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Instructions Napoleon Fireplaces, Modèle GE38NT-M

Fabricant : Napoleon Fireplaces
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VERTICAL RISE IN FEET V T HORIZONTAL VENT RUN PLUS OFFSET IN FEET H T The shaded area within the lines represents acceptable values for H and V TT . For vent configurations requiring more than one 90° elbow, the following formulas apply: Formula 1: H < V T T Formula 2: H T Example 1: FIGURE 6 V 1 =3 ft V 2 =8 ft V T = V 1 + V 2 = 3 + 8 = 11 ft H 1 = 2.5 ft H 2 =2 ft H R = H 1 + H 2 = 2.5 + 2 = 4.5 ft H O =.03(three 90° elbows - 90°) = .03(270° - 90°) = 5.4 ft H T = H R + H O = 4.5 + 5.4 = 9.9 ft + V T < 40 feet 90° 90° 90° H 1V2H2V1 H T + V T = 9.9 + 11 = 20.9 ft Formula 1: H T < V T 9.9 < 11 Formula 2: H T + V T < 40 feet 20.9 < 40 Since both formulas are met, this vent configuration is acceptable. W415-0429 / C / 11.04.05 TOP EXIT / HORIZONTAL TERMINATION TOP EXIT / HORIZONTAL TERMINATION when (Ht) > (Vt) 90° elbow) tical rise VTrun HT. REQUIRED VERTICAL RISE IN INCHES V T HORIZONTAL VENT RUN PLUS OFFSET IN FEET H T The shaded area within the lines represents acceptable values for H and V TT . For vent configurations requiring more than one 90° elbow the following formulas apply: Simple venting configuration (only one See graph to determine the required verfor the required horizontal FIGURE 7 H2H1V190° Formula 1: H T < 4.2 V T Formula 2: HT + VT < 24.75 feet Example 2: FIGURE 8 V 1 = V T = 6 ft H 1 =3 ft H 2 =5 ft H R = H 1 + H 2 = 3 + 5 = 8 ft H O = .03(two 90° elbows - 90°) = .03(180° - 90°) = 2.7 ft H T = H R + H O = 8 + 2.7 = 10.7 ft H T + V T = 10.7 + 6 =16.7 Formula 1: H < 4.2 V T T 4.2 V T = 4.2 x 6 = 25.2 ft 10.7 < 25.2 Formula 2: H + V < 24.75 feet T T 16.7 < 24.75 Since both formulas are met, this vent configuration is acceptable. Example 3: FIGURE 9 H2H4V290° H3H1V190° V 1 =4 ft V 2 = 1.5 ft V = V + V = 4 + 1.5 = 5.5 ft T 1 2 H 1 =2 ft H 2 =1 ft H 3 =1 ft H 4 = 1.5 ft H = H + H + H + H = 2 + 1 + 1 + 1. 5 = 5.5 ft R 12 34 H O = .03(four 90° elbows - 90°) = .03(360° - 90°) = 8.1 ft H T = H R + H O = 5.5 + 8.1 = 13.6 ft H T + V T = 13.6 + 5.5 = 19.1 ft Formula 1:H < 4.2 V T T 4.2 V T = 4.2 x 5.5 = 23.1 ft 13.6 < 23.1 Formula 2:H + V < 24.75 feet T T 19.1 < 24.75 Since both formulas are met, this vent configuration is acceptable. W415-0429 / C / 11.04.05 TOP EXIT VERTICAL TERMINATION TOP EXIT VERTICAL TERMINATION when (Ht) < (Vt) Simple venting configurations FIGURE 10 See graph to determine the required vertical rise V T for the required horizontal run H T . REQUIRED VERTICAL RISE IN FEET V T HORIZONTAL VENT RUN PLUS OFFSET IN FEET H T The shaded area within the lines represents acceptable values for H and V TT . For vent configurations requiring elbows, the following formulas apply: Formula 1: H < V T T Formula 2: H + V < 40 feet T T Example 4: FIGURE 11 90° 90° 90° V90° H1V1H3H2 V 1 =5 ft V 2 =10 ft V T = V 1 + V 2 = 5 + 10 = 15 ft H1 =1 ft H 2 =3 ft H 3 =2.5 ft H = H + H + H = 1 + 3 + 2.5 = 6.5 ft R 123H O = .03(four 90° elbows - 90°) =.03(90 + 90 + 90 + 90 - 90) = 8.1 ft H T =H R + H O = 6.5 + 8.1 = 14.6 ft H T + V T = 14.6 + 15 = 29.6 ft Formula 1: H T < V T 14.6 < 15 Formula 2: H T + V T < 40 feet 29.6 < 40 Since both formulas are met, this vent configuration is acceptable. W415-0429 / C / 11.04.05 11 W415-0429 / C / 11.04.05 Formula 1: HT < 3VT Formula 2: HT + VT < 40 feet Example 5: V1 =2 ft V2 =1 ft V3 =1.5 ft See graph to determine the required vertical rise VT for the required horizontal run HT. For vent configurations requiring elbows, the following formulas apply: FIGURE 13 VT =V1 + V2 + V3 = 2 + 1 + 1.5 = 4.5 ft H1 =6 ft H2 =2 ft HR =H1 + H2 = 6 + 2 = 8 ft HO = .03(four 90° elbows - 90°) = .03(90 + 90 + 90 + 90 - 90) = 8.1 ft HT =HR + HO = 8 + 8.1 = 16.1 ft HT + VT = 16.1 + 4.5 = 20.6 ft Formula 1: HT < 3VT 3VT = 3 x 4.5 = 13.5 ft 16.1 > 13.5 Since this formula is not met, this vent configuration is unacceptable. Formula 2: HT + VT < 40 feet 20.6 < 40 Since only formula 2 is met, this vent configuration is unacceptable and a new fireplace location or vent configuration will need to be established to satisfy both formulas. Example 6: V1 =1.5 ft V2 =5 ft VT =V1 + V2 = 1.5 + 5 = 6.5 ft H1 =1 ft H2 =1 ft H3 =1 ft H4 =10.75 ft HR =H1 + H2 + H3 + H4 = 1 + 1 + 1 + 10.75 = 13.75 ft HO = .03(four 90° elbows + one 45° elbow - 90°) = .03(90 + 90 + 90 + 90 + 45 - 90) = 9.45 ft HT =HR + HO = 13.75 + 9.45 = 23.2 ft HT + VT = 23.2 + 6.5 = 29.7 ft Formula 1: HT < 3VT 3VT = 3 x 6.5 = 19.5 ft 19.5 = 19.5 Formula 2: HT + VT < 40 feet 29.7 < 40 Since both formulas are met, this vent configuration is acceptable. when (Ht) > (Vt) Simple venting configurations FIGURE 14 FIGURE 12 H 1 H 2 V 3 V 1 90° 90° 90° V 2 90° H 4 V 1 V 2 90° 45° 90° H 1 90° H 3 H 2 TOP EXIT VERTICAL TERMINATION MAXIMUM VERTICAL RISE IN FEET VT HORIZONTAL VENT RUN PLUS OFFSET IN FEET H T The shaded area within the lines represents acceptable values for HT and VT . INSTALLATION WALL AND CEILING PROTECTION FOR SAFE AND PROPER OPERATION OF THE FIREPLACE, FOLLOW THE VENTING INSTRUCTIONS EXACTLY. NOTE: Only a clearance to combustibles of 1...