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]. 
View Answer Report Discuss 5050! 
Answer: (d).km = k0 [1 + β (t1 + t2)/2].
