Heat Transfer Test Questions Set - 8

1)   A hollow sphere of internal diameter ID = 20 cm and outer diameter OD = 30 cm contains hot fluid. What should be the critical radius of insulation for maximum rate of heat transfer? Thermal conductivity k = 0.86 W/mK and convection heat transfer coefficient  of outer fluid ho = 20 W/m2K.

a. 0.086 cm
b. 0.86 m
c. 8.6 mm
d. 8.6 cm
Answer  Explanation  Related Ques

ANSWER: 8.6 cm

Explanation:
No explanation is available for this question!


2)   Consider a plane wall of finite thickness having energy generation in it. The sum of the rate of heat energy generation in the wall and rate of energy transfer by conduction into the wall is equal to

a. rate of heat transfer by conduction out of the wall
b. rate of heat transfer by convection out of the wall
c. both a. and b.
d. none of the above
Answer  Explanation  Related Ques

ANSWER: rate of heat transfer by conduction out of the wall

Explanation:
No explanation is available for this question!


3)   The maximum temperature difference between heat generation temperature (T) and the temperature of outer surface (To) of a plane wall of thickness 2l is given by

Consider,

Heat is generated at the rate of at the centre of wall being both the outer surfaces are at distance l from the heat generation point

k  is thermal conductivity of the wall


a. (T – To)max = q̇l2 / k
b. (T – To)max = 2q̇l2 / k
c. (T – To)max = 4q̇l2 / k
d. (T – To)max = q̇l2 / 2k
Answer  Explanation  Related Ques

ANSWER: (T – To)max = q̇l2 / 2k

Explanation:
No explanation is available for this question!


4)   What is the formula for maximum temperature (Tmax) at the center of a long and solid cylindrical electric wire of radius r?

Where,

is rate of heat generation into the electric wire
k is thermal conductivity of the electric wire
and To is the surface temperature of the wire


a. (Tmax)=(To) +(q̇r2/2k)
b. (Tmax) =(To) +(q̇r2/4k)
c. (Tmax) =(q̇r2/2k)
d. Tmax) =(To) +(4q̇r2/k)
Answer  Explanation  Related Ques

ANSWER: (Tmax) =(To) +(q̇r2/4k)

Explanation:
No explanation is available for this question!


5)   What is the purpose of using fins in a particular heat transfer system?

a. to decrease rate of heat transfer
b. to increase rate of heat transfer
c. to maintain rate of heat transfer at a constant rate
d. cannot say
Answer  Explanation  Related Ques

ANSWER: to increase rate of heat transfer

Explanation:
No explanation is available for this question!


6)   Temperature at the end tip of the fin having uniform cross-sectional area is

a. maximum
b. minimum
c. similar to the heat generation temperature
d. unpredictable
Answer  Explanation  Related Ques

ANSWER: minimum

Explanation:
No explanation is available for this question!


7)   Consider that heat transfer is taking place through a fin having circular cross-sectional area, one dimensionally as shown in figure.
The rate of heat transfer by conduction into a section at x is equal to


Heat-Transfer-through-Fin with-Uniform-Cross-sectional-Area.png
a. sum of rate of heat transfer by convection out of the element (x+dx) and heat transfer by convection from the surface between x to (x+dx)
b. sum of rate of heat transfer by conduction out of the element (x+dx) and heat transfer by conduction from the surface between x to (x+dx)
c. sum of rate of heat transfer by conduction out of the element (x+dx) and heat transfer by convection from the surface between x to (x+dx)
d. none of the above
Answer  Explanation  Related Ques

ANSWER: sum of rate of heat transfer by conduction out of the element (x+dx) and heat transfer by convection from the surface between x to (x+dx)

Explanation:
No explanation is available for this question!


8)   In the process of heat transfer through extended surfaces or fins, the entire surface area is at

a. the same constant temperature
b. different temperatures
c. maximum base temperature
d. minimum temperature
Answer  Explanation  Related Ques

ANSWER: different temperatures

Explanation:
No explanation is available for this question!


9)   An engine is fitted with pin fins having thermal conductivity k = 200 W/mK. The diameter and length of the fin is 2 cm and 50 cm respectively. Calculate the temperature at 10 cm from the fin base if fine base temperature is 500 0C and fin is in contact with air at 50 0C. Take h = 12 W/m2K
Consider that the fin is infinitely long


a. 80.67 0C
b. 100 0C
c. 85.67 0C
d. 185.67 0C
Answer  Explanation  Related Ques

ANSWER: 185.67 0C

Explanation:
No explanation is available for this question!


10)   What is the relation between the intensity of radiation and the thickness of the layer, in case of gas?

a. the intensity of radiation increases with increase in the thickness of gas layer
b. the intensity of radiation decreases with increase in the thickness of gas layer
c. the intensity of radiation remains unaffected with change in the thickness of gas layer
d. unpredictable
Answer  Explanation  Related Ques

ANSWER: the intensity of radiation decreases with increase in the thickness of gas layer

Explanation:
No explanation is available for this question!


11)   90% of the incident radiation on glass in the visible range is

a. absorbed
b. reflected
c. transmitted
d. none of the above
Answer  Explanation  Related Ques

ANSWER: transmitted

Explanation:
No explanation is available for this question!


12)   In greenhouse effect, the only infrared radiations

a. are allowed to exit from the interior surface
b. are not allowed to exit from the interior surface
c. cannot absorb heat energy
d. none of the above
Answer  Explanation  Related Ques

ANSWER: are not allowed to exit from the interior surface

Explanation:
No explanation is available for this question!


13)   How is the greenhouse effect experienced on earth?

a. global worming
b. pollution
c. both a. and b.
d. none of the above
Answer  Explanation  Related Ques

ANSWER: global worming

Explanation:
No explanation is available for this question!


14)   Which of the following gases is/are responsible for global warming?

a. Carbon dioxide (CO2)
b. Water vapour (H2O)
c. both a. and b.
d. none of the above
Answer  Explanation  Related Ques

ANSWER: both a. and b.

Explanation:
No explanation is available for this question!


15)   In desert areas, there is large difference between day and night temperatures mainly because of

a. presence of carbon dioxide in air as it acts as barrier for emanating infrared radiation from the earth surface
b. presence of water vapour in air as it acts as barrier for emanating infrared radiation from the earth surface
c. absence of carbon dioxide in air as it acts as barrier for emanating infrared radiation from the earth surface
d. absence of water vapour in air as it acts as barrier for emanating infrared radiation from the earth surface
Answer  Explanation  Related Ques

ANSWER: absence of water vapour in air as it acts as barrier for emanating infrared radiation from the earth surface

Explanation:
No explanation is available for this question!


16)   How is the cooling at night in coastal areas?

a. the day temperature reduces rapidly at night in coastal areas
b. the day temperature reduces slowly at night in coastal areas
c. the day temperature and a night temperature in coastal areas are similar
d. unpredictable
Answer  Explanation  Related Ques

ANSWER: the day temperature reduces slowly at night in coastal areas

Explanation:
No explanation is available for this question!


17)   The radiation energy from the sun is produced by

a. fission reaction
b. fusion reaction
c. both a. and b.
d. none of the above
Answer  Explanation  Related Ques

ANSWER: fusion reaction

Explanation:
No explanation is available for this question!


18)   The rate at which the solar energy reaching the earth's atmosphere is called as

a. solar constant
b. radiation constant
c. electromagnetic constant
d. atmospheric constant
Answer  Explanation  Related Ques

ANSWER: solar constant

Explanation:
No explanation is available for this question!


19)   The radiation from fusion of hydrogen

a. can be achieved on earth's surface
b. cannot be achieved on earth's surface
c. can be achieved on earth's surface at high maintained temperature
d. unpredictable
Answer  Explanation  Related Ques

ANSWER: cannot be achieved on earth's surface

Explanation:
No explanation is available for this question!


20)   What is the purpose of using baffles in shell-and-tube heat exchangers?

a. to maintain uniform spacing between tubes
b. to enhance heat transfer
c. both a. and b.
d. none of the above
Answer  Explanation  Related Ques

ANSWER: both a. and b.

Explanation:
No explanation is available for this question!


21)   Large sized and heavy shell-and-tube heat exchangers are suitable for

a. automotive applications
b. aircraft applications
c. marine applications
d. none of the above
Answer  Explanation  Related Ques

ANSWER: none of the above

Explanation:
No explanation is available for this question!


22)   How can the total thermal resistance of a heat exchanger, in which two fluids are separated by a plane wall of surface area A, thickness t and thermal conductivity k, be calculated?
h1 and h2 are the convection heat transfer coefficients on each side of wall.


a. (t / h1 A) + (t / h2 A)
b. (1 / h1 A) + (1 / h2 A)
c. (1 / h1 A) + (1 / h2 A) + (1 / k A)
d. (1 / h1 A) + (1 / h2 A) + (t / k A)
Answer  Explanation  Related Ques

ANSWER: (1 / h1 A) + (1 / h2 A) + (t / k A)

Explanation:
No explanation is available for this question!


23)   Which side is suitable to provide fins on to enhance heat transfer, when heat exchanges between gas and liquid?

a. Gas side
b. Liquid side
c. None of the sides

Answer  Explanation  Related Ques

ANSWER: Gas side

Explanation:
No explanation is available for this question!


24)   How is the effective surface area of finned tube surface calculated?
Where,
Afin = the area of tube surface on which fines are provided
Aunfinned = the area of tube surface on which fines are not provided
ηfin = fin efficiency


a. A = Aunfinned + Afin
b. A = ηfin (Aunfinned + Afin)
c. A = Aunfinned + (ηfin x Afin)
d. A = (ηfin x Afin)
Answer  Explanation  Related Ques

ANSWER: A = Aunfinned + (ηfin x Afin)

Explanation:
No explanation is available for this question!


25)   The scales form in heat exchangers after a period of operation and provide additional resistance to heat transfer with some heat transfer coefficient. The reciprocal of this scale heat transfer coefficient is called as

a. scaling factor
b. fouling factor
c. forming factor
d. resisting factor
Answer  Explanation  Related Ques

ANSWER: fouling factor

Explanation:
No explanation is available for this question!


26)   Fouling factor of a heat exchanger can be calculated by

a. [Thermal resistance of heat exchanger without scaling] – [Thermal resistance of heat exchanger with scaling]
b. [Thermal resistance of heat exchanger with scaling] – [Thermal resistance of heat exchanger without scaling]
c. [Thermal resistance of heat exchanger without scaling] x [Thermal resistance of heat exchanger with scaling]
d. none of the above
Answer  Explanation  Related Ques

ANSWER: [Thermal resistance of heat exchanger with scaling] – [Thermal resistance of heat exchanger without scaling]

Explanation:
No explanation is available for this question!


27)   How is the logarithmic mean temperature difference (LMTD) calculated for heat exchangers?Where,
ΔTi= temperature difference between hot and cold fluid at inlet of heat exchanger
ΔTe = temperature difference between hot and cold fluid at exit of heat exchanger


a. ln (ΔTi– ΔTe)
b. ln (ΔTe – ΔTi)
c. (ΔTi– ΔTe) / ( ln (ΔTe / Δti))
d. (ΔTi– ΔTe) / ( ln (ΔTi / Δte))
Answer  Explanation  Related Ques

ANSWER: (ΔTi– ΔTe) / ( ln (ΔTi / Δte))

Explanation:
No explanation is available for this question!


28)   Which type of flow in heat exchanger is represented in below diagram?

Temperature-Profile-of-Fluids-in-Heat-Exchanger.png
a. Parallel flow heat exchanger
b. Counter flow heat exchanger
c. Cross flow heat exchanger
d. none of the above
Answer  Explanation  Related Ques

ANSWER: Parallel flow heat exchanger

Explanation:
No explanation is available for this question!


29)   The arithmetic mean temperature difference for parallel flow heat exchanger is given as

a. ΔTam = (ΔTi– ΔTe)
b. ΔTam = (ΔTi+ ΔTe)
c. ΔTam = (ΔTi– ΔTe) / 2
d. ΔTam = (ΔTi+ ΔTe) / 2
Answer  Explanation  Related Ques

ANSWER: ΔTam = (ΔTi+ ΔTe) / 2

Explanation:
No explanation is available for this question!