# 2.2: Derived Units

• • R.L. Cerro, B. G. Higgins, S Whitaker
• Professors (Chemical Engineering) at University of Alabama at Huntsville & University of California at Davis

In addition to using some alternative basic units for length, time, mass and electric charge, we make use of many derived units in the SI system and a few are listed in Table $$\PageIndex{1}$$. Some derived units are sufficiently notorious so that they are named after famous scientists and represented by specific symbols. For example, the unit of kinematic viscosity is the stokes (St), after the British mathematician, Sir George G. Stokes (Rouse and Ince)7, while the equally important molecular and thermal diffusivities are known only by their generic names and represented by a variety of symbols. The key point to remember concerning units is that the basic units represented in Table $$2.1.1$$ are sufficient to describe all physical phenomena, while the alternate units illustrated Table $$2.1.2$$ and the derived units listed in Table $$\PageIndex{1}$$ are used as a matter of convenience.

Table $$\PageIndex{1}$$: Derived SI Units
Physical Quantity Unit (Symbol) Definition
force newton (N) $$kg$$ $$m/s^{ 2}$$
energy joule (J) $$kg$$ $$m^{2}/s^{ 2}$$
power watt (W) J/s
electrical potential volt (V) J/(A s)
electric resistance ohm ($$\Omega$$) V/A
frequency hertz (Hz) cycle/second
pressure pascal (Pa) $${ N/m}^{ 2}$$
kinematic viscosity stokes (St) $${ cm}^{ 2} /s$$
thermal diffusivity square meter/second $${ m}^{ 2} /s$$
molecular diffusivity square meter/second $${ m}^{ 2} /s$$

While the existence of derived units is simply a matter of convenience, this convenience sometimes leads to confusion. This is particularly true in the case of Newton’s second law which can be stated as

$\left\{\begin{array}{c} {\text{ force acting}} \\ {\text{ on a body}} \end{array}\right\}=\left\{\begin{array}{c} {\text{ time rate of change}} \\ {\text{ of linear momentum}} \\ {\text{ of the body}} \end{array}\right\} \label{2-5}$

In terms of mathematical symbols, we express this axiom as

$\mathbf{f}=\frac{d}{dt} \left(m\mathbf{v}\right) \label{2-6}$

Here we adopt a nomenclature in which a lower case, boldface Roman font is used to represent vectors such as the force and the velocity. Force and velocity are quantities that have both magnitude and direction and we need a special notation to remind us of these characteristics.

Let us now think about the use of Equation \ref{2-6} to calculate the force required to accelerate a mass of 7 kg at a rate of 13 $$m/s^2$$. From Equation \ref{2-6} we determine the magnitude of this force to be

$f=(7 \ kg )(13 \ m/s^{2})= 91 \ kg \ m/s^{2} \label{2-7}$

where $$f$$ is used to represent the magnitude of the vector $$\mathbf{f}$$. Note that the force is expressed in terms of three of the four fundamental standards of measure, i.e., mass, length and time. There is no real need to go beyond Equation \ref{2-7} in our description of force; however, our intuitive knowledge of force is rather different from our knowledge of mass, length and time. Consider for example, pushing against a wall with a “force” of 91 $$kg$$ $$m/s^2$$. This is simply not a satisfactory description of the event. What we want here is a unit that describes the physical nature of the event, and we obtain this unit by defining a unit of force as

$1 \text{ newton} = \left(1 \ kg \right)\times \left(1 \ m/s^{2} \right) \label{2-8}$

When pushing against the wall with a force of 91 $$kg$$ $$m/s^2$$ we feel comfortable describing the event as

\begin{align}\text{ force} &= 91 \ kg \ m/s^{2} \nonumber\\[4pt] &= \left(91 \ kg \ m/s^{2} \right) \times \left(1\right) \nonumber\\[4pt] &= \left(91 \ kg \ m/s^{2} \right) \times \left(\frac{1 \text{ newton}}{ kg \ m/s^{2}} \right) \nonumber\\[4pt] &= 91 \text{ newtons} \label{9} \end{align}

Here we have arranged Equation \ref{2-8} in the form

$1=\frac{\text{ newton}}{kg \ m/s^{2}} =\frac{ N}{\ kg \ m/s^{2}} \label{10}$

and multiplied the quantity 91 $$kg$$ $$m/s^2$$ by one in order to effect the change in units. Note that in our definition of the unit of force we have made use of a one-to-one correspondence represented by Equation \ref{2-8}. This is a characteristic of the S.I. system and it is certainly one of its attractive features. One must keep in mind that Equation \ref{2-8} is nothing more than a definition and if one wished it could be replaced by the alternate definition given by8

$1 \text{ euler} = \left(17.07 \ kg\right) \times \left(1 \ m/s^{2} \right) \label{11}$

However, there is nothing to be gained from this second definition and it is clearly less attractive than that given by Equation \ref{2-8}.

In the British system of units the one-to-one correspondence is often lost and confusion results. In the British system we choose our standards of mass, length and time as

Table $$\PageIndex{2}$$

 mass $$lb_m$$ length ft time s

and we define the pound-force according to

$1 \ lb_f =\left( 1 \ lb_m \right) \times \left(32.17 \ ft/s^{2} \right) \label{2-12}$

This definition was chosen so that mass and force would be numerically equivalent when the mass was acted upon by the earth’s gravitational field. While this may be a convenience under certain circumstances, the definition of a unit of force given by Equation \ref{2-12} is certainly less attractive than that given by Equation \ref{2-8}.

In summary, we note that there are only four standards needed to assign a numerical values to all observables, and the choice of standard is arbitrary, i.e. the standard of length could be a foot, an inch or a centimeter. Once the standard is chosen, i.e. the meter is the standard of length, other alternate units can be constructed such as those listed in Table $$2.1.2$$. In addition to a variety of alternate units for mass, length and time, we construct for our own convenience a series of derived units. Some of the derived units for the SI system are given in Table $$\PageIndex{1}$$. Finally, we find it convenient to tabulate conversion factors for the derived units for the various different systems of units and some of these are listed in Table $$\PageIndex{3}$$. An interesting history of the SI system is available from the Bureau International des Poids et Mesures9.

 Length Mass 1 inch (in) = 2.54 centimeter (cm) 1 pound mass ($$lb_m$$) = 453.6 gram (g) 1 angstrom $$(\mathop{\mathrm{ A}}\limits^{o}) = 10^{-8}$$ centimeter (cm) 1 g = $$10^{-3}$$ kilogram (kg) 1 foot (ft) = 0.3048 meter (m) 1 ton (short) = 2000 $$lb_m$$ 1 mile (mi) = 1.609 kilometer (km) 1 ton (long) = 2240 $$lb_m$$ 1 yard (yd) = 0.914 meter (m) 1 nanometer (nm) = $$10^{-9}$$ meter (m) Time 1 minute (min) = 60 second (s) Force 1 hour (h) = 60 minute (min) pound-force ($$lb_f$$) = 32.17 ft $$lb_m$$ $$s^{-2}$$ 1 day = 24 hours (h) dyne (dyn) = g cm $$s^{-2}$$ 1 newton (N) = $$10^{5}$$ dyne (dyn) Density 1 dyne (dyn) $$= 2.248 \times 10^{-6}$$ pound force ($$lb_f$$) 1 $$g/cm^{3} = 10^{3}$$ $$kg/m^{3}$$ 1 poundal = $$3.108 \times 10^{-2}$$ $$lb_f$$ 1 $$lb_m/ft^{3} = 16.018$$ $$kg/m^3$$ 1 pound-force ($$lb_f$$) = 4.448 newton (N) 1 $$lb_m/$$gal (US) = $$119.83 \ kg/m^{3}$$ Volume Pressure 1 $$in^3$$ = 16.39 $$cm^3$$ 1 atmosphere (atm) = 14.7 $$lb_f$$ /in$$^2$$ 1 $$ft^{3} = 2.83 \times 10^{-2}$$ $$m^{3}$$ 1 $$lb_f/in^{ 2} = 6.89 \times 10^{ 3}$$ $$N/m^{ 2}$$ 1 gallon (US) (gal) = 231 $$in^3$$ 1 atm = $$1.013 \times 10^{ 6}$$ dyne/$$cm^{ 2}$$ 1 quart (liquid) (qt) = 0.25 gal (US) 1 pascal (Pa) = 1 N/m$$^{ 2}$$ 1 barrel = 31.5 gal (US) 1 atmosphere (atm) = 1.01325 bar 1 gal (US) = 0.003785 $$m^{3}$$ 1 liter (L) = $$10^3$$ $$cm^3$$ Power 1 horsepower (hp) = 745.7 watt Area 1 hp = 42.6 Btu/min 1 $$in^2$$ = 6.452 $$cm^2$$ 1 watt = $$9.51 \times 10^{ -4}$$ Btu/s 1 $$ft^{2} =9.29 \times 10^{-2}$$ $$m^{2}$$ 1 ft $$lb_f/s$$ = 1.356 watt (W) 1 acre $$= 4.35 \times 10^{4}$$ $$ft^{ 2}$$ Temperature 1 C = 1.8 F Energy K = C + 273.16 1 calorie (cal) = 4.186 joule (J) R = F + 459.60 1 British Thermal Unit (Btu) = 252 cal F = 1.8C + 32 1 erg = $$10^{-7}$$ joule (J) 1 Btu = 1055 watt sec Flow Rate 1 ft $$lb_f$$ = 1.356 joule (J) 1 gal (US)/min = gpm = $$6.309 \times 10^{-5}$$ $$m^{ 3}/s$$ 1 $$ft^{ 3}$$/min = $$4.719 \times 10^{-5}$$ $$m^{ 3}/s$$ Viscosity 1 poise (P) = 1 g/cm s Kinematic Viscosity 1 poise (P) = 0.10 N $$s/m^2$$ 1 $$m^{ 2}/s$$ = $$3.875 \times 10^{ 4}$$ $$ft^{ 2}$$/hr 1 poise (P) = $$6.72\times 10^{-2}$$ $$lb_m$$/ft s 1 $$cm^{ 2}/s$$ = $$10^{ -4}$$ $$m^{ 2}/s$$ 1 centipoise (cP) = $$10^{-3}$$ kg/m s 1 stokes (St) = 1 $$cm^{2}/s$$