Napoleon Fireplaces BGD36NTR, BGD36PTR, GD36NTR, GD36PTR when, FEET, FEET, 16.1, 19.5, =2 ft, =1 ft, =6 ft, =2 ft, =5 ft, =1 ft, =1 ft, 26 <40, MAXIMUM, RISE IN, =1.5 ft, =1.5 ft, HT + VT, VERTICAL, =10.75 ft

14

TOP OR REAR EXIT VERTICAL TERMINATION

when (HT) > (VT)

Simple venting configurations

FIGURE 17

See graph to determine the required vertical rise VT for the required horizontal run HT.

MAXIMUM

VERTICAL

RISE IN

FEET

VT

HORIZONTAL VENT RUN PLUS OFFSET IN FEET HT

The shaded area within the lines represents acceptable values for HT and VT .

For vent configurations requiring more than two 90° elbow (top exit) or one 90° elbow (rear exit), the following formulas apply:

Formula 1: HT < 3VT

Formula 2: HT + VT < 40 feet

Example 7:

90°		90°
			V3
V		V	H2
1
	H1	2
			90°
		90°

FIGURE 18

V1	=2 ft
V2	=1 ft
V3	=1.5 ft
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 ftFormula 1: HT < 3VT3VT = 3 x 4.5 = 13.5 ft16.1 > 13.5Since 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 8:
	90°		H	V2
				45°
			2
	V1			90°
	H1			H3
		90°
				FIGURE 19
V1	=1.5 ft
V2	=5 ft
VT	= V1 + V2 = 1.5 + 5 = 6.5 ft
H1	=1 ft
H2	=1 ft
H3	=10.75 ft
HR	= H1 + H2 + H3 = 1 + 1 + 10.75 = 12.75 ft
HO	= .03(three 90° elbows + one 45° elbow - 90°)
	= .03(90 + 90 + 90 + 45 - 90) = 6.75 ft
HT	=HR + HO = 12.75 + 6.75 = 19.5 ft
HT + VT	= 19.5 + 6.5 = 26 ft

Formula 1: HT < 3VT

3VT = 3 x 6.5 = 19.5 ft19.5= 19.5

Formula 2: HT + VT < 40 feet

26 < 40

Since both formulas are met, this vent configuration is ac- ceptable.

W415-0210 / E / 06.25.03