Specifications

40
Component resistance
ASHRAE has set up a range of k-values for different
types of components in a chimney system. Chimney
and stack manufacturers have set specic values for
their own products. A list of component k-values can
be found in Appendix C.
Other resistance
Resistance from negative building pressure, external
building pressures, eddies etc. must also be taken
into consideration. This could actually be measured
by using a pressure gauge in the replace chimney
before starting a re. It is not that unusual to see
a negative pressure of 0.05-0.1”WC for problem
replaces.
Estimating natural draft
The theoretical draft of a gravity chimney or vent
is the difference in weight (mass) between a given
column of warm ue gas and an equal column of
colder ambient air. The theoretical can be derived
from the following formula:
The formula indicates that the draft increases with the
height. However, as the height increases the mean
ue gas temperature decreases, so adding height to
create more draft only works up to a certain height.
Example: With a barometric pressure of 29.92 inHg,
a 25’ chimney, 70°F ambient temperature and a
mean ue gas temperature of 250°F, the theoretical
draft will be:
D
t
= .2554 x 29.92 x 25 x (1/530-1/710) = 0.091 “WC
According to the formula a chimney without draft
does not exist.
Estimating available draft
To derive the available draft (D
a
) for a chimney
system, the static pressure loss is deducted from the
theoretical draft:
D
a
= -D
t
+ P
s
Notice that draft is always negative, while static
pressure is always positive.
Example:
Theoretical draft, D
t
- 0.138 “WC
Static Pressure Loss, P
s
+ 0.100 “WC
Available draft, D
a
- 0.038 “WC
Analyzing a replace venting system
The following is an example on how to analyze and
estimate replace systems. The data is as follows:
One-sided replace for wood: 24” x 36”
Ambient temperature = 60°F
Mean chimney temperature = 300°F
Chimney = single wall steel chimney
Flue size = 8”
Chimney height = 25’
Off-sets = one 30° off-set (= 2x30° elbows)
d
m
= 0.075 lbs/ft3 @ 60°F
= 0.047 lbs/ft3 @ 300°F
1. Determine ow
Q
t
= q x A
inlet
x V
inlet
/ 144
= 1.6 x 24 x 36 x 48 = 460 ACFM
144
* V
in
= see table in Chapter 5.1
2. Determine k-values
Pipe:
K
L
= F x L / D
i
= 0.32 x 25 / 8 = 1.0
Components:
2 x 30 ° elbows = 2 x 0.15 = 0.30
Inlet from rebox = 2.00
Total k-value = 2.30
Σk = 1.0 + 2.30 = 3.3
3. Determine system pressure loss
P
s
= 0.003 x d
m
x (Q
t
/ A
pipe
)
2
x Σk
= 0.003 x 0.047 x (460 / 50)
2
x 3.3
= 0.039”WC
4. Determine natural draft
D
t
= 0.2554 x B x H x (1 / T
o
– 1 / T
m
)
= 0.2554 x 29.92 x 25 x (1/530 – 1/710)
= 0.091”WC
D
t
= .2554 x B x H x (1/T
o
-1/T
m
)
D
t
= Theoretical draft (“WC)
B = Local barometric pressure (inHg)
H = Height of chimney (feet)
T
o
= Absolute ambient temperature(°F) – Ambient(°F) +460 (°R)
T
m
= Absolute mean ue gas temperature(°F) – Mean ue
gas(°F) + 460 (°R)