Heat Of Rejection and The Flash Steam Engine.

 " 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.

The percentage of flash steam can be calculated by finding the difference in heat content between the high and low pressure waters, then dividing by the latent heat of the steam at the lower pressure. To convert this decimal answer to a percentage, multiply by 100. For example, a steam trap in a 100 psi system discharges condensate to atmospheric pressure (0psi). The steam table (Table 1-1)  shows that 100 psi water contains 309 Btu/lb, 0 psi water contains 180 Btu/lb and that the latent heat of 0 psi steam is 970.3 Btu/lb. What is the percentage of water flashed into steam? 

Wasting 13% of the water of the water is bad enough, but now calculate how much heat is lost by letting the flash steam get away. What percentage of heat does the flash steam contain compared to the saturated water at 100 psi?

"end quote from book"

   Now, lets say in the case above 13.29% flash steam was generated. And released about 2.5 times that same percentage of heat, say 41.7%

    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).

    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, The amount of flash steam produced, doubled in this situation, but it's useless.

   Since so much heat was transferred nearly all of the flash 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 get 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.

Why?

   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 into the engine block at a temperature of 75°f. So, the engine block has a 39btu/lb capacity and this is the discharge area.

 The injector has a internal pressure of 1784.4psi and 620°f, the discharged water, has 638.3btu/lb. The percentage of flash steam produced is 61.7% The percentage of heat transfer is 93.8%

 That's nearly 100% heat transfer rate, and this is not good.

 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, as the result there was no positive pressure produced in the cylinder.

 But, at a block temp of 180°f(147.9btu/lb) the engine turned over, there was positive pressure developed from the injected water, since heat was conserved or rejected by the system.

    In this case, the equation yielded 50% flash steam discharge and 77% heat transfer rate, that's 20% less (heat transfer) than with the cold engine block.

 It may be noticed that as heat transfer decreases. That the resulting steam produced, performs more work. This occurs with less percentage of flash steam than would occur at colder temperatures according to the formula.

 This is what is meant by Heat Of Rejection.