What is the difference between 100V 30A and 300A 10V ?

  • Thread starter Panthera Leo
  • Start date
  • Tags
    Difference
In summary: So the effective temperature would be much lower than the theoretical value. So, for the resistor in your example, if you wanted to heat it up to 2000K, you would need a voltage of 10.4V and a current of 568.8A.
  • #1
Panthera Leo
109
0
What is the difference between 100V 30A and 300A 10V ?

Hello, I am very new at this so please guide me on the following issue;

Lets say I have an electrical resistor with a resistivity of 1.5*10^-5 ohm m (graphite)

and let's say the total resistance of this particular graphite heating element is 0.15ohm and I need it to heat up to 2000 kelvin.

Considering the power required to radiate 2000 K is 3000 watt, how should I determine the required voltage and current...? I do know that graphite heating elements are only used with dc step down transformers, but how low should the voltage go?!

if I hook this up to a 100V 30Amp or 10V 300A transformers the result is 3000 Watt in either case, so what is the difference?! I think I am missing something very fundamental here... PLEASE help...

The way I got 3000 watt as the required power is from the black-body radiation's formula :
Power=(surface area of blackbody)*(boltzman constant)*emissivity*(Temperature)^4

is this a good method for a rough approximation (+/- 500 Kelvin) ?
or correct at all?

Thanks in advance.
 
Engineering news on Phys.org
  • #2
You have the resistance. So compute :

P=I2 R
 
  • #3
The difference in the two powers is this:

Start with Ohm's Law.

If your heating element has a total resistance of 0.15 ohm then if you hook it up to a 10V source you will consume 66.6A and generate about 665W of heat.

If you hook it up to a 100V, 30A source then the one of those two values is no longer valid: at 100V and 0.15ohm the circuit wants 6666.66A and at 0.15ohm and 30A the circuit wants 4.5V.

So if you tried the 100V, 30A source and you had a current limiting power supply the power supply voltage would drop to 3.5V with the current pegged at 30A which only gives you 105W. If you have a completely unregulated system something will probably start to smoke.How to find what you need:

You have to balance this with P = I^2*R = V^2/R.

So with R = 0.15ohms you need 141.42A and 21.213V
 
Last edited:
  • #4
Floid said:
The difference in the two powers is this:

Start with Ohm's Law.

If your heating element has a total resistance of 0.15 ohm then if you hook it up to a 10V source you will consume 66.6A and generate about 665W of heat.

If you hook it up to a 100V, 30A source then the one of those two values is no longer valid: at 100V and 0.15ohm the circuit wants 6666.66A and at 0.15ohm and 30A the circuit wants 4.5V.

So if you tried the 100V, 30A source and you had a current limiting power supply the power supply voltage would drop to 3.5V with the current pegged at 30A which only gives you 105W. If you have a completely unregulated system something will probably start to smoke.


How to find what you need:

You have to balance this with P = I^2*R = V^2/R.

So with R = 0.15ohms you need 141.42A and 21.213V


Thaaaaaaaaaaaaank you very much Floid, this clarifies everything...

What is your take on my assumption that I will need 3000Watt to reach 2000K ?
Is this formula applicable in this context?
Power=(surface area of blackbody)*(boltzman constant)*emissivity*(Temperature)^4

I got this from the following wikipedia page:

http://en.wikipedia.org/wiki/Stefan–Boltzmann_law


Thanks again. :D
 
  • #5
Now you are adding the thermal performance into the equation, things are much harder. Did you want a particular temperature? 2000K is quite high and would require a specialist heating element, I think. Imo, you would be best to find a manufacturer and ask their advice. The temperature achieved would depend as much on the enclosure you keep it in as the characteristics of the resistor.
 
  • #6
Yes. A specialist Heating element must be used. Fortunately such high grade graphites are available and already used in Atomic Absorption Spectroscopy (Graphite AAS) by Buck and other manufacturers.

The unfortunate part is they don't provide any information regarding the heating behavior of the graphite, other than their resistivity (1.5*10^-5)

But what what if we don't consider effect of temperature due to enclosure, and just consider the sole temperature of the graphite rod... How to calculate it?
 
  • #7
Panthera Leo said:
Yes. A specialist Heating element must be used. Fortunately such high grade graphites are available and already used in Atomic Absorption Spectroscopy (Graphite AAS) by Buck and other manufacturers.

The unfortunate part is they don't provide any information regarding the heating behavior of the graphite, other than their resistivity (1.5*10^-5)

But what what if we don't consider effect of temperature due to enclosure, and just consider the sole temperature of the graphite rod... How to calculate it?

That would give you a value for the notional temperature you could reach - just applying the formula in your first post. But the element is not in outer space and it is connected by conducting wires, held in a support, surrounded by air and, presumably, it is being used to 'heat' something up. All those factors will significantly affect the actual temperature reached by the element (in both directions). You would also, of course, need to consider the thermal coefficient of the resistivity, which is significant and negative (leading to possible thermal runaway in carbon filament lamps, for instance). If this has a serious application then you would need to consider some thermostatic control. Also, I can't imagine you would actually want to be operating the element out in the open - even if only from the running cost consideration.
 

1. What is the difference between voltage and amperage?

Voltage and amperage are both measures of electricity, but they represent different aspects of it. Voltage is the measure of the force or pressure of electricity, while amperage is the measure of the amount of electrical current flowing through a circuit.

2. How is electricity affected by changes in voltage or amperage?

Changes in voltage or amperage can impact the flow of electricity in a circuit. Higher voltage means that the electricity is under greater pressure and can potentially cause damage to electronic devices. Higher amperage means that there is more electricity flowing through a circuit, which can also cause damage or overheating.

3. What does 100V 30A mean?

100V 30A means that the voltage in a circuit is 100 volts and the amperage is 30 amps. This could refer to a specific appliance or device that requires this amount of electricity to function properly.

4. How does 300A 10V differ from 100V 30A?

In general, 300A 10V means that there is a higher amperage (300 amps) and a lower voltage (10 volts) compared to 100V 30A. This could indicate a larger or more powerful appliance that requires more electricity to function.

5. What are the practical implications of the difference between 100V 30A and 300A 10V?

The main practical difference between these two measurements is the amount of electricity being used. 300A 10V would require more power and potentially be used for larger or more heavy-duty appliances, while 100V 30A would be suitable for smaller devices. Additionally, the higher amperage and lower voltage of 300A 10V could result in a higher electric bill.

Similar threads

Replies
14
Views
3K
Replies
12
Views
1K
  • Electrical Engineering
2
Replies
68
Views
10K
  • Electrical Engineering
Replies
7
Views
2K
  • Introductory Physics Homework Help
2
Replies
41
Views
4K
Replies
1
Views
1K
Replies
9
Views
640
  • General Engineering
2
Replies
67
Views
4K
  • Electromagnetism
Replies
5
Views
3K
Back
Top