Monday, December 6, 2010

How do you calculate using Dimensional Analysis?

Actually, dimensional analysis isn't a tool for
calculating; rather, it helps you show if a formula you're using won't work.  What it
shows is whether one expression uses the same units of measurement (called dimensions)
as another expression.


For example: suppose we're asked to
calculate how long it takes to fill a tank of water, given the rate at which water
enters the tank and the volume of the tank, which starts out
empty.


We want to find the time,
t, measured in seconds.  We're
given the volume of water coming into the tank in gallons per second, or
gal/sec.


Here's the
dimensional analysis: we want to find t, the number of
seconds, and we're given the flow rate in
gallons/sec.  These are the units of measurement which are
called "dimensions".


We can do algebra with these
dimensions just as we can with numeric values.  We write the beginning of a
formula:


seconds = gallons/second *
-- what do we write next to make both sides
equivalent?


We have to rewrite our trial
formula:


seconds (time to fill
the tank) =



seconds/gallon
(reciprocal of flow rate) *
gallons
(tank volume)


Look at the units of
measurement (dimensions) in the formula and ignore for the moment what's being measured.
 We have


seconds = seconds/gallon *
gallons


Rearranging,


seconds
= seconds * (gallons /
gallons)


Algebraically, the gallons cancel
out, leaving us with


seconds =
seconds


which at least shows us our formula
is self-consistent.


So we have, finally, t =
tank volume / flow rate


Now, here's an
example of how this analysis can detect an inconsistency.  Suppose we had been given
tank volume in cubic meters
(m^3).


Now our units of
measurement are:


t (seconds) = tank volume
(m^3) / flow rate
(gallons/sec)


Dimensionally,


seconds
= m^3 * (seconds / gallon)


Clearly, the
units don't match.  We either have to convert volume in m^3 to volume in gallons, or
vice versa.  Then we might have


seconds = m^3
* (seconds / m^3)


Now we use algebra to show
that this formula is
self-consistent.


--> Note
that you have to consider units of measurement rather than
name of the quantity being measured.  If we just used the
quantity names we would have


time = volume *
(time / volume)


which would appear correct
but would be wrong if volume was measured in m^3 and flow rate was measured in
gallons/second.


For a more technical discussion of units of
measurement, see the reference below, in the section "Units".

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