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

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Discussion Overview

The discussion centers on the energy required to evaporate 1 liter of water using a Scheffler dish system, particularly in the context of a large-scale operation to evaporate wastewater. Participants explore the calculations involved, including the latent heat of vaporization and the heat needed to raise the water temperature to 100°C.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant estimates that it takes around 0.62 kWh to evaporate 1 liter of water based on the latent heat of vaporization.
  • Another participant suggests that the total energy requirement includes both the heat of vaporization and the heat needed to raise the water to 100°C, but does not provide a specific energy value.
  • A different participant claims that the energy required to raise the water to 100°C contributes only 5% of the total energy requirement.
  • One participant proposes a calculation method involving calories, BTUs, and kilowatt hours to determine the energy needed based on the initial temperature of the wastewater.

Areas of Agreement / Disagreement

Participants express differing views on the contribution of the heat required to raise the water temperature to 100°C, with some asserting it is minimal while others do not quantify it. The discussion remains unresolved regarding the exact energy requirements.

Contextual Notes

Participants reference specific calculations and assumptions regarding temperature and energy conversion factors, but these are not universally agreed upon, and the initial temperature of the wastewater is not specified.

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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