Question

7.39 A long, thin metal plate is hung vertically after a heat treating process. The plate, of width W= 0.2 m, ver- tical dimension of 3 m, and an initial temperature of T-320°C, is cooled by atmospheric air of velocity V=4 m/s and Toe-30°C. Determine the initial rate of heat loss from the plate if the air is in parallel flow over the plate. Account for heat loss from both surfaces of the thin plate. Determine the initial rate of heat loss if the plate is turned so that its width is perpendicular to the air flow. Account for heat loss from both the front surface and the back surface of the plate. Which orien tation will maximize the cooling rate?

Answer 1

1. Initially width is parallel to the flow of air Calculate the Reynolds number. Calculate the air properties at mean temperature T.+T, 320 +30 At meanttemperature, kinematic viscosity of air V Substitute the values in the given equation. 3.168x10-m2/s - 4x0.2 Re3168x10 25252.5 Here the flow is laminar Calculate the nusselt number. 033 Nu0.664 (Re) Pr) 033 0.664 (25252.50.684) 93.0869 Calculate the heat transfercoefficient. hl 930869 3,110x10 h-17.2676W/m2 k

Calculate the heat loss through the plate. -17.2676x2x0.2x3x(320-30) 3.004kW 2. If the width is perpendicular and length is parallel to the air flow. Write the expression for the Reynolds number. Calculate the 3.168x 10 378787.8 Nu-0.664(Re)(Pr) s:1/81.8 V 3.168x10 Calculate the nusselt number. 05 0.664x(378787.8) (0.684) 360.068 Calculate the heat transfer coefficient. 360.068hx3 3.710x10 4.4528 W/m-K

Calculate the heat loss. Q, =Id(7-5) -4.4528x2x0.2x3x(320-30) = 1.549kW Here the cooling rate is high in case when the plate is parallel to the air flow. Hence the plate should be placed in such a way that width should be parallel to the air flow.

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