A furnace door, 1.5m high and 1m wide is insulated from inside and has an outer surface temperature of $70^0C$. If the surrounding ambient air is at $30_0C$. Calculate steady state heat loss from the door. The properties at film temperature $50^0C$ are $\rho= 1.093kg/m^3, C_p = 1.005kJ/kg-K, v = 17.95 × 10^{-6}m^2/s, k = 0.02826W/m-K, Pr= 0.698$. Use the relation $Nu= 0.13(Ra)^{1/3}$.

Given,

$A=1.5×1=1.5m^2 \ (Surface \ Area \ of \ furnace \ door) \\ ts=70℃=343K \ (Door \ surface \ temperature) \\ tb=30℃=303K \ (Surrounding \ temperature) \\ tf=50℃=323K \ (Mean \ film \ temperature) \\ ρ=1.093kg/m^3 \ (Density \ of \ air) \\ Cp=1.005kJ/kgK \ (Specific \ heat \ of \ air) \\ ν=17.95×10^{-6}m^2/s \ (Kinematic \ viscosity) \\ K=0.02826W/mK \ (Thermal \ conductivity)\\ Pr=0.698 \ (Prandtl \ number) \\ Nu=0.13(Ra)^{1/3}$

Find: (1) Q (Heat loss from door)

Solution:

Characteristic Length

l= height of door

l = 1.5m

Grashoff’s Number

Gr = ρ2.l3(β.g.∆t)μ2 = ρ2.l3(β.g.∆t)(ρ.ν)2

Gr = l3(β.g.∆t)ν2 …(1)

Temperature Coefficient

β = 1tf = 1323

$β = 3.0959 × 10^{-3}$

∆t=|ts-tb|=|343-303|

∆t=40

∴From(1)

Gr=1.53×(3.0959×10-3×9.81×40)(17.95×10-6)2

Gr=1.272×1010

Nusselt’s Number

Nu=0.13(Ra)1/3= 0.13(Gr.Pr)1/3=0.13(1.272×1010×0.698)1/3

Nu=269.189

Also,

Nu=h.lK

269.189=h×1.50.02826

∴h=5.0715 W/m2K

Heat Transfer Rate

Q=h.As.(|ts-tb|)= 5.0715×1.5×(|343-303|)

Q=304.29 W