I often say that when you can measure what you are speaking about, and
express it in numbers, you know something about it; but when you cannot measure
it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory
kind; it may be the beginning of knowledge, but you have scarcely, in your
thoughts, advanced to the stage of science, whatever the matter may be
– William Thompson (Lord Kelvin)
All engineered systems require
measurements for specifying the size, weight, speed, etc. of objects as well as
characterizing their performance. Understanding the application of these units
is the single most important objective of this textbook because it applies to
all forms of engineering and everything that one does as an engineer.
Understanding units is far
more than simply being able to convert from feet to meters or vice versa;
combining and converting units from different sources is a challenging topic.
For example, if building insulation is specified in units of BTU inches per hour
per square foot per degree Fahrenheit, how can that be converted to thermal
conductivity in units of Watts per meter per degree C? Or can it be converted?
Are the two units measuring the same thing or not? (For example, in a new
engine laboratory facility that was being built for me, the natural gas flow
was insufficient… so I told the contractor I needed a system capable of
supplying a minimum of 50 cubic feet per minute (cfm) of natural gas at 5
pounds per square inch (psi). His response was “what’s the conversion between
cfm and psi?” Of course, the answer is that there is no conversion; cfm is a
measure of flow rate and psi a measure of pressure.) Engineers must struggle
with these misconceptions every day.
Engineers in the United States
are burdened with two systems of units and measurements: (1) the English or
USCS (US Customary System) ☺
and (2) the metric or SI (Système International d’Unités) ☺. Either
system has a set of base units, that is, units which are defined based on a standard
measure such as a certain number of wavelengths of a particular light source.
These base units include:
• Length (meters (m), centimeters
(cm), millimeters (mm); feet (ft), inches (in), kilometers (km),miles (mi))
• 1 m = 100 cm =
1000 mm = 3.281 ft = 39.37 in
• 1 km = 1000 m
• 1 mi = 5280 ft
• Mass (lbm, slugs, kilograms);
(1 kg = 2.205 lbm = 0.06853 slug) (lbm = “pounds mass”)
• Time (seconds; the standard
abbreviation is “s” not “sec”) (same units in USCS and SI!)
• Electric current (really
electric charge in units of coulombs [abbreviation: ‘coul’] is the base unit
and the derived unit is current = charge/time) (1 coulomb = charge on 6.241506
x 1018 electrons) (1 ampere [abbreviation: amp]= 1 coul/s)
Moles are often reported as a
fundamental unit, but it is not; it is just a bookkeeping convenience to avoid
carrying around factors of 1023 everywhere. The choice of the number
of particles in a mole of particles is completely arbitrary; by convention
Avogadro’s number is defined by N A = 6.0221415 x 1023 ,
the units being particles/mole (or one could say individuals of any kind, not
limited just to particles, e.g. atoms, molecules, electrons or students).
Temperature is frequently
interpreted as a base unit but again it is not, it is a derived unit, that is, one
created from combinations of base units. Temperature is essentially a unit of
energy divided by Boltzman’s constant. Specifically, the average kinetic energy
of an ideal gas particle in a 3-dimensional box is 1.5kT, where k is Boltzman’s
constant = 1.380622 x 10-23 J/K (really (Joules/particle)/K; every
textbook will state the units as just J/K but you’ll see below how useful it is
to include the “per particle” part as well). Thus, 1 Kelvin is the temperature
at which the kinetic energy of an ideal gas (and only an ideal gas, not any other material) molecule is 1.5kT
=2.0709 x 10-23 J.
The ideal gas constant (ℜ) with which are you are very familiar is simply Boltzman’s
constant multiplied by Avogadro’s number, i.e.
In the above equation, note that
we have multiplied and divided units such as Joules as if they were numbers;
this is valid because we can think of 8.314 Joules as 8.314 x (1 Joule) and
additionally we can write (1 Joule) / (1 Joule) = 1. Extending that further, we
can think of (1 Joule) / (1 kg m2 /s2 ) = 1, which will
be the basis of our approach to units conversion – multiplying and dividing by
1 written in different forms.
There’s also another type of gas
constant R = ℜ/M, where M = molecular mass of the gas;
R depends on the type of gas whereas ℜ is the
“universal” gas constant, i.e., the same for any gas. Why does this
discussion apply only for an ideal gas? By definition, ideal gas particles have
only kinetic energy and negligible potential energy due to inter-molecular
attraction; if there is potential energy, then we need to consider the total
internal energy of the material (E, units of Joules) which is the sum of the
microscopic kinetic and potential energies, in which case the temperature for
any material (ideal gas or not) is defined as
where S is the entropy of the
material (units J/K) and V is the volume. This intimidating-looking definition
of temperature, while critical to understanding thermodynamics, will not be
needed in this course.
Derived units are units created
from combinations of base units; there are an infinite number of possible
derived units. Some of the more important/common/useful ones are:
• Area = length2 ; 640
acres = 1 mile2 , or 1 acre = 43,560 ft2
• Volume = length3 ; 1
ft3 = 7.481 gallons = 28,317 cm3 ; also 1 liter = 1000 cm3
= 61.02 in3
• Velocity = length/time
• Acceleration = velocity/time =
length/time2 (standard gravitational acceleration on earth = g =
32.174 ft/s2 = 9.806 m/s2 )
• Force = mass * acceleration =
mass*length/time2
o 1 kg m/s2
= 1 Newton = 0.2248 pounds force (pounds force is usually abbreviated lbf and
Newton N) (equivalently 1 lbf = 4.448 N)
• Energy = force x length = mass
x length2 /time2
o 1 kg m 2
/s 2 = 1 Joule (J)
o 778 ft lbf = 1
British thermal unit (BTU)
o 1055 J = 1 BTU
o 1 J = 0.7376
ft lbf
o 1 calorie =
4.184 J
o 1 dietary
calorie = 1000 calories
• Power (energy/time = mass x length2
/time3 )
o 1 J/s = 1 kg m2
/s3 = 1 Watt
o 746 W = 550 ft
lbf/sec = 1 horsepower
• Heat capacity = J/moleK or
J/kgK or J/mole˚C or J/kg˚C (see note below)
• Pressure = force/area
o 1 N/m 2
= 1 Pascal
o 101325 Pascal
= 101325 N/m2 = 14.696 lbf/in2 = 1 standard atmosphere
• Current = charge/time (1 amp =
1 coul/s)
• Voltage = energy/charge (1 Volt
= 1 J/coul)
• Capacitance = amps / (volts/s)
(1 farad = 1 coul2 /J)
• Inductance = volts / (amps/s)
(1 Henry = 1 J s 2 / coul2 )
• Resistance = volts/amps (1 ohm
= 1 volt/amp = 1 Joule s / coul2 )
• Torque = force x lever arm
length = mass x length2 /time2 – same as energy but one
would usually report torque in Nm (Newton meters), not Joules, to avoid
confusion.
• Radians, degrees, revolutions –
these are all dimensionless quantities, but must be converted between each
other, i.e. 1 revolution = 2π radians = 360 degrees.
By far the biggest problem with
USCS units is with mass and force. The problem is that pounds is both a unit of
mass AND force. These are distinguished by lbm for pounds (mass) and lbf for pounds
(force). We all know that W = mg where W = weight, m = mass, g = acceleration
of gravity.
So
1 lbf = 1 lbm x g = 32.174 lbm
ft/s2 (Equation
2)
Sounds ok, huh? But wait, now we
have an extra factor of 32.174 floating around. Is it also true that
1 lbf = 1 lbm ft/s2
which is analogous to the SI unit
statement that
1 Newton = 1 kg m/s2 (Equation
3)
No, 1 lbf cannot equal 1 lbm ft/s2
because 1 lbf equals 32.174 lbm ft/sec2 . So what unit of mass satisfies
the relation
1 lbf = 1 (mass unit) ft/s2
?
This mass unit is called a “slug”
believe it or not. With use of equation (2) it is apparent that
1 slug = 32.174 lbm = 14.59 kg (Equation
4)
Often when doing USCS
conversions, it is convenient to introduce a conversion factor called gc
; by rearranging Equation 2 we can write
Since Equation 2 shows that gc
= 1, one can multiply and divide any equation by g c as many times as necessary
to get the units into a more compact form (an example of “why didn’t somebody
just say that?”). Keep in mind that any units conversion is simply a matter of
multiplying or dividing by 1, e.g.
For some reason 32.174 lbm ft/
lbf s2 has been assigned a special symbol called gc even
though there are many other ways of writing 1 (e.g. 5280 ft / mile, 1 kg m / N
s2 , 778 ft lbf / BTU) all of which are also equal to 1 but none of
which are assigned special symbols.
If this seems confusing, I don’t
blame you. That’s why I recommend that even for problems in which the givens
are in USCS units and where the answer is needed in USCS units, first convert everything
to SI units, do the problem, then convert back to USCS units. I disagree with
some authors who say an engineer should be fluent in both systems; it is
somewhat useful but not necessary. The first example below uses the approach of
converting to SI, do the problem, and convert back to USCS. The second example
shows the use of USCS units employing gc :
Example 1
What is the weight (in lbf) of
one gallon of air at 1 atm and 25˚C? The molecular mass of air is 28.97 g/mole
= 0.02897 kg/mole.
Ideal gas law:
PV = nℜT
(P = pressure, V
= volume, n = number of moles, ℜ =
universal gas constant, T = temperature)
Mass of gas (m)
= moles x mass/mole = nM (M = molecular mass)
Weight of gas (W)
= mg, where g = acceleration of gravity = 9.81 m/s2
Combining these
3 relations: W = PVMg/ℜT
Note that it’s easy to write down
all the formulas and conversions. The tricky part is to check to see if you’ve
actually gotten all the units right. In this case I converted everything to the
SI system first, then converted back to USCS units at the very end – which is a
pretty good strategy for most problems. The tricky parts are realizing (1) the
temperature must be an absolute temperature, i.e. Kelvin not ˚C, and (2) the
difference between the universal gas constant ℜ and the
mass-specific constant R = ℜ/M. If in
doubt, how do you know which one to use? Check the units!
Example 2
A car with a mass of 3000 lbm is
moving at a velocity of 88 ft/s. What is its kinetic energy (KE) in units of ft
lbf? What is its kinetic energy in Joules?
Now what can we do with lbm ft2
/s2 ??? The units are (mass)(length)2 /(time)2
, so it is a unit of energy, so at least that part is correct. Dividing by gc
, we obtain
Note that if you used 3000 lbf
rather than 3000 lbm in the expression for KE, you’d have the wrong units – ft
lbf2 /lbm, which is NOT a unit of energy (or anything else that I
know of…) Also note that since gc = 1, we COULD multiply by gc
rather than divide by gc ; the resulting units (lbm2 ft3
/lbf s4 ) is still a unit of energy, but not a very useful one!
Many difficulties also arise with
units of temperature. There are four temperature scales in “common” use: Fahrenheit,
Rankine, Celsius (or Centigrade) and Kelvin. Note that one speaks of “degrees
Fahrenheit” and “degrees Celsius” but just “Rankines” or “Kelvins” (without the
“degrees”).
T (in units of ˚F) = T (in units
of R) - 459.67
T (in units of ˚C) = T (in units
of K) - 273.15
1 K = 1.8 R
T (in units of˚C) = [T (in units
of ˚F) – 32]/1.8,
T (in units of ˚F) = 1.8[T (in
units of ˚C)] + 32
Water freezes at 32˚F / 0˚C,
boils at 212˚F / 100˚C
Special note (another example of
“that’s so easy, why didn’t somebody just say that?”): when using units involving
temperature (such as heat capacity, units J/kg˚C, or thermal conductivity,
units Watts/m˚C), one can convert the temperature in these quantities these
to/from USCS units (e.g. heat capacity in BTU/lbm˚F or thermal conductivity in
BTU/hr ft ˚F) simply by multiplying or dividing by 1.8. You don’t need to add
or subtract 32. Why? Because these quantities are really derivatives with
respect to temperature (heat capacity is the derivative of internal energy with
respect to temperature) or refer to a temperature gradient (thermal
conductivity is the rate of heat transfer per unit area by conduction divided
by the temperature gradient, dT/dx). When one takes the derivative of the
constant 32, you get zero. For example, if the temperature changes from 84˚C to
17˚C over a distance of 0.5 meter, the temperature gradient is (84-17)/0.5 =
134˚C/m. In Fahrenheit, the gradient is [(1.8*84 +32) – (1.8*17 + 32)]/0.5 =
241.2˚F/m or 241.2/3.281 = 73.5˚F/ft. The important point is that the 32
cancels out when taking the difference. So for the purpose of converting
between ˚F and ˚C in units like heat capacity and thermal conductivity, one can
use 1˚C = 1.8˚F. That doesn’t mean that one can just skip the + or – 32
whenever one is lazy.
Also, one often sees thermal
conductivity in units of W/m˚C or W/mK. How does one convert between the two?
Do you have to add or subtract 273? And how do you add or subtract 273 when the
units of thermal conductivity are not degrees? Again, thermal conductivity is
heat transfer per unit area per unit temperature gradient. This gradient could
be expressed in the above example as (84˚C-17˚C)/0.5 m = 134˚C/m, or in Kelvin
units, [(84 + 273)K – (17 + 273)K]/0.5 m = 134K/m and thus the 273 cancels out.
So one can say that 1 W/m˚C = 1 W/mK, or 1 J/kg˚C = 1 J/kgK. And again, that
doesn’t mean that one can just skip the + or – 273 (or 460, in USCS units) whenever
one is lazy.
Example 3
The thermal conductivity of a
particular brand of ceramic insulating material is 0.5 x (BTU in/ft2
hr °F) (I’m not kidding, these are the units commonly reported in commercial
products!) where the standard abbreviations in = inch and hr = hour are used.
What is the thermal conductivity in units of W/m˚C? (Here “W” = Watt, not
weight.)
Note that the thermal
conductivity of air at room temperature is 0.026 Watt/m˚C, i.e. about 3 times lower
than the insulation. So why don’t we use air as an insulator? We’ll discuss
that in future articles.
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