## Heat Transfer Miscellaneous

#### Heat-Transfer

1. A slender rod of length L, diameter d (L >> d) and thermal conductivity k1 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

1. As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
(Rth) eq = R1 + R2

 2L = L + L Akeq AK1 AK2

 2 = 1 + 1 Keq K1 K2

 ⇒ Req = 2K1K2 K1 + K2

##### Correct Option: A

As per question, heat transfer in radial direction and contact resistance are negligible. In series connection,
(Rth) eq = R1 + R2

 2L = L + L Akeq AK1 AK2

 2 = 1 + 1 Keq K1 K2

 ⇒ Req = 2K1K2 K1 + K2

1. 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

1. From wein’s displacement law λmax T = constant
max)1 T1 = (λmax)2 T2
1.45 × 2000 = 2.90 × T2

 T2 = 1.45 × 2000 2.90

T2 = 1000 K

##### Correct Option: B

From wein’s displacement law λmax T = constant
max)1 T1 = (λmax)2 T2
1.45 × 2000 = 2.90 × T2

 T2 = 1.45 × 2000 2.90

T2 = 1000 K

1. A balanced counter flow heat exchanger has a surface area of 20 m2 and overall heat transfer coefficient of 20 W/m2K. Air (cp = 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

1. Counter flow heat exchanged
Surface Area A = 20 m2, mass flow rate = 0.4 kg/s

 u = 20W ; m2

Temperature Tci = 280K
 Cp of air = 1000 1 kg K

Since m is same for both flow = 0.4 kg/s
Assume Cp is same = 100 J/kg.K
Hence ∆T1 = Tci – Tco = ∆T2 = Th0 – Tci
∆T1 = 300 – Tco = Th0 – 280
∆Tm = ∆T1 = ∆T2
uA∆Tm = mCp (Tco – Tci)
20 × 20 × (300 – Tco) = 0.4 × 1000(Tco – 280)
2Tco = 300 + 280
 Tco = 580 = 290 K 2

##### Correct Option: A

Counter flow heat exchanged
Surface Area A = 20 m2, mass flow rate = 0.4 kg/s

 u = 20W ; m2

Temperature Tci = 280K
 Cp of air = 1000 1 kg K

Since m is same for both flow = 0.4 kg/s
Assume Cp is same = 100 J/kg.K
Hence ∆T1 = Tci – Tco = ∆T2 = Th0 – Tci
∆T1 = 300 – Tco = Th0 – 280
∆Tm = ∆T1 = ∆T2
uA∆Tm = mCp (Tco – Tci)
20 × 20 × (300 – Tco) = 0.4 × 1000(Tco – 280)
2Tco = 300 + 280
 Tco = 580 = 290 K 2

1. Consider a parallel-flow heat exchanger with area Ap and a counter-flow heat exchanger with area Ac. 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 Ac /Ap is _____.

1. ##### Correct Option: A

1. In a counter-flow 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/m2K. The required heat transfer surface area (in m2) is

1. Heat exchanger is a counterflow type.

∆Ti = 120 – 80 = 40°C (θ1)
∆Te = 60 – 40 = 20°C (θ2)

 LMTD = θ1 - θ2 In θ1 θ2

 = 40 - 20 = 28.86°C In 40 20

Q = mc Cpc(Tce - Tci)
= 1.5 × 4.2 × 103 (80 – 40) watt
Q = 252000 W
Q = UA∆Tm .
 A = Q = 1.5 × 4.2 × 103 × 40 U∆Tm 400 × 28.86

= 21.83 m2.

##### Correct Option: D

Heat exchanger is a counterflow type.

∆Ti = 120 – 80 = 40°C (θ1)
∆Te = 60 – 40 = 20°C (θ2)

 LMTD = θ1 - θ2 In θ1 θ2

 = 40 - 20 = 28.86°C In 40 20

Q = mc Cpc(Tce - Tci)
= 1.5 × 4.2 × 103 (80 – 40) watt
Q = 252000 W
Q = UA∆Tm .
 A = Q = 1.5 × 4.2 × 103 × 40 U∆Tm 400 × 28.86

= 21.83 m2.