Kilojoules per kilo to watts per second

[Ripcrow]

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.

 

[russ_watters]

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.

 

[Baluncore]

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.

 

[Ripcrow]

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.

 

[russ_watters]

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?

 

[Ripcrow]

got it thanks. For some reason I kept getting 31 kw. Must have been interpreting the answer wrong.

 

[Ripcrow]

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.

 

[russ_watters]

got it thanks. For some reason I kept getting 31 kw. Must have been interpreting the answer wrong.

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.

 

[Ripcrow]

So is it 31 watts or 31 kw

 

[Baluncore]

31 thousand watts = 31 kW.

 

[Ripcrow]

That’s a tremendous amount of power

 

[Baluncore]

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.

 

[Ripcrow]

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.

 

[DaveE]

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

 

[Ripcrow]

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.

 

[Baluncore]

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

 

[Ripcrow]

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