W
Wayne Farmer
Guest
"paul" <reetix@gmail.com> wrote in message
news:1115246621.068110.165600@f14g2000cwb.googlegroups.com...
course at my local community college), I believe there are 3 ways to measure
the power drawn by a load from an alternating voltage source:
1. Apparent power. This is RMS AC Volts x RMS AC Amps. It is reported not
in Watts, but in "VA" = Volt-Amperes.
2. True power. This is the resistive component of Apparent Power, and is
reported in Watts. If the load has no capacitive or inductive elements,
then all the load is resistive. True power is the power that is dissipated
in heat through your load.
3. Reactive power. This is the inductive and capacitive (= reactive)
component of Apparent Power, and is reported in "VAR" = Volt-Amperes
Reactive. Reactive power isn't dissipated in heat, but cycles back and
forth between you and the power company as alternating current flow.
Apparent power is the vector sum of True Power and Reactive Power: the
square root of (True Power**2 + Reactive Power**2).
The ratio of True power to Apparent power is the Power Factor.
Motors (hard drives, fans, etc.) are typically inductive loads.
Now let's look at the figures your meter is reporting:
51.2 VA = the 31 Watts true power being consumed by the resistive loads in
your computer and hard drive. At the same time, there is a reactive
(inductive and capacitive) power draw of sin(arccos 0.60) x 51.2 VA = 41
VAR. The square root of (true power**2 + reactive power**2) is indeed 51.4
VA, which matches your figure of 51.4 VA.
of that apparent power is due to the resistive loads in your computer and
hard drive; the true power is 42.5 Watts, or approximately the 41 Watts you
reported. There is little or no inductive and capacitive load. That makes
sense, because the hard drive motor was a major part of the inductive load
when it was connected.
So, you can see that the apparent power draw does decrease, from 51.2 VA to
42.5 VA, when you disconnect the hard drive.
Can somebody else explain why the true power (resistive load) -increases-
from 31 Watts to 41 Watts when the hard drive is disconnected?
Wayne
news:1115246621.068110.165600@f14g2000cwb.googlegroups.com...
Based on my limited knowledge (having just finished an AC circuit theoryHello,
Today I plugged in a power meter I bought a while ago, and decided to
check out how much power my old Pentium 233MMX was using, here are the
results (UK Power):
Normal Operation
244 Volts, 0.21 Amps, 31 Watts, Power Factor = 0.60
But if I disconnect the hard drive, I get:
250 Volts, 0.17 Amps, 41 Watts, Power Factor = 1.00
How could it be using more power when the hard drive is disconnected?
41 Watts is how much power it is using, right? Which one (if any) is
accurate?
Thanks!
paul
course at my local community college), I believe there are 3 ways to measure
the power drawn by a load from an alternating voltage source:
1. Apparent power. This is RMS AC Volts x RMS AC Amps. It is reported not
in Watts, but in "VA" = Volt-Amperes.
2. True power. This is the resistive component of Apparent Power, and is
reported in Watts. If the load has no capacitive or inductive elements,
then all the load is resistive. True power is the power that is dissipated
in heat through your load.
3. Reactive power. This is the inductive and capacitive (= reactive)
component of Apparent Power, and is reported in "VAR" = Volt-Amperes
Reactive. Reactive power isn't dissipated in heat, but cycles back and
forth between you and the power company as alternating current flow.
Apparent power is the vector sum of True Power and Reactive Power: the
square root of (True Power**2 + Reactive Power**2).
The ratio of True power to Apparent power is the Power Factor.
Motors (hard drives, fans, etc.) are typically inductive loads.
Now let's look at the figures your meter is reporting:
244 V x .21 A = 51.2 VA apparent power. Of that, the 0.60 Power Factor xNormal Operation
244 Volts, 0.21 Amps, 31 Watts, Power Factor = 0.60
51.2 VA = the 31 Watts true power being consumed by the resistive loads in
your computer and hard drive. At the same time, there is a reactive
(inductive and capacitive) power draw of sin(arccos 0.60) x 51.2 VA = 41
VAR. The square root of (true power**2 + reactive power**2) is indeed 51.4
VA, which matches your figure of 51.4 VA.
250 V x .17 A = 42.5 VA apparent power. Since the Power Factor is 1.00, allBut if I disconnect the hard drive, I get:
250 Volts, 0.17 Amps, 41 Watts, Power Factor = 1.00
of that apparent power is due to the resistive loads in your computer and
hard drive; the true power is 42.5 Watts, or approximately the 41 Watts you
reported. There is little or no inductive and capacitive load. That makes
sense, because the hard drive motor was a major part of the inductive load
when it was connected.
So, you can see that the apparent power draw does decrease, from 51.2 VA to
42.5 VA, when you disconnect the hard drive.
Can somebody else explain why the true power (resistive load) -increases-
from 31 Watts to 41 Watts when the hard drive is disconnected?
Wayne