9.18
KNOWN: PROBLEM During a winter day, the window of a patio door with a height of 1.8 m and width of 1.0 m shows a frost line near its base. FIND: (a) Explain why the window would show a frost layer at the base of the window, rather than at the top, and (b) Estimate the heat loss through the window due to free convection and radiation. If the room has electric baseboard heating, estimate the daily cost of the window heat loss for this condition based upon the utility rate of 0.08 $/kWh. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Window has a uniform temperature, (3) Ambient air is quiescent, and (4) Room walls are isothermal and large compared to the window. PROPERTIES: Table A-4, Air (Tf = (Ts + T)/2 = 280 K, 1 atm): = 14.11 10 m /s, k = 0.0247 -5 2 W/mK, = 1.986 10 m /s, Pr = 0.710. ANALYSIS: (a) For these winter conditions, a frost line could appear and it would be at the bottom of the window. The boundary layer is thinnest at the top of the window, and hence the heat flux from the warmer room is greater than compared to that at the bottom portion of the window where the boundary layer is thicker. Also, the air in the room may be stratified and cooler near the floor compared to near the ceiling. (b) The heat loss from the room to the window having a uniform temperature Ts = 0C by convection and radiation is
-6 2
q loss = q cv + q rad
4 4 q loss = As h L ( T Ts ) + Tsur Ts properties evaluated at Tf = (Ts + T)/2.
(1)
(
)
(2)
The average convection coefficient is estimated from the Churchill-Chu correlation, 9.26, Eq. using
0.387 Ra1/ 6 h L L Nu L = L = 0.825 + 8 / 27 k 1 + ( 0.492 / Pr )9 /16
2 (3)
Ra L = g T ( T Ts ) L3 /
Substituting numerical values in the correlation expressions, find
(4)
Ra L = 1.084 1010
Nu L = 258.9
h L = 3.6 W / m2 K
Continued ..
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PROBLEM 9.18 (Cont.)
Using Eq. (2), the heat loss with = 5.67 10 W/m K is
-8 2 4
q loss = (1 1.8 ) m 2 3.6 W / m 2 K (15 K ) + 0.940 2884 2734 K 4
(
)
qloss = ( 96.1 + 127.1) W = 223 W
The daily cost of the window heat loss for the gi...