The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is 225∘C and the change in plate temperature with time (dT/dt) is -0.022 K/s. The ambient air temperature is 25∘C and the plate measures 0.3×0.3with a mass of 3.75 kg and a specific heat of 2770J/kg⋅K.

h = 6.35 W/m².k

Explanation:

In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:

Rate of Heat Transfer Through Convection = Energy Stored in Plate

- h A (Ts - T₀) = m C dT/dt

where,

h = convection heat transfer coefficient = ?

A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²

Ts = Surface Temperature of hot thin plate = 225⁰C

T₀ = Ambient Temperature = 25°C

m = mass of plate = 3.75 kg

C = Specific Heat = 2770 J/kg. k

dT/dt = rate of change in plate temperature = - 0.022 K/s

Therefore,

- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)

h = (- 228.525 W)/(- 36 m².k)

h = 6.35 W/m².k

answer:

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