The wall of an oven used to cure plastic parts is of thickness L = 0.05 m and is exposed to large surroundings and air at its outer surface. The air and the surroundings are at 300 K. If the temperature of the outer surface is 400K and its convection coefficient and emissivity are h = 20 W/m 2 .K and e = 0.8, respectively, what is the temperature of the inner surface if the wall has a thermal conductivity of k = 0.7 W/m . K?

the temperature of the inner surface is 599.55 K

Explanation:

first, let Am²  represent area of the oven surface

now heat transfer rate due to convection

qconv = hAΔT

where ΔT is temperature difference in kelvin (To -Ts) temperature of outer surface - temperature of surroundings

ΔT = 400K - 300k = 100k

and h = 20 W/m².K

so we substitute

qconv = 20 × A × 100 = 2000A W

Now heat transfer due to emission

qemi = ∈α(To⁴ - Ts⁴)A

∈ is emissivity =  0.8, α is boltzman constant = 5.67×10⁻⁸ W/m⁻²K⁻⁴

so we substitute

qemi = 0.8 × 5.67×10⁻⁸ (400⁴ - 300⁴)A

qemi = 4.536×10⁻⁸ × 1.75×10¹⁰ × A

qemi =  793.8A W

NOW total heat loss rate = Qconv + Qemi

2000A + 793.8A = 2793.8A W

so since outer surface is to be maintained at a temp of 400K then heat must be transferred from inner surface to outer surface at the same rate.

Therefore

Heat transfer rate to outer surface by conduction = kA(Ti-To)/L

k is thermal conductivity  0.7 W/m.K, Ti is temperature of inner surface, L is the thickness of plate  0.05m

so Heat transfer rate = 0.7× A × ( Ti - 400 ) / 0.05

Now

This heat transfer rate equals heat loss rate

so  0.7× A × ( Ti - 400 ) / 0.05  =  2793.8 × A

divide both side by A

THEREFORE

Ti = 400 + 199.55

Ti = 599.55 K

∴  the temperature of the inner surface is 599.55 K

yes indeed

mayonaise pizza

good

dad gone me