B.J. Newlin
2014-12-04 20:55:02 UTC
To balance the demand pressure with an overhead system isn't all I need is
to create a k-factor at the point the in-racks tie into the overhead system?
I have done this, but the pressure demand for the in-rack system is still 15
psi less than the pressure at that junction point for the overhead system.
Is that ok, or did I do something incorrectly?
For example 203 is the junction node. At that point the overhead system
demand is 99.09 psi with ~1450 gpm flowing.
I come up with a k factor of 145.66 (1450/sqrt of 99.09).
When I insert this K at the node and run my calcs, I end up with only 84.2
psi at node 203 for the in-racks. I assume this is acceptable since I think
I'm doing everything properly, but I can't figure out how to explain it to
the FM rep.
Any help would be appreciated. (Below is what i believe is the applicable
code text)
23.4.2.4 Hydraulic Junction Points.
23.4.2.4.1 Pressures at hydraulic junction points shall balance
within 0.5 psi (0.03 bar).
23.4.2.4.2 The highest pressure at the junction point, and
the total flows as adjusted, shall be carried into the calculations.
23.4.2.4.3 Pressure balancing shall be permitted through the
use of a K-factor developed for branch lines or portions of
systems using the formula in 23.4.2.5.
23.4.2.5 K-Factor Formula. K-factors, flow from an orifice, or
pressure from an orifice shall be determined on the basis of
the following formula:
K
Q
P n =
where:
Kn = equivalent K at a node
Q = flow at the node
P = pressure at the node
B.J. Newlin
Service Sales
Aegis Fire Protection LLC
***@aegisfirepro.com
P (913) 825-0343
F (913) 322-4475
C (913) 238-0035
"Whoever is careless with the truth in small matters cannot be trusted with
important matters"
― Albert Einstein
to create a k-factor at the point the in-racks tie into the overhead system?
I have done this, but the pressure demand for the in-rack system is still 15
psi less than the pressure at that junction point for the overhead system.
Is that ok, or did I do something incorrectly?
For example 203 is the junction node. At that point the overhead system
demand is 99.09 psi with ~1450 gpm flowing.
I come up with a k factor of 145.66 (1450/sqrt of 99.09).
When I insert this K at the node and run my calcs, I end up with only 84.2
psi at node 203 for the in-racks. I assume this is acceptable since I think
I'm doing everything properly, but I can't figure out how to explain it to
the FM rep.
Any help would be appreciated. (Below is what i believe is the applicable
code text)
23.4.2.4 Hydraulic Junction Points.
23.4.2.4.1 Pressures at hydraulic junction points shall balance
within 0.5 psi (0.03 bar).
23.4.2.4.2 The highest pressure at the junction point, and
the total flows as adjusted, shall be carried into the calculations.
23.4.2.4.3 Pressure balancing shall be permitted through the
use of a K-factor developed for branch lines or portions of
systems using the formula in 23.4.2.5.
23.4.2.5 K-Factor Formula. K-factors, flow from an orifice, or
pressure from an orifice shall be determined on the basis of
the following formula:
K
Q
P n =
where:
Kn = equivalent K at a node
Q = flow at the node
P = pressure at the node
B.J. Newlin
Service Sales
Aegis Fire Protection LLC
***@aegisfirepro.com
P (913) 825-0343
F (913) 322-4475
C (913) 238-0035
"Whoever is careless with the truth in small matters cannot be trusted with
important matters"
― Albert Einstein