Question 1
A cylindrical rod of length $h$ and diameter $d$ is placed inside a cubic enclosure of side length $L$. $S$denotes the inner surface of the cube. The view-factor $F_{S-S}$ is
 A $0$ B $1$ C $\frac{(\pi d h+\pi d^2/2)}{6L^2}$ D $1-\frac{(\pi d h+\pi d^2/2)}{6L^2}$
GATE ME 2023   Heat Transfer
Question 1 Explanation:

Surface area of centrifugal rod $=\pi d h+\frac{2 \pi d^{2}}{4}$
Or $A_{1}=\pi\left(d h+\frac{d^{2}}{2}\right)$
Surface area of cube $A_{2}=6 \times \mathrm{L}^{2}$

For surface 1 (cylinder), $\mathrm{F}_{11}=0$
So $\mathrm{F}_{12}=1 \quad \text {...(i) }$

For surface 2(cube)
\begin{aligned} \mathrm{F}_{21}+\mathrm{F}_{22} & =1 \\ \mathrm{F}_{22} & =1-\mathrm{F}_{21} \\ \mathrm{F}_{21} & =1-\mathrm{F}_{22}\;\;...(ii) \end{aligned}
By reciprocity theorem for surface 1 and 2
\begin{aligned} A_{1} F_{12} & =A_{2} F_{21} \\ F_{21} & =\frac{A_{1}}{A_{2}} \end{aligned}
$\left[F_{12}=1\right.$ from equation (i)]
So, By equation (ii)
$F_{22}=F_{s S}=1-\left(\frac{\pi \mathrm{dh}+\frac{\pi d^{2}}{2}}{6 L^{2}}\right)$
 Question 2
Wien's law is stated as follows: $\lambda _mT=C$, where $C$ is 2898 $\mu mK$ and $\lambda _m$ is the wavelength at which the emissive power of a black body is maximum for a given temperature $T$. The spectral hemispherical emissivity $(\varepsilon _ \lambda)$ of a surface is shown in the figure below ($1\mathring{A}=10^{-10}m$). The temperature at which the total hemispherical emissivity will be highest is __________ K (round off to the nearest integer).

 A 4210 B 6520 C 3650 D 4830
GATE ME 2022 SET-2   Heat Transfer
Question 2 Explanation:
From the figure we can say that at $6000 \mathring{A}$ wavelength total hemispherical emissivity is maximum
$\lambda _m=6000 \mathring{A} =0.6\mu m$
Wein's law
$\lambda _m T=C$
$\lambda _m T=2898 \mu mK$
$T=\frac{2898}{0.6}=4830K$
Note: In this question, we need to find temperature at which total hemispherical emissivity will be highest.
But to calculate total hemispherical emissivity we require additional data like function chart and emissivity value at particular wavelength.
With the use of Wein's law we can find out temperature at which spectral emissivity is highest but in question it is asked temperature for total hemispherical emissivity.

 Question 3
A flat plate made of cast iron is exposed to a solar flux of $600 W/m^2$ at an ambient temperature of $25 ^{\circ}C$. Assume that the entire solar flux is absorbed by the plate. Cast iron has a low temperature absorptivity of 0.21. Use Stefan-Boltzmann constant $= 5.669 \times 10^{-8} W/m^2-K^4$. Neglect all other modes of heat transfer except radiation. Under the aforementioned conditions, the radiation equilibrium temperature of the plate is __________ $^{\circ}C$ (round off to the nearest integer).
 A 491.34 B 583.36 C 965.24 D 218.34
GATE ME 2022 SET-1   Heat Transfer
Question 3 Explanation:

Equilibrium Temperature $=T_S=?$
In this it is mentioned that entire flux is absorbed by the plate it means for solar flux absorptivity is 1. Kirchoff's law $\alpha =\varepsilon =0.21$
Surface of cast iron and surrounding fluid temperature difference is small due to this we can use Kirchoff's law
At equilibrium condition
Energy absorbed = Energy leaving
\begin{aligned} \alpha GA&=\varepsilon A\sigma (T_S^4-T_\infty ^4)\\ 1 \times 600 &=0.21 \times 5.669 \times 10^{-8}(T_S^4-298^4)\\ T_S&=491.34K\\ T_S&=218.34^{\circ}C\\ T_S&=218^{\circ}C \end{aligned}
 Question 4
A solid sphere of radius 10 mm is placed at the centroid of a hollow cubical enclosure of side length 30 mm. The outer surface of the sphere is denoted by 1 and the inner surface of the cube is denoted by 2. The view factor $F_{22}$ for radiation heat transfer is ________ (rounded off to two decimal places).
 A 0.12 B 0.45 C 0.88 D 0.77
GATE ME 2021 SET-1   Heat Transfer
Question 4 Explanation:

\begin{aligned} r_{1} &=10 \mathrm{~mm} \\ A_{1} &=4 \pi r_{1}^{2} \\ A_{2} &=6 \times\left(30^{2}\right) \\ F_{12} &=1 \\ A_{1} F_{12} &=A_{2} F_{21} \quad \Rightarrow \quad F_{21}=\frac{A_{1}}{A_{2}}=\frac{4 \pi \times(10)^{2}}{6 \times(30)^{2}}=0.2327 \\ F_{22} &=1-F_{21}=0.7672 \simeq 0.77 \end{aligned}
 Question 5
The spectral distribution of radiation from a black body at $T_1$=3000 K has a maximum at wavelength$\lambda _{max}$. The body cools down to a temperature $T_2$. If the wavelength corresponding to the maximum of the spectral distribution at $T_2$ is 1.2 times of the original wavelength $\lambda _{max}$, then the temperature $T_2$ is ________ K (round off to the nearest integer).
 A 3000 B 4500 C 2500 D 1800
GATE ME 2020 SET-2   Heat Transfer
Question 5 Explanation:
From Wien's Displacement law,
\begin{aligned} \lambda_{m} T &=\text { constant } \Rightarrow \lambda_{m 1} T_{1}=\lambda_{m 2} T_{2} \\ \lambda_{m 1} \times 3000 &=1.2 \lambda_{m 1} \times T_{2} \\ T_{2} &=\left(\frac{3000}{1.2}\right) K=2500 K \end{aligned}

There are 5 questions to complete.