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Old 24th February 2009, 10:18 PM   #1
Fat80y Thread Starter
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Default How to calculate power required to heat water?

Hey all,

I'm in the planning stage of building a parts washer. What I need to do is calculate the size of the heating element I need to reach the temperatures required.

Eg 200l water @ 45 degrees will need xxxx watt heating element to reach 90 degrees in 30 minutes.

OR

200l of water @ 45 degrees and 2000w heating element will take xxx minutes to heat to 90 degrees.

Assume insulated container if that will affect things...
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Old 24th February 2009, 11:45 PM   #2
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Yay, physics!

Specific heat of water is 4.186 joules/gram degree C

So to heat up 200L of water, we need:

Q = cm(dT), where c is the specific heat, m is the mass, and dT is the change in temperature in Kelvin (Always better to work in Kelvin for thermodynamics), and Q is how much energy we need to do that.

Q = 4.186 J/g K *200,000g * 45K = 37,674,000 joules required to heat 200L of water from 45 deg to 90 deg

1 Watt is 1 joule/second, and hence:

Quote:
Eg 200l water @ 45 degrees will need xxxx watt heating element to reach 90 degrees in 30 minutes.
37,674,000 J / (30*60)seconds = 20 930 watts, so you'd need a 21kW element to do it, given perfect conditions, i.e., the element converts electricity to heat perfectly (it won't), nothing is lost though the tank (it is, the tank is heated as well, and the air outside the tank, and if there's a breeze blowing past it, that cools it further...)

Quote:
200l of water @ 45 degrees and 2000w heating element will take xxx minutes to heat to 90 degrees.
This is a bit more realistic given that 240V @ 10A gives 2400W of power.

So we know the energy requirement (37,674,000 J), so at 2000W:

37,674,000 J/2000(J/s) = 5.23 hours.

At 2400W = 4.36 hours

At 3600W (15A line, possible) = 2.9 hours.
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Old 25th February 2009, 12:40 AM   #3
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^ Looks good except you assumed the water was at 0 degrees celcius
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Old 25th February 2009, 10:03 AM   #4
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Quote:
Originally Posted by Eddyah View Post
^ Looks good except you assumed the water was at 0 degrees celcius
Except all we care about is the delta.
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Old 25th February 2009, 10:29 AM   #5
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Quote:
Originally Posted by Eddyah View Post
^ Looks good except you assumed the water was at 0 degrees celcius
dT = (T2 - T1) = (90 - 45)= 45
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Old 26th February 2009, 12:03 AM   #6
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Quote:
Originally Posted by amano.ginji View Post
dT = (T2 - T1) = (90 - 45)= 45
Ahh i'm a silly goose. I thought he wanted to heat a body of water to 45 degrees
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Old 26th February 2009, 12:10 AM   #7
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This isn't completely right as C is not constant, but it's definitely close enough to answer the question.
I was actually going to dig out the old steam tables, but this works too.
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Old 27th February 2009, 9:07 AM   #8
Fat80y Thread Starter
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Quote:
Originally Posted by amano.ginji View Post
Yay, physics!

Specific heat of water is 4.186 joules/gram degree C

So to heat up 200L of water, we need:

Q = cm(dT), where c is the specific heat, m is the mass, and dT is the change in temperature in Kelvin (Always better to work in Kelvin for thermodynamics), and Q is how much energy we need to do that.

Q = 4.186 J/g K *200,000g * 45K = 37,674,000 joules required to heat 200L of water from 45 deg to 90 deg

1 Watt is 1 joule/second, and hence:



37,674,000 J / (30*60)seconds = 20 930 watts, so you'd need a 21kW element to do it, given perfect conditions, i.e., the element converts electricity to heat perfectly (it won't), nothing is lost though the tank (it is, the tank is heated as well, and the air outside the tank, and if there's a breeze blowing past it, that cools it further...)



This is a bit more realistic given that 240V @ 10A gives 2400W of power.

So we know the energy requirement (37,674,000 J), so at 2000W:

37,674,000 J/2000(J/s) = 5.23 hours.

At 2400W = 4.36 hours

At 3600W (15A line, possible) = 2.9 hours.
Thanks amano, a very informative reply
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Old 27th February 2009, 3:24 PM   #9
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Quote:
Originally Posted by spludgey View Post
This isn't completely right as C is not constant, but it's definitely close enough to answer the question.
I was actually going to dig out the old steam tables, but this works too.
Yea, it does change - but I'd be willing to be that loss due to the environment and inefficiencies in the heating element would have a larger effect on the answer. Also, without a stiring mechanism in the water heater, depending on the number of elements, the water will heat unevenly, meaning you get hotter water around the element and cooler further way - and you'll get a longer heating time as a result.

Quote:
Originally Posted by Fat80y View Post
Thanks amano, a very informative reply
NP, got to use this physics degree I'm studying for something :-)
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