Heat Transfer Miscellaneous
 Biot number signifies the ratio of

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Biot  number:
Biot number provides a way to compare the conduction resistance within a solid body to the convection resistance external to the body (offered by the surrounding fluid) for heat transfer:Bi = hs ; s = Volume of the body k Surface area
Where ‘s’ is a characteristic dimension of the solid
‘h’ is convective heat transfer coefficient
‘k’ is thermal conductivity of the body.Correct Option: B
Biot  number:
Biot number provides a way to compare the conduction resistance within a solid body to the convection resistance external to the body (offered by the surrounding fluid) for heat transfer:Bi = hs ; s = Volume of the body k Surface area
Where ‘s’ is a characteristic dimension of the solid
‘h’ is convective heat transfer coefficient
‘k’ is thermal conductivity of the body.
 A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030°C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30°C, with convective heat transfer coefficient h = 20 W/m²K. The thermophysical properties of steel are: density ρ = 7800 kg/m², conductivity k = 40 W/mK and specific heat c  600 J/kgK. The time required in seconds to cool the steel ball in air from 1030°C to 430°C is

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Diameter of steel ball,
d = 0.06 m
Initial temperature,
T_{i} = 1030°C
Ambient temperature
T_{0} = 30°C
Convective heat transfer coefficienth = 20 W m²K ρ = 7800 kg m³ k = 40 W mK C = 600 J kg
Final temperature of steel ball, T = 430°c
Time: t =?T_{i}  T_{0} = e^{ht/ρCl}.......(i) T_{i}  T_{0} l = V = πd³ = d = 0.6 = 0.1 m A 6 × πd² 6 6
From (i)430  30 = e^{20×t/7800×600×0.1} 1030  30
t = 2144.16 secondsCorrect Option: D
Diameter of steel ball,
d = 0.06 m
Initial temperature,
T_{i} = 1030°C
Ambient temperature
T_{0} = 30°C
Convective heat transfer coefficienth = 20 W m²K ρ = 7800 kg m³ k = 40 W mK C = 600 J kg
Final temperature of steel ball, T = 430°c
Time: t =?T_{i}  T_{0} = e^{ht/ρCl}.......(i) T_{i}  T_{0} l = V = πd³ = d = 0.6 = 0.1 m A 6 × πd² 6 6
From (i)430  30 = e^{20×t/7800×600×0.1} 1030  30
t = 2144.16 seconds
 Consider steadystate heat conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides.
Given: h_{i} = 20 W/m²K; h_{0} = 50 W/m²K; T_{∞i} = 20°C; T_{*infin;,0} = –2°C; k_{1} = 20 W/mK; k_{2} = 50 W/mK; L_{1} = 0.30 m and L_{2} = 0.15 m.
Assuming negligible contact resistance between the wall surfaces, the interface termperature, T(in °C), of the two walls will be

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R_{eq} = 1 + l_{1} + l_{2} + 1 h_{2}A h_{1}A h_{2}A h_{2}A = 1 1 + 0.30 + 0.15 + 1 = 0.088 A 20 20 50 50 A Now Q_{2} = (t_{1} + t_{2}) = t_{1}  t R_{eq} 1 + L_{1} h i_{A} h 1_{A} ∴ t_{1}  t = t_{1}  t 0.8 0.065 A A ⇒ 2.95 × 0.065 = 293  t 0.088
⇒ t = 3.75°C
Alternately
Under steady state conditions(20  t_{i}) = t_{1}  (2) 1 + l_{1} l_{2} + 1 h_{i}A k_{1}A k_{2}A h_{0}A ∴ 20  t_{i} = t_{1} + 2 1 + 0.3 0.15 + 1 20 20 50 50 or 20  t_{i} = T_{1} + 2 1.3 1.15 20 50 or 20  t_{i} = 1.3 × 50 {t_{i} + 2} 1.05 20
or 20 – t_{i} = 2.826 (t_{i} + 2)
or 3.826 t_{i} = 20 – 2 × 2.826
or t_{i} = 3.75°CCorrect Option: C
R_{eq} = 1 + l_{1} + l_{2} + 1 h_{2}A h_{1}A h_{2}A h_{2}A = 1 1 + 0.30 + 0.15 + 1 = 0.088 A 20 20 50 50 A Now Q_{2} = (t_{1} + t_{2}) = t_{1}  t R_{eq} 1 + L_{1} h i_{A} h 1_{A} ∴ t_{1}  t = t_{1}  t 0.8 0.065 A A ⇒ 2.95 × 0.065 = 293  t 0.088
⇒ t = 3.75°C
Alternately
Under steady state conditions(20  t_{i}) = t_{1}  (2) 1 + l_{1} l_{2} + 1 h_{i}A k_{1}A k_{2}A h_{0}A ∴ 20  t_{i} = t_{1} + 2 1 + 0.3 0.15 + 1 20 20 50 50 or 20  t_{i} = T_{1} + 2 1.3 1.15 20 50 or 20  t_{i} = 1.3 × 50 {t_{i} + 2} 1.05 20
or 20 – t_{i} = 2.826 (t_{i} + 2)
or 3.826 t_{i} = 20 – 2 × 2.826
or t_{i} = 3.75°C
 The logarithmic mean temperature difference (LMTD) of a counter flow heat exchanger is 20°C. The cold fluid enters at 20°C and the hot fluid enters at 100°C. Mass flow rate of the cold fluid is twice that of the hot fluid. Specific heat at constant pressure of the hot fluid is twice that of the cold fluid. The exit temperature of the cold fluid

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θ_{1} = θ_{2}
θ_{m} = 20°C
T_{h1} – T_{c2} = 20°C
100 – T_{c2} = 20°C
T_{c2} = 80°CCorrect Option: C
θ_{1} = θ_{2}
θ_{m} = 20°C
T_{h1} – T_{c2} = 20°C
100 – T_{c2} = 20°C
T_{c2} = 80°C
 In a counterflow heat exchanger, hot fluid enters at 60°C and cold fluid leaves at 30°C. Mass flow rate of the hot fluid is 1 kg/s and that the cold fluid is 2 kg/s. Specific heat of the hot fluid is 10 kJ/kgK and that of the cold fluid is 5 kJ/kgK. The Log Mean Temperature Difference (LMTD) for the heat exchanger in °C is

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Heat capacity of hot fluid
= 1 × 10 = 10 kJ/ k – s
Heat capacity of cold fluid
= 2 × 5 = 10 kJ/ k – s
Since heat capacity is same, so LMTD is difference of temperature at either end
i.e. LMTD = 60° – 30° = 30° CCorrect Option: B
Heat capacity of hot fluid
= 1 × 10 = 10 kJ/ k – s
Heat capacity of cold fluid
= 2 × 5 = 10 kJ/ k – s
Since heat capacity is same, so LMTD is difference of temperature at either end
i.e. LMTD = 60° – 30° = 30° C