Question 1
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 1 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 2
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 2 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}
 Question 3
Three sets of parallel plates LM, NR and PQ are given in Figures 1, 2 and 3.The view factor $F_{IJ}$ is defined as the fraction of radiation leaving plate-I that is intercepted by plate-J. Assume that the values of $F_{LM} \; and \; F_{NR}$ are 0.8 and 0.4, respectively. The value of $F_{PQ}$(round off to one decimal place) is ______. A 0.2 B 0.4 C 0.6 D 0.8
GATE ME 2019 SET-2   Heat Transfer
Question 3 Explanation: $\begin{array}{l} \mathrm{F}_{\mathrm{LM}}=\mathrm{F}_{\mathrm{L}-\mathrm{a}}+\mathrm{F}_{\mathrm{L}-\mathrm{b}}+\mathrm{F}_{\mathrm{L}-\mathrm{c}}=0.8-\mathrm{c}\ldots(1) \\ \mathrm{F}_{\mathrm{N}-\mathrm{R}}=0.4 \end{array}$
From the figure,
$\begin{array}{l} \mathrm{F}_{\mathrm{L}-\mathrm{a}}=\mathrm{F}_{\mathrm{L}-\mathrm{c}}=\mathrm{F}_{\mathrm{P}-\mathrm{e}}(\text { similar }) \\ \mathrm{F}_{\mathrm{L}-\mathrm{b}}=\mathrm{F}_{\mathrm{N}-\mathrm{R}}(\text { similar }) \end{array}$
From equation (1)
$\begin{array}{l} \mathrm{F}_{\mathrm{L}-\mathrm{a}}+\mathrm{F}_{\mathrm{L}-\mathrm{b}}+\mathrm{F}_{\mathrm{L}-\mathrm{c}}=0.8 \\ 2 \times \mathrm{F}_{\mathrm{L}-\mathrm{a}}+\mathrm{F}_{\mathrm{N}-\mathrm{R}}=0.8 \\ \mathrm{F}_{\mathrm{L}-\mathrm{a}}=\frac{0.8-0.4}{2}=0.2 \end{array}$
From the fig -3
\begin{aligned} \mathrm{F}_{\mathrm{P}-\mathrm{Q}}&=\mathrm{F}_{\mathrm{P}-\mathrm{d}}+\mathrm{F}_{\mathrm{P}-\mathrm{e}} =\mathrm{F}_{\mathrm{N}-\mathrm{R}}+\mathrm{F}_{\mathrm{L}-\mathrm{a}} \\ (\mathrm{F}_{\mathrm{P}-\mathrm{d}}&=\mathrm{F}_{\mathrm{N}-\mathrm{R}} \text{ and } \mathrm{F}_{\mathrm{P}-\mathrm{e}}=\mathrm{F}_{\mathrm{L}-\mathrm{a}}(\text { due to similarity })) \\ &=0.4+0.2\\ F_{P-Q}&=0.6 \end{aligned}
 Question 4
Sphere-1 with a diameter of 0.1 m is completely enclosed by another sphere-2 of diameter 0.4 m. The view factor $F_{12}$ is
 A 0.0625 B 0.25 C 0.5 D 1
GATE ME 2019 SET-2   Heat Transfer
Question 4 Explanation: Given data:
$\mathrm{d}_{1}=0.1 \mathrm{m}$
$\mathrm{d}_{2}=0.4 \mathrm{m}$
$\mathrm{F}_{1-1}=0$ (from the geometry)
From the additive rule
$\mathrm{F}_{1-1}+\mathrm{F}_{1-2}=1$
$\mathrm{F}_{1-2}=1$
 Question 5
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 $\mu$m. If the peak wavelength of emitted radiation changes to 2.90 $\mu$m, then the temperature (in K) of the black body is
 A 500 B 1000 C 4000 D 8000
GATE ME 2018 SET-2   Heat Transfer
Question 5 Explanation:
From Wein's displacement law
For black body,
\begin{aligned} \lambda_{M} T &=\text { constant } \\ \lambda_{M 1} T_{1} &=\lambda_{M 2} T_{2} \\ 1.45 \times 2000 &=\lambda_{M 2} T_{2}=2.90 \times T_{2} \\ \therefore \quad T_{2} &=\left(\frac{1.45}{2.90} \times 2000\right)=1000 \mathrm{K} \end{aligned}
 Question 6
The emissive power of a blackbody is P. If its absolute temperature is doubled, the emissive power becomes
 A 2P B 4P C 8P D 16P
GATE ME 2017 SET-2   Heat Transfer
Question 6 Explanation:
$E_{b} \propto T^{4}$(Stefan Boltzmann Law of Radiation)
$\frac{E_{b_{2}}}{E_{b_{1}}}=\left(\frac{2 T}{T}\right)^{4} \Rightarrow E_{b_{2}}=16 P$
 Question 7
Two black surfaces, AB and BC, of lengths 5m and 6m, respectively, are oriented as shown. Both surfaces extend infinitely into the third dimension. Given that view factor $F_{12}=0.5$,$T_{1}=800K$,$T_{2}=600K$ , $T_{surrounding} =300K$, and Stefan Boltzmann constant, $\sigma =5.67\,\times\,10^{-8}W/(m^{2}K^{4})$, the heat transfer rate from Surface 2 to the surrounding environment is ____________ kW. A 12.85 B 13.77 C 15.23 D 17.33
GATE ME 2017 SET-1   Heat Transfer
Question 7 Explanation:
F^{12} =0.5 \begin{aligned} T_{1} &=800 \mathrm{K} \\ T_{2} &=600 \mathrm{K} \\ T_{\text {surounding }} &=300 \mathrm{K} \end{aligned}
Let (3) denote the surrounding environment
(Assume unit width perpendicular to plane of Fig.)
From summation rule:
\begin{aligned} F_{21}+F_{\beta 2}^{0}+F_{23} &=1 \\ A_{1} F_{12} &=A_{2} F_{21}=6 \times 1 \times 0.5 \\ &=(5 \times 1) \times F_{21} \\ F_{21} &=\frac{6 \times 0.5}{5}=0.6\\ \therefore \quad F_{23}&=1-F_{21}=0.4 \end{aligned}
since surrounding environment being large,
No surface has a surface resistance. Heat transfer rate from (2) and (3)$=\frac{E_{b_{2}}-E_{b_{3}}}{\frac{1}{A_{2} F_{23}}}$
\begin{aligned} &=\frac{\sigma\left(T_{2}^{4}-T_{3}^{4}\right)}{1} \\ &=\frac{5.67 \times 10^{-8}\left(600^{4}-300^{4}\right)}{A_{2} F_{23}} \text { watt/metre }\\ &=13.778 \mathrm{kW} / \mathrm{metre} \\ \end{aligned}
Assume that heat exchange is per unit width perpendicular to plane of figure
 Question 8
Two large parallel plates having a gap of 10 mm in between them are maintained at temperatures T1 = 1000 K and T2 = 400 K. Given emissivity values, $\varepsilon{1}=0.5$ , $\varepsilon_{2}=0.25$ and Stefan-Boltzmann constant $\sigma= 5.67 \times 10^{-8} W/m^{2}-K^{4}$, the heat transfer between the plates (in kW/$m^{2}$) is __________ _________
 A 11.05 B 56.54 C 45.54 D 65.32
GATE ME 2016 SET-3   Heat Transfer
Question 8 Explanation: \begin{aligned} Q &=\frac{\sigma\left[T_{1}^{4}-T_{2}^{4}\right]}{\frac{1}{\epsilon_{1}}+\frac{1}{\epsilon_{2}}-1}=\frac{5.67 \times 10^{-8} \times\left(1000^{4}-400^{4}\right)}{\frac{1}{0.5}+\frac{1}{0.25}-1} \\ &=11049.7 \mathrm{W} / \mathrm{m}^{2}=11.05 \mathrm{kW} / \mathrm{m}^{2} \end{aligned}
 Question 9
Consider the radiation heat exchange inside an annulus between two very long concentric cylinders. The radius of the outer cylinder is $R_{0}$ and that of the inner cylinder is $R_{i}$ . The radiation view factor of the outer cylinder onto itself is
 A $1-\sqrt{\frac{R_{i}}{R_{0}}}$ B $\sqrt{1-\frac{R_{i}}{R_{0}}}$ C $1-( \frac{R_{i}}{R_{0}} )^{\frac{1}{3}}$ D $1-\frac{R_{i}}{R_{0}}$
GATE ME 2016 SET-2   Heat Transfer
Question 9 Explanation: To find: Radiation view factor of the outer cylinder onto itself.
Hence $F_{11}$ is required.
\begin{aligned} \Rightarrow A_{1} F_{12}&=A_{2} F_{21} \qquad[\text { Reciprosity }]\\ F_{11}+F_{12}&=1 \qquad \text { [Summation] }\\ F_{21}+F_{22}&=1\\ \therefore \quad F_{11}&=1-F_{12} \\ F_{22}&=0 \text { [No radiation leaving }\\ &\text{surface 2 reaches back to 2]}\\ \therefore \quad F_{21} &=1 \\ F_{12} &=\frac{A_{2}}{A_{1}} \times F_{21}=\frac{A_{2}}{A_{1}} \times 1 \\ &=\frac{2 \pi R_{i} L}{2 \pi R_{0} L} \times 1=\frac{R_{i}}{R_{0}} \\ F_{11} &=1-F_{12}=1-\frac{R_{i}}{R_{0}} \end{aligned}
 Question 10
An infinitely long furnace of 0.5 m x 0. 4 m cross-section is shown in the figure below. Consider all surfaces of the furnace to be black. The top and bottom walls are maintained at temperature $T_{1}=T_{3}=927^{\circ}C$ while the side walls are at temperature $T_{2}=T_{4}=527^{\circ}C$ . The view factor, $F_{1-2}$ is 0.26. The net radiation heat loss or gain on side 1 is_________ W/m.
Stefan-Boltzmann constant = $5.67\times10 ^{-8}W/m^{2}-k^{4}$ A 24800.95W/m B 24530.68W/m C 45856.25W/m D 45897.25W/m
GATE ME 2016 SET-1   Heat Transfer
Question 10 Explanation: \begin{aligned} \text { Given: } F_{12}&=0.26 \\ F_{12}=F_{14}&=0.26 \qquad [symmetry]\\ \text{Given}\quad \sigma&=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \mathrm{K}^{4} \end{aligned}
side (1) and (3) will not exchange and heat being at same temp.
side (2) and (4) will not exchange any heat being at same temp.
Hence side (1) and side (2) and side (1) and side
(4) will exchange heat
\begin{aligned} Q_{1-2} &=A_{1} F_{12} \sigma\left[T_{1}^{4}-T_{2}^{4}\right] \\ &=0.5 \times 0.26 \times 5.67 \times 10^{-8} x \\ &=12265.34 \mathrm{W} / \mathrm{m} \\ Q_{1-4} &=0.5 \times 0.26 \times 5.67 \times 10^{-8} \mathrm{x} \\ &=12265.34 \mathrm{W} / \mathrm{m} \\ Q_{\text {net }} &=Q_{1-2}+Q_{1-4}=12265.34+\\ &=24530.68 \mathrm{W} / \mathrm{m} \end{aligned}
There are 10 questions to complete.

### 2 thoughts on “Radiation”

1. Question 3 has incorrect Diagram !

2. 