Document EGzvaG5BpQpXQNZZ8Qm6vRKj

622 CHAPTER 31 1949 Guide - The two inside walls are assumed to separate rooms, which are filled with 68 F air, so that both surfaces of each wall are in close contact with 68 F air. The surface tem- with the heated ceiling, they are exposed to the heat radiation from the ceiling. In the following example, 70 F will be selected as the mean surface tempera ture of the inside walls. For the two Outside walls, the mean inside surface temperature can be calculated with fair accuracy. For a heat transmission coefficient of 0.25 and a temperature difference of 68 deg, heat flows through the wall at the rate of 17 Btuh per square foot. Fig. 8. Chabt fob Estimating Inside Subface Tempebatubes of Outside Walds* . *Note: .The value of /. the over-all coefficient of heat transmission, cannot exceed 1.29 if the inside and. outride film coefficients are 1.65 and 6.0 respectively (... y - 55 + j' 0jf}- . Therefore 'TT. b -I M nutrimiiTn). f If the indoor, film coefficient is 1.65, the temperature difference, indoor air to inside wall surface, is 17/1.65 or 10 deg and the wall surface temperature is 68 -- 10, or ,58 F, This value may be taken directly from Fig. 8. However, the film coefficient 1.65 was rdetermined to represent the sum of the heat flow into the wall, by conduction from the air in contact with the wall, and by radiation from the warmer surfaces seen by the wall surface. In a panel-heated room, the rate.of heat flow.into the outside wall by radiation is greater than it is in a radiator-heated room; consequently', the film coefficient is higher, and the temperature difference, air to wall surface, is smaller, and'therefore, the wall surface temperature is higher than the calculated 58 F. It is "impossible to determine accurately how much higher than 58 F the temperature of 'the wall surface will be until the corresponding indoor air film coefficient has .been determined accurately. . : .i 'iii the following example, 60 F will be selected as the probable mean inside surface temperature of the outside walls. . The probable mean inside surface temperatures of the floor and the glass may be 'determined by calculations and by reasoning similar to that, employed to determine the inside surface temperature of the outside walls. ; Panel Heating and Radiant Heating 623 Table 1. Calculated Heat Loss of Room Surface Area Sq Ft U Calculation - Heat Loss Btuh Outside Walls.................. Infiltration....................... 360 216 480 480 480 0.25 1.13 . 0.10 360 x 0.25 x 68 216 x 1.13 x 68 6,120 . 16,597 480 x 0 10 x 38 5,760 cu ft x 1.50 x 68 x 0.018 Total......................................... 1 9A 10/576 35,117 In the following example, SO F and 70 F will be selected as the probable inside surface temperatures of the glass and floor, respectively. 4: Determine the heat loss of the room. In the following example, the heat loss calculation will be based on an outdoor air temperature of 0 F. Since the functioning of a panel-heating system differs very little from that of a radiator-type heating system, the heat loss shown in Table 1 may be calculated according to Chapter 14. The heat loss through the outside walls and through the glass is probably a little greater than calculated because the calculation is based on an indoor air film coef ficient of 1.65 Btuh, whereas, for a panel-heated room, this coefficient is a little higher, but the difference is probably not sufficiently large to be considered in design calcu lations for a heating system. 5. Estimate the Mean Radiant Temperature. The Mean Radiant Temperature of the surfaces enclosing the room but not includ ing the heating panels may be estimated as follows: . Surface Interior Walls.................................... Exterior Walls.................................... Glass.................................................... Floor................................ .................. Area 480 360 216 480 1,636 Fahb Deo 70 ' 60 30 70 Product 33,600 21,600 . 6,480 33,600 95,280 The sum of these products divided by the sum of the surface areas is: 95,280/1,536 or 62.03 F, the required mean surface temperature. In the following example, 62 F will be selected as the MRT of walls, glass and floors. 6. Determine the temperature of the ceiling panel. Determine the temperature of the ceiling so that the ceiling panel will deliver heat to the room at a rate equal to the rate at which the room is calculated to lose heat, namely, 35,117 Btuh. When a room is heated by means of a panel, air convection currents are developed in the room similar to those which are developed when the room is heated by means of a free-standing radiator. Consequently, the heating panel delivers heat to the room partly by radiation and partly Dy convection. The proportion of the total heat flow delivered by convection varies with the location of the heating panel, with the height of the ceiling, and with the size, number, and location of pieces of furniture and other articles which interfere with the free flow of air along the floor and along the walls.. It is generally sufficiently accurate to assume that a ceiling panel will deliver 70 per cent of its heat by radiation and 30 per cent by convection; a floor panel 55 per cent by radiation and 45 per cent by convection; and a wall panel 65 per cent by radiation and 35 per cent by convection. ..