Heat and Mass Transfer MCQs

Question 1

Unit of thermal conductivity in M.K.S. units is

Accepted Answer

K cal m/hr m2 °C

Answer

K cal/kg m2 °C

Answer

K cal/hr m2 °C

Answer

K calm/hr °C

Question 2

In heat exchangers, degree of approach is defined as the difference between temperatures of

Accepted Answer

Hot medium outlet and cold water outlet

Answer

Cold water inlet and outlet

Answer

Hot medium inlet and outlet

Answer

Hot medium outlet and cold water inlet

Question 3

Fouling factor is used

Accepted Answer

In heat exchanger design as a safety factor

Answer

In case of Newtonian fluids

Answer

When a liquid exchanges heat with a gas

Answer

None of the above

Question 4

The natural convection air cooled condensers are used in

Accepted Answer

All of these

Answer

Domestic refrigerators

Answer

Water coolers

Answer

Room air conditioners

Question 5

The concept of overall coefficient of heat transfer is used in heat transfer problems of

Accepted Answer

Conduction and convection

Answer

Conduction

Answer

Convection

Answer

Radiation

Question 6

An electric cable of aluminium conductor (k = 240 W/mK) is to be insulated with rubber (k = 0.15 W/mK). The cable is to be located in air (h = 6 W/m2). The critical thickness of insulation will be

Accepted Answer

25 mm

Answer

40 mm

Answer

160 mm

Answer

800 mm

Question 7

The logarithmic mean temperature difference ($${{	ext{t}}_{	ext{m}}}$$) is given by (where $$\Delta {{	ext{t}}_1}$$ and $$\Delta {{	ext{t}}_2}$$ are temperature differences between the hot and cold fluids at entrance and exit)

Accepted Answer

$${{	ext{t}}_{	ext{m}}} = \frac{{\Delta {{	ext{t}}_1} - \Delta {{	ext{t}}_2}}}{{{{\log }_{	ext{e}}}\left( {\frac{{\Delta {{	ext{t}}_1}}}{{\Delta {{	ext{t}}_2}}}} ight)}}$$

Answer

$${{	ext{t}}_{	ext{m}}} = \frac{{{{\log }_{	ext{e}}}\left( {\frac{{\Delta {{	ext{t}}_1}}}{{\Delta {{	ext{t}}_2}}}} ight)}}{{\Delta {{	ext{t}}_1} - \Delta {{	ext{t}}_2}}}$$

Answer

$${{	ext{t}}_{	ext{m}}} = \left( {\Delta {{	ext{t}}_1} - \Delta {{	ext{t}}_2}} ight){\log _{	ext{e}}}\left( {\frac{{\Delta {{	ext{t}}_1}}}{{\Delta {{	ext{t}}_2}}}} ight)$$

Answer

$${{	ext{t}}_{	ext{m}}} = \frac{{{{\log }_{	ext{e}}}\left( {\Delta {{	ext{t}}_1} - \Delta {{	ext{t}}_2}} ight)}}{{\frac{{\Delta {{	ext{t}}_1}}}{{\Delta {{	ext{t}}_2}}}}}$$

Question 8

Emissivity of a white polished body in comparison to a black body is

Accepted Answer

Lower

Answer

Higher

Answer

Same

Answer

Depends upon the shape of body

Question 9

The heat of sun reaches to us according to

Accepted Answer

Radiation

Answer

Conduction

Answer

Convection

Answer

None of these