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Topic: adapters (6 msgs / 354 lines)
1) From: Spencer W. Thomas
Ok, so I made an arithmetic error, and the perpetual motion machine
won't work.  Darn!
0. Watts are watts.  AC / DC / burning paper.  Doesn't matter where
they come from, they're the same amount of power.  You can't create
new energy with a transformer or converter.  In fact, any *real*
device will lose energy (and get warm in the process). So if it's
putting out 300 watts, it's gotta be taking in MORE THAN 300 watts.
1. Even though your power supply is rated at "5 amps" it's not
actually drawing that much.  If it did, you'd be wasting about 250
watts as heat in the power supply, which I think you would notice
pretty quickly.  It's probably drawing somewhere under 400 watts, or
less than 3.3 amps on the input.
2. I think the current draw question for an inverter is answered
pretty well by the calculator at the URL someone else posted earlier
http://www.skingcompany.com/portable_power/calc2.htm). These guys
are in the business of selling inverters, and I have to assume they
know their stuff.  Their calculator claims that an inverter that puts
out 300(real) watts will draw 30 amps from the battery.  (Just for
grins (check my math!) 30 amps x 12 volts = 360 watts (got it right
this time! :-) so the inverter will be putting out about 60 watts as
heat, too.)
3. (for geeks only) Why does the Volts * Amps = Watts thing work for
AC too?  Well, it's because the "120" volt AC current is really
delivered as a 60Hz sine wave with a peak voltage of about 170 volts.
If we pretend for the moment that it's a DC current at 120 V and 2.5A,
then it would dissipate 300W in a 48ohm resistor.  If, instead, we
apply our 170V peak sine wave to the 48 ohm resistor, we get current
varying in a sine wave from -3.5A to +3.5A. Multiplying the current by
the voltage, we get a power curve in the shape of cos^2 with a peak
power of 600W. (Precisely, we get P(t) = 600*cos^2(t*60*2*pi).)
Averaging over a full cycle, we get 300W.
(The average power over a period DT is equal to
integral(t=0,DT)(600W * cos^2(t*60*2pi)dt) / DT =
  600W * (DT/2 + 1/(4*60*2pi) * sin(2 * DT * 60 * 2pi) / DT
If DT = 1/60 this reduces to 600W * (1/120 + 1/(240*2pi) * sin(2*2pi))
/ (1/60), but sin(240*2pi) is zero, so we get 600W * (1/120) / (1/60)
= 300W.)
=Spencer
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2) From: Ryuji Suzuki -- JF7WEX
From: "Spencer W. Thomas" 
Subject: +Re: adapters
Date: Wed, 28 Mar 2001 17:05:07 -0500
Although we have more important coffee issues than electrical
engineering...
<Snip>
This is right, but there is another factor in AC case. What
if the load is not purely dissipative? That is, what if the load
has inductance and/or capacitance in addition to resistor. This
is typical of motor devices. Then, there will be a phase shift
between voltage and current, and therefore multiplying RMS values
of voltage and current will overestimate the actual (real) power
deilvered to the dissipative (resistive) part of the load!
If you look at such motor deviecs, your UPS and switching power
unit used for computers, etc., you're more likely to see a unit
called VA or Volt-Ampere. This is the product of RMS voltage and
RMS current, which is never less than Watt, but usually larger.
So what happens to the difference, VA value - real Watt value?  That
amount of power is sucked in and spit out during the AC cycle.  Your
electricity delivering company will not like it if there is too much
of that, like factories do. (There are circuits to compensate for the
effect though.)
<Snip>
This is also right. If your circuit is drawing 5A through a 5A fuse,
the fuse will slowly blow off. There is no such thing as ideal fuse,
so that the fuse blows off at the instant when the circuit current
slightly exceeds the rated current but never blows off if the circuit
current is slightly below the rated current. There is also
manufacturing tolerance factor.
More importantly, when you turn on the switch to a DC power supply for
your amateur radio rigs, the DC power supply draws much more current
(rush current) in the first few cycles than its steady state load,
even if the DC load draws a constant current from the
supply. Therefore it is not uncommon to use a 5A fuse when the device
draws only 3A at steady state to give sufficient margin for rush
current.
<Snip>
Although being considered a geek is much better than being considered
a loser (sorry :-) this discussion applies only if the load is purely
resistive. In real devices there are some deviation from that. However,
as you noted, real components are lossy. So in any case real Watt is
not more than Volt times Ampere in sinusoidal power supply.
Ryuji Suzuki
# radio amateur since 1983
# JF7WEX
--
Ryuji Suzuki
"I can't believe I'm here.
People always say that I'm a long way from normal."
(Bob Dylan, Normal, Illinois, 13 February 1999)
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3) From: Alex McGregor
Ahh this brings me back to days spent studying power distribution systems
during college. (painful memories I might add)
The AC discussion below reminds me of the reason power companies force users
who burden the network with highly inductive loads to install large
capacitors.
Now there is a thought!  A few farads of capacitance would do the trick.
Solve the current spike issues as well.
Alex

4) From: JCAbbott Postoffice
"Transmission Theory"  a class designed to make you appreciate all others -
but it didn't help with my feelings about Chem 200+ YUK!

5) From: Ryuji Suzuki -- JF7WEX
From: "Alex McGregor" 
Subject: RE: +Re: adapters
Date: Wed, 28 Mar 2001 16:42:17 -0800
<Snip>
THe key is that you probably didn't roast your own coffee to
appreciate all important applications of science and engineering.
--
Ryuji Suzuki
Q. What is your real message?
A. Keep a good head and always carry a light bulb. (Bob Dylan 1965)
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6) From: Alex McGregor
Suzuki-san,
You are absolutely correct! My priorities were wrong when I was in college.
I now make time for roasting and it has aided my journey.
Reminds me of the following quote.
" Like the sharp edge of a razor, the sages say, is the path, Narrow it is,
and difficult to tread."
Alex


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