Question 1 |

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).

4210 | |

6520 | |

3650 | |

4830 |

Question 1 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.

\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 2 |

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).

491.34 | |

583.36 | |

965.24 | |

218.34 |

Question 2 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 3 |

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).

0.12 | |

0.45 | |

0.88 | |

0.77 |

Question 3 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 4 |

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).

3000 | |

4500 | |

2500 | |

1800 |

Question 4 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}

\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 5 |

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 ______.

0.2 | |

0.4 | |

0.6 | |

0.8 |

Question 5 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 6 |

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

0.0625 | |

0.25 | |

0.5 | |

1 |

Question 6 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 7 |

The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 \mum. If the peak wavelength of emitted radiation changes to 2.90 \mum, then the temperature (in K) of the black body is

500 | |

1000 | |

4000 | |

8000 |

Question 7 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}

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 8 |

The emissive power of a blackbody is P. If its absolute temperature is doubled, the emissive power becomes

2P | |

4P | |

8P | |

16P |

Question 8 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

\frac{E_{b_{2}}}{E_{b_{1}}}=\left(\frac{2 T}{T}\right)^{4} \Rightarrow E_{b_{2}}=16 P

Question 9 |

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.

12.85 | |

13.77 | |

15.23 | |

17.33 |

Question 9 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

\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 10 |

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 __________ _________

11.05 | |

56.54 | |

45.54 | |

65.32 |

Question 10 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}

There are 10 questions to complete.

Question 3 has incorrect Diagram !

Q.3 has wrong figure given , pls correct it.