## Heat Transfer Miscellaneous

#### Heat-Transfer

1. Hot oil is cooled from 80 to 50°C in an oil cooler which uses air as the coolant. The air temperature rises from 30 to 40°C. The designer uses a LMTD value of 26°C. The type of heat exchanger is

1.  θ = θ1 - θ2 ln θ1 θ2

θ1 = 50° , θ2 = 10°
θ1 = 40° , θ2 = 20°
As given value is between these two, so cross flow is the best option.

##### Correct Option: D

 θ = θ1 - θ2 ln θ1 θ2

θ1 = 50° , θ2 = 10°
θ1 = 40° , θ2 = 20°
As given value is between these two, so cross flow is the best option.

1. In a heat exchanger, it is observed that ∆T1 = ∆T2 where ∆T1 is the temperature difference between the two single phase fluid streams at one end and ∆T2 is the temperature difference at the other end. This heat exchanger is

1. NA

##### Correct Option: C

NA

1. In certain heat exchanger, both the fluids have identical mass flow rate specific heat product. The hot fluid enters at 76°C and leaves at 47°C, and the cold fluid entering at 28°C leave at 55°C, the effectiveness of the HE is

1.  ε = Ch(Th1 - Th2) Cmin(Th1 - Tc1)

Here , Ch = Cc'
 ε = Th1 - Th2 = 76 - 47 = 0.604 Th1 - Tc1 76 - 28

##### Correct Option: B

 ε = Ch(Th1 - Th2) Cmin(Th1 - Tc1)

Here , Ch = Cc'
 ε = Th1 - Th2 = 76 - 47 = 0.604 Th1 - Tc1 76 - 28

1. Two rods, one of length L and the other of length 2L are made of the same material and have the same diameter. The two ends of the longer rod are maintained at 100°C. One end of the shorter rod is maintained at 100°C while the other end is insulated. Both the rods are exposed to the same environment at 40°C. The temp at the insulated end of the shorter rod is measured to be 55°C. The temperature at the mid point of the longer rod would be.

1. NA

##### Correct Option: C

NA

1. A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 Wm–1K–1. One end of the fin is maintained at 130°C and its remaining surface is exposed to ambient air at 30°C. If the convective heat transfer coefficient is 40 Wm–2K–1, the heat loss (in W) from the fin is

1. Fin diameter, d = 5 mm
Fin length, L = 100 mm
k = 400 W/mK
T0 = 130°C
T = 30°C
h = 40 W/m2K
∴ Heat loss from fin Q = √kAhP (T0 - T) tanh ml
where P = πd = 3.14 × 5 × 10– 3 = 0.015

 A = π d2 = π × (5 × 10– 3)2 = 1.96 × 10– 5 m 2 4 4

∴ m = √hp / kA = √(40 × 0.0157) / (400 × 1.96 × 10– 5) = 8.949
∴ ml = 8.949 × 0.100 = 0.895
∴ Q = √40 × 0.0157 × 400 × 1.96 × 10– 5(T0 - T) tanh (0.895) = 5.0 W

##### Correct Option: B

Fin diameter, d = 5 mm
Fin length, L = 100 mm
k = 400 W/mK
T0 = 130°C
T = 30°C
h = 40 W/m2K
∴ Heat loss from fin Q = √kAhP (T0 - T) tanh ml
where P = πd = 3.14 × 5 × 10– 3 = 0.015

 A = π d2 = π × (5 × 10– 3)2 = 1.96 × 10– 5 m 2 4 4

∴ m = √hp / kA = √(40 × 0.0157) / (400 × 1.96 × 10– 5) = 8.949
∴ ml = 8.949 × 0.100 = 0.895
∴ Q = √40 × 0.0157 × 400 × 1.96 × 10– 5(T0 - T) tanh (0.895) = 5.0 W