Shell and Multi Tube Heat Exchanger?
545704
Well, get out your heat transfer text book and start calculating. You should have all the info you should need.
553982
Fan Controller, Temperature Gauges, Floppy, Card Reader, Hard Drive if you can get it to fit..thats where I have mine so the water cooling pump had more room..Or just save a spot for all of your wires to hide.\n\nNewegg.com\nTigerDirect.com\nXoxide.com\nPcToys.com
860530
visit\nhttp://www.diytrade.com/directory/global/buy/products-list/0902-t-i-1/Wire_Mesh.html
1096290
I suspect you may have to install seperate systems and in respect of central heating you would have to have isolation valves to allow transfer from pressurised to gravity fed system - it would be complx but technically acheivable - I currently run two seperate heating systems in a now combined house (formerly two) and have considered this.\n heating of mains pressure water via wood stove boiler and or solar would be difficult and therefore a gravity fed system would be required.\nThorough planning and installation of switching options is required
301282
decreasing the heat to avoid degredation of the PVC
329557
Try this....\n\nhttp://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm
221103
There is water/gas cooling heat exchanger are started first to avoid overheating. There is some checking for Hydrogen which can mean a water leak into the core. The 3mile island problem is online. It explains what the operators did to worsen the problem and what could have been done in detail.\nTry Am Institute of Physics, Physics Today site.
774136
http://www.fike.com/products_overview_products.html\nhttp://www.ansul.com\n\nIf I knew even more specifics about the application, I could make a recommendation.  (Tank Farm?  Holding area?  Process Area?  Tanker Filling Station?)
738632
What do you mean by the outlet temperature not being uniform. Do you mean not the same over time? Or are there multiple outlets and the temperature from each outlet is not the same?\n\nIf the inlet and outlet channels or headers are not designed carefully to allow for good fluid dynamics then there can certainly be portions of the exchanger that is not getting the same flow as other portions.\n\nWithout more information, I can not see how an energy balance will help solve the problem.
256725
I agree that I believe this to be a heat transfer equation however it seems that something is missing.  From the gut I would think that it is a way to calculate heat loss through something like a double-paned window.
381553
You should use high velocity air conditioning, sometimes called mini ducts.  This uses 2" flexible tubing, and is really ideal for retro fits.
174163
thermodynamics?  heat transfer?  physics?
107612
This is potentially a massive and complex task.\nSearch the web for "computational fluid dynamics".  Here's what I found in a couple of minutes:\nThe 1st ref. below has some interesting simulation ideas (but unfortunately for a liquid, not air).\nThe next discusses things you should know before even deciding whether to do this.\nThe last gives links to software downloads.
37324
http://www.hvacplus.com/Bryant-471.html\n\ntry this
795677
cooling it or pressurized it
898274
Yes, you can do it, but it will require iterations.  \n\n1) Convert all units to standard engineering units (feet, pounds, seconds).\n\n2) Set up the mechanical energy equation:\n(p2 - p1)/gamma = z2 - z1 + (f*L*v^2)/(2*D*g)\nwhere p2 -p1 is the pressure across the pipe (20 psi, but convert to lb/ft^2), z2 - z1 is the rise in the pipe (if any), gamma is the specific weight of water (62.4 lb/ft^3), L is the length of the pipe (50 ft), D is the I.D. of the pipe (ft), and g is the acceleration of gravity.  \n\n3) Now you have a problem: v is the velocity of the water, and f is the friction factor, which depends on v.  So, you guess a value for v, and calculate f from it.  You look up f on a chart (it's called a "Moody chart").  But you'll need the Reynolds number and relative roughness of the pipe surface to do that.  Then you try another v, find f, etc.  Or, you can try this:\n\nhttp://www.fluidmech.net/jscalc/vmd03.htm\n\n4) Just keep plugging in different velocities until the head loss (right side of the equation) equals the left side.  Then convert to volumetric flow rate.\n\nHey, I get 14.2 ft/s, 0.0774 ft^3/s = 34.7 gpm
618003
You're going to need containers to move stuff around in which fit closely to the tubes, the tubes themselves, and then an air compressor or vacuum pump to blow them around.\n\nThe link below is for a company which will build it for you. I suspect that not many people build these themselves.\n\nGood luck!