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)

подпись: 
winter (t0<ti)
Buoyancy pressure (Aps)

Natural convection

Neutral

Level

(O) STACK ACTION ONLY WITH NEUTRAL PRESSURE LEVEL AT MID-HEIGHT

подпись: 
(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

подпись: 
i b j wind action only with pressures of equal magnitude on winowaro ano leeward si0e5

( c 1 WIND AND STACK ACTION COMBINED

подпись: 
( c 1 wind and stack action combined
Fig. 3 Distribution of Inside and Outside Pressures over Height of Building

2Ap

подпись: 2apAirflow through large openings:

Q = cda

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

подпись: 0.5Flow 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


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