Amount of energy required to evaporate 1 liter if water through scheffler dish

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To evaporate 1 liter of water using a Scheffler dish, approximately 0.62 kWh is needed, primarily due to the latent heat of vaporization. Additionally, the energy required to heat the water to 100°C contributes only about 5% to the total energy requirement. The calculation for this energy includes raising the temperature from the initial temperature of the wastewater, which can be converted from calories to BTUs and then to kilowatt hours. The discussion highlights the importance of considering both heating and vaporization energy in the overall energy assessment. Understanding these energy requirements is crucial for effectively planning the evaporation of large volumes of wastewater.
ashishgourav
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Hi
Our company is planning to setup a scheffler system to evaporate 9 lakh liters of "waste water" per day.

I just want to know that how much energy would it take to evaporate 1 liter of water.

According to "latent heat of vaporization of water" it would take around 0.62 kWh/liter but I'm not sure...please help!
 
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From an ideal standpoint, it will be both heat of vaporization plus the heat required to raise the water to 100°C. The only google hit I can find on Scheffler water systems is this thread, so I can't help you any more.
 
I know that "both heat of vaporization plus the heat required to raise the water to 100°C" but the energy required to raise the water to 100°C contributes just 5% of the total requirement
 
ashishgourav said:
I know that "both heat of vaporization plus the heat required to raise the water to 100°C" but the energy required to raise the water to 100°C contributes just 5% of the total requirement

If that is true then it will take 20(1000 calories) times the number of degrees (100-T beginning)
or
20000 (100 - temp of waste water in celsius) in calories
If you take that number and multiply by 3.968 you will have BTUs
You can multiply BTUs by 0.000293 for kilowatt hours.
 

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