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 }
Question 6
A cylindrical steel rod, 0.01 m in diameter and 0.2 m in length is first heated to 750^{\circ}C and then immersed in a water bath at 100^{\circ}C . The heat transfer coefficient is 250 W/m^{2}-K. The density, specific heat and thermal conductivity of steel are \rho = 7801 kg/m^{3}, c = 473 J/kg-K, and k = 43 W/m-K, respectively. The time required for the rod to reach 300^{\circ}C is ________ seconds.
A
25.326
B
43.65
C
25.68
D
12.56
GATE ME 2016 SET-3   Heat Transfer
Question 6 Explanation: 
Given data:
\mathrm{D}=0.01 \mathrm{m} ; \quad L=0.2 \mathrm{m} ; \quad T_{i}=750^{\circ} \mathrm{C}
T_{\infty}=100^{\circ} \mathrm{C} ; \quad h=250 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} ; \quad \rho=7801 \mathrm{kg} / \mathrm{m}^{3}
\mathrm{c}=473 \mathrm{J} / \mathrm{kgK}
\begin{aligned} \frac{T-T_{\infty}}{T_{i}-T_{\infty}}&=\mathrm{e}^{-\frac{h A t}{\rho V c}}\\\text{ Where }\quad \frac{V}{A}&=\frac{\pi D^{2} L}{4 \times \pi D L}=\frac{D}{4}=\frac{0.01}{4} \\ &=2.5 \times 10^{-3} \mathrm{m} \\ \therefore \frac{300-100}{750-100}&=e^{\frac{-250 \times t}{7801 \times 2.5 \times 10^{-3} \times 473}} \\ \frac{200}{650}&=e^{-0.0271 t} \\ 0.30769&=e^{-0.0271 t}\end{aligned}
Taking \log _{e} both sides, we get
\begin{aligned} \log _{e} 0.30769&=-0.0271 t \log _{e} e \\ &\qquad\left(\because \log _{\theta} e=1\right) \\ t&=\frac{1.1786}{0.027}=43.65 \mathrm{s} \end{aligned}
Question 7
Two cylindrical shafts A and B at the same initial temperature are simultaneously placed in a furnace. The surfaces of the shafts remain at the furnace gas temperature at all times after they are introduced into the furnace. The temperature variation in the axial direction of the shafts can be assumed to be negligible. The data related to shafts A and B is given in the following Table.

The temperature at the centerline of the shaft A reaches 400^{\circ} C after two hours. The time required (in hours) for the centerline of the shaft B to attain the temperature of 400^{\circ} C is _______
A
4.5hrs
B
3hrs
C
2.5hrs
D
1.5hrs
GATE ME 2016 SET-2   Heat Transfer
Question 7 Explanation: 
\begin{aligned} T_{(r, t)} &=T_{0}+\left(T_{\infty}-T_{0}\right) \operatorname{erf}(z) \\ \frac{T_{(r, t)}-T_{0}}{T_{\infty}-T_{0}} &=\operatorname{erf}(z) \\ \frac{T_{(r, t)}-T_{0}}{T_{\infty}-T_{0}} &=\operatorname{erf}(z) \end{aligned}
For shaft A \quad \frac{T_{0.2, t}-T_{i}}{400-T_{i}}
For shaft B \quad \frac{T_{0.05, t}-T_{i}}{400-T_{i}}
Hence error function of both shafts will be same.
\therefore \qquad z_{A}=z_{B}\begin{aligned} \frac{\Gamma_{A}}{2 \sqrt{\alpha_{A} t_{A}}} &=\frac{T_{B}}{2 \sqrt{\alpha_{B} t_{B}}} \\ \frac{\Gamma_{A}}{\sqrt{\alpha_{A} t_{A}}} &=\frac{\Gamma_{B}}{\sqrt{\alpha_{B} t_{B}}} \\ \frac{\Gamma_{A}^{2}}{\alpha_{A} t_{A}} &=\frac{r_{B}^{2}}{\alpha_{B} t_{B}} \\ t_{B} &=\left(\frac{r_{B}}{r_{A}}\right)^{2} \times \frac{\alpha_{A}}{\alpha_{B}} \times t_{A} \\ \alpha_{A} &=\frac{k}{\rho c_{p}}=\frac{40}{2 \times 10^{6}}=\frac{20}{10^{6}} \mathrm{m}^{2} / \mathrm{s} \\ \alpha_{B} &=\frac{k}{\rho c_{p}}=\frac{20}{2 \times 10^{7}}=\frac{1}{10^{6}} \mathrm{m}^{2} / \mathrm{s} \\ t_{B} &=\left(\frac{r_{B}}{r_{A}}\right)^{2} \times \frac{\alpha_{A}}{\alpha_{B}} \times t_{A} \\ &=\left(\frac{0.1}{0.4}\right)^{2} \times \frac{20}{10^{6} \times \frac{1}{10^{6}}} \times 2 \\ &=\frac{1}{16} \times 20 \times 2=2.5 \mathrm{hrs} \end{aligned}
Question 8
A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in a water environment at 300 K. The convective heat transfer coefficient is 1000 W/m^{2}-K Thermal conductivity of steel is 40 W/m-K. The time constant for the cooling process \tau is 16 s. The time required (in s) to reach the final temperature is __________
A
56.24s
B
42.23s
C
25.36s
D
15.26s
GATE ME 2016 SET-1   Heat Transfer
Question 8 Explanation: 
\begin{aligned} \text { Diameter of ball } &=10 \mathrm{mm} \\ T_{\text {initial }} &=1000 \mathrm{K} \\ T_{\text {final }} &=350 \mathrm{K} \\ T_{\infty} &=300 \mathrm{K} \\ h &=1000 \mathrm{W} / \mathrm{m}^{2} \mathrm{K} \\ k &=40 \mathrm{W} / \mathrm{mK} \\ \text { Time constant } &=16 \mathrm{s} \\ \text { Time required } &=t \mathrm{s} \\ \frac{T-T_{\infty}}{T_{i}-T_{\infty}} &=e^{-\frac{h A t}{\rho v c}} \\ &=16 \mathrm{second} \\ \text { But } \quad \frac{\rho v c}{h A} &=\text { Time constant } \\ \frac{350-300}{1000-300} &=e^{-\frac{1}{16}} \times t \\ 0.07143 &=e^{-t/ 16} \end{aligned}
Taking \log _{e} both sides, we get
\begin{aligned} \log _{e} 0.07143 &=\frac{-t}{16} \log _{e} e & &\left(\because \log _{e} e=1\right) \\ -2.639 &= \frac{-t}{16} \\ \text{or}\quad t &=42.23 \mathrm{s} \end{aligned}
Question 9
Biot number signifies the ratio of
A
convective residtance in the fluid to conductive resistance in the solid
B
conductive resistance in the solid to convective resistance in the fluid
C
inertia force to viscous force in the fluid
D
buoyancy force to viscous force in the fluid
GATE ME 2014 SET-1   Heat Transfer
Question 9 Explanation: 
Biot number
\mathrm{Bi}=\frac{h l}{k}=\frac{\frac{l}{k A}}{\frac{1}{h A}} \\ \mathrm{Bi}=\frac{\text { Conductive resistance of solid }}{\text { Convective resistance of fluid }}
Question 10
A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030^{\circ}C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30^{\circ}C, with convective heat transfer coefficient h=20W/m^{2}K. The thermo-physical properties of steel are: density \rho =7800 kg/m^{3}, conductivity k = 40 W/mK and specific heat c = 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030^{\circ}C to 430^{\circ}C is
A
519
B
931
C
1195
D
2144
GATE ME 2013   Heat Transfer
Question 10 Explanation: 
Diameter of ball,
d=60 \mathrm{mm}=0.06 \mathrm{m}
Initial temperature,
T_{i}=1030^{\circ} \mathrm{C}
Ambient temperature,
T_{0}=30^{\circ} \mathrm{C}
Convective heat transfer coefficient
h=20 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}
Density: \quad \rho=7800 \mathrm{kg} / \mathrm{m}^{3}
Conductivity: k=40 \mathrm{W} / \mathrm{mK} Specific heat: c=600 \mathrm{J} / \mathrm{kgK}
Final temperature of steel ball,
T=430^{\circ} \mathrm{C}
\text{Time:}\qquad t=?
The temperature distribution relation is given by
\frac{T-T_{0}}{T_{i}-T_{0}}=e^{\frac{h t}{\rho c l}} \cdots(i)
where \quad l=\frac{V}{A} characteristic length of balls
\begin{aligned} &=\frac{\pi}{6} \frac{d^{3}}{\pi d^{2}}\\ \because \qquad V&=\frac{\pi}{6} d^{3}, A=\pi d^{2} \\ \therefore \qquad l&=\frac{d}{6}=\frac{0.06}{6}=0.01 \mathrm{m} \end{aligned}
Substituting the value of T, T_{0}, T_{i}, h, \rho, c and l in Eq. (i), we get
\begin{aligned} \frac{430-30}{1030-30} &=e^{\frac{-20 x t}{7800 \times 600 \times 0.01}} \\ 0.4 &=e^{-4.273 \times 10^{-4} t} \end{aligned}
Taking \log _{\mathrm{e}} both sides, we get
\log _{e} 0.4=-4.273 \times 10^{-4} \times t \times 1
\text{or }\quad t=\frac{0.9162}{4.273 \times 10^{-4}}=2144.16 \mathrm{s}
There are 10 questions to complete.

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