Info

1.7442 1.7568 1.7596 S-Entropy Btu/lb °F

Figure 33. Enthalpy/entropy diagram.

In order to cool the vapor to its condensing temperature of 220°F, 11.2 Btu/lb of heat is removed. This can be done by introducing condensate at 220°F from the steam side of the evaporator. The 11.2 Btu/lb of heat removed from the vapor is absorbed into the condensate, some of which will vaporize and give off what is known as "excess vapor." For every pound of vapor cooled, 11.2 Btu of heat is absorbed in the condensate, which requires 965.2 Btu/lb to boil. Therefore, for each point of vapor leaving the compressor, 11.2/965.2 or 0.0116 lb of excess vapor is available.

This excess vapor is used in several ways. Since there is a slight difference in latent heat between the steam and the vapor, slightly more steam is required than the vapor generated. Other excess vapor is used to cover losses due to radiation and venting, is made available in some instances for preheating, or is sent to a condenser. It is significant to note that the condenser on an MVR evaporator is responsible only for vent and excess vapors. This results in a much lower cooling requirements than is necessary for steam evaporators.

It is possible to calculate an equivalent steam economy for an MVR system. In this example, for every pound of water evaporated, 970.3 Btu is absorbed. The compressor supplies 14.1 Btu but with motor and gear losses, probably requires 14.5 Btu of energy. The equivalent economy (970.3/14.5) is 67 to 1. Since one horsepower is equivalent to 2545 Btu/h, the compressor in the example requires 14.5/2545 or 0.0057 hp/lb • h of evaporation.

It should be noted that pressure losses through the evaporator that must be absorbed by the compressor have not been considered in this example. These losses would be taken into account by either higher compressor horsepower or lower AT over the heat-transfer surface.

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