Calculating Heat Loss for Improved Efficiency in a Steam Generator Design

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SUMMARY

The discussion focuses on redesigning a steam generator with specific requirements, including a minimum water capacity of 2.25 liters and a maximum heat loss of 75W at -3°C. The design includes a fan capacity of 0.07 to 0.10 m³/min at 0.5 mbar static pressure and six steam outlet holes. The user seeks guidance on the necessary calculations to determine heat loss for improving efficiency, noting that the device functions more like a vaporizer than a traditional steam generator. The current method of increasing insulation is insufficient without proper calculations.

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  • Thermodynamics principles related to heat transfer
  • Understanding of heat loss calculations
  • Familiarity with fluid dynamics for fan capacity
  • Basic knowledge of insulation materials and their thermal properties
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Engineers, product designers, and anyone involved in the design and optimization of heating devices, particularly those focused on steam generation and thermal efficiency.

Jt2015
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At current I am trying to redesign a steam generator as seen in the picture attached, the requirements for this are as follows:
· The water contain must have a minimum capacity of 2.25 litres
· The heat loss as boiling point shall not exceed 75W in ambient of -3 C +/- 1 C
· The fan shall have a capacity of 0.07 to 0.10m^3/min at 0.5 mbar static pressure
· Six steam outlets holes shall be positioned at the top of the generator around the perimeter spaced at equal distances
· The generator should be calibrated at -3 C +/- 1 C to give reading for each 70 +/- 5g/h output up to a maximum of n times this figure where n is the maximum number of seating positions designated by the manufacturer.

Having not studied physics I am struggling to find what calculations would be required in order to calculate the heat loss and therefore cannot try improving it.
The dimensions for the base are not set and can be altered, only the nozzle and dispersion chamber have set dimensions.

If somebody could point me in the right direction to what equations I will need that would be great,

Thanks
 

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Jt2015 said:
70 +/- 5g/h output
Grams? Or, gallons?
 
Bystander said:
Grams? Or, gallons?

Sorry, it's grams per hour
 
Here's the requirements for the nozzle and chamber too.
 

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  • Nozzle & Chamber.png
    Nozzle & Chamber.png
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Jt2015 said:
Here's the requirements for the nozzle and chamber too.
I don't think this device is a steam generator in the conventional sense, more like a vaporizer or a tea kettle. 70 g an hour is equivalent to converting about 5 tablespoons of water to vapor.
 
SteamKing said:
I don't think this device is a steam generator in the conventional sense, more like a vaporizer or a tea kettle. 70 g an hour is equivalent to converting about 5 tablespoons of water to vapor.

Yeh that's a problem I've had as regards to research into it, there's not much information on them. I've literally called it that because that's what it is listed as in the guidelines.

Effectively it is just a kettle heating element in an aluminium (Previous machine) casing lined with insulator, with the nozzle, chamber and fan on top.
I just have little knowledge on what calculations I need to work out the heat loss through it in order to then improve the efficiency for the design as currently the method for this has been just add more insulator.
 

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