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

0 | |

1 | |

\frac{(\pi d h+\pi d^2/2)}{6L^2} | |

1-\frac{(\pi d h+\pi d^2/2)}{6L^2} |

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).

4210 | |

6520 | |

3650 | |

4830 |

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.

\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).

491.34 | |

583.36 | |

965.24 | |

218.34 |

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).

0.12 | |

0.45 | |

0.88 | |

0.77 |

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).

3000 | |

4500 | |

2500 | |

1800 |

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}

\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.

Question 3 has incorrect Diagram !

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