## Heat Transfer Miscellaneous

#### Heat-Transfer

1. In the laminar flow of air (Pr = 0.7) over a heated plate, if δ and δT denote, respectively, the hydrodynamic and thermal boundary layer thicknesses, then

1. When Pr < 1 ⇒ δT > δ
Pr > 1 ⇒ δT < δ
Pr = 1 ⇒ δT = δ

##### Correct Option: C

When Pr < 1 ⇒ δT > δ
Pr > 1 ⇒ δT < δ
Pr = 1 ⇒ δT = δ

1. If a foam insulation is added to a 4 cm outer diameter pipe as shown in the figure, the critical radius of insulation (in cm) is ________.

1.  Kfoam = 0.1 W mK

 h0 = 2 W m²K

 Critical radius, rc = Kfoam for a pipe h0

 = 0.1 = 0.05 m = 5 cm 2

##### Correct Option: C

 Kfoam = 0.1 W mK

 h0 = 2 W m²K

 Critical radius, rc = Kfoam for a pipe h0

 = 0.1 = 0.05 m = 5 cm 2

Direction: Consider steady one-dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 80 MW/m³. The left and right faces are kept at constant temperatures of 160°C and 120°C respectively. The plate has a constant thermal conductivity of 200 W/mK.

1. The maximum temperature within the plate in °C is

1. Tmax = 165º [putting x = 5 × 10–3 is equation (1)]

##### Correct Option: B

Tmax = 165º [putting x = 5 × 10–3 is equation (1)]

1. With an increase in the thickness of insulation around a circular pipe, heat loss to surroundings due to

1. With an increases in the thickness of insulation around a circular pipe, heat loss to surroundings due to convection increases and due to conduction decreases.

##### Correct Option: A

With an increases in the thickness of insulation around a circular pipe, heat loss to surroundings due to convection increases and due to conduction decreases.

1. A fluid (Prandt/number, Pr = 1) at 500 K flows over a flat plate of 1.5 m length, maintained at 300 K. The velocity of the fluid is 10 m/s. Assuming kinematic viscosity. μ = 30 × 10–6 m²/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is ______.

1.  Rex=0.5 = Ux = 10 × 0.5 = 166666.67 30 × 10-6

SinceRex=0.5 ≤ 5 × 105
Then flow will be laminar.
 ∴ δ = 5x = 5 × 0.5 = 6.1237 × 10-3 √Rex √166666.67

 δt = (pr)1/3 ⇒ δt = (1)1/3 × δ = 1 × 6.1237 × 10-3 m = 6.1237 mm K

##### Correct Option: A

 Rex=0.5 = Ux = 10 × 0.5 = 166666.67 30 × 10-6

SinceRex=0.5 ≤ 5 × 105
Then flow will be laminar.
 ∴ δ = 5x = 5 × 0.5 = 6.1237 × 10-3 √Rex √166666.67

 δt = (pr)1/3 ⇒ δt = (1)1/3 × δ = 1 × 6.1237 × 10-3 m = 6.1237 mm K