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MCQs

Total Questions : 75 | Page 7 of 8 pages
Question 61. Find the value of corrected length for rectangular fin?
Where, b is width and t is length of the fin
  1.    L C = L + b t/2 (b + t)
  2.    L C = L + b t/ (b + t)
  3.    L C = L + 2 (b + t)
  4.    L C = L + b t
 Discuss Question
Answer: Option A. -> L C = L + b t/2 (b + t)
Answer: (a).L C = L + b t/2 (b + t)
Question 62. In convection from the tip what is the value of correction length?
  1.    L C = A C/P
  2.    L C = L + A C
  3.    L C = L + P
  4.    L C = L + A C/P
 Discuss Question
Answer: Option D. -> L C = L + A C/P
Answer: (d).L C = L + A C/P
Question 63. Let us assume there are two pieces of copper wire 0.1625 cm in diameter with a device that melts it at 195 degree Celsius. The wires are positioned vertically in air at 24 degree Celsius and the heat transfer coefficient of the wire is 17 W/ square m K. Let us say k = 335W/m K i.e. of wire. Find out the energy input?Let Us Assume There Are Two Pieces Of Copper Wire 0.1625 Cm ...
  1.    1.234W
  2.    2.652W
  3.    4.562W
  4.    9.435W
 Discuss Question
Answer: Option B. -> 2.652W
Answer: (b).2.652W
Question 64. Which one is true for the spine?
  1.    A C = π d²/4 and P = 4 π d
  2.    A C = π d²/4 and P = 3 π d
  3.    A C = π d²/4 and P = π d
  4.    A C = π d²/4 and P = 2 π d
 Discuss Question
Answer: Option C. -> A C = π d²/4 and P = π d
Answer: (c).A C = π d²/4 and P = π d
Question 65. The figure shows a 5 cm diameter rod, 90 cm long, which is having its lower face grinded smooth. The remainder of the rod is exposed to 32 degree Celsius room air and a surface coefficient heat transfer equal to 6.8 W/m² degree exists between the rod surface and the room air. The grinder dissipates mechanical energy at the rate of 35 W. If thermal conductivity of rod material is 41.5 W/m degree, find the temperature of the rod at the point where the grinding is taking placeThe Figure Shows A 5 Cm Diameter Rod, 90 Cm Long, Which Is H...
  1.    161.45 degree celsius
  2.    151.45 degree celsius
  3.    141.45 degree celsius
  4.    131.45 degree celsius
 Discuss Question
Answer: Option B. -> 151.45 degree celsius
Answer: (b).151.45 degree celsius
Question 66. If heat dissipation for one fin is given by 377.45 k J/hour, then what is the heat dissipation for 12 fins?
  1.    7529.4 k J/hour
  2.    6529.4 k J/hour
  3.    5529.4 k J/hour
  4.    4529.4 k J/hour
 Discuss Question
Answer: Option D. -> 4529.4 k J/hour
Answer: (d).4529.4 k J/hour
Question 67. If fin is sufficiently thin, so heat flows pertains to
  1.    One dimensional heat conduction
  2.    Two dimensional heat conduction
  3.    Three dimensional heat conduction
  4.    No heat flow is there
 Discuss Question
Answer: Option A. -> One dimensional heat conduction
Answer: (a).One dimensional heat conduction
Question 68. In order to achieve maximum heat dissipation the fin should be designed in such a way that has a
  1.    Maximum lateral surface towards the tip side of fin
  2.    Minimum lateral surface near the center line
  3.    Maximum lateral surface at the root side of fin
  4.    Maximum lateral surface near the center of fin
 Discuss Question
Answer: Option C. -> Maximum lateral surface at the root side of fin
Answer: (c).Maximum lateral surface at the root side of fin
Question 69. A steel rod (k = 30 W/m degree) 1 cm in diameter and 5 cm long protrudes from a wall which is maintained at 10 degree Celsius. The rod is insulated at its tip and is exposed to an environment with h = 50 W/m² degree and t a = 30 degree Celsius. Calculate the rate of heat dissipation
  1.    2.658 W
  2.    3.658 W
  3.    4.658 W
  4.    5.658 W
 Discuss Question
Answer: Option B. -> 3.658 W
Answer: (b).3.658 W
Question 70. On a heat transfer surface, fins are provided to
  1.    Increase turbulence in flow for enhancing heat transfer
  2.    Increase temperature gradient so as to enhance heat transfer
  3.    Pressure drop of the fluid should be minimized
  4.    Surface area is maximum to promote the rate of heat transfer
 Discuss Question
Answer: Option D. -> Surface area is maximum to promote the rate of heat transfer
Answer: (d).Surface area is maximum to promote the rate of heat transfer

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