Question

9.21 A household oven door of 0.5-m height and 0.7-m width reaches an average surface temperature of 32℃ during operation. Estimate the heat loss to the room with ambient air at 22℃. If the door has an emissivity of 1.0 and the surroundings are also at 22℃, comment on the heat loss by free convection relative to that by radiation.

Answer 1

Q convection = 11.361 W

Q radiation = 21.438 W

Step-by-step explanation:

Given data:

`v = 1.524 * 10^(-5)m^2/s`

Pr = 0.7

Isobaric volume expansion is calculated as

`\beta = (1)/(v) {\partial v}{\partial T}`

`\beta = (1)/(T_(ambient)) = (1)/(22+273) = 3.39* 10^(-3) K^(-1)`

L = 0.5 m Since door is vertical

grasshoff number is

`Gr = (g \beta(T_S - T_A)L^3)/(v^2)`

`Gr = (9.81 * 3.39* 10^(-3) (32.22) * 0.5^3)/((1.524* ^(-5))^2)`

Gr = 1.79\times 10^8

Rayleigh number

Ra = GrPr

`Ra = 1.253* 10^8 * 0.7 = 1.253* 10^8< 10^8`

therefore given flow is laminar

Nusselt number is

`Nu = 0.5 (Ra)^(1/4)`

`Nu= 0.59(1.253* 10^8)^(1/4))`

Nu = 62.4223

Heat transfer coefficient

`h = (NuK_(air))/(L)`

`h = 62.4223 * 0.026}{0.5}`

h = 3.246 W/m^2 K

convection heat transfer

`Q = hA(T_S - T_A)`

`Q = 3.246* 0.5* 0.7(32-22)`

Q = 11.361 W

Radiation heat transfer

`Q_R = \sigma \epsilon A(T_s^4 - T_A^4}`

`Q_R = 5.67 * 10^(-8) * 1* 0.5* 0.7(305^4 - 294^4)`

`Q_R = 21.438 W`