Question

A4: A refrigerator door has a height and width of H = 1 m and W = 0.65 m, respectively, and is situated in a large room for which the air and walls are at T. = Tu = 25°C. The

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Answer 1

Son Given that - HIMA = 1X0.65 W = 0.65 m A = 065m2 K = 0.03 W/m ok Two = Tsus = 25 = 298 K Ts, - 5°C = 298k &= 0.6 So from table of air a T = 2881) N = 14. 82X10-6 m²ls Lo = 20.92 x 10-6 m2 15 K = 0.0053 wm. Pr=0.71 0= 5.67810 -8 W/m²Kt B=0.00 3 47K-1 we know that the formula of heat gain - Two = has (To-Tsi) + E0 AS (TW - T we also find h So, ñ = NUH (KH) - 0

Nun = 5 0.825 + 0-381|2.19X10JY [1+LD1492 Judaler Nun = 156.6 Also find 20.92x10 6 x 14.82 X10-6 Rayleigh numer - Rom- g B To-Ts, He Rom - 9.81X0.00347/298-278) * 13 RaH = 2-19 810 9 So from egno - 5 = Nur ( x ) - 156.6 X 0.0259 ħ = 3.961 W/m2K 1 h = 4 W/ m² K/

Also from egn.o - 2.0 - 4 X 0.651298 - 298 + 0.625.67X168 x0.65 x/1998(278)4] 9w.o=152 +42.80) 190,0 = 94.30 w AS we have to from energy balanced Con + had = "cond ñ | To-Ts.o) +€0 | 134-8%) = Ki 160-ts.) So, 9w = Kids (T5.0 Tsi) 9w = 0.0253X0-65/298-2.78) 0.025 - 3289 = 13.156 W 9. aw - 250 1960 - 19.2W 40 No warm to feel.