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About LinkBackssimple question, how much electricity (volts/ampere, not sure which one heats up the metal) would you have to run through a block of steel to heat it up so much that it melts?
July 5th, 2009, 05:56 AM
simple question, how much electricity (volts/ampere, not sure which one heats up the metal) would you have to run through a block of steel to heat it up so much that it melts?
The resistance, and so the voltage across the steel, depends on the cross-sectional area of the steel. A wire of 5mm diameter will melt before a 150mm girder.
thanks, i got that much.Originally Posted by drowsy turtle
The resistance, and so the voltage across the steel, depends on the cross-sectional area of the steel. A wire of 5mm diameter will melt before a 150mm girder.
but what is the multiplier?
that formula only tells me that more area equals more resistance, which isn't really answering my question.
more directly:
how many watts are 1350 degrees celcius? am i completely off track?
maybe i am:
http://www.xtremesystems.org/forums/...d.php?t=130637
'Steel' varies in composition. You would need to know the alloy of steel you wish to melt....otherwise your answer will be 'about' x degrees celsius.
But it is: the voltage/current you need to melt it, depends on the length, cross-sectional area and composition of the steel.thanks, i got that much.
The resistance, and so the voltage across the steel, depends on the cross-sectional area of the steel. A wire of 5mm diameter will melt before a 150mm girder.
but what is the multiplier?
that formula only tells me that more area equals more resistance, which isn't really answering my question.
If you want to know a value, please define the above variables.
I'm afraid it doesn't really work like that. This will vary with size, shape, etc.
jesus christ... yes, thats what i want, "about" x degrees is perfectly fine.
not to be rude, but you do realize by answering like that, you're giving me more questions, instead of giving me an answer(any answer that gives a result)But it is: the voltage/current you need to melt it, depends on the length, cross-sectional area and composition of the steel.thanks, i got that much.
The resistance, and so the voltage across the steel, depends on the cross-sectional area of the steel. A wire of 5mm diameter will melt before a 150mm girder.
but what is the multiplier?
that formula only tells me that more area equals more resistance, which isn't really answering my question.
If you want to know a value, please define the above variables.
I'm afraid it doesn't really work like that. This will vary with size, shape, etc.
the best way to answer this would be "here's one specific instance that might answer part of what you're asking about but keep in mind results will vary", instead of having me guessing what i need to ask in order for you to give me an answer i can live with, and have the thread go on for 3 pages with people disagreeing before i have an answer.
again, sorry for being rude, you're the knowledgeable master, and i'm just a frustrated noob. i know you got things that are more important than answering some silly question at a science forum.
so specific values: 6cm diameter ball of steel, made of an alloy of.. say , 10% carbon, 90% iron?
specific heat capacity of steel: 420 JKg-1C-1
mass of the ball: 118.5g
0.1185x420x1500=74655J
so, you'd be thinking 75MW+ to melt it quickly, but I'm not sure about the rate at which it would radiate the heat so this is only a wild estimate. This is also assuming the resistance of the rest of the circuit is negligable (which it wouldn't be).
awesome! thanks! i completely forgot about time. what is the 1500 value? and how fast is quickly?
The 1500 is the change in temperature required to reach the estimated boiling point (couldn't find one for the exact composition).
Quickly in this case is between 1-2 seconds. But more likely, the thinnest part of the metal will melt first and break the circuit, so the rest is heated but not molten.
awesome! thanks! i completely forgot about time. what is the 1500 value? and how fast is quickly?
The 1500 is the degrees you have to heat the steel to in order to raise it to the melting point.
Though drowsy turtle missed a step in his calculation as he only calculated the amount of energy needed to get the steel up to 1500C. You also have to add the latent heat energy involved in making the phase transition from solid to liquid. This would be in the order of 272,000J per kg or another 32,000 J in his example.
The steel ball at 1500C radiates at about 85 watts. So it would take 85 watts just to hold the ball at melting temp.
So, about 106740W for 1 sec should melt your 118.5 g steel ball(1 Joule= 1 watt-sec).
At the other end, 86 watts for 1.25 days should also do the job. (Assuming the ball is isolated in a vacuum. and doesn't lose heat via contact with anything else. )
Of course in reality, the ball will be in contact with the conductors delivering the electricity...
July 5th, 2009, 04:20 PM
It is not a simple question, and it has no answer with only the information that you provided.
This is because it is a matter of heat transfer and the steel will be radiating strongly long before it melts. The heat loss due to radiation will depend on the temperature of the environment and the geometry of the piece of steel itself.
What you need to determine is the cuirrent required to provide an equilibrium temperature that is equal to the melting temperature of the particular allowy that you have. The equilibrium temperature will be the temperature at which energy loss due to radiation is equal to the energy being applied through ohmic heating. This in turn depends on the geometry of the steel, the path provided for conduction of the current (the amperage will drive the heating profile, but the amperage will be determined by the applied voltage and the resistivity of the steel, which is dependent on temperature). The heat loss is governed by the general law for radiation, the available surface area and it view factor with regard to the environment and the temperature of the environment.
http://en.wikipedia.org/wiki/Electric_arc_furnaceA mid-sized modern steelmaking furnace would have a transformer rated about 60,000,000 volt-amperes (60 MVA), with a secondary voltage between 400 and 900 volts and a secondary current in excess of 44,000 amperes.
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