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 NA = 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).
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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.
ℜ = kNA = 1.38×10−23 J
particle K
⎛
⎝
⎜ ⎞
⎠
⎟ 6.02×1023 particle
mole
⎛
⎝
⎜ ⎞
⎠
⎟ = 8.314 J
mole K =1.987 cal
mole K
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
T ≡ ∂U
∂S

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(
V=const.
(Equation 1)
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 m2
/s2 = 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/m2 = 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 s2 / 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)
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Often when doing USCS conversions, it is convenient to introduce a conversion factor called gc
; by
rearranging Equation 2 we can write
gc = 32.174 lbm ft
lbf s
2 =1 (Equation 5).
Since Equation 2 shows that gc = 1, one can multiply and divide any equation by gc 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.
5280 ft
mile = 1;
1 kg m
N s
2 = 1;
778 ft lbf
BTU = 1; etc.
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 second example below uses the approach of converting to SI, do the problem, and
convert back to USCS. The third example shows the use of USCS units employing gc:
Example 1
An object has a weight of 300 lbf at earth gravity. What is its mass in units of lbm?
F = ma ⇒ m = F
a = F
a (1) = F
a g( c ) = 300 lbf
32.174 ft
s
2
32.174 lbm ft
lbf s
2
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟ = 300 lbm
This shows that an object that weighs 300 lbf at earth gravity has a mass of 300 lbm. At any other
gravity level, its mass would still be 300 lbm but its weight would be different, but in all cases this
weight would still be calculated according to F = ma.
Example 2
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
W = PV Mg
ℜT =
1atm
101325N/m2
atm
"

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3
7.481gal
m
3.281ft
"

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&
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3 "

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'
0.02897kg
mole
"

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&
' 9.81m
s
2
"

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8.314J
moleK (25+273)K
= 0.0440
N
m2
"

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' m3
( ) kg
mole
"

$ %
&
' m
s
2
"

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J
mole
= 0.0440
(N)(m)(kg) m
s
2
"

$ %
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J = 0.0440
(Nm) kg
m
s
2
"

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J
= 0.0440 N 0.2248 lbf
N = 0.00989 lbf ≈ 0.01 lbf
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 3
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?
KE = 1
2
(mass)(velocity)
2
= 1
2
(3000 lbm) 88 ft
s
!
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2
=1.16 ×107 lbm ft
2
s
2
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
KE =1.16×107 lbm ft
2
s
2
×
1
gc
= 1.16×107 lbm ft
2
s
2
⎛
⎝
⎜ ⎞
⎠
⎟ lbf s
2
32.174 lbm ft
⎛
⎝
⎜ ⎞
⎠
⎟ = 3.61×105 ft lbf
€
KE = 3.61×105
( ft lbf) 1 J
0.7376 ft lbf

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( = 4.89 ×105
J
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 s
4
) 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 [(1.884 +32) – (1.817 + 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 4
The thermal conductivity of a particular brand of ceramic insulating material is 0.5 BTU in
ft
2 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.)
0.5 BTU in
ft
2 hr °F × 1055 J
BTU × ft
12 in
× 3.281 ft
m × hr
3600 s
× 1 W
1 J/s
× 1.8˚F
˚C = 0.0721 W
m˚C
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?