be zero it means $T_S=T_ \infty$
It indicate that $T_S$ is independent of heat transfer coefficient.
So it means temperature of body any where is independent of heat transfer coefficient.
So by increasing heat transfer coefficient there is no change in the temperature at mid location.
 Question 2
An uninsulated cylindrical wire of radius 1.0 mm produces electric heating at the rate of 5.0 W/m. The temperature of the surface of the wire is 75 $^{\circ}C$ when placed in air at 25 $^{\circ}C$. When the wire is coated with PVC of thickness 1.0 mm, the temperature of the surface of the wire reduces to 55 $^{\circ}C$. Assume that the heat generation rate from the wire and the convective heat transfer coefficient are same for both uninsulated wire and the coated wire. The thermal conductivity of PVC is ______W/m.K (round off to two decimal places).
 A 0.45 B 0.32 C 0.11 D 0.05
GATE ME 2021 SET-1   Heat Transfer
Question 2 Explanation:
For uninsulated wire:
Given: $T_{\infty}=25^{\circ} \mathrm{C}, R_{1}=1 \mathrm{~mm}, \dot{q}_{\text {gen }}=5 \mathrm{~W} / \mathrm{m}, T_{s_{1}}=7.5^{\circ} \mathrm{C}$ \begin{aligned} q &=\dot{q}_{g e n} \times L=\frac{T_{s 1}-T_{\infty}}{\frac{1}{h \times 2 \pi R_{1} L}} \\ 5 \times L &=h \times(2 \pi \times 0.001) L \times(75-25) \\ h &=15.91 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} \end{aligned}
For Insulated wire: $R_{2}=2 \mathrm{~mm}$
Heat transfer rate after insulation kept on wire,
$q=q_{\text {gen, wire }} \times L=5 \times L$ \begin{aligned} q &=5 \times L=\frac{55-25}{\frac{\ln (2 / 1)}{2 \pi k_{\text {PVC }} L}+\frac{1}{15.91 \times 2 \times \pi \times 0.002 \times L}} \\ k_{P V C} &=0.1103 \mathrm{~W} / \mathrm{m}-\mathrm{K} \\ &=0.11 \mathrm{~W} / \mathrm{m}-\mathrm{K} \end{aligned}

 Question 3
An infinitely long pin fin, attached to an isothermal hot surface, transfers heat at a steady rate of $Q_1$ to the ambient air. If the thermal conductivity of the fin material is doubled, while keeping everything else constant, the rate of steady- state heat transfer from the fin becomes $Q_2$. The ratio $Q_2/Q_1$ is
 A $\sqrt{2}$ B 2 C $\frac{1}{\sqrt{2}}$ D $\frac{1}{2}$
GATE ME 2021 SET-1   Heat Transfer
Question 3 Explanation:
Fin problem: $q=\sqrt{h P k A}\left(T_{o}-T_{\infty}\right) \text { wait }$
If k gets doubled q increases by $\sqrt{2}$ times.
 Question 4
A metal ball of diameter 60mm is initially at $220\, ^{\circ}C$. The ball is suddenly cooled by an air jet of $20\, ^{\circ}C$. The heat transfer coeffient is $200 \: W/m^{2}$. The specific heat,thermal conductivity and density of the metall ball are $400\:J/kg\! \cdot\! K$ ,$400\:W/m\! \cdot\! K$ and $9000\,kg/m^{3}$, The ball temperature (in $^{\circ}C$) after 90 seconds will be approximately.
 A 141 B 163 C 189 D 210
GATE ME 2017 SET-2   Heat Transfer
Question 4 Explanation: \begin{aligned} c_{p} &=400 \mathrm{J} / \mathrm{kgK} \\ k &=400 \mathrm{W} / \mathrm{mK} \\ \rho &=9000 \mathrm{kg} / \mathrm{m}^{3} \\ \text { Time }(\tau) &=90 \mathrm{sec} \end{aligned}
since K being high and size of ball being small,
lumped heat analysis is valid.
\begin{aligned} \Rightarrow \quad e^{\left(\frac{h A}{\rho V C_{p}}\right)^{\tau}}&=\frac{T_{i}-T_{\infty}}{T-T_{\infty}} \\ \Rightarrow \quad\left(\frac{h_{A}}{\rho v c_{p}}\right) \tau&=\ln \left(\frac{T_{i}-T_{\infty}}{T-T \infty}\right) \\ \text{Put }\quad \frac{A}{V}&=\frac{3}{R}=\frac{3}{\left(\frac{30}{1000}\right)} \\ \left(\frac{h A}{\rho v c_{p}}\right) \tau &=\left(200 \times \frac{3}{0.03} \times \frac{1}{9000} \times \frac{1}{400}\right) \times 90 \\ &=0.5 \\ \frac{220-20}{T-20} &=e^{0.5}=1.6487 \\ T &=141.3^{\circ} \mathrm{C} \end{aligned}
 Question 5
The heat loss from a fin in 6W. The effectiveness and efficiency of the fin are 3 and 0.75, respectively. The heat loss (in W) from the fin, keeping the entire fin surface at base temperature, is ________.
 A 7 B 9 C 11 D 8
GATE ME 2017 SET-2   Heat Transfer
Question 5 Explanation:
\begin{aligned} q_{\text {actual }} \text { from fin } &=6 \text { watt } \\ \eta_{\text {fin }} &=\frac{q_{\text {actual }}}{q_{\text {max possible }}} \text { i.e. when entire } \end{aligned}
fin at base temperature
$\therefore$q from fin if entire fin at base temp
$=\frac{q_{\text {act }}}{0.75}=\frac{6}{0.75}=8 \text { watt }$

There are 5 questions to complete.

### 2 thoughts on “Fins and Unsteady Heat Transfer”

1. Question 18 ) Explanation is correct but answer is showing in correct once look at this que

2. 