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?

\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?

T_A+\frac{\dot{q}L}{2h} | |

2T_A | |

T_A | |

T_A \left (1-\frac{\dot{q}L}{4 \pi h} \right )+\frac{\dot{q}L}{4 \pi h}T_\infty |

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).

0.45 | |

0.32 | |

0.11 | |

0.05 |

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}

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

\sqrt{2} | |

2 | |

\frac{1}{\sqrt{2}} | |

\frac{1}{2} |

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.

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.

141 | |

163 | |

189 | |

210 |

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 ________.

7 | |

9 | |

11 | |

8 |

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 }

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.

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

Que. 10’s Ans. is option A