Heat Transfer Miscellaneous
 A slender rod of length L, diameter d (L >> d) and thermal conductivity k_{1} i s joined with another rod of identical dimensions, but of thermal conductivity k 2, to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is

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As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
(R_{th}) _{eq} = R_{1} + R_{2}2L = L + L Ak_{eq} AK_{1} AK_{2} 2 = 1 + 1 K_{eq} K_{1} K_{2} ⇒ R_{eq} = 2K_{1}K_{2} K_{1} + K_{2} Correct Option: A
As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
(R_{th}) _{eq} = R_{1} + R_{2}2L = L + L Ak_{eq} AK_{1} AK_{2} 2 = 1 + 1 K_{eq} K_{1} K_{2} ⇒ R_{eq} = 2K_{1}K_{2} K_{1} + K_{2}
 The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 μm. If the peak wavelength of emitted radiation changes to 2.90 μm, then the temperature (in K) of the black body is

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From wein’s displacement law λ_{max} T = constant
(λ_{max})_{1} T_{1} = (λ_{max})_{2} T_{2}
1.45 × 2000 = 2.90 × T_{2}T_{2} = 1.45 × 2000 2.90
T_{2} = 1000 KCorrect Option: B
From wein’s displacement law λ_{max} T = constant
(λ_{max})_{1} T_{1} = (λ_{max})_{2} T_{2}
1.45 × 2000 = 2.90 × T_{2}T_{2} = 1.45 × 2000 2.90
T_{2} = 1000 K
 A balanced counter flow heat exchanger has a surface area of 20 m^{2} and overall heat transfer coefficient of 20 W/m2K. Air (c_{p} = 1000 J/kgK) entering at 0.4 kg/s and 280 K is to be preheated by the air leaving the system at 0.4 kg/s and 300 K. The temperature (in K) of the preheated air is

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Counter flow heat exchanged
Surface Area A = 20 m^{2}, mass flow rate = 0.4 kg/su = 20W ; m^{2}
Temperature T_{ci} = 280KC_{p} of air = 1000 1 kg K
Since m is same for both flow = 0.4 kg/s
Assume C_{p} is same = 100 J/kg.K
Hence ∆T_{1} = T_{ci} – T_{co} = ∆T_{2} = T_{h0} – T_{ci}
∆T_{1} = 300 – T_{co} = T_{h0} – 280
∆T_{m} = ∆T_{1} = ∆T_{2}
uA∆Tm = mC_{p} (T_{co} – T_{ci})
20 × 20 × (300 – T_{co}) = 0.4 × 1000(T_{co} – 280)
2T_{co} = 300 + 280T_{co} = 580 = 290 K 2 Correct Option: A
Counter flow heat exchanged
Surface Area A = 20 m^{2}, mass flow rate = 0.4 kg/su = 20W ; m^{2}
Temperature T_{ci} = 280KC_{p} of air = 1000 1 kg K
Since m is same for both flow = 0.4 kg/s
Assume C_{p} is same = 100 J/kg.K
Hence ∆T_{1} = T_{ci} – T_{co} = ∆T_{2} = T_{h0} – T_{ci}
∆T_{1} = 300 – T_{co} = T_{h0} – 280
∆T_{m} = ∆T_{1} = ∆T_{2}
uA∆Tm = mC_{p} (T_{co} – T_{ci})
20 × 20 × (300 – T_{co}) = 0.4 × 1000(T_{co} – 280)
2T_{co} = 300 + 280T_{co} = 580 = 290 K 2
 Consider a parallelflow heat exchanger with area A_{p} and a counterflow heat exchanger with area A_{c}. In both the heat exchangers, the hot stream flowing at 1 kg/s cools from 80°C to 50°C. For the cold stream in both the heat exchangers, the flow rate and the inlet temperature are 2 kg/s and 10°C, respectively. The hot and cold streams in both the heat exchangers are of the same fluid. Also, both the heat exchangers have the same overall heat transfer coefficient. The ratio A_{c} /A_{p} is _____.

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Correct Option: A
 In a counterflow heat exchanger, water is heated at the rate of 1.5 kg/s from 40°C to 80°C by an oil entering at 120°C and leaving at 60°C. The specific heats of water and oil are 4.2 kJ/ kgK and 2 kJ/kgK, respectively. The overall heat transfer coefficient is 400 W/m^{2}K. The required heat transfer surface area (in m^{2}) is

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Heat exchanger is a counterflow type.
∆T_{i} = 120 – 80 = 40°C (θ_{1})
∆T_{e} = 60 – 40 = 20°C (θ_{2})LMTD = θ_{1}  θ_{2} In θ_{1} θ_{2} = 40  20 = 28.86°C In 40 20
Q = m_{c} C_{pc}(T_{ce}  T_{ci})
= 1.5 × 4.2 × 10^{3} (80 – 40) watt
Q = 252000 W
Q = UA∆Tm .A = Q = 1.5 × 4.2 × 10^{3} × 40 U∆T_{m} 400 × 28.86
= 21.83 m^{2}.Correct Option: D
Heat exchanger is a counterflow type.
∆T_{i} = 120 – 80 = 40°C (θ_{1})
∆T_{e} = 60 – 40 = 20°C (θ_{2})LMTD = θ_{1}  θ_{2} In θ_{1} θ_{2} = 40  20 = 28.86°C In 40 20
Q = m_{c} C_{pc}(T_{ce}  T_{ci})
= 1.5 × 4.2 × 10^{3} (80 – 40) watt
Q = 252000 W
Q = UA∆Tm .A = Q = 1.5 × 4.2 × 10^{3} × 40 U∆T_{m} 400 × 28.86
= 21.83 m^{2}.