Kilojoules per kilo to watts per second

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Discussion Overview

The discussion revolves around the conversion of energy measurements, specifically kilojoules per kilogram to watts, and the implications of these conversions in practical applications, such as vaporizing water. Participants explore the relationship between power and energy, clarifying definitions and calculations related to watts, kilowatts, and joules.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that a watt is defined as a joule per second, emphasizing that "watts per second" is not a valid term.
  • One participant calculates that to vaporize 0.0129 kg of water per second, given a requirement of 2405 kJ per kilogram, would yield a power requirement of 31 kW.
  • Another participant expresses confusion over interpreting the power output, initially believing it to be 31 kW but later clarifying it as 31,000 watts.
  • Some participants discuss the cost implications of using 31 kW, noting it would be significant at a rate of 20 cents per kW⋅hr.
  • There are calculations presented regarding energy output using kinetic energy formulas, with one participant noting that the energy output is only 1690 joules, which raises questions about the efficiency of the process.
  • Several participants reference dimensional analysis as a helpful tool for understanding unit conversions in STEM fields.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the definitions of watts and joules, but there is disagreement regarding the interpretation of power requirements and the implications of the calculated values. The discussion remains unresolved on the efficiency and practicality of the power calculations presented.

Contextual Notes

Some calculations depend on specific assumptions about flow rates and energy requirements, and there are unresolved questions about the efficiency of energy conversion in the discussed context.

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TL;DR
I have a system that requires 2405 kilojoules to vaporise 1 kilogram of water. I have a mass flow rate of 0.0129 kg per second. How many watts are required per second to sustain the required vaporisation rate.
My understanding is that electrical devices are rated in watts which is really watts per second. For instance if a peltier module is rated at 5 watts it will use 5 watts per second. I had thought that it should be simple to convert kilojoules to watts but everywhere I’ve looked i can only find kilojoules to watt hour conversion yet I can find calculations to convert joules to watts in turbines ( power output ). Is it as simple as converting kilojoules to watt hours and then dividing by 3600 to get back to watts required per second.
 
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A Watt is a joule per second. Watts is power, Joules is energy. There's no such thing as "watts per second". You just want Watts.
 
1 kg of water will require 2405 kJ.
At a flow rate of 0.0129 kg per second, 1 kg will take 1 / 0.0129 = 77.5 sec.
That is 2405 kJ / 77.5 sec = 31.0 kJ/sec = 31.0 kW.
 
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russ_watters said:
A Watt is a joule per second. Watts is power, Joules is energy. There's no such thing as "watts per second". You just want Watts.

yes. How many watts do I need to vaporise 0.0129 kg of water per second given I need 2405 kJ per kilo.
 
Ripcrow said:
yes. How many watts do I need to vaporise 0.0129 kg of water per second given I need 2405 kJ per kilo.
You know kg per second and kJ per kilo, so how many kJ do you need per second?
 
got it thanks. For some reason I kept getting 31 kw. Must have been interpreting the answer wrong.
 
russ_watters said:
You know kg per second and kJ per kilo, so how many kJ do you need per second?

thanks. The answer above here confirmed my maths. I was interpreting the answer as kw and not as watts.
 
Ripcrow said:
got it thanks. For some reason I kept getting 31 kw. Must have been interpreting the answer wrong.
Ripcrow said:
thanks. The answer above here confirmed my maths. I was interpreting the answer as kw and not as watts.
Just to be clear; a kW is 1000 watts. 31 kW is a correct answer. If you just want plain watts, you multiply by 1000: 31,000 W.
 
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So is it 31 watts or 31 kw
 
  • #10
31 thousand watts = 31 kW.
 
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  • #11
That’s a tremendous amount of power
 
  • #12
Ripcrow said:
That’s a tremendous amount of power
It is not trivial.
At 20 cents per kW⋅hr it will cost you about $6.20 per hour to run the plant.
 
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  • #13
And energy out is only 1690 joules using 1/2 mass times velocity squared where mass is 0.0129 divided by 2 = 0.00645 kg multiplied by 512 metres per second squared = 1690 joules. Or it it kilojoules.
 
  • #15
DaveE said:
I think you will find these videos about working with dimensional units really helpful in your
future problems. This "dimensional analysis" is a great tool to make all STEM work easier.
https://www.khanacademy.org/math/al...on/v/dimensional-analysis-units-algebraically
I had the right answer as 31 kw but I didn’t believe it. Seems excessive. Just needed to clarify it. 31 watts made this very possible but 31 kilowatts destroys it.
 
  • #16
Ripcrow said:
And energy out is only 1690 joules using 1/2 mass times velocity squared where mass is 0.0129 divided by 2 = 0.00645 kg multiplied by 512 metres per second squared = 1690 joules. Or it it kilojoules.
0.5 * 0.0129 kg * (512 m/sec)2 = 1690.8 joule = 1.6908 kJ
 
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  • #17
Baluncore said:
0.5 * 0.0129 kg * (512 m/sec)2 = 1690.8 joule = 1.6908 kJ
Sad output for the amount of energy going in. That equals 1690 nm. About 2 hp out for about 25 hp input
 

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