Heat Transfer Miscellaneous
 Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different temp is valid when

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For lumed system analysis, Bi < 0.1
Correct Option: A
For lumed system analysis, Bi < 0.1
 When the fluid velocity is doubled the thermal time constant of a thermometer used of measuring the fluid temperature reduces by a factor of 2 (T/F)

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It is not necessary that h would become a half if velocity becomes double.
Correct Option: D
It is not necessary that h would become a half if velocity becomes double.
 Biot number signifies

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Biot number = R_{conduction}/R_{convective}
Correct Option: D
Biot number = R_{conduction}/R_{convective}
 The heat transfer process between body and its ambient is governed by an Intemal Conductive Resistance (ICR) and an External Convective Resistance (ECR). The body can be considered to be a lumped heat capacity system is

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Lumped heat capacity can be applied when Bi < 0.1
B_{i} hl/k, for desired condition i.e. Bi < 0
k > h, ICR is smallCorrect Option: D
Lumped heat capacity can be applied when Bi < 0.1
B_{i} hl/k, for desired condition i.e. Bi < 0
k > h, ICR is small
 In a concentric counter flow heat exchange, water flows through the inner tube at 25°C and leaves at 42°C. The engine oil enters at 100°C and flows in the annular flow passage. The exit temperature of the engine oil is 50°C. Mass flow rate of water and the engine oil are 1.5 kg/s and 1 kg/s, respectively. The specific heat of water and oil are 4178 J/kgK and 2130 J/kgK, respectively. The effectiveness of this heat exchange is

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T_{c,i} = 25°C;
T_{c,o} = 42°C;
m_{c} = 1.5 kg/s;
C_{c} = 4.178 kJ/kg.k
T_{h,i} = 100°C;
T_{h,o} = 50°C;
m_{h} = 1 kg/s;
C_{h} = 2.130 kJ/kg.k
m_{c}C_{c} = 1.5 × 4.178
= 6.267 kW/°C = C_{max}
m_{h}C_{h} = 1 × 2.130
= 2.130 kW/°C = C_{min}effectiveness = ε = C_{h}(T_{h,i}  T_{h,o}) C_{min}(T_{h,i}  T_{c,i}) = 100  50 = 0.666 100  25 Correct Option: A
T_{c,i} = 25°C;
T_{c,o} = 42°C;
m_{c} = 1.5 kg/s;
C_{c} = 4.178 kJ/kg.k
T_{h,i} = 100°C;
T_{h,o} = 50°C;
m_{h} = 1 kg/s;
C_{h} = 2.130 kJ/kg.k
m_{c}C_{c} = 1.5 × 4.178
= 6.267 kW/°C = C_{max}
m_{h}C_{h} = 1 × 2.130
= 2.130 kW/°C = C_{min}effectiveness = ε = C_{h}(T_{h,i}  T_{h,o}) C_{min}(T_{h,i}  T_{c,i}) = 100  50 = 0.666 100  25