One more calculation. Using equation given in previous message, if you are using pipes with one cm diameter and a differential between in and out temperatures of 60 degrees, you need 12 km worth of pipes to get rid of 1 MW.
Please check my calculations but again if this is true it gives a good mental picture of why any real world physical solution of "where the heat go" is completely absurd. On Mon, Aug 15, 2016 at 12:41 PM, Giovanni Santostasi <[email protected] > wrote: > So actually the solution for the differential equation of a heat exchanger > is the following: > > dQ/dt=h*D*L*(T1-T2)/log( (T1-T0)/(T2-T0)) > > where Q is the heat exchanged, h is the heat transfer coefficient (if you > have a pipe made of steel and water is the cooling material h=400 W/m^2 K), > D is the size of the pipe, L is the length of the pipe, T1 is the incoming > temperature, T2 final temperature and T0 is the constant temperature of the > pipe. > You can play with this equation and envision the size of the pipe, initial > and final temperature and you will have really hard time to get rid of 1 MW > with any reasonable size pipe and temperatures. > > > > On Mon, Aug 15, 2016 at 11:57 AM, Giovanni Santostasi < > [email protected]> wrote: > >> Peter: >> *very simple in principle and the drain carries so much warm water with >> ease in an industrial area.* >> >> It is NOT an industrial area. And that is an essential part of what Jed >> and I are communicating over and over. >> >> >> >> On Mon, Aug 15, 2016 at 11:53 AM, Giovanni Santostasi < >> [email protected]> wrote: >> >>> Jed calculations is an order of estimate but actually on the >>> conservative side: >>> >>> 1 MJ/s = 239,000 calories/s. 80°C - 20°C = 60°C. 239,000 calories / 60°C >>> = 3,983 g/s = 239 L/min (61 gallons), which far exceeds the capacity of the >>> entire building. >>> >>> It would take time to heat up the water to 80 degrees, then it will take >>> time to exchange heat with the 20 degree water and so on, this will reduce >>> the efficiency of the exchange. The right solution is probably a >>> differential equation and I bet you get a even higher value of water flow >>> needed. >>> Unless you had enormous pools of water involved (cold and hot) you >>> cannot think of this as a thermo-equilibrium situation. >>> >>> For an order of magnitude estimate I prefer to think that you will need >>> a ton of water with 1 degree difference coming in an out of the building >>> every second to deal with 1 MW poer (similar considerations about time >>> needed to heat up and cool down the water applies but it is easier to think >>> about 1 m^3 of water with a slight higher temperature involved to represent >>> a solution closer to the non-equilibrium realistic one). >>> >>> In any case even if one adopted the conservative Jed estimate one can >>> see easily the heat exchange idea is nonsense. >>> >>> >>> >>> >>> >>> >>> >>> On Mon, Aug 15, 2016 at 10:17 AM, Jed Rothwell <[email protected]> >>> wrote: >>> >>>> Peter Gluck <[email protected]> wrote: >>>> >>>> >>>>> Both, Jed and goiovanni are trying to show tht they are not >>>>> understanding >>>>> what I have told about the energy: the cut the Gordian solution of >>>>> consuming the energy is to pass the steam pipe through a system of heat >>>>> exchangers where it heats water. that is discraded to the drain channel as >>>>> water at 40-50 C. being, (exactly as this discussion diluted to >>>>> insignificance) >>>>> >>>> >>>> The heat exchangers would produce waste heat, which would be readily >>>> apparent. It would heat up the entire warehouse. >>>> >>>> More to the point, the heat exchangers would not serve any purpose. >>>> They do not make heat vanish, they merely transfer it. The total amount of >>>> water you need is the same with or without the exchangers (except for the >>>> waste heat lost from the exhangers). So, you might as well leave out the >>>> exchangers and use the water to cool the main loop. This, however, is >>>> impossible. A normal commercial building in Florida has a 2" water supply >>>> pipe, which is not large enough for the flow of water you need to keep the >>>> temperature below the legal maximum of 80°C, with 1 MW of heat. >>>> >>>> 1 MJ/s = 239,000 calories/s. 80°C - 20°C = 60°C. 239,000 calories / >>>> 60°C = 3,983 g/s = 239 L/min (61 gallons), which far exceeds the capacity >>>> of the entire building. >>>> >>>> Your plan makes no sense. Evidently you do not understand the >>>> conservation of energy or thermodynamics. >>>> >>>> - Jed >>>> >>>> >>> >> >
