Closing the door
The problem of opening
and shutting doors plagues the refrigeration and air conditioning businesses.
How many times per hour can one allow those pesky customers open a door to an
AC area without adding to the heat load too much? Or those people who complain
the refrigerator is not cooling effectively, when they keep opening the fridge
door wide? But the problem that bothers me is: how much air pollution (PM2.5)
is getting in to my living room each time the door is opened to the balcony
whose air is not purified? In my friend Chandan’s living room the PM2.5 level
immediately responds, going up by easily 10-20%.
A strip of the door of height h and width dx has
an area dA (Fig.a):
dA= h dx
When the
door is being shut it has an angular velocity w and if r is
the distance from the pivot point, the velocity v at that point is given
by (Fig.b):
v = wr
where r can be calculated from the distance x
and the perpendicular distance y (see part b) of above diagram):
r = (x2 + y2)1/2
Let no be the number of
particles of PM2.5 in the room initially. Then the rate of change of n is given
by:
dn/dt = nov
dA
But the velocity v is
independent of time, and so it is also independent of y. Setting y = 0, r = x:
dn/dt = now òh x dx = now h d2/2
So that, since door area
A = hd and taking Dq = wDt:
Dn = A (Dq) d n0
where
x is integrated across the width d of the door, and x = 0 at the pivot of the door. Of course, this
assumes that the time period when the door is being accelerated to velocity v,
and when it is decelerated to zero velocity, is negligible. Also, the
temperature is being assumed to be the same in both rooms 1 & 2, as is the
density. This may not be a very realistic assumption and, indeed, others do not make it (see below).
Take
n0 = 3x1025 m-3 (1 atm at normal temperature
& pressure), h = 2 m. d = 1 m, and Dq = w(Dt) = p/4 (45° door opening).
so
that Dn is independent of the time taken, and only dependent on the
angle Dq traversed.
So:
Dn = 2.36 x 1025
atoms
A
room of volume 22.4 m3 contains 6x1026 atoms (1000 mols x
Avogadro number, at NTP), so the Dn = 2.36 x 1025 atoms corresponds to 0.88 m3.
That is, opening the door by this much allows about 4% of the outside PM2.5
contamination to get in: so the PM2.5 level increases by about 4% of the
outside PM2.5 level (for every 45° opening). For a 90° opening, it should be ~8%.
As
a reality check: if we assume a 45° angle (which is what we did), the volume pushed out by the door
while closing is (pr2/8)(h) = 0.78 m3.
The
latter volume is 88% of the former, so a 12% difference can be attributed to
the fact that the velocity v is not constant across the door (in the
x-direction).
The
amount of 880 L is probably an overestimate because the angular velocity w is assumed to be constant while the door is
moving, whereas it is more likely that it starts from zero and increases.
But
if it is only dependent on the angle Dq traversed then the time
variation of the angular velocity will not matter.
This
analysis will work if the door is closing: moving CW about its hinge so that
air is being forced from Room2 to Room1.
However,
if it is opening (moving CCW), the air from room 1 is not being pushed, but
pulled (while the air already in room 2 is being pushed CCW).
The
analysis for that case is different: so opening and closing the door gives rise
to different effects - and I shall not go there.
I
found 4 papers in the literature that tackled a similar topic albeit with
different assumptions.
Foster et al [1] refers to the case of an air curtain above the door so it is not directly
applicable, and it has an analytical solutions as well as a computational fluid
dynamics (3D CFD) study.
Lagus
[2] considers a pressure difference between both rooms with buoyancy effects
driven by the difference in densities in the two rooms. But, for the case of
equal density, no buoyancy effects occur, and the volume displaced by the door
is given by:
Vp
= A(Dq)d/2
which
is similar to that derived above for Dn.
Hendiger
et al [3] also consider the flow rate due to a pressure difference between the
two rooms, and obtain the flow rate Q as:
Q (m3/s) = 0.067 (Dp)0.63
Where
Dp is in Pa, and the
exponent n is usually between 0.6 and 0.7. They also conclude that: “quick door
swing causes a greater transfer of contaminants, regardless of the width of the
door opening, which shows that it is necessary to open the door more slowly.”
[3].
Marshall
et al [4] have derived an expression – using very similar logic to that gven
above – but specifically trying to calculate the force required to close a door,
overcoming air resistance. They show that the force required is pretty low,
until the point where it is almost shut, and there it increases sharply for the
last few degrees of shutting it:
“When the windows in a small room are shut, the effect of
displacing air as the door is pushed makes the door more diffcult to push when
it is nearly closed.”
But they add that air flow
under the door will reduce this effect somewhat at small angles.
A more complete analysis would take into account temperature and pressure profiles (as has been done [1,2,3] by other authors. But simply the effect of door swing has also been considered [2.4], although the latter is more rigorous since it takes acceleration into account.
1 1) “Three-dimensional effects of an air curtain
used to restrict cold room infiltration” A.M. Foster et al Applied Mathematical
Modelling 31 (2007) 1109–1123
2 2) “Air inleakage due to door opening and closing”
P.L.Lagus (26th Nuclear Air Cleaning & Treatment Conference,
Sep.2000)
3 3) “Influence of the Pressure Difference and Door
Swing on Heavy Contaminants Migration between Rooms” Jacek Hendiger et al PloS One 11(5): e0155159.
doi:10.1371/journal.pone.0155159 (11th May 2016)
4 4) “Slamming doors due to open windows”,
D.A.Marshall et al J.Phys. Special Topics A2.7 (23Nov.2011)
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