Engineering, 21.04.2020 19:34 torresnoemi899

Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while the motorcycle is moving at 80 km/h in air at 27 C. The air is in parallel flow over both surfaces of the fin, and turbulent flow conditions may be assumed to exist throughout. What is the rate of heat removal per unit width of the fin?

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Answer from: amycressey1970

q' = 5826 W/m

Explanation:

Given:-

- The length of the rectangular fin, L = 0.15 m

- The surface temperature of fin, Ts = 250°C

- The free stream velocity of air, U = 80 km/h

- The temperature of air, Ta = 27°C

- Parallel flow over both surface of the fin, assuming turbulent conditions through out.

Find:-

What is the rate of heat removal per unit width of the fin?

Solution:-

- Assume steady state conditions, Negligible radiation and flow conditions to be turbulent.

- From Table A-4, evaluate air properties (T = 412 K, P = 1 atm ):

Dynamic viscosity , v = 27.85 * 10^-6 m^2/s

Thermal conductivity, k = 0.0346 W / m.K

Prandlt number Pr = 0.69

- Compute the Nusselt Number (Nu) for the - turbulent conditions - the appropriate relation is as follows:

$Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}$

Where, Re_L: The average Reynolds number for the entire length of fin:

$Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909$

Therefore,

$Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378$

- The convection coefficient (h) can now be determined from:

$h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}$

- The rate of heat loss q' per unit width can be determined from convection heat transfer relation, Remember to multiply by (x2) because the flow of air persists on both side of the fin:

$q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}$

- The rate of heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The heat loss per unit width (q') due to radiation:

q' = 2*a*T_s^4*L

Where, a: Stefan boltzman constant = 5.67*10^-8

$q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}$

- We see that radiation loss is not negligible, it account for 20% of the heat loss due to convection. Since the emissivity (e) of the fin has not been given. So, in the context of the given data this value is omitted from calculations.

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Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a...

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