Natural convection
Driving force of natural convection:
• pressure difference caused by wind
• air density difference due to buoyancy (stack effect)
• pressure difference caused by appliance operations (combustion devices, hood, etc.)
Ap = Po + Pw — Pir + Aps
Where Ap = pressure difference between outdoors and indoors at location (Pa) p0 = static pressure at reference height of undisturbed flow (Pa) pw = wind pressure at location (Pa)
Pir = interior static pressure at reference height (Pa)
Aps = pressure difference due to buoyancy (Pa)
O
Where Cp = surface pressure coefficient p = air density (kg/m3)
V = wind speed (m/s)
Cp = In [1.248 — 0.703 sin (a/2) — 1.175 sin2 (a)
+ 0.131 sin3 (2aG) + 0.769 cos (a/2)
+ 0.07 G2 sin2 (a/2) + 0.717 cos2 (a/2)]
Where a = angle between wind direction and outward normal of wall under consideration G = natural log of ratio of wall width under consideration to adjacent wall
O
Where g = gravity (9.81 m/s )
HjsjpL = height of natural pressure level
Winter (T0<Ti) |
Buoyancy pressure (Aps)
|
Neutral Level |
(O) STACK ACTION ONLY WITH NEUTRAL PRESSURE LEVEL AT MID-HEIGHT |
I b J WIND ACTION ONLY WITH PRESSURES Of EQUAL MAGNITUDE ON WINOWARO ANO LEEWARD SI0E5 |
( c 1 WIND AND STACK ACTION COMBINED |
Fig. 3 Distribution of Inside and Outside Pressures over Height of Building
2Ap |
Airflow through large openings:
Where Q = airflow rate through the opening (in ‘/s)
Cd = discharge coefficient (-)
A = cross sectional area of the opening (m2) p = air density (kg/m3)
Ap = pressure difference across opening (Pa)
• Flow caused by wind:
Q = Cv A V
Where Cv = effectiveness of openings (-)
(0.5 to 0.6 for perpendicular winds and 0.25 — 0.35 for diagonal winds) A = opening area (m2)
0.5 |
Flow caused by thermal forces: Q = k A [ 2g AhNPL (Ti — Tc)/Ti ]
Where k = discharge coefficient (k = 0.40 + 0.0045 |T — Td| )
AhNPL = height from midpoint of lower opening to NPL (m)
Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning