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4.411 Building Technology Laboratory
Buoyancy-driven flows
Consider conservation of mass, whereby the mass flow entering the single zone through
the lower window equals that leaving through the upper window:
P ,,, C dA,
2
2 APowUflo.
A P f lo w
P out
Al2 ou APinf low
i,,Cd A2
1~fkN
P n
A2P
Al'2 APinf low - A22 A ,utflow
,)
A
Tu!
T,"
[(ou
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(P
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i,)= (Po
-
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,)gz2
=
in i-PingZ
T
T
2
)- (Pou-Pout gZ 2 )]
.
With the pressure difference calculated, we can now determine the volumetric flow in
through the lower window:
Vinflow = C
[(Pou-
tPi )--(Pou
]
in)gZ
0.5
=
CA, 2
-
gz 2
i2
1
Zl
Z2
2) Tr,,,
0.5
= CAl
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TIN
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g(z 2 -z,
1
( A 0
T) ,
If z is set to zero and, for simplicity z 2 is defined as z,
1
4.411 Building Technology Laboratory
U.5
Vinf low = Cd
A1
2 AT
1
T (2
(~,
Note that if z equals zero, there is no buoyancy-driven flow, as expected. If Tout = Tin,
there is also no flow, again as expected.
The above equation is exact. Introducing our first approximation, let
T,,,,
Note that the temperature difference must be retained. Then
A\ r
U.,
AT
CdAI
inf low
1
2- gzA
+
L2
::
05
FI0.5
r0.5
AT
______
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Al + A
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Till
AT
Till1Z~
]Y
The effective area is sometimes defined with a factor of two, as follows:
2A12 A2
Aeff
=
]05
A2 + A2
in which case Aeff'= Al = A2 = A if the two windows are equal in area and
Vinf/ow =
CdAfe
dflflol= gZ
gz AT]
This equation takes a slightly different form when the height is defined in terms of a
neutral plane, where the indoor and outdoor pressures are equal:
Vinflow = Cd Al [ 2g AhNPL (T - To)/Ti ]0.5
2
4.411 Building Technology Laboratory
where Cd = discharge coefficient
AhNPL = height from midpoint of lower opening to NPL (m)
Cd = 0.40 + 0.0045 [T -To
if there is bidirectional flow in a single opening and 0.65
otherwise.
However, this form is somewhat difficult to use because the location of the neutral plane
must be determined. Comparing expressions above we find that
0.5
Vinflow
AT
CdAI 2..gz-
1
g
= CdA [2 AT gAhNp
TLr,,l
J0.5
[ ,A2) j
AhNPL = Z
1]
1
If there are two windows and they are of equal area, neutral pressure is half-way between
the two windows, as intuitively expected. If the lower window is much larger, the
neutral-pressure point shifts toward the lower window. If the upper window is larger, the
neutral-pressure point shifts upward. The neutral-pressure height is more difficult to
determine with multiple openings.
3
4.411 Building Technology Laboratory
I
or, large upper openings
height
ndoor, large lower openings
tdoor
pressure
Example
Determine the volumetric inflow for the following case:
z
= 30 °C
= 20 °C
=3m
Aeff
= 2 m2
Tin
Tout
Solution:
inflow
= CdA·ffX
[
]
= 06 ·2*98
9
'
3
3AT10
Tll303j
1.2
S
Note that absolute temperature is required, because the ideal gas law was used to replace
densities with temperatures.
Wind-driven flows
We start with the same expression of conservation of mass as for buoyancy-driven flows.
For wind-driven flows, we could choose to let Tin equal Tout, which implies that pin
equals pout. Flows are driven by wind and not temperature differences. However, if
there are heat sources inside the building, Tin will be higher than Tout and we will
therefore retain distinct indoor and outdoor temperatures and densities.
4
4.411 Building Technology Laboratory
Poo
Cd A 1
2
APi f ' v n = i'd
A2 1
2
oflow
A1 Poupinflow = A2 Pi,,l AP,tflw
A2Atnf/w
1
A
,i,,W
[
A2,,,,,~,w
I,.fW
= A22
Til
p+
2o luvl]ulU
O
[1
~
A
(Po
2
C
( )Al
V
eOouxCpl
- pi )= h-2
)U
>
2Til
,
Al
2
Cp Pou VW2
p
Ivw212]
2
Till
If the inflow and outflow windows are at different heights, the wind speed associated with
these openings may vary, according to
V =VG
where vG = gradient or free-stream velocity
8 = gradient or boundary layer height
= height exponent
With the pressure difference calculated, we can now determine the volumetric flow in
through the windward window:
Vinfow = C-A
1
inf low
=
Cd
1
[(Po_/
AFout~
KL
iP,,,,pL ,
P,,
2 l)-C 2p
O° v 2
]]
'
U.)
= CA
CvVwl
= C,Alvwl
where Cv represents the effectiveness of the opening area
5
4.411 Building Technology Laboratory
C,
= Cd
This expression is exact, within the framework of the model. To simplify, we will equate
the indoor and outdoor temperatures and the heights of the windward and leeward
openings, such that the wind velocities are equal. In this case
A.
0.5
lCp
2
CpA
inflow = CdAI Cp v
(A 2
= C,,AC v
where
,
[
Ae =L
C,
A2A2
0.5
A, 2]
Cd
2C
l
­
Finally, if the leeward wind-pressure coefficient is equal in magnitude to the windward
coefficient and opposite in sign,
C, = CdCp
C, equals 0.5 to 0.6 for perpendicular winds and 0.25 - 0.35 for diagonal winds.
Combined flows
Again, let's start with conservation of mass, this time including both buoyancy and wind
terms in the indoor and outdoor pressures.
Note
6
4.411 Building Technology Laboratory
Po,,, Cd A,
Pi"Cd A 2
i
P¥oul
A21 out\
A
2
inflfow= A
Apinf low
I
T.1
Pin,
2
2 APoultflo
w
Tl
l2
((1
TI ,,,
PAP,,outflow
-Po.,gzI )+ 2
, (p) ,, z
CPIV1
)-
-P
=
Z
Cp2V2
[(A
)2/
Till
(Phu.-Pn) =(P,,.,-Pn)gz2
z'
A2
)1
+1]
1
1
2
[LAYL2:,i2
2
,,,)+
1
L
2ILI
|.To u2
Al
2
Till+
CP2
V2
]
Note that we have two terms on the right side of the equation: a buoyancy-driven term
followed by a wind-driven term. Both are exactly as we calculated earlier. In other
words, pressure differences, NOT flows, add.
The inflow can be computed as
Vinflow =
CdAI
2
AT
-
g(z - z.
2
~A2
rout)
The first term under the radical is what we had before for buoyancy-driven flow and the
second is as before for wind-driven flow.
If we again set the temperature ratio to one
7
12
4.411 Building Technology Laboratory
Vinf low = CAeff
gh + CplVI
Cj
1
1
1
Ae~J=
1
IA
A2
Using an alternative formulation for the wind-driven term we have
Vinflow = CdAeftT,,
gh+
CplvI2-Cp22
1
The windows must be at different heights for the buoyancy force to work. If, however,
the height differential is such that the wind velocities are essentially the same, and if the
leeward wind-pressure coefficient is equal in magnitude and opposite and sign to the
windward coefficient,
iinflow =
CdAfft
gh + 2C V
= CdAf
gh + CP
Complex building geometries
The expressions developed above are adequate for single-zone buildings with no internal
obstructions and simple arrangements of windows. For more complex configurations,
such computer programs as CONTAM are very useful. CONTAM, freeware from NIST,
accounts for forced and natural ventilation, with multiple zones and multiple openings.
Infiltration
Infiltration is from the combined effect of buoyancy and wind. It can be estimated as an
air-change rate on the basis of experience or calculated from leakage areas and the orifice
equation, using the crack method:
Q = A CAp"
where A = effective leakage area of cracks (m 2 )
C = flow coefficient
Ap = pressure difference between outside and inside
Infiltration in houses from Sherman-Grimsrud model
8
4.411 Building Technology Laboratory
The Sherman-Grismrud model, presented in Chapter 26 of the 2001 ASHRAE Handbook
of Fundamentals, is a version of the crack method that is often used for residential
buildings:
Q = L (Cs AT + Cw V 2 )0 5
where Q = infiltration rate (L/s)
L = effective leakage area (cm 2)
Cs = stack coefficient (L2 s 2 cm4'
K- l)
AT = average indoor-outdoor temperature difference (K)
Cw = wind coefficient (L 2 s-4 cm-4 K- 1 m - 2)
V = average wind speed measured at local weather station (m/s)
The stack coefficient depends solely on house height, in stories, and the wind coefficient
is a function of height and the extent to which the building is sheltered.
Example
Estimate the infiltration at design conditions for a two-story house in Boston. The house
has an effective leakage area of 500 cm 2 , a volume of 340 m 3 , and is surrounded by a
thick hedge (shielding class 3).
Solution:
Under winter design conditions: To = -13 C, V = 8 m/s, Ti = 22 °C.
From Tables 7 and 9 from Chapter 26 of the 2001 ASHRAE Handbook of
Fundamentals, Cs = 0.000290 and Cw = 0.000231.
Q = L (Cs AT + Cw, V2) 5
= 500 (0.000290 x (22 - (-13)) + 0.00231 x 82)05
= 79 L/s = 284 m3 /h
Air change rate = 284 / 340 = 0.84 ACH.
Air exchange method for houses:
extremely low
low
normal
high
extremely high
-
0.1 ACH (air change rate per hour)
0.5 ACH
1 ACH
2 ACH
3 ACH
9
4.411 Building Technology Laboratory
Combined Energy and Ventilation Analysis
Consider what is usually of more interest than airflow: indoor temperature. What
happened when we open the windows, with a given indoor heat load? Nature adjusts:
*
*
Indoor pressure to conserve mass
Indoor temperature to conserve energy.
Let's look at the two conservation laws:
Q= PC CpVnfflow (Tin -,,u ) ±
Vinflow =d
I-
Aet fAT
gh +
UA, (T,,, - Tt )
w
Substituting for Vinflow in the first equation yields a cubic equation in AT.
Q = p,CpCdAcf/[
gh APw+]
UiA, (T,, -To,,
Solve for the temperature difference, then substitute and solve for the flow.
Alternatively, solve the following equation for the flow and then substitute to obtain the
temperature difference.
+APwj
Either way, recognize that cubic equations have three solutions, more than one of which
may be physically realizable. One such case concerns a wind that opposes the flow
direction established by buoyancy-driven flows. One solution has a relatively high indoor
temperature and a net flow out of the top window. Another has a lower temperature,
smaller buoyancy force, and flow out the lower window. The time history of the flows
determines which solution is present under given conditions. An interesting discussion is
found in Li et al., Building andEnvironment 36(2001):851-58.
One way of looking at multiple solutions is shown below. This figure graphs the
conservation-of-mass relations for various wind speeds and the conservation of energy
relations. Intersection points indicate that both conservation laws are satisfied.
10
4.411 Building Technology Laboratory
Volumetric flows
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