The decay equation
Although from the point of view of a purist it might be better to derive the decay equation in general terms, it is certainly easier to understand a derivation phrased with reference to a particular application. Accordingly, the equation is derived in terms of the rate at which a contaminant decays in a ventilated room under the influence of a diluting influx of fresh air. The contaminant chosen is carbon dioxide.
Consider a room, as shown in Figure 16.2, having a volume V m3, in which the concentration of carbon dioxide is c, expressed as a fraction (e. g. parts of carbon dioxide per million parts of air). Suppose that during time At, a small quantity of air, Aq, entirely free of carbon dioxide (for simplicity), enters the room. A similar small quantity, Aq, of contaminated air is forced out of the room. The concentration within the room is therefore reduced by an amount (Aq/V)c. This reduction in concentration can be expressed by Ac, defined as follows:
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(16.6)
The negative sign is used because the concentration is decreasing.
It follows that the rate of change of concentration is given by Ac/At, defined as follows:
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(16.7) |
Ac _ cAq
At VAt
Influx of air Aq
Entirely——— ►
Free of contaminant
Room of volume V in which the concentration is c
Aq Efflux of
Contaminated air
Fig. 16.2 A notional room, used as a model for the derivation of a decay equation.
But Aq/At is the rate of influx of ventilating air and is a constant, Q, say. Thus the rate of change of the concentration with respect to time may be written as
Dc cQ
The physical problem has now been phrased as a simple differential equation, and a solution to this will be of practical value in determining the answers to real problems. By integration, the solution to equation (16.8) is
, Qt, ,
Loge c = — y + loge A
Where loge A is a constant of integration.
Hence,
Loge C — loge A = — y
And
A t~(Q,/V) = c
The value of the constant A is established by considering the boundary condition c — c0 (the initial concentration in the room) at t = 0 (the instant that the ventilation began). The solution to equation (16.8) is therefore
C = c0 e+Qt/V) (16.9)
If it is observed that in a ventilated room, where V is in m3, Q is in m3 s-1 and t is in seconds, Qt/V is the number of air changes, then
C = c0e~n (16.10)
Where n is the number of times the cubical content of the room is changed.
The graph of equation (16.10) is an exponential curve, as shown in Figure 16.3. It can be seen that concentration of the contaminant decays rapidly if the ventilating air is entirely free of the contaminant. After one air change it is 36.8 per cent of its initial value and after three air changes it is only 5 per cent.
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Air changes Fig. 16.3 Exponential decay. |
EXAMPLE 16.5
If the air in a room has an initial concentration of 1000 ppm (parts per million) of hydrogen, how many air changes are required to reduce the value of this to 50 ppm?
Answer
From equation (16.10) we can write
N = loge(c0/c) = loge 20 = 3
Suppose a more practical case where, with carbon dioxide as a contaminant, the fresh air used for ventilating purposes also contains some carbon dioxide. Suppose further, that there are people in the room who top up the level of carbon dioxide continually by respiratory activity.
Let c = the concentration of carbon dioxide in the room at any instant, expressed as parts per million of air,
Q’ = the rate of fresh air supply in m3 s-1 per person,
V = the volume of the room in m3 per person, t = the time in seconds after the beginning of occupancy and ventilation, ca = the concentration of carbon dioxide present in the ventilating air, expressed in parts per million of air, and Vc = the volume of carbon dioxide produced by breathing on the part of the occupants, in m3 s“1 per person.
The volume of air entering the room in time At is Q’At, expressed in m3 per person. Hence, the increase of carbon dioxide in time At, due to the contamination of the ventilating air, is (j2/A0(ca/106), in m3 per person. (This is because even if the air in the room were initially free of the contaminant the fraction of the entering airstream which is carbon dioxide is ca/106 and, therefore, this fraction of the entering volume is an addition to the contamination in the room in m3.)
The volume of air forced out of the room by the entering airstream is also Q’At and so, in a similar way, the amount of carbon dioxide leaving the room in m3 per person is (G’ArXc/lO6).
Thus, a balance can be drawn up for the net change of carbon dioxide, in time At, expressed in m3 per person:
Net change of C02 = VcAt + (Q’AtcJIO6) — (Q’AtcllO6)
In this equation, VcAt is the volume of C02 produced in time At by one person breathing, and is expressed in m3 per person.
Since the concentration is the volume of C02 divided by the volume of the room, we can write the change in concentration, per person, as
Net change of C02 in m3 per person volume of the room in m3 per person
And this can be defined by
Ac = [VcAt + (Q’AtcJIO6) — (Q’Atc/106)]/V’
Expressed as a fraction.
Hence
= +(Q, ca/106)-({2’c/106)]/r (16.11)
Also expressed as a fractional change in concentration per unit time. Equation (16.11) may be rearranged in a form which is recognisable as amenable to solution:
Dc Q’c _ 106Vc + Q’ca nf-
Dt T7- ~ V’
This is the differential equation which expresses the physical problem in mathematical terms, just as was equation (16.8) for the first simple case dealt with. Equation (16.12) is obtained from equation (16.11) by multiplying throughout by 106, thus giving the rate of change of concentration in parts of C02 per 106 parts of air, in unit time. It may be solved by multiplying throughout by an integrating factor, e(C t/v’
Then
, rr _ { ia6t/ , n> „
AQ’tfl/’)
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A_ J |
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Ev |
Dc (gw’t, Q’c e(gw’) _ (106Vc + Q c dt 6 V’ ~{ V’
The left-hand side of this is the derivative of a product and so, by integrating, we get
A + r e(Q’l/v’) — 1Q6^ + ^
— y, where A is a constant of integration.
Write B = AQ’, then
B e(_eW) = (106VC + Q’cA — Q’c)
The boundary condition is c = cq when t = 0, and so
B = 106VC + Q’c, — Q’c0 The solution to equation (16.12) is then
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106VC Q’ |
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C = |
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Ca (1 — e-") + c0e-" |
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Where n is the number of air changes after the passage of time t and is equal to Q’t/V’. Reference to Figure 16.4 shows a graphical interpretation of this solution.
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(1) C=Coen (2) c = ca(1 — e-n) |
(3) c.^ + ca)(1-e-)
(4) C=(-^F^ + Ca)(1-e~"> + C°e-"
Fig. 16.4 Graphical illustration of the build up of equation (16.13).
If the room were initially free of C02, the concentration in the room would change along curve (2), following the law c = ca(l — e~"), and attaining an ultimate value of ca.
If people were present but the room was initially free of C02, the law would be
And the curve (3) would be followed. The ultimate concentration attained would be (106VC/Q’) + ca. When the room volume is very small the same ultimate concentration is approached more rapidly. This would be shown in Figure 16.4 by another curve, starting
at the origin and lying between curve 3 and the upper broken line. When the room volume is very large the approach is slower and the curve would then lie beneath 3 in Figure 16.4.
If the initial concentration in the room were c0, people were present, and the ventilating air had a concentration of ca, then curve (4) would be followed according to equation (16.13), and the concentration in the room would approach a value of (106VC/Q’) + ca.
In the steady state, when n is very large (—> °°), e“" in equation (16.13) approaches zero and equation (16.13) degenerates to
C = (UtVJQ’) + ca
Which becomes
Q’ = 106Ve/(c — ca) (16.14)
And is the same as a dilution equation given in the CIBSE Guide. The ventilation effectiveness factor, ev, from Table 16.1, could be applied to the value of the air change rate, n.
EXAMPLE 16.6
If local government regulations stipulate that the minimum amount of fresh air which may be supplied to a place of public entertainment is 8 litres s-1 per person and that the minimum amount of space allowable in the room is 12 m3 per person, calculate the concentration of carbon dioxide present after one hour, expressed as a percentage. Assume that fresh air contains 0.032 per cent of carbon dioxide and that human respiration produces 4.72 x 10“3 litres s“1 of carbon dioxide per person. Also that the supply and exhaust points are above the occupied zone and that (ts — tr) < 0. Then, from Table 16.1, es = 0.9 to 1.0. Take this as 0.95.
Answer
Q’ = 0.008 m3 s"1 per person, ca = 320 parts per million,
Vc = 4.72 x 10-6 m3 s-1 per person, n = evQt/V’ (for minimum ventilation),
V = 12 m3 per person.
Hence,
N = 0.95 x 0.008 x 3600/12 = 2.28 air changes From equation (16.13)
C = [1Q6 0^)081Q6 + 32°) (1 " e~2’28} + 320 e~2 28
= 849.7 ppm = 0.085 per cent
EXAMPLE 16.7
A garage measures 60 m x 30 m x 3 m high and contains a number of motor cars which produce a total of 0.0024 m3 s~’ of carbon monoxide.
(a) If the maximum permissible concentration is to be 0.01 per cent of carbon monoxide, what number of air changes per hour are required if the garage is in continual use?
(b) If the garage is in use for periods of 8 hours only, and if at the start of any such period the concentration of carbon monoxide is zero, what number of air changes per hour is needed if the concentration is to reach 0.01 per cent only by the end of the 8-hour period?
(c) What is the concentration after the first 20 minutes of an 8-hour period?
(d) If at the end of an 8-hour period the concentration is 0.01 per cent, for how long should the ventilation plant be run in order to reduce the concentration to 0.001 per cent?
Assure ev = 1.0 for the case of supply at high level and exhaust at low level, with (ts — tr) > 0.
Answer
(a) Cmax = 100 ppm
Since the garage is in continual use, c0 also equals one hundred ppm.
The number of persons present is irrelevant but, for ease of understanding, it may be convenient to assume an occupancy of one person.
Q’ = W3600, since t is one hour
V = 5400 m3
Ca = 0 and Vc = 0.0024 m3 s-1 Hence, from equation (16.13),
100 = pP— x ^24 x 3600 + o j (i _ e- ) + 100 e~n
1 = — (1 — e~n ) + e " n
(1 — e ") = —^ (1 — e-") n
Hence n = 16
(b) From equation (16.13)
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100 = |
‘lO6 x 0.0024 x 3600
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5400 x n |
+ 0(1 — e-B) + c0e-"
In this equation, n is the total number of air changes that has occurred after 1 hour, if the other terms of the equation are for one hour. It follows that a factor of 8 must be introduced in the appropriate places:
100 = f — X0’^X3600X8)(l-e-)
^ 5400 xn J
Where n is the total number of air changes that has occurred after 8 hours.
1 = 118 (1 _ e_„ j n
Therefore,
If n equals 128, e_" is almost zero and so we can conclude that the required air change rate is 16 per hour.
(c) From equation (16.13)
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X e |
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-5.33 |
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= 99 ppm
(In the above solution, 0.0008 is the production of CO in m3 in twenty minutes.)
(d) Equation (16.9) is used because there is no production of carbon monoxide after the eight-hour period and the fresh air used to ventilate contains no contaminant of this sort.
Observing that 0.001 per cent is a concentration of 10 ppm, we can write
10 = 100 e~n because c0 is 100 ppm.
Hence,
N = 2.3
At a rate of 16 air changes an hour, it takes 8.6 minutes to effect 2.3 air changes.
Posted in Engineering Fifth Edition