ÿþ<HTML> <HEAD> <TITLE>SUPER CHIMNEY</TITLE> <meta name="robots" content="index,follow"> <link rel="stylesheet" type="text/css" href="../allCSS.css"> </HEAD> <BODY text="#000000" link="#1846a3" vlink="#800080" alink="#ff0000" bgcolor="#ffffff" leftmargin="0" topmargin="0" marginheight="0" marginwidth="0" style="BACKGROUND: url(../image/back1.jpg) #ffffff no-repeat"> <table width="100%" height="100%" border="0" cellpadding="0" cellspacing="0"> <tr> <td height="150" colspan="4" align="left" valign="center"><A href="default.html"><IMG height=150 alt="SUPER CHIMNEY" src="../image/00.gif" width=400 border=0></a></td> </tr> <tr> <td width="250" align="left" valign="top"> <p class="menu_left" style="MARGIN-TOP: 20px"><A class=a_menu_left href="./default.html">Summary</a></p> <p class="menu_left"><A class=a_menu_left href="./principle.html">How the invention works. Principle</a></p> <p class="menu_left"><A class=a_menu_left href="./electricity.html">How the super-chimney will produce electricity</a></p> <p class="menu_left"><A class=a_menu_left href="./water.html">How the super-chimney will produce water</a></p> <p class="menu_left"><A class=a_menu_left href="./co2.html">How the super-chimney will trap CO<sub>2</sub> and promote agriculture</a></p> <p class="menu_left"><A class=a_menu_left href="./atmosphere.html">How the super-chimney will cool the atmosphere</a></p> <p class="menu_left"><A class=a_menu_left href="./build.html">How we can build the super-chimney</a></p> <p class="menu_left"><A class=a_menu_left href="./calculation.html">Calculations</a></p> <p class="menu_left"><A class=a_menu_left href="./faq.html">FAQ</A></p> <p class="menu_left"><A class=a_menu_left href="./Volunteer.html">Be a Volunteer!</A></p> <p class="menu_left"><A class=a_menu_left href="./references.html">References</a></p> <p class="menu_left"><A class=a_menu_left href="./download.html">Downloads</a></p> <p>&nbsp;</p> <p class="menu_left"><A class=a_menu_left_red href="./contact.html">Contact</a></p> <br> <IMG height=1 src="../image/00.gif" width=250 border=0> </td> <td width="15"><IMG height=1 src="../image/00.gif" width=15 border=0></td> <td width="65%" align="left" valign="top"><!-------------------------------------------------------------------------------------------------------------> <H1 style="MARGIN: 12pt 0in 3pt"><FONT size=5><FONT face=Arial>Calculations<o:p></o:p></FONT></FONT></H1> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal">Example for calculations.<o:p></o:p></B></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">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 <SUP>2</SUP>.<o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal">1. Air Flow Calculations, according to Natural Draft Pressure Calculator <SUP>1 </SUP><SPAN style="mso-spacerun: yes">&nbsp;&nbsp;</SPAN><o:p></o:p></B></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><EM><SPAN style="FONT-FAMILY: Arial">q</SPAN></EM><FONT face=Arial> </FONT><I><FONT face=Arial>= À d<SUB>h</SUB><SUP>2</SUP> /4 [ (2 g (Á<SUB>o</SUB> - Á<SUB>r</SUB>) h ) / (<EM><SPAN style="FONT-FAMILY: Arial"> » (l Á<SUB>r </SUB>/ d<SUB>h</SUB>)</SPAN></EM> + </FONT><EM><SPAN style="FONT-FAMILY: Arial">£ ¾ Á<SUB>r</SUB> ) ]<SUP>1/2</SUP><SPAN style="mso-spacerun: yes">&nbsp; </SPAN></SPAN></EM></I><o:p></o:p></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><EM><SPAN style="FONT-FAMILY: Arial"><FONT size=2>where<o:p></o:p></FONT></SPAN></EM></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><EM><SPAN style="FONT-FAMILY: Arial">d<SUB>h</SUB></SPAN></EM><I><FONT face=Arial> </FONT><EM><SPAN style="FONT-FAMILY: Arial">= <A href="http://www.engineeringtoolbox.com/hydraulic-equivalent-diameter-d_458.html"><SPAN style="COLOR: windowtext">hydraulic diameter</SPAN></A> (m, ft)<o:p></o:p></SPAN></EM></I></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><EM><SPAN style="FONT-FAMILY: Arial">»</SPAN></EM><I><FONT face=Arial> </FONT><EM><SPAN style="FONT-FAMILY: Arial">= <A href="http://www.engineeringtoolbox.com/colebrook-equation-d_1031.html"><SPAN style="COLOR: windowtext">D'Arcy-Weisbach friction coefficient</SPAN></A></SPAN></EM></I><o:p></o:p></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><EM><SPAN style="FONT-FAMILY: Arial">l</SPAN></EM><I><FONT face=Arial> <EM><SPAN style="FONT-FAMILY: Arial">= length of duct or pipe (m, ft)</SPAN></EM></FONT></I><o:p></o:p></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><FONT face=Arial><I>g = gravity - 9.81 (m/s<SUP>2</SUP>)</I><o:p></o:p></FONT></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><EM><SPAN style="FONT-FAMILY: Arial"><SPAN style="mso-spacerun: yes">&nbsp;</SPAN>q = air volume (m<SUP>3</SUP>/s)</SPAN></EM><o:p></o:p></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><FONT face=Arial><I>Á<SUB>o</SUB> = density outside air (kg/m<SUP>3</SUP>)</I><o:p></o:p></FONT></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><FONT face=Arial><I>Á<SUB>r</SUB> = density inside air (kg/m<SUP>3</SUP>)</I><o:p></o:p></FONT></FONT></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><I><FONT size=2><FONT face=Arial>h = height between outlet and inlet air (m)<o:p></o:p></FONT></FONT></I></P> <P class=NormalWeb1 style="MARGIN: auto 0in 3.75pt"><FONT size=2><EM><SPAN style="FONT-FAMILY: Arial">£ ¾</SPAN></EM><I><FONT face=Arial> = </FONT><A href="http://www.engineeringtoolbox.com/minor-pressure-loss-ducts-pipes-d_624.html"><SPAN style="COLOR: windowtext"><FONT face=Arial>minor loss coefficient</FONT></SPAN></A><FONT face=Arial> (summarized and taken as 1)</FONT></I><o:p></o:p></FONT></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">We can expect the natural draft of air flow at the speed of 139.2 m/s. This translates into 109,357,369 m<SUP>3</SUP>/s of air going through the super-chimney. As the air density equal 1.164 kg/m<SUP>3</SUP><SPAN style="mso-spacerun: yes">&nbsp; </SPAN>at 30°C <SUP>2</SUP>, we can calculate that 127,292,000 kg/s of air moves through the super-chimney.<o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal">2. Producing electric power.<o:p></o:p></B></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">Calculate amount of electric energy produced.<o:p></o:p></P> <P class=subpagecopy style="MARGIN: auto 0in"><FONT color=#333333><FONT face=Verdana><STRONG><I><SPAN style="FONT-WEIGHT: normal; FONT-SIZE: 12pt; mso-bidi-font-weight: bold">Wind Turbine Power <SUP>7</SUP>:</SPAN></I></STRONG><SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: 'Times New Roman'"><o:p></o:p></SPAN></FONT></FONT></P> <P class=subpagecopy style="MARGIN: auto 0in"><FONT color=#333333><FONT face=Verdana><STRONG><SPAN style="FONT-WEIGHT: normal; FONT-SIZE: 12pt; mso-bidi-font-weight: bold">P = 0.5 x rho x A x Cp x V<SUP>3</SUP> x Ng x Nb </SPAN></STRONG><SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: 'Times New Roman'"><o:p></o:p></SPAN></FONT></FONT></P> <P class=subpagecopy style="MARGIN: auto 0in"><EM><SPAN style="FONT-SIZE: 12pt; mso-bidi-font-weight: bold"><FONT face=Verdana color=#333333>where:</FONT></SPAN></EM><SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: 'Times New Roman'"><BR><FONT color=#333333>P = power in watts (746 watts = 1 hp) (1,000 watts = 1 kilowatt)<BR>rho = air density (about 1.225 kg/m<SUP>3</SUP> at sea level, less higher up)<BR>A = rotor swept area, exposed to the wind (m<SUP>2</SUP>) count as a whole super-chimney area<SPAN style="mso-tab-count: 2">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </SPAN>(3.14 * 500<SUP>2</SUP>)<BR>Cp = Coefficient of performance (.59 {Betz limit} is the maximum theoretically possible, .35 for a good design) <BR>V = wind speed in meters/sec (139.2 m/s)<BR>Ng = generator efficiency (50% for car alternator, 80% or possibly more for a permanent magnet generator or grid-connected induction generator)<BR>Nb = gearbox/bearings efficiency (depends, could be as high as 95% if good)<o:p></o:p></FONT></SPAN></P> <P class=subpagecopy style="MARGIN: auto 0in"><SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: 'Times New Roman'"><FONT color=#333333>P= 0.5 *1.164 * (3.14)*(500)<SUP>2</SUP>*0.35*(139.2)<SUP>3</SUP>* 0.8 * 0.95<o:p></o:p></FONT></SPAN></P> <P class=subpagecopy style="MARGIN: auto 0in"><SPAN style="FONT-SIZE: 12pt; FONT-FAMILY: 'Times New Roman'"><FONT color=#333333>P=327,787,194,991.65696 watt=327,786 Mega Watt<o:p></o:p></FONT></SPAN></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal">3.</B> <B style="mso-bidi-font-weight: normal">Calculation of amount of water condensate.<I style="mso-bidi-font-style: normal"><o:p></o:p></I></B></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"> 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 &nbsp;kg of water condensate per second.<o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal">4. Calculation of CO<SUB>2 </SUB><SPAN style="mso-spacerun: yes">&nbsp;&nbsp;</SPAN>uptake by irrigated desert.<o:p></o:p></B></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">There are different estimates of CO<SUB>2</SUB> 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 CO<SUB>2</SUB> which will be fixed in soil. Thus, one super-chimney will allow to trap:<o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><I style="mso-bidi-font-style: normal"><o:p>&nbsp;</o:p></I></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><I style="mso-bidi-font-style: normal">300* 640 Sq.Acres/ sq.mile* 2.1=403, 200 tons of carbon,<o:p></o:p></I></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><I style="mso-bidi-font-style: normal">1 ton Carbon equivalent = 3.667 ton CO<SUB>2</SUB>,<o:p></o:p></I></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal"><I style="mso-bidi-font-style: normal">thus one<SPAN style="mso-spacerun: yes">&nbsp; </SPAN>super-chimney will allow to<SPAN style="mso-spacerun: yes">&nbsp; </SPAN>trap 1,478,354 tons of CO<SUB>2</SUB> per year.<o:p></o:p></I></B></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><B style="mso-bidi-font-weight: normal">5. Calculation of number of super-chimneys needed to cool the atmosphere.</B> <o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">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 Kiehl<SUP><SPAN lang=EN style="mso-ansi-language: EN">3</SPAN></SUP>, annually Earth receives 492 W/m<SUP><SPAN lang=EN style="mso-ansi-language: EN">2</SPAN></SUP> of radiation combined (direct solar and due to the green house effect) . Global Warming is attributed to the fact <SPAN lang=EN style="mso-ansi-language: EN">that the Earth is presently absorbing 0.85 ± 0.15 W/m<SUP>2</SUP> more than it emits into space<SUP>4</SUP> (Hansen et al. 2005). So the planet absorbs approximately 0.2% of radiation energy more than it should to maintain constant temperature.</SPAN><o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">According to <SPAN lang=EN style="mso-bidi-font-weight: bold; mso-ansi-language: EN">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.</SPAN><SUP><SPAN lang=EN style="mso-ansi-language: EN"> 5</SPAN></SUP><SPAN lang=EN style="mso-bidi-font-weight: bold; mso-ansi-language: EN"><SPAN style="mso-spacerun: yes">&nbsp; </SPAN>The super-chimney will emit air at the 5000m, </SPAN>which is roughly the point in the atmosphere where half the amount of air<SPAN style="mso-spacerun: yes">&nbsp; </SPAN>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. <o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><o:p>&nbsp;</o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">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 4x10<SUP><SPAN lang=EN style="mso-ansi-language: EN">15</SPAN></SUP>kg . According to the<st1:place><st1:placename>National</st1:PlaceName> <st1:placetype>Center</st1:PlaceType></st1:place> for Atmospheric Research, "The total mean mass of the atmosphere is 5.1480×10<SUP><SPAN lang=EN style="mso-ansi-language: EN">18</SPAN></SUP> kg... .<o:p></o:p></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt">Therefore, annually 7.7 ×10<SUP><SPAN lang=EN style="mso-ansi-language: EN">-4 </SPAN></SUP>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<SUP><SPAN lang=EN style="mso-ansi-language: EN">-4<SPAN style="mso-spacerun: yes">&nbsp; </SPAN></SPAN></SUP>or 2.31×10<SUP><SPAN lang=EN style="mso-ansi-language: EN">-2</SPAN></SUP>% more heat than it would otherwise. Since the <SPAN lang=EN style="mso-ansi-language: EN">planet absorbs approximately 0.2% of radiation energy more than it should to maintain constant temperature, we can aproximate <B style="mso-bidi-font-weight: normal">that 10 super-chimneys will offset Global</B></SPAN><B style="mso-bidi-font-weight: normal"> Warming</B><SPAN lang=EN style="mso-ansi-language: EN">.<o:p></o:p></SPAN></P> <P class=MsoNormal style="MARGIN: 0in 0in 0pt"><SPAN lang=EN style="mso-ansi-language: EN"><o:p>&nbsp;</o:p></SPAN></P> <P><SPAN lang=EN style="FONT-SIZE: 12pt; FONT-FAMILY: 'Times New Roman'; mso-ansi-language: EN; mso-fareast-font-family: 'Times New Roman'; mso-fareast-language: EN-US; mso-bidi-language: AR-SA"><BR style="PAGE-BREAK-BEFORE: always; mso-special-character: line-break" clear=all></SPAN><!--- Begin BOTTOM -----------------------------------------------------------------------------------------> <IMG height=1 src="/image/00.gif" width=550 border=0> <p>&nbsp;</p> </td> <td width="35%"><IMG height=1 src="image/00.gif" width=20 border=0></td> </tr> <tr> <td colspan="5"> <table width="100%" border="0" cellpadding="0" cellspacing="0" background="image/pesok_back.jpg"> <tr> <td><IMG height=230 src="image/pesok_rostok.jpg" width=248></td> <td>&nbsp;</td> </tr> </table> </td> </tr> </table> </BODY> </HTML>