Calculations
Example for calculations.
If we have super-chimney 5000 m high and diameter 1000m or 20m , 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 (po - pr) h ) / ( λ (l pr / dh) + ∑ξ pr ) ]½
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)
po = density outside air (kg/m3)
pr = density inside air (kg/m3)
h = height between outlet and inlet air (m)
∑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 (50,916 m3/s for 20m diamterer). As the air density equal 1.164 kg/m3 at 30°C2, we can calculate that 127,292,000 kg/s of air moves through the super-chimney(50,916 kg/s for 20m diameter).
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 (131.1 Mega Watt for 20m diamterer)
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(371 kg for 20m diamterer).
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(592tons for 20m diamterer)
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 theNational Center for Atmospheric Research, "The total mean mass of the atmosphere is 5.1480×1018 kg...".
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(25,000 for 20m diamterer).
6. Calculation of building a flexible chimney
There will be three limitations to consider:
1. Whether flexible chimney will hold its shape.
2. Whether it will be standing upright.
3. Whether it will withstand side winds.
1. Whether a flexible chimney will hold its shape.
In order to guarantee that chimney holds its shape, its theoretical mass needs to be less than pressure multiplied by inner area surface.
In order to guarantee that chimney holds its shape, its theoretical mass needs to be less than pressure multiplied by inner area surface.
where (using SI units):
= static pressure in Pascals,
= density in kg/m3
R= specific gas constant (287.05 J/(kg·K) for air),
T= absolute temperature in Kelvin (K),
M= Mach number (non-dimensional),
= ratio of specific heats (non-dimensional) (1.4 for air at sea level conditions),
= fluid velocity in m/s,
a= speed of sound in m/s
http://en.wikipedia.org/wiki/Dynamic_pressure#cite_note-5
Thus we can estimate the additional pressure due to the flow of air in the chimney. It will be 8400Pa. It means that as long as each square meter of the surface weighs less then 840 kg, the chimney will hold its shape and can be standing upright.
2. Whether it will be standing upright.
There are three forces which will be pushing the chimney up: reaction force at the mushroom cap, buoyancy and drag force along the surface.The sum of these forces will be indicative of the maximum allowed weight of the chimney.
a. Reaction force.
In order to overcome reactive force of the exiting air the chimney will need to be equipped with "Mushroom cap" where exits for air will be on the down facing surface of the cap. Such assembly will ensure that exiting air will be pushing chimney upwards.
The force will be equal to
F =p *v2 *A .
- density at the exit 5000m =0.74
(see http://www.denysschen.com/catalogue/density.aspx)
V- Speed of air at exit
A = ΠR2 area of the opening
For 10 m A =78.5 m2 Fr=0.74* (139)2* 78.5=1,122,356 N
For 20m A =314m2= Fr=0.74* (139)2* 314=4,489,427N
b. Buoyancy force
The air inside the chimney is warmer than the outside air. Yet, pressure is almost the same. Thus air inside is less dense. Therefore, chimney will act like a hot air balloon and support itself by buoyancy force. In order to average up our estimates for the whole system, we will calculate the buoyancy force for the chimney at the mid level altitude of 2500 m.
F = V*g*(pcool - phot)
V=Area*height m3 is volume
g = 9.81 m/s3
pcool=0.929 kg/m3 is air density outside the chimney at 2500 m and 5° C
phot= 0.838 kg/m3 air density inside the chimney at 2500 m and 30° C
(see http://www.denysschen.com/catalogue/density.aspx)
For 10 m A=78.5 m2 Fb=78.5*5000*9.81*(0.929-0.838) =350,388 N
For 20m A=314 m2 Fb=314*5000*9.81*(0.929-0.838) =1,401,555 N
c. Air Friction
It is difficult to estimate air resistance as it will greatly depend on the inner surface property of the chimney. Since our objective to have a higher air flow, we will always try to keep friction force at minimum by providing smooth polished surface inside.
For the purpose of our calculations we will give it 0 value.
Ff=0
Total Force:
Total Upward Force
Ft= Fr + Fb + Ff
For 10 m A =78.5 m2 Ft=1,122,356 +350,388+0=1,472,744 N=147 tons
For 20m A =314 m2 Ft=4,489,427+1,401,555 +0=5,890,982 N=589 tons
7. Erecting the chimney. Whether resulting buoyancy will be enough to support such structure.
In order to average up our estimates for the whole system we will calculate that buoyancy force for the chimney at the mid level altitude 2500 m and inside air at 70°C.
F = V g (pcool - phot)
V=Area*height=78.5*5000=392500
g = 9.81 m/s2
pcool=0.929 kg/m3 is air density outside the chimney at 2500 m and 5° C
phot= 0.637 kg/m3air density inside the chimney at 2500 m and 70° C
For 10 m A =78.5 m2 Fb=78.5*5000*9.81*(0.929- 0.637) = 1,124,324N
For 20m A =314 m2 Fb=314*5000*9.81*(0.929-0.637) = 4,497,296N
(see http://www.denysschen.com/catalogue/density.aspx)
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