Homework #4 Solution

Transcription

1 Homework #4 Solution Problem 6.4 Determine the design winter heat loss tough each of the following components of a building located in Minneapolis, Minnesota: (a) Wall having 648 ft 2 of area and construction of 4-in. face brick; 3/4-in. plywood sheathing; 2-/2-in. glass fiber insulation in 2 by 4 stud space (6-in. on centers); /2-in. plasterboard interior wall. (b) A 285 ft 2 ceiling topped by a 2622 ft2 hip roof. The ceiling consists of /2-in acoustical tile with R-9 insulation between the 2 by 6 (6-in on centers) ceiling joists. The roof has asphalt shingles on 3/4-in. plywood sheathing on the roof rafters. The attic is unvented in winter. (c) Two 4 ft by 6 ft single pane glass windows with storm windows. (a) For combined parallel resistance of insulation/dead air space and 2 by 4 studs, ft2 F U R studs studs 0.3 R studs ft 2 F 2.5 R fglass 3.5 ft F R airspace A studs ft2 F A fglass U fglass 0.27 R fglass ft 2 F U airspace 0.99 R airspace A studs ft 2 F U fgairspace R fgairspace R fglass + R airspace ft 2 F U eq A studs U studs + U fgairspace ft 2 F R eq 7.66 U eq ft 2 F Now for entire wall, rest of resistances add in series R outsideair 0.7 R brick R plywood 0.93 A fglass From Table 5-2 From Table 5-5 From Table 5-5 R plaster 0.45 R insideair 0.68 From Table 5-5 From Table 5-2

2 R tot R outsideair + R brick + R plywood + R eq + R plaster + R insideair ft2 F R tot U tot R tot ft 2 F q wall U A T i T o A 648 ft 2 T i 72 F Suitable indoor design temperture T o 6 F q wall ft ft 2 [ 72 F ( 6 F) ] 5, 520 F (b) ceiling/roof Determine Uc and Ur: R roof R outsideair + R shingle + R plywood + R atticair ( ) Table 4-7A, 0.4% design condition was selected since wall is not very well insulated. answer (a) R outsideair 0.7 From Table 5-2 R shingle 0.44 From Table 5-5 R plywood 0.93 From Table 5-5 R atticair 0.62 From Table 5-2 R roof for ceiling, first combine R-9 insulation with ceiling joists in parallel: A joist A ins A joist U joist R joist U ins R ins ft 2 F ft 2 F

3 U eq A joist R eq U eq A ins U joist + U ins ft 2 F Now, sum other resistance in series with equivalent resistance of insulation and ceiling joists: ft2 F R ceil R atticair + R eq + R tile + R insideair U ceil R ceil ft 2 F Now, combine Uceil & Uroof into an overall U based upon ceiling area: U o U ceil + A ceil A roof U roof U o ft 2 F q ceilroof U o A ceil ( T i T o ) ( 0.053) ( 285) [ 72 ( 6) ] 9, 86 answer (b) (c) window Table 5-6 does not include effect of storm windows so estimate window + storm window as two single glass windows (U.27 from Table 5-6) with a 0.5-in. dead air space between them (R0.9 from Table 5-4) R eq R window + R space + R storm U eq R eq ft 2 F q window U eq A ( T i T o ) ( 0.402) ( 4) ( 6) [ 72 ( 6) ] 850 answer (c) Problem 6.8 Estimate the heat loss from the uninsulatedslabe floor of a frame house having dimensions of 8 m by 38 m. The house is maintained at 22C. Outdoor design temperature is -5C in a regio with 5400 Kelvin days. Problem 6.9 Repeat problem 6.8 for the case where insulation (R0.9 m 2 -K/W) is applied to the slab edge and extended below grade to the frost line. for both problems: q F 2 P T i T o F to 0.93 /-ft-f for an uninsulated slab floor, pg. 6.5 of textbook second paragraph F to 0.54 /-ft-ffor an insulated slab floor (R0.95 m 2 -K/W), pg 6.5

4 so.73 W m K q noinsul ( 0.86) ( ) [ 22 ( 5) ] 6, 65 W ft F 6.65 kw answer W m K q winsul ( 0.50) ( ) [ 22 ( 5) ] 3, 585 W ft F kw answer 6.9 Problem 6.20 To preclude attic condensation, an attic ventilation rate of 59 L/s is provided with outside air at -3C. Roof area is 244 m 2 and Uroof 2.7 W/m 2 -K. Ceiling area is 203 m 2 and Uclg0.30 W/m 2 -K. Inside design temperature is 22C. Determine the ceiling heat loss W with ventilation and compare to the loss if there had been no ventilation. without ventilation: U o q novent U clg + A clg A roof U roof U o ft 2 F U o A clg T i T o ( 0.275) ( 203) [ 22 ( 3) ], 954 W.95 kw answer with ventilation: Q in substitute for Qin and Qout. Also, note eqtn 3-22 can be used to calculate mc p t for ventilation air U clg A clg T i T u solve for Tu: Q out ( ) m o c p T u T o U roof A roof T u T o energy balance on attic m o c p T i T o.23 Q vent T u T o T u U clg A clg T i.23 Q vent T o + U roof A roof +.23 Q vent + U roof A roof + U clg A clg T o ( 0.30) ( 203) ( 22) + ( 2.7) ( 244) ( 3) +.23 ( 59) ( 3) T u 0.3 C.23 ( 59) + ( 2.7) ( 244) + ( 0.30) ( 203) q vent U clg A clg T i T u ( 0.30) ( 203) [ 22 ( 0.3) ], 967 W.97 kw answer

5 Problem 7.9 A small parts assembly area in a factory has a working force of 25 men and occupies a space of 27.4 m by 9. m with a 3-m ceiling. Smoking is not allowed. Determine: (a) the sensible heat load from the occupants (b) the latent heat load from the occupants (c) the moisture added from the occupants (d) the minimum volume of outdoor air for ventilation (e) a suitable summer design inside dry-bulb temperature For sensible and latent heat loads from occupants, use Tale 7-4, use values for light bench work at a factory, qs 275 //person, ql 475 //person (a) (b) q s , 875 person q l , 875 person (c) (d) q l m o h fg W m w h fg m w q l h fg, lb lb Should use Table 5-9 to determine required cfm/person, but there is not an obvious match between any of the applications listed. At a minimum 5 cfm/person (8 L/s-person) must be provided, but from comparing other applications with similar activity (e.g., office space) 20 cfm/person (0 L/s-person) is probably a more appropriate choice. Q vent cfm 500 cfm 250 L person s (e) use figures 4-2, should also adjust temperature for the fact their is moderate activity going on (~ 2.0 met) T(sedentary) 75 to 78 F use eqtn 4-5 to adjust for 2.0 met activity level (-5.4*(+0.5)*(2.0-.2) -6.5 T(active) 69F to 72F Problem 7.20 The west wall of a school building in Abilene, TX is 250 ft by 9 ft overall and includes 0 windows, each 6 ft by 4 ft. The wall is 2-in. concrete (k2 -in/-ft 2 -F). followed by 8-in. cinder aggregate concrete block. The windows are single pane, 0.25-in. reglar plate glass, with light venetian blinds. For design conditions in late July at 6 pm, determine the cooling load for the complete wall. Desgin conditions (%) for Abilene, TX (closest to Dallas) are 98F dbt, 74 wbt, 20.3 daily range, and located at 32.9 N LAT dailyrange 20.3 T om T o

6 assume indoor design temp. is 78F, then CLTD correction factor is: wall A wall ( 250) ( 9) 0 ( 4) ( 6) 200 ft 2 R wall R outside + R concrete + R block + R inside R-values taken from table 7-23 R concrete R wall Most massive wall material is 2-in concrete with code number C, but this is not listed as one of the primary materials in Table Of principal wall materials listed in Tables 7-24, C8 is most massive so it would be the best to select as approximating the actual wall. using Table 7-24B since there is no insulation (mass is evenly distributed with respect to insulation), closest matching secondary material is face brick so wall type is 6 From Table 7-22-J32, for wall #6 at 6 pm (8 solar time) and a west facing wall: CLTD 20 R outside 0.33 R inside 0.69 R block 2.00 L 2 in k in 2.0 ft 2 F q wall window U window.4 ft 2 F ( ) CLTD corr ( 78 78) + T om ( ) 2.9 I used 8-in. lightweight concrete block U wall R wall ft 2 F U A ( CLTD) ( 0.249) ( 200) ( ), 460 From Table 5-6 (single glaze, /4-in.) CLTD 2 From Table 7-25 so CLTD corr SC 0.67 From Table 7-8 (clear with light venetian blinds) insufficient information to determine zone type in Table 7.27-J32, look at all 4 and choose one with the highest SCL so SCL 40 (zone A) q window U win A win CLTD corr + SC SCL A (.4) ( 240) ( 4.9) + ( 0.67) ( 40) ( 240) 26, 590 q total q wall + q window, , , 050 answer