A refrigerated truck moving at 100km/h has a trailer that is 3 m wide by 15 m lo

Question

A refrigerated truck moving at 100km/h has a trailer that is 3 m wide by 15 m long and 3 m high. The refrigeration system is providing cooling at 2000 W. The top, sides, front, and rear surfaces exchange thermal energy convectively at an ambient temperature of 40 degree C. Determine the equilibrium surface temperature of all surfaces of the trailer. Radiation heat transfer Is small to the surroundings. Calculate Re_L for the length of the trailer and assume it applies to the entire surface of the trailer. Assume film temperature equal to ambient temperature and verify at end of solution.

Explanation / Answer

* Here, We have

Truck Velocity, V=100 km/h = 27.778 m/s

Truck Width, W=3m

Truck length, L=15m

Truck height, H=3m

ambient Temperature, Ta = 40 oC

Cooling, Q = 2000W

Total Surface area, As = 2 * (3 x 15 + 3 x 15 + 3 x 3) = 198 m2

*Air Properties at 40 deg.C are as below

Thermal Condutivity of air, k = 0.02732 W/moC

Kinematic Viscosity, v = 16.768 x 10-6 m2/s

Prendlt Number, Pr = 0.7

*Calculation :

Reynolds No., ReL = V * L / v

ReL = 27.778 x 15 / (16.768 x 10-6)

ReL = 24.85 x 106 > 5 x 105

Here ReL > Recr , Hence Flow is Terbulance,

So Correlation for find out convective heat transfer coefficient is Nu = hL/k = 0.0375 * ReL0.8 * Pr0.333

h = k / L * 0.0375 * ReL0.8 * Pr0.333

h = 0.02732 / 15 * 0.0375 * (24.85 x 106)0.8 * (0.7)0.333

h = 50.007 W/m2oC ~ 50 W/m2oC

Convective Heat transfer, Q = h x As x (Ta - Teq), Where Teq is film temperature

Teq = Ta - Q / (h x As)

Teq = 40 - 2000 / (50 x 198)

Teq = 39.79 oC

By comparing value of Teq and Ta, Film temperature is equal to ambient temperature.

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