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

Answer:

Q convection = 11.361 W

Q radiation = 21.438 W

Explanation:

Given data:

$v = 1.524 \times 10^{-5}m^2/s$

Pr = 0.7

Isobaric volume expansion is calculated as

$\beta = \frac{1}{v} {\partial v}{\partial T}$

$\beta = \frac{1}{T_{ambient}} = \frac{1}{22+273} = 3.39\times 10^{-3} K^{-1}$

L = 0.5 m Since door is vertical

grasshoff number is

$Gr = \frac{g \beta(T_S - T_A)L^3}{v^2}$

$Gr = \frac{9.81 \times 3.39\times 10^{-3} (32.22) \times 0.5^3}{(1.524\times ^{-5})^2}$

Gr = 1.79\times 10^8

Rayleigh number

Ra = GrPr

$Ra = 1.253\times 10^8 \times 0.7 = 1.253\times 10^8< 10^8$

therefore given flow is laminar

Nusselt number is

$Nu = 0.5 (Ra)^{1/4}$

$Nu= 0.59(1.253\times 10^8)^{1/4})$

Nu = 62.4223

Heat transfer coefficient

$h = \frac{NuK_{air}}{L}$

$h = 62.4223 \times 0.026}{0.5}$

h = 3.246 W/m^2 K

convection heat transfer

$Q = hA(T_S - T_A)$

$Q = 3.246\times 0.5\times 0.7(32-22)$

Q = 11.361 W

Radiation heat transfer

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

$Q_R = 5.67 \times 10^{-8} \times 1\times 0.5\times 0.7(305^4 - 294^4)$

$Q_R = 21.438 W$