Calculating heat lost (transmission lines)? (1 Viewer)

[QZP]

I'm pretty bad with electricity and this whole section is confusing me. Power loss in transmission lines is given by P = I^2 R
But isn't this calculating power? Help please!

 

[Fizzy_Cyst]

This calculates power dissipated (essentially, heat loss) by a conductor.

 

[QZP]

This calculates power dissipated (essentially, heat loss) by a conductor.

... yeah I basically stated that. So why does it calculate power loss and NOT power? I must be confused by something.

 

[anomalousdecay]

... yeah I basically stated that. So why does it calculate power loss and NOT power? I must be confused by something.

So in a conductor, the higher the current flow, the more the electrons are flowing and thus the more kinetic collisions occur in the metallic lattice of the transmission lines. The lower the current flow, the less kinetic collisions that occur.

Now, if the Voltage is high, there is a high potential difference, and hence the energy flows at a high rate per unit of charge (Out of scope of the syllabus, this is defined as V= E/Q).

By increasing the Voltage supplied, there is less current flowing through the transmission lines, as:



So we get:



The energy rate is constant (E/T is constant) so the only thing changing is the current flow. So by decreasing the current, the Voltage is increased (by simple maths) and then we get less kinetic collisions occurring in the metallic lattice of the transmission lines. Hence, the total power is defined by:

 

[EpaX]

so we know the derivation of P_total = VI, but how about P_loss = I^2R?

 

[anomalousdecay]

so we know the derivation of P_total = VI, but how about P_loss = I^2R?

Its behind the increased kinetic collision in the lattice.

So in a conductor, the higher the current flow, the more the electrons are flowing and thus the more kinetic collisions occur in the metallic lattice of the transmission lines. The lower the current flow, the less kinetic collisions that occur.

In HSC they tend to hide the details about how power is the rate of work (or change in kinetic energy) done.

 

[EpaX]

Its behind the increased kinetic collision in the lattice. So in a conductor, the higher the current flow, the more the electrons are flowing and thus the more kinetic collisions occur in the metallic lattice of the transmission lines. The lower the current flow, the less kinetic collisions that occur.

I understand the qualitative explanation, but not the quantitative derivation. From what I've seen, P = I^2R is derived from P = VI and V = IR (Ohm's Law). Substituting Ohm's Law into P = VI gives P = I^2R, but I don't understand why P now gives you 'power loss' instead of 'total power'.

 

[QZP]

I understand the qualitative explanation, but not the quantitative derivation. From what I've seen, P = I^2R is derived from P = VI and V = IR (Ohm's Law). Substituting Ohm's Law into P = VI gives P = I^2R, but I don't understand why P now gives you 'power loss' instead of 'total power'.

Yeah this was my initial question but no one answered it (satisfactorily). I have a solution but I'll tell you after school today.

 

[hit patel]

Its behind the increased kinetic collision in the lattice.



In HSC they tend to hide the details about how power is the rate of work (or change in kinetic energy) done.

Wait doessnt that come in I2I?

I'm pretty bad with electricity and this whole section is confusing me. Power loss in transmission lines is given by P = I^2 R
But isn't this calculating power? Help please!

Oh if this helps lemme clear it:
P (transmitted)=VI is the formula that shows the initial relationship between power transmitted. Note: This is not the power lost. So you can calculate either of the quantities from this. Then but V(drop)=IR and note resistance means energy loss and the voltage in this equation is the voltage drop. So when you calculate the voltage drop you can calculate P(loss)=V(drop)^2/R or P(loss)=I^2R. And to directly answer your question again P=VI, again repeating is the POWER TRANSMITTED however resistance means energy loss, therefore in Ohms law the voltage refers to the VOLTAGE DROP (since voltage is Joules/Coloumb since energy decreases voltage must decrease) due to Resistance. So when you sub in to find P=V^2/R, the P represent the POWER LOSS since the V represents the VOLTAGE DROP over the TRANSMISSION LINES. And you know that the Voltage DROP occurs therefore the current must also change since RESISTANCE IS CONSTANT and when you sub in terms of current in the Power Equation with the current transmitted you are taking into account VOLTAGE DROP OVER TRANSMISSION.
Hope that Helps.

 

[QZP]

Wait doessnt that come in I2I?



Oh if this helps lemme clear it:
P (transmitted)=VI is the formula that shows the initial relationship between power transmitted. Note: This is not the power lost. So you can calculate either of the quantities from this. Then but V(drop)=IR and note resistance means energy loss and the voltage in this equation is the voltage drop. So when you calculate the voltage drop you can calculate P(loss)=V(drop)^2/R or P(loss)=I^2R. And to directly answer your question again P=VI, again repeating is the POWER TRANSMITTED however resistance means energy loss, therefore in Ohms law the voltage refers to the VOLTAGE DROP (since voltage is Joules/Coloumb since energy decreases voltage must decrease) due to Resistance. So when you sub in to find P=V^2/R, the P represent the POWER LOSS since the V represents the VOLTAGE DROP over the TRANSMISSION LINES. And you know that the Voltage DROP occurs therefore the current must also change since RESISTANCE IS CONSTANT and when you sub in terms of current in the Power Equation with the current transmitted you are taking into account VOLTAGE DROP OVER TRANSMISSION.
Hope that Helps.

Yep, this was what I was going to say to EpaX. The V in P = IV is different to the V in V = IR. By subbing in the voltage drop V = IR across the transmission line into P = IV, you are effectively calculating the energy "drop" (loss) across that line.

 

[anomalousdecay]

Wait doessnt that come in I2I?

Its year 9 general knowledge.

Wait doessnt that come in I2I?

Oh if this helps lemme clear it:
P (transmitted)=VI is the formula that shows the initial relationship between power transmitted. Note: This is not the power lost. So you can calculate either of the quantities from this. Then but V(drop)=IR and note resistance means energy loss and the voltage in this equation is the voltage drop. So when you calculate the voltage drop you can calculate P(loss)=V(drop)^2/R or P(loss)=I^2R. And to directly answer your question again P=VI, again repeating is the POWER TRANSMITTED however resistance means energy loss, therefore in Ohms law the voltage refers to the VOLTAGE DROP (since voltage is Joules/Coloumb since energy decreases voltage must decrease) due to Resistance. So when you sub in to find P=V^2/R, the P represent the POWER LOSS since the V represents the VOLTAGE DROP over the TRANSMISSION LINES. And you know that the Voltage DROP occurs therefore the current must also change since RESISTANCE IS CONSTANT and when you sub in terms of current in the Power Equation with the current transmitted you are taking into account VOLTAGE DROP OVER TRANSMISSION.
Hope that Helps.

Just be careful there as current does not change in series meaning the current does not change across the resistor. I know you were saying that if voltage changes, then the current changes, but it wasn't very clear.

The explanation is in the voltage drop (which in turn is proportional to the rate of change in energy).

Power is the rate of change of energy.



Energy is conserved in a way such that:



So what happens is the energy lost is in the Voltage drop.

By stepping up the voltage to a maximum, the current will decrease. Hence if the current decreases (which is constant throughout the series circuit which transmission lines are relative to a reference point when "simplifying" (this is not HS stuff) the circuit), then the voltage drop as indicated below decreases:



So hence the energy lost through the voltage drop is decreased (as the constant of current is decreased). This voltage drop is where the energy converts into heat. Hence, by stepping up the source voltage, the current flowing is decreased in the transmission lines, which decreases the voltage drop across the lines (which have a resistance).

If you are still confused, then keep asking.

Don't wait 3 months later like how QZP did.

 

[QZP]

Don't wait 3 months later like how QZP did.

Well it didn't really help when you derived P = IV and then the next line stated Ploss = I^2R, completely unanswering the question. So I just buckled down and figured it out myself 20 minutes later.

 

[anomalousdecay]

Well it didn't really help when you derived P = IV and then the next line stated Ploss = I^2R, completely unanswering the question. So I just buckled down and figured it out myself 20 minutes later.

Yeah my friends complain I'm bad at explaining things. So I tend to not be very clear.

 

[hit patel]

Its year 9 general knowledge.


It is used extensively in I2I calculations. For the question asked, I said what u said.

 

[anomalousdecay]

It is used extensively in I2I calculations. For the question asked, I said what u said.

Not really. Its a concept introduced in I2I as a recap of what internal resistance in wiring is.


There were a few points that were somewhat not clear. But then again who is to say that what I said is clear

 

[EpaX]

Okay guys, so based on the info given, I have deduced the stuff below. Can you guys just verify that it's all correct? Thanks

Ohm’s Law: V = IR
where V = voltage (drop in EPE per Coulomb of charge due to resistance)
I = current flowing through resistor
R = resistance of resistor

Ohm’s Law considers the relationship between voltage drop, current and resistance within a particular resistor.


Electrical Power: P = VI

Where:
P = rate at which electrical potential energy is transmitted through the circuit [Evident from formula P = ∆EPE/t]
V = voltage (EPE given to each Coulomb of charge that is being transmitted through the circuit)
I = current being transmitted through the circuit


When you want to sub in Ohm’s Law into the electrical power formula, you must change the definition of ‘V’ in ‘P = VI’ to be the drop in EPE per Coulomb of charge due to resistance. Since the definition of ‘V’ changes, so does the definition of ‘P’ and ‘I’:


P = VI

Where:
P = rate at which electrical potential energy is being dissipated due to resistance
V = voltage (drop in EPE per Coulomb of charge due to resistance)
I = current flowing through the resistor (which is the same as the current being transmitted through the circuit since it is a series circuit)

Subbing in Ohm’s Law:

P = I^2R

Where:
R = resistance of resistor, noting that the resistor is the entire circuit.

We eliminate ‘V’ instead of ‘I’ because ‘I’ remains the same regardless of the change in definition and is thus more easier to calculate.

 

[anomalousdecay]

Energy is conserved in a way such that:

If you understand that equation then pretty much you can get why it occurs. The definition of voltage and current doesn't change so be careful with how you interpret it. Its just a voltage drop (hence energy drop) which we get through Ohm's law. Nothing more than that.

The supplied voltage is different to the voltage drop, due to the resistor having an assumed constant resistance which will dissipate energy at a rate at which the voltage drop is defined in terms of Ohm's law.

The below is not HSC stuff and you don't need to know it, but it might help you understand it if you understand the maths and application behind it

Of course, we must assume here that the current is instantaneous and appears everywhere in the circuit at the same time.

However, of all things, this is never the case for real power transmission lines. Its safe to assume it for small scale electronic or lab circuits, but for actual transmission lines, realistically the answer would be in terms of a differential equation in terms of energy. Then you get this whole answer in terms of an exponential function where things get really interesting.

 

[EpaX]

uhh, so am I correct?

 

[anomalousdecay]

uhh, so am I correct?

Generally yes. Its not the fact that you are incorrect, its just that some wording was a little "misplaced", but its perfectly fine.

 

[EpaX]

Okay thanks for the help guys!!