Talk:Vapor-compression evaporation

message: I am reshaping this article. It will maybe be necessary to separate the MVR and thermocompression items, but for the time being this seem premature - at least until the page reaches a decent length and completeness. I will probably be obliged to amend some points, which are clearly referred only to MVR.--UbUb (talk) 20:56, 20 April 2008 (UTC)
replies: text: Well, it's done, at least as a first phase. The name should probably be changed to V. C. evaporator rather than evaporation; I will leave that to the next contributor.--UbUb (talk) 22:07, 20 April 2008 (UTC)

text: I agree and I have added a clarify tag. Biscuittin (talk) 15:05, 15 January 2015 (UTC)
thread_title: Comments

message: These were used in fleet-type submarine boats for producing fresh water?  Jack Waugh (talk) 16:49, 29 July 2012 (UTC)
replies: text: See Evaporator (marine) Andy Dingley (talk) 17:20, 29 July 2012 (UTC)
thread_title:  Usage 

message: Quote: "The compression ratio in a MVR unit does not usually exceed 1.8. This means that, if evaporating at atmospheric pressure (0.101 MPa and 100 °C), the condensation pressure at the heat exchanger will be 116.9 °C".  Comment: 116.9 °C is a temperature, not a pressure, and how is the figure calculated? It would be sensible to use absolute temperatures, rather then Celsius temperatures. Biscuittin (talk) 18:23, 14 January 2015 (UTC)
replies: text: When dealing with saturated vapor, the pressure defines the saturation temperature (and vice versa). Thus while the above statement is technically incorrect, it's not completely inaccurate. I clarified that list item and showed how the 2nd temperature is calculated. Since we're dealing with water vapor, Celsius is more correct to use here than Kelvin. You'll notice that standard SI steam tables and charts are always given in C, not K.  --84.94.83.131 (talk) 18:32, 9 June 2017 (UTC)
thread_title:  Error 

message: Quote: "This theoretical value shall be increased by the efficiency, usually in the order of 30 to 60%". Comment: Suppose the theoretical energy input is 300 kJ and the efficiency is 30%. The actual energy input would be 300 x 100/30 = 1,000 kJ. However, the sentence is unclear and might be interpreted as 300 + 30% = 390 kJ. This needs clarification. Biscuittin (talk) 10:22, 15 January 2015 (UTC)
replies: text: I have re-written the sentence to clarify it. Biscuittin (talk) 10:35, 15 January 2015 (UTC)
thread_title:  Energy input 

message: I can't see any reason for using Vapor-compression evaporation. Surely, it would always be more efficient to use direct heating? Biscuittin (talk) 15:13, 15 January 2015 (UTC)
replies: text: Heat transfer efficiency may well be higher, but total system efficiency will be lower. Direct heating requires an external heat source (fuel). The higher the energy input to the drying system, the lower the system efficiency. Vapor-compression evaporation uses the latent heat of the evaporated water as a 'free' heat source. By investing energy in vapor compression, we create a temperature difference that lets heat flow back in to the evaporation chamber, and allows us to reuse most of the energy we invested in evaporation in the first place. (Cost-effectiveness is a different question... that typically depends on the scale of the system.) --84.94.83.131 (talk) 18:42, 9 June 2017 (UTC)
thread_title:  Efficiency