Air of temperature T[infinity] = 27 °C is being used to cool an engine. Consider a single cooling fin at a temperature of Ts = 523 K that can be approximated as a rectangular plate that is 150 cm long by 10 cm high. Air is flowing parallel over both surfaces of the plate. The air flow across the plate at 50 miles per hour may be assumed turbulent throughout. What is the rate of heat removal q in Watts from the cooling fin?

q = 4870.0524 W

Explanation:

Given that:

From Table A4; Thermophysical Properties of Gases at Atmospheric Temperature; Properties of air at film Temperature are as follows:

density

kinematic viscosity (v)  =

thermal conductivity K =

Prandtl number (Pr) = 0.689

Given that :

the length of the plate is = 150 cm = 1.5 m

the width of the plate = 10 cm = 0.1 m

Then the Area = W×L = 1.5  × 0.10 = 0.15 m²

Air flow velocity

The Reynolds number is calculated by using the formula :

turbulent flow.

Thus, we can say that the turbulent flow is throughout the entire plate. Therefore, the appropriate correlation is addressed via the Nusselt number;

where;

and  

Then;

A = 871

Now;

We can now determine the convention coefficient since the is known. By using the equation;

Finally the rate of heat removal q from both  cooling surface of the plate  (fin) is calculated as:

q = 4870.0524 W

c) up is the direction the magnetic force is acting on the charge.

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