E-PolyLearning

51. For the same amount of fabrication material and same inside capacity, the heat loss is lowest in
a. Sphere
b. Cylinder
c. Rectangle
d. Cube
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Answer: (b).Cylinder

52. With variable thermal conductivity, Fourier law of heat conduction through a plane wall can be expressed as
a. Q = -k0 (1 + β t) A d t/d x
b. Q = k0 (1 + β t) A d t/d x
c. Q = – (1 + β t) A d t/d x
d. Q = (1 + β t) A d t/d x
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Answer: (a).Q = -k0 (1 + β t) A d t/d x

53. The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation
K = (1.45 + 0.5 * 10¯⁵ t²) KJ/m hr deg

Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
a. 1355.3 kJ/m² hr
b. 2345.8 kJ/m² hr
c. 1745.8 kJ/m² hr
d. 7895.9 kJ/m² hr
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Answer: (c).1745.8 kJ/m² hr

54. A plane wall of thickness δ has its surfaces maintained at temperatures T₁ and T₂. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k₀ T². Find the expression to work out the steady state heat conduction through the wall?
a. Q = 2A k₀ (T₁³ – T₂³)/3 δ
b. Q = A k₀ (T₁³ – T₂³)/3 δ
c. Q = A k₀ (T₁² – T₂²)/3 δ
d. Q = A k₀ (T₁ – T₂)/3 δ
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Answer: (b).Q = A k₀ (T₁³ – T₂³)/3 δ

55. The mean thermal conductivity evaluated at the arithmetic mean temperature is represented by
a. km = k0 [1 + β (t1 – t2)/2].
b. km = k0 [1 + (t1 + t2)/2].
c. km = k0 [1 + β (t1 + t2)/3].
d. km = k0 [1 + β (t1 + t2)/2].
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Answer: (d).km = k0 [1 + β (t1 + t2)/2].