Heat Of Rejection

Heat Of Rejection And The Flash Steam Engine.

I want to go out on a limb here and discuss a factor that I have not seen covered in any thermodynamics texts. I call it heat of rejection. And it specifically addresses the phenomenon of flash steam formation, and residual pressure of such flash steam. It is advantageous to form high pressure flash steam for the purposes of running an engine. And current flash steam equations may be utilized to understand this. Now I want to reference another very significant book. -The Boiler Operator's Exam Preparation Guide- by Theodore B. Sauselein, P.E. McGraw-Hill -ISBN-O-07-057968-7- Anyway the sub chapter titled FLASH STEAM pg 8, is very interesting. It goes something like this. Nature will not permit water to remain in liquid form at temperatures higher than 212°f, and to contain more than 180 btu/lb at atmospheric pressure, in liquid form. And that saturated water at 0 psi at 212°f contains 180 btu/lb. Saturated water at 150 psi at 366°f contains 339 btu/lb. In the latter case, the btu/lb exceeding 180 btu/lb must be jettisoned. Nature takes care of this surplus by converting a fraction of the water to flash steam. Live steam is generated in a boiler, while Flash steam is produced when hot water at its saturated temperature is released to a lower pressure. And say in the case above you generated 16.3% flash steam. And released 3 times that same percentage of heat, say 46.9%. Now take the example of the engine block, it is at 75°f ( assume 0 psi), and with a btu/lb energy of 43.09 (substitute 180 btu/lb). now run the flash steam equation, what do you get if you discharge water at 150psi and 366°f, containing 339 btu/lb. The percentage of flash steam is 30.4% and a heat transfer of 87.2% According to the formula, flash steam doubled in this situation, but it's useless. Since so much heat transferred all steam was absorbed. You could say it was un-rejected. The heat transfer almost doubled from 46% to 87%. In the next example take water at 1784.4psi and 620°f, it has 638.3 btu/lb. What percent will flash when it is released to a 180 btulb area? 47.2% flash steam and 71.8% heat transfer. In this situation the engine will turn over. But at 638btu/lb you got 61% flash steam from your hot water with a 43 btu/lb release area (cold block 75°f) while releasing 93% of the heat. The engine did not turn over. Even though 10% more flash steam was produced.

Lets explore this further, remember in the situation with the cold engine block temp. The discharge from the injector would not push the piston. Then I heated the block to 180°f, and then the discharge from the injector pushed the piston. now lets run the equation with a discharge from the injector to the engine at 75°f. so the engine block has a 39btu/lb capacity and this is the discharge area. Now the injector has 1784.4psi and 620°f so the discharge water has 638.3btu/lb. the percentage of flash steam produced is 61.7% The percentage of heat transfer is 93.8% That's almost 100% heat transfer rate. Remember the engine did not turn over under these conditions. All the heat from the flash steam was transferred into the engine block. None of it was rejected. And there was no positive pressure. But at a block temp of 180°f(147.9btu/lb) the engine turned over, there was positive pressure. In this case the equation yielded 50% flash steam discharge. And 77% heat transfer rate, 20% less than with the cold engine block. It may be noticed that as heat transfer decreases. That the resulting steam performs more work. even with less percentage of flash steam than would occur at colder temperatures. This is what is meant by Heat Of Rejection.

J.W. 11/06/2002