Calculations
Example for
calculations.
If we have super-chimney 5000 m
high and 1000m in diameter, wherein the temperature of air at the bottom is 30°C
and at 5000 m it is -20°C 2.
1. Air Flow Calculations, according to
Natural Draft Pressure Calculator 1
q = π dh2 /4 [ (2 g (ρo -
ρr) h ) / ( λ (l ρr /
dh) + Σ ξ
ρr ) ]1/2
where
dh
=
hydraulic diameter (m,
ft)
λ = D'Arcy-Weisbach friction
coefficient
l = length of duct or pipe (m,
ft)
g = gravity - 9.81
(m/s2)
q = air
volume (m3/s)
ρo = density outside air
(kg/m3)
ρr = density inside air
(kg/m3)
h = height between outlet and inlet air
(m)
Σ ξ = minor loss
coefficient (summarized and taken as
1)
We can expect the natural draft
of air flow at the speed of 139.2 m/s. This translates into 109,357,369
m3/s of air going through the super-chimney. As the air density equal
1.164 kg/m3 at 30°C
2, we can calculate that 127,292,000 kg/s of air moves through the
super-chimney.
2. Producing electric
power.
Calculate amount of electric
energy produced.
Wind
Turbine Power 7:
P = 0.5
x rho x A x Cp x V3 x Ng x Nb
where:
P = power in watts (746 watts = 1 hp) (1,000 watts = 1
kilowatt)
rho = air density (about 1.225 kg/m3 at sea level, less
higher up)
A = rotor swept area, exposed to the wind (m2) count as
a whole super-chimney area
(3.14 * 5002)
Cp = Coefficient of performance (.59 {Betz
limit} is the maximum theoretically possible, .35 for a good design)
V =
wind speed in meters/sec (139.2 m/s)
Ng = generator efficiency (50% for
car alternator, 80% or possibly more for a permanent magnet generator or
grid-connected induction generator)
Nb = gearbox/bearings efficiency
(depends, could be as high as 95% if good)
P=
0.5 *1.164 * (3.14)*(500)2*0.35*(139.2)3* 0.8 *
0.95
P=327,787,194,991.65696 watt=327,786 Mega
Watt
3. Calculation of amount of water
condensate.
Suppose the relative air
humidity at the bottom of the super-chimney is 30%, which is typical for deserts.
Thus, the air contains roughly 9g of water per kg of air. At elevation of 5000m
the temperature is roughly -20C and the pressure is about 55,000 Pa (and this
is roughly half of the pressure existing at sea level). At such conditions, air
can hold at most 1.7 g of water per kg of air (point of saturation). Therefore,
once the air from the super-chimney is expelled out, water from the air
will condensate in the amount of 7.3g per kg of air. In the given system,
where 127,292, 000 kg/s of air is going through the super-chimney, this means
that 929,231 kg of water condensate per second.
4. Calculation of CO2 uptake by irrigated
desert.
There are different estimates of
CO2 uptake capacity given by the EPA. For example, it is estimated
that each acre of reforested land will absorb up to 2.1 tons of carbon per year
for period of 120 years. In fact, this number does not count additional
CO2 which will be fixed in soil. Thus, one super-chimney will allow
to trap:
300* 640 Sq.Acres/ sq.mile* 2.1=403, 200
tons of carbon,
1 ton Carbon equivalent = 3.667 ton
CO2,
thus
one super-chimney will allow
to trap 1,478,354 tons of
CO2 per year.
5. Calculation of number of super-chimneys
needed to cool the atmosphere.
There are many factors to
analyze. However, it is clear that in principle, the super-chimney will
facilitate heat exchange and, given the enormous amount of air coming through
the super-chimney, it will have effect on the heat balance of the Earth
atmosphere. According to Kiehl3, annually Earth receives 492
W/m2 of radiation
combined (direct solar and due to the green house effect) . Global Warming is
attributed to the fact that the
Earth is presently absorbing 0.85 ± 0.15 W/m2 more than it emits into
space4 (Hansen et al. 2005). So the planet absorbs approximately 0.2%
of radiation energy more than it should to maintain constant
temperature.
According to Wikipedia, Earth
surface heat captured by the atmosphere. More than 75% can be attributed to the
action of greenhouse gases that absorb thermal radiation emitted by the Earth's
surface. The atmosphere in turn transfers the energy it receives both into space
(38%) and back to the Earth's surface (62%), where the amount transferred in
each direction depends on the thermal and density structure of the
atmosphere.
5 The super-chimney will emit air at the
5000m, which is roughly the point in the atmosphere where half the amount
of air is below and half is above.
Thus, it can be assumed that re-absorption will be cut in half. So, instead of
normal distribution we will have ~70 % of energy lost to space and only 30%
reabsorbed into atmosphere. In other words, the air, which will go through the
super-chimney, will loose 30% more heat than a normal air.
According to the above
calculation 127,292, 000 kg of air will go through the super-chimney every
second. In a year it comes to 4x1015kg . According to the
Therefore, annually 7.7
×10-4 of the whole
atmosphere will go through the super-chimney. As shown above, that air will lose
30% more heat than normal air. Thus, the whole atmosphere will lose
2.31×10-4 or 2.31×10-2% more heat than it would
otherwise. Since the planet absorbs
approximately 0.2% of radiation energy more than it should to maintain constant
temperature, we can aproximate that 10
super-chimneys will offset Global Warming.
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