btw altho many of u have mentioned water heaters. I've assumed these questions were referring to electric baseboard heaters. Which I always wondered why they were running on 240v. That part makes sense now sort of. I mean the cost of 240v vs 120v still to me seems like they would cost the same but you would have to double the amps on the 120v or put in 2 baseboard heaters to equal 1 240v baseboard heater right?
Help understanding meters vs electric heaters
- Thread starter jtomara37
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It seems to me like the Question in #3 seems like a trick question or something. If you answer them the questions in succession 1,2,3 it seems like the power will remain the same (watts) seem constant. The resistance will stay the same if you double the amps like in question 2. From what we have done out of the book so far it didnt expect a trick question in #3, unless they are figuring amperes to be the measurement of cost?
I'll be going back to school Monday night @ 6pm and going over our answers so I'll report back with what the instructor has to say about this.
Thank you all for your input.
Jerry
I'll be going back to school Monday night @ 6pm and going over our answers so I'll report back with what the instructor has to say about this.
Thank you all for your input.
Jerry
LarryFine
Master Electrician Electric Contractor Richmond VA
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The meter is a motor whose speed depends on both the voltage and the current. If the PF is poor, the voltage and current peaks don't occur simultaneously, and the motor spins slower.
This is similar to the way a system supplying a low-PF load must be sized to carry the unproductive, reactive, out-of-phase current, as well as the productive, useable, in-phase current.
This is similar to the way a system supplying a low-PF load must be sized to carry the unproductive, reactive, out-of-phase current, as well as the productive, useable, in-phase current.
winnie
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I know that the below repeats things that others have said....
1) A KWH meter reports 2 things:
a) power: the speed that the disk turns is proportional to the power being delivered to the load
b) energy: by counting the turns of the wheel, the meter reports energy being used.
The relationship between power and energy is analogous to the relationship between speed and distance. The more power you use (the faster you go) the more energy you use (the greater the distance traveled).
2) You don't need to measure power factor if you are taking instantaneous measurements. Power factor is a tool that you have to use when you are dealing with voltage and current measurements that are averaged over time. If you measure RMS current, and you measure RMS voltage, and you want to figure out average power, then you need to know the power factor. But if you have a tool that somehow measures the moment to moment produce of voltage and current, and then averages that product over time, then this tool is reporting average power directly.
In a common spinning disk wattmeter, the torque applied to the spinning disk is proportional to the instantaneous voltage and current, so the torque is proportional to the instantaneous power. The disk is arranged with system that applies drag torque proportional to speed, so the disk will turn at the speed that makes the drag torque equal to the power measurement torque. (If the drag torque is less than the power measurement torque, then the disk accelerates; greater and the disk slows down. So the power torque equals the drag torque, and the drag is proportional to speed thus the speed is proportional to the power.) The disk acts like a flywheel, smoothing out small variations, net result is that the speed is proportional to the average power.
-Jon
1) A KWH meter reports 2 things:
a) power: the speed that the disk turns is proportional to the power being delivered to the load
b) energy: by counting the turns of the wheel, the meter reports energy being used.
The relationship between power and energy is analogous to the relationship between speed and distance. The more power you use (the faster you go) the more energy you use (the greater the distance traveled).
2) You don't need to measure power factor if you are taking instantaneous measurements. Power factor is a tool that you have to use when you are dealing with voltage and current measurements that are averaged over time. If you measure RMS current, and you measure RMS voltage, and you want to figure out average power, then you need to know the power factor. But if you have a tool that somehow measures the moment to moment produce of voltage and current, and then averages that product over time, then this tool is reporting average power directly.
In a common spinning disk wattmeter, the torque applied to the spinning disk is proportional to the instantaneous voltage and current, so the torque is proportional to the instantaneous power. The disk is arranged with system that applies drag torque proportional to speed, so the disk will turn at the speed that makes the drag torque equal to the power measurement torque. (If the drag torque is less than the power measurement torque, then the disk accelerates; greater and the disk slows down. So the power torque equals the drag torque, and the drag is proportional to speed thus the speed is proportional to the power.) The disk acts like a flywheel, smoothing out small variations, net result is that the speed is proportional to the average power.
-Jon
I believe:
I believe:
I believe that the watt-meter is based on the equation,
p(t) = v(t)*i(t)
I believe gar's statement about the in-phase component is not literally true, but it is true in effect.
In a reactive circuit, p(t) is negative when energy is flowing out of the load. The watt-hour meter integrates p(t) with respect to time which results in real energy used in KWH. Does the disk turn backward? No, but it would stop with a PF of zero.
BTW, instantaneous values can be represented by rotating phasors of the form,
v(t) = Vpk[cos(wt) + jsin(wt)]
I believe:
I believe that the watt-meter is based on the equation,
p(t) = v(t)*i(t)
I believe gar's statement about the in-phase component is not literally true, but it is true in effect.
In a reactive circuit, p(t) is negative when energy is flowing out of the load. The watt-hour meter integrates p(t) with respect to time which results in real energy used in KWH. Does the disk turn backward? No, but it would stop with a PF of zero.
BTW, instantaneous values can be represented by rotating phasors of the form,
v(t) = Vpk[cos(wt) + jsin(wt)]
gar
Senior Member
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Instantaneous power is p=v*i. Pave = average of v*i over some period.
If we are measuring a steady DC voltage and current within an averaging interval, then p = v*i and Pave = Vave*Iave. Vave and Iave are what a DC meter reads.
If we have sine wave AC voltage source of constant amplitude and a constant resistance load, then instantaneous power is
p = Vpeak*sin(wt)*Ipeak*sin(wt)
These two components are in phase as I stated in previous post.
This equation can be rewritten as
p = Vpeak*Ipeak*(sin(wt))^2
From trigonometric identities
(sin(wt))^2 = 1/2 ( 1 - cos(2wt) )
Note there is an instantaneous double frequency component and constant (DC) component. Thus
p = Vpeak*Ipeak*(1 - cos(2wt))/2
If you average this over an integral number of cycles, then the result is
Pave = Vpeak*Ipeak/2
For a sine wave Vpeak = 2^-2*Vrms, and Ipeak = 2^-2*Irms.
Note: 2^-2 is the sq-root of 2 = 1.414. Also the sq-root of 2 times the sq-root of 2 = 2.
Thus, the average power of a steady-state AC voltage applied to a constant resistance is
Pave = Vrms*Irms.
If the averaging time is very short and not over an integral number of cycles, then there will be a random variation in the value because of the double frequency component.
.
Instantaneous power is p=v*i. Pave = average of v*i over some period.
If we are measuring a steady DC voltage and current within an averaging interval, then p = v*i and Pave = Vave*Iave. Vave and Iave are what a DC meter reads.
If we have sine wave AC voltage source of constant amplitude and a constant resistance load, then instantaneous power is
p = Vpeak*sin(wt)*Ipeak*sin(wt)
These two components are in phase as I stated in previous post.
This equation can be rewritten as
p = Vpeak*Ipeak*(sin(wt))^2
From trigonometric identities
(sin(wt))^2 = 1/2 ( 1 - cos(2wt) )
Note there is an instantaneous double frequency component and constant (DC) component. Thus
p = Vpeak*Ipeak*(1 - cos(2wt))/2
If you average this over an integral number of cycles, then the result is
Pave = Vpeak*Ipeak/2
For a sine wave Vpeak = 2^-2*Vrms, and Ipeak = 2^-2*Irms.
Note: 2^-2 is the sq-root of 2 = 1.414. Also the sq-root of 2 times the sq-root of 2 = 2.
Thus, the average power of a steady-state AC voltage applied to a constant resistance is
Pave = Vrms*Irms.
If the averaging time is very short and not over an integral number of cycles, then there will be a random variation in the value because of the double frequency component.
.
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
080928-1359 EST
Basically the rotating disk KWH meter is like a two phase motor, and it will run backwards when you invert the phase of one of the two components. If you supply power from your house to the grid, then the meter rotates backwards.
.
Basically the rotating disk KWH meter is like a two phase motor, and it will run backwards when you invert the phase of one of the two components. If you supply power from your house to the grid, then the meter rotates backwards.
.
For gar:
For gar:
Gar, look at this way:
Assume,
v(t) = Vp*sin(wt)
i(t) = Ip*sin(wt + phi)
then normalizing Vp and Ip to unity,
p(t) = sin^2(wt)*cos(phi) + sin(wt)*cos(wt)*sin(phi)
Clearly, the first term contains the power factor, cos(phi), and the second term is alternately postive and negative and averages to zero over a period.
Clearly, sin(wt) is in phase with itself, and sin(wt) is out of phase with cos(wt), but this term is not excluded from the product. Cos(phi) and sin(phi) are constants.
For gar:
Gar, look at this way:
Assume,
v(t) = Vp*sin(wt)
i(t) = Ip*sin(wt + phi)
then normalizing Vp and Ip to unity,
p(t) = sin^2(wt)*cos(phi) + sin(wt)*cos(wt)*sin(phi)
Clearly, the first term contains the power factor, cos(phi), and the second term is alternately postive and negative and averages to zero over a period.
Clearly, sin(wt) is in phase with itself, and sin(wt) is out of phase with cos(wt), but this term is not excluded from the product. Cos(phi) and sin(phi) are constants.
gar
Senior Member
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rattus post #30:
I do not disagree. cos(phi) is simply extracting the in phase component magnitude of the current.
Besoeker postg #31:
Why not describe it as the DC component of the instantaneous power? Where DC doesn't mean precisely that we are referencing current, but rather the low frequency component (base band) of a demodulator.
Many times DC and AC are extended to mean a moderately steady signal vs a rapidly varying signal. It is a loose useage of the two terms.
When you apply an AC voltage or current to a resistor you have a rapidly varying component that you usually do not notice because of the long thermal time constant of the resistor relative to the excitation frequency. Apply a 0.1 Hz current to a small wire and you will be able to feel this double frequency component. Do it at 60 Hz and you won't feel it, but you will feel the steady average component.
I have heard people referring to alternating voltage, but more often to AC voltage, and AC current. So AC is more linked to an abrevation for alternating than specifically to current.
.
rattus post #30:
I do not disagree. cos(phi) is simply extracting the in phase component magnitude of the current.
Besoeker postg #31:
Why not describe it as the DC component of the instantaneous power? Where DC doesn't mean precisely that we are referencing current, but rather the low frequency component (base band) of a demodulator.
Many times DC and AC are extended to mean a moderately steady signal vs a rapidly varying signal. It is a loose useage of the two terms.
When you apply an AC voltage or current to a resistor you have a rapidly varying component that you usually do not notice because of the long thermal time constant of the resistor relative to the excitation frequency. Apply a 0.1 Hz current to a small wire and you will be able to feel this double frequency component. Do it at 60 Hz and you won't feel it, but you will feel the steady average component.
I have heard people referring to alternating voltage, but more often to AC voltage, and AC current. So AC is more linked to an abrevation for alternating than specifically to current.
.
Bob NH
Senior Member
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It wasn't a trick question. It was a question to determine your understanding of the relationship between voltage, current, resistance, and power.jtomara37 said:
The examination you will be taking will have similar questions that won't look exactly the same, but that will require your understanding of those principles. Unless you are able to examine the questions and frame the problem in the proper way to apply the principles you may not pass the exam.
weressl
Esteemed Member
iwire said:
Watt: power (magnitude) spedometer
watthour: energy (quantity) odometer
Just like looking at your odometer you are unable to determine weather your kid was speeding or not, you can't determine the watts from your kilowatthour-meter either.
Last edited:
LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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- Electrical Contractor
In a way, it was, if you read it literally: "Is it less expensive to operate the electric heating unit in question 1 on 240v or 120v?"Bob NH said:
If we're talking about the same heater, and not modifying it for the voltages, the instantaneous power at 120v is 1/4 that at 240v, which means it costs less per hour to operate.
But, because of the lower power, the heater must run four times as long to produce the same heat, which means that, over time, both heaters cost the same to heat the area.
The point of it all is really that it takes a given amount of energy to produce a given amount of heat in a given amount of time. The power determines the overall heating time.
gar
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- EE
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rattus:
Add some harmonic content to either or both signals and the PF is still the ratio of real power to apparent power. Calcultate phi from arc cos of PF and phi is not very meaningful in terms of any waveform phase shift.
.
rattus:
Add some harmonic content to either or both signals and the PF is still the ratio of real power to apparent power. Calcultate phi from arc cos of PF and phi is not very meaningful in terms of any waveform phase shift.
.
winnie
Senior Member
- Location
- Springfield, MA, USA
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- Electric motor research
weressl said:
If you have an accurate odometer and a clock, you can figure out speed.
As commonly used, watthour meters can reasonably be used to figure power consumption as well as total energy used. Most modern digital 'watthour' meters directly report power as well as energy.
-Jon
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