NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/rec/boats/8839

Internet Message Format | 1992-12-31 | 6.9 KB

Path: sparky!uunet!gatech!concert!rutgers!cmcl2!calvin!mchip00.med.nyu.edu!roy
From: [e-mail] (Roy Smith)
Newsgroups: rec.boats
Subject: Re: inverter and battery use
Keywords: inverter dc battery life
Message-ID: <[e-mail]>
Date: 31 Dec 92 16:30:46 GMT
References: <[e-mail]>
Organization: New York University, School of Medicine
Lines: 120
NNTP-Posting-Host: mchip00.med.nyu.edu

> My brother just took delivery of a Catalina 36 with 3 deep cycle 900 A/H
> batteries and an inverter.

    OK, folks, pet peeve time.  There is no such unit as "A/H".  Or
rather, there is, but it's not what you think it is.  A/H means "amps per
hour", i.e. current divided by time.  What you really mean is AH (or A*H),
conventionally read as "amp-hours", which is current multiplied by time.
This may sound like nit-picking, but the error is so pervasive in
discussions of battery life and it leads to such confusion that it's worth
complaining about.

    Well, at least you did the most important thing right, you looked at
your answer and realized it doesn't make any sense.

> running his 600W microwave for 1 hr at 110V (which takes approx 6A/H,
> right),

    This is where you start to go wrong.  AH only makes sense as a unit
of energy if you state the voltage you're talking about.  AH by itself is a
unit of charge, i.e. how many electrons you have.  It says nothing about how
much energy each electron has, which is what the voltage means.  You may
have heard the unit eV, electron-volt, used in particle physics, as a unit
of energy.  One eV is a horribly tiny unit of energy. A VAH (volt-amp-hour)
is a more reasonable amount of energy.  For example, your typical small-boat
bow light draws about an amp at 12 volts, so it takes 12 VAH of energy to
run a bow light for an hour.

> if we estimate the efficiency of his inverter at 50% (pretty conservative)
> that he should be able to run his microwave for 37 hours, even using
> only half the charge of his batteries:
>
>         microwave takes: 6A/H
>         inverter 50% efficiency: 12A/H
>         battery (50% of total): 450A/H
>         so, you can get 37 hours.

    OK, lets start with the microwave; it draws 600W (watts).  1W is one
volt-amp (it's a unit of power, energy divided by time).  Actually, what it
probably does is put 600W into the cooking chamber, which means it draws
more than 600W from the mains due to less than 100% efficient conversion,
but let's ignore that for now.  You say the inverter is 50% efficient, so
that means you need 1200W (1200VA) into the inverter.  1200VA / 12V = 100A.
That's what you're going to draw from your battery bank.  You say you've got
3 900 AH batteries, so you add their capacities to get a total of 2700 AH,
but you say you only want to draw out half that capacity, so you've
effectively got 1350 AH available.  1350AH / 100A = 13.5H.  You can run the
microwave for 13.5 hours.  That's a lot of popcorn.

    I'm assuming you meant 3 batteries, each of which is 900AH.  If what
you meant was 3 batteries totalling 900AH, you obviously have to divide the
13.5H running time by 3.  The key is dimensional analysis.  Every quantity
you are dealing with has some unit attached to it.  You need to make sure
the units work out; it they don't, then you've divided something where you
should have multiplied.  That's why the "A/H" notation is such a horrible
thing.  It makes it look like your dividing current by time, when in reality
you're multiplying.

    Here's some electrical units for your calculating pleasure:

    C = Coulomb, a unit of charge.  This is the fundamental electrical
        unit, from which all other electrical units can be derived in
        combination with the other fundamental MKS units of length
        (meter, m), mass (kilogram, kg), and time (second, s).  In most
        electrical problems, you'll never actually use this unit, but
        it's there.  The real fundamental unit of charge is the
        electron.  Each electron has 1 unit of negative charge (and each
        proton has one unit of positive charge, but since it's the
        electrons that move in common electrical circuits involving
        copper wires, we can ignore the protons for now).  Each Coulomb
        is some very large number of electrons (6.24 x 10^18).

    A = Amp, a unit of current.  Current is the amount of charge that
        passes through a wire per unit time.  1A = 1C/s.

    V = Volt, a unit of potential difference (or, more conversationally,
        voltage).  Looking at it another way, it's the amount of energy
        each electron has.  This is not really correct, since there is
        no absolute potential (which is why it's technically called
        potential difference), but it's good enough to get a handle on
        what's going on.  When you charge up a 12V battery, you pull
        electrons out of the positive terminal, give each one 12eV worth
        of energy, and shove them back into the negative terminal.  When
        you use the battery, those electrons flow out of the negative
        terminal back to the positive terminal through your load, giving
        back the 12eV you put into each one.  12ev is a very small amount
        of energy, but you get a whole hell of a lot of electrons.

    AH = Amp-Hour.  Hey, if A = C/s, then isn't AH just C*H/s and since
        1H = 3600s, isn't AH the same as C, except for a constant factor
        of 3600.  Yup, that's why AH is a unit of charge.

    J = Joule, a unit of energy.  1J = 1CV (coulomb-volt), i.e. it's the
        amount of energy you would pump into a coulomb of charge (i.e.
        6.24 x 10^18 electrons) if you raised it's potential by 1 volt.
        Since 1A = 1C/s, if you charge a 12V battery at a charging
        current of 1A for 1s, you have pumped 12J of energy into the
        battery, so you can see 1J is a smallish emount of energy, but
        at least something you can get a handle on.  You rarely see
        joules used as a unit in typical problems, but it's there, and
        it's important to understand the concept, since what were're
        really talking about with battery charging and discharging is
        storing and moving around energy, i.e. joules.

    W = Watt, a unit of power, which is rate of flow of energy.  1W =
        1J/s.  That 600W microwave uses 600J of energy every second.
        It's also 1VA (volt-amp).  You should be able to work out why;
        1W = 1J/s = 1(CV)/s = 1V(C/s) = 1VA.  1W is a fairly small
        amount of power; that little bulb in your bow light draws about
        1A at 12V, so it draws about 12W.  If you've got a 60A
        alternator, it generates 60A * 12V = 72W, which is to say, 72
        joules per second.

    That pretty much does it for everything you need to know to figure
out problems like this.  Just make sure that everytime you see "A/H" written
you say to yourself "this is wrong, it probably really means "AH".
--
Roy Smith <[e-mail]>
Hippocrates Project, Department of Microbiology, Coles 202
NYU School of Medicine, 550 First Avenue, New York, NY 10016
"This never happened to Bart Simpson."