9.2: Appendix B- Dimensions and Units
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B.1 Dimensions and Units
Consider the following mathematical statement: \[L=3.00 \mathrm{~m} \nonumber \] In words you might say, "The length \(L\) is \(3.00\) meters." Implicit in the mathematical expression is an indication of both a physical quantity and its size.
The nature or description of a physical quantity is known as its dimension. The size of the physical quantity is expressed as a certain number of standard reference amounts. These reference amounts are known as units. In the example above, the dimension is length and the units are meters. (Note that the unit symbol "\(m\)" is a mathematical entity, not an abbreviation.) For any given dimension, there are an infinite number of possible units.
B.2 Systems of Units and Dimensions
Experience has shown that there is a set of independent dimensions that can be used to represent all physical quantities. The members of this set are known as the fundamental (primary) dimensions. Once the fundamental dimensions have been selected, all other physical quantities are described in terms of derived (secondary) dimensions.
Once the fundamental dimensions have been chosen, it is then possible to select their corresponding units - the base (primary) units. And as you might expect the derived dimensions have a corresponding set of units — the derived (secondary) units.
There are many different systems of units and dimensions based upon the choice of fundamental dimensions. Two of the most common systems of units are the SI system and the American Engineering System. Table B.1 shows the base units for these two systems.
| Quantity | SI Base Units | AES Base Units | ||
|---|---|---|---|---|
| Name | Symbol | Name | Symbol | |
| length | meter | \(\mathrm{m}\) | foot | \(\mathrm{ft}\) |
| mass | kilogram | \(\mathrm{kg}\) | pound-mass | \(\mathrm{lbm}\) |
| time | second | \(\mathrm{s}\) | second | \(\mathrm{s}\) |
| electric current | ampere | \(\mathrm{A}\) | ampere | \(\mathrm{A}\) |
| thermodynamic temperature | kelvin | \(\mathrm{K}\) | Rankine | \({ }^{\circ}\) R |
| amount of substance | mole | \(\mathrm{mol}\) | pound-mole | \(\mathrm{lbmol}\) |
| luminous intensity | candela | \(\mathrm{cd}\) | candela | \(\mathrm{cd}\) |
B.3 Calculations with Dimensions and Units
Units would not provide significant problems if we did not have to use them in calculations. Unfortunately, unit errors are one of an engineer's worst enemies. Although they always seem like trivial mistakes in school, in practice the consequences can be catastrophic.
B.3.1 Dimensional Homogeneity
All theoretically derived equations that describe physical phenomena must be dimensionally homogeneous. An equation is dimensionally homogeneous if the dimensions of both sides of the equation are the same and all additive terms have the same dimensions.
B.3.2 Converting Units
The most common errors in using units occur when converting a physical quantity from one set of units to another set. When you convert units you are not changing the size of the physical quantity, only the numerical value associated with the units in which it is measured.
The relationship between two units for the same dimension are typically found in a handbook as an equivalence relation, such as \(1 \mathrm{~ft}=12 \mathrm{~in}\). Note again that the unit symbols are mathematical entities and cannot be neglected.
The key to converting units is to recall that multiplying a mathematical expression by unity (1) does not change the magnitude of the mathematical expression. A unit conversion factor equals unity and can be constructed from an equivalence relation. Example B.1 shows how to convert equivalence statements into unit conversion factors.
Convert the given equivalence relations into unit conversion factors.
\[ \begin{aligned} 1 \mathrm{~ft} = 12 \mathrm{~in} \quad\quad &\Rightarrow \quad\quad 1 = 12 \ \frac{\mathrm{in}}{\mathrm{ft}} \\ 1 \mathrm{~slug} = 32.174 \mathrm{~lbm} \quad &\Rightarrow \quad 1 = 32.174 \ \frac{\mathrm{lbm}}{\mathrm{slug}} \\ 1 \mathrm{~mol} = 0.001 \mathrm{~kmol} \quad &\Rightarrow \quad 1 = 0.001 \ \frac{\mathrm{kmol}}{\mathrm{mol}} \\ 1 \mathrm{N} = 1 \ \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2} \quad &\Rightarrow \quad 1 = 1 \ \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2} \end{aligned} \nonumber \]
The left-hand column shows the equivalence relations from a handbook and the right-hand column shows the resulting unit conversion factors. Notice how it would be mathematically incorrect to just drop the unit symbols.
Example B.2 illustrates how to perform a simple unit conversion for pressure, now that we have the unit conversion factors.
Given a pressure of \(13.0 \mathrm{~lbf} / \mathrm{in}^{2}\), convert the pressure to \(\mathrm{lbf} / \mathrm{ft}^{2}\).
\[ p = 13.0 \ \frac{\mathrm{lbf}}{\mathrm{in}^{2}} = 13.0 \ \frac{\mathrm{lbf}}{\mathrm{in}^{2}} \times \left(12 \ \frac{\mathrm{in}}{\mathrm{ft}}\right)^{2} = 1872.0 \ \frac{\mathrm{lbf}}{\mathrm{ft}^{2}} \nonumber \]
Now convert the pressure value to \(\mathrm{N} / \mathrm{m}^{2}\).
\[p = 13.0 \ \frac{\mathrm{lbf}}{\mathrm{in}^{2}} \times \underbrace{\left(4.448 \ \frac{\mathrm{N}}{\mathrm{lbf}}\right)}_{1} \times \underbrace{\left(\frac{1 \mathrm{~in}}{0.0254 \mathrm{~m}}\right)^{2}}_{1} = 89,627 \ \frac{\mathrm{N}}{\mathrm{m}^{2}} \nonumber \]
If done, correctly the intermediate units should cancel. Check this out by drawing lines through the units that cancel.
Example B.2 is relatively straightforward. Sometimes, however, you are faced with a unit conversion that appears to be both a unit and a dimension conversion. However, it is impossible to convert dimensions. Example B.3 demonstrates this type of unit conversion.
The kinetic energy per unit mass for a baseball can be described by the expression \(k e=(1 / 2) V^{2}\). Calculate the kinetic energy per unit mass in kilojoules per kilogram if the speed of the baseball is \(10 \mathrm{~m} / \mathrm{s}\).
\[\begin{aligned} ke &= \frac{\mathrm{V}^{2}}{2} = \frac{\left(10 \ \dfrac{\mathrm{m}}{\mathrm{s}}\right)^{2}}{2} = 50 \ \frac{\mathrm{m}^{2}}{\mathrm{~s}^{2}} \\[5pt] &= 50 \frac{\mathrm{m}^{2}}{\mathrm{~s}^{2}} \times \underbrace{\left[\frac{\left(\dfrac{\mathrm{kJ}}{\mathrm{kg}}\right)}{\left(\dfrac{\mathrm{kJ}}{\mathrm{kg}}\right)}\right]}_{1} = \left[50 \frac{\mathrm{kJ}}{\mathrm{kg}}\right] \times \left[\frac{\dfrac{\mathrm{m}^{2}}{\mathrm{~s}^{2}}}{\dfrac{\mathrm{kJ}}{\mathrm{kg}}}\right] = 50 \ \frac{\mathrm{kJ}}{\mathrm{kg}} \times \frac{\left[\dfrac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}}\right]}{\mathrm{kJ}} \\[5pt] &= 50 \ \frac{\mathrm{kJ}}{\mathrm{kg}} \times \frac{\left[\dfrac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}}\right]}{\mathrm{kJ}} \times \underbrace{\left[ \dfrac{\mathrm{kJ}}{1000 \mathrm{~J}} \right]}_{1} \times \underbrace{\left[\frac{\mathrm{J}}{\mathrm{N} \cdot \mathrm{m}}\right]}_{1} \times \underbrace{\left[ \frac{\mathrm{N}}{\dfrac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2}} \right]}_{1} = 0.050 \ \frac{\mathrm{kJ}}{\mathrm{kg}} \end{aligned} \nonumber \]
The procedure here was to multiply the calculated quantity in the second line by a unit conversion factor formed from the desired final units. Then it was just a matter of massaging the units until we eliminated all but the desired combinations. If it had not been possible to achieve this, then the proposed conversion is really a conversion of dimensions and that is impossible.
B.4 Handling Units in Equations
Unit errors can also occur within equations as you substitute numerical values into symbolic equations. To prevent this type of error you should follow three steps:
- always write down the units with a number when you substitute in a numerical value,
- write down your unit conversion factors and show the algebra that cancels out the units, and
- show the unit conversions as a separate step.
The last step is not always necessary, but it is always a good idea when the conversions are complicated. In addition, it is a useful step for novices. This approach is demonstrated in Example B.4.
A tank contains \(15 \mathrm{~mol}\) of an ideal gas. The pressure in the tank is \(1500 \mathrm{~kPa}\) and the volume of the tank is \(10 \mathrm{~m}^3\). The ideal gas constant is \(8.314 \mathrm{~kJ} / ( \mathrm{kmol} \cdot \mathrm{K})\). Determine the temperature of the gas in the tank.
\[ \begin{aligned} & \text{The ideal gas equation is } pV = n \bar{R} T. \\ & \text{We solve for } T = \frac{pV}{n \bar{R}} \\ &\therefore T = \frac{pV}{n \bar{R}} = \frac{(1500 \mathrm{~kPa}) \times \left(10 \mathrm{~m}^{3}\right)}{(15 \mathrm{~kmol}) \times \left(8.314 \ \dfrac{\mathrm{kJ}}{\mathrm{kmol} \cdot \mathrm{K}}\right)} = \frac{15000 \ \mathrm{kPa} \cdot \mathrm{m}^{3}}{124.71 \ \dfrac{\mathrm{kJ}}{\mathrm{K}}} \times \underbrace{\left[ \frac{\mathrm{kJ}}{\mathrm{kPa} \cdot \mathrm{m}^{3}} \right]}_{1} = 120.3 \mathrm{~K} \end{aligned} \nonumber \]
Again, the check is to see if the appropriate units cancel out.
B.5 Weight and Mass
People frequently confuse the terms weight and mass. The weight of an object is the force exerted by the earth's gravitational field on the object. Mathematically, \(W=m g\), where \(m\) is the mass of the object and \(g\) is the local gravitational field strength. The local gravitational field strength is also referred to as the local acceleration of gravity.
Standard values for the local gravitational field strength are \[g = 9.80665 \mathrm{~N} / \mathrm{kg} = 32.174 \mathrm{~lbf} / \mathrm{slug} = 1.000 \mathrm{~lbf} / \mathrm{lbm}. \nonumber \]
Standard values for the local acceleration of gravity are \[g = 9.80665 \mathrm{~m} / \mathrm{s}^{2} = 32.174 \mathrm{ft} / \mathrm{s}^{2} . \nonumber \]
TEST YOURSELF: Why do these two interpretations for \(g\) come up with similar numbers but different units?
Much of the confusion about mass and weight can be directly attributed to the fact that the mass and force units in the American Engineering System are both called "pounds." To eliminate this problem, it is highly recommended that you only talk about pound-force \((\mathrm{lbf})\) or a pound-mass \((\mathrm{lbm})\). You would never confuse a newton with a kilogram, but then they have different names. Unfortunately, you will still find "pound" and "\(\mathrm{lb}\)" used frequently to mean both mass and weight. Always approach "pounds" with caution when doing calculations. Remember that the weight of an object is always a function of the local gravitational field strength, but its mass is independent of the gravitational field.
B.6 Problems
Calculate the magnitude of the physical quantities in the indicated units. Show all of your work, i.e. explicitly show the use of unit conversion factors and how the units cancel out. If it is impossible to make the conversions indicated, please state the reasons why you believe this to be the case. [Hint: It is frequently necessary to break a secondary unit down into its primary units or other secondary units in the process of converting to the desired secondary units. See Example B.3 in Appendix B of the ES201 notes.]
(a) Pressure: \(100 \mathrm{~lbf} / \mathrm{in}^{2} = \text{_______________ } \mathrm{N} / \mathrm{cm}^{2} = \mathrm{bar}\)
b) Energy per unit mass: \(2000 \mathrm{~Btu} / \mathrm{slug} = \text{_______________ } \mathrm{ft} \cdot \mathrm{lbf} / \mathrm{lbm} = \text{_______________ } \mathrm{kJ} / \mathrm{kg} \)
c) Product of pressure and volume: ["\(\text{psi}\)" equals "pounds-force per square inch," \(\mathrm{lbf} / \mathrm{in}^{2}\)]
\(3000 \mathrm{~psi} \cdot \mathrm{ft} = \text{_______________ } \mathrm{ft} \cdot \mathrm{lbf} = \text{_______________ } \mathrm{N} / \mathrm{kg} \)
On one of the moon landings, astronauts collected moon rocks for study back on earth. The weight of the rocks measured on the moon was \(50 \mathrm{~lbf}\). The strength of the gravity on the moon is \(1 / 6\) the value on earth. (Many times this is stated as "The acceleration of gravity on the moon is \(1 / 6\) the value on earth.")
(a) Determine the mass of the rocks collected on the moon, in \(\mathrm{lbm}\) and in slugs.
(b) Determine the mass and the weight of the rocks back on earth, in \(\mathrm{lbm}\) and \(\mathrm{lbf}\), respectively.
(c) A newspaper report from a later moon landing reports that the astronauts collected "200 pounds of rocks." What, if any, additional information would you need in order to actually specify the amount of rocks they brought back? (Be careful of making implicit assumptions.)
The units for a physical quantity can often seem to be at odds with the description used for them. A good example is the common units for "energy per unit mass." Showing all the steps, develop and verify the following conversion factor: \(1 \mathrm{~ft} \cdot \mathrm{lbf} / \mathrm{lbm} = 32.174 \mathrm{~ft}^{2} / \mathrm{s}^{2}\).
Note that both \(\mathrm{ft} \cdot \mathrm{lbf} / \mathrm{lbm}\) and \(\mathrm{ft}^{2} / \mathrm{s}^{2}\) are correct units for \([\mathrm{energy}] / [\mathrm{mass}]\).
Calculate the magnitude of the physical quantities in the indicated units. Show all of your work. [Do not just look up one single unit conversion factor in a table.] If it is impossible to make the conversions indicated, please state the reasons why you believe this to be the case.
a) Dynamic Viscosity: \(15 \mathrm{~kg} /(\mathrm{m} \cdot \mathrm{s}) = \text{________________ } \mathrm{Pa} \cdot \mathrm{s} = \text{_______________ } \mathrm{slug} / \mathrm{ft} \cdot \mathrm{s} \)
b) Pressure: \(100 \mathrm{~lbf} / \mathrm{in}^{2} = \text{________________ } \mathrm{lbf} / \mathrm{ft}^{2} = \text{________________ } \mathrm{bar}\)
c) Energy per unit mass: \(2000 \mathrm{~ft} \cdot \mathrm{lbf} / \mathrm{slug} = \text{______________ } \mathrm{ft}^{2} / \mathrm{s}^{2} = \text{______________ } \mathrm{kJ} / \mathrm{kg} \)
d) Product of pressure and specific volume:
\(3000 \mathrm{~bar} \cdot \mathrm{m}^{3} / \mathrm{kg} = \text{______________ } \mathrm{kJ} / \mathrm{kg} = \text{________________ } \mathrm{Btu} / \mathrm{s}\)
The following unit equivalence factors for moment of inertia were copied from a standard controls textbook: \[1 \mathrm{~lb} \cdot \mathrm{in} \cdot \mathrm{s}^{2}=386 \mathrm{~lb} \cdot \mathrm{in}^{2} \quad \text{ and } \quad 1 \mathrm{~g} \cdot \mathrm{cm} \cdot \mathrm{s}^{2}=980
\mathrm{~g} \cdot \mathrm{cm}^{2} \nonumber \]
A quick examination seems to indicate some strange results that are dimensionally inconsistent, e.g. \(1 \mathrm{~in} \cdot \mathrm{s}^{2} = 386 \(\mathrm{~in}^{2}\) and \(1 \mathrm{~cm} \cdot \mathrm{s}^{2}=980 \mathrm{~cm}^{2}\). How can this be? What if the author had distinguished between \(\mathrm{lbf}\) (pound-force) and \(\mathrm{lbm}\) (pound-mass) and between \(\mathrm{g}_{\mathrm{f}}\) (gram-force) and \(\mathrm{g}_{\mathrm{m}}\) (gram-mass)? Would that make the expressions above dimensionally correct? Please explain.
A moon landing craft has a mass of \(5000 \mathrm{~lbm}\) on the surface of the earth.
(a) Determine the following information when the object is on the surface of the earth:
- weight of the object in \(\mathrm{lbf}\)
- mass of the object mass in slugs
(b) Determine the following information for the object when it sits on the surface of the moon if the strength of gravity on the moon is \(1 / 6\) of the value on earth:
- weight of the object in \(\mathrm{lbf}\)
- mass of the object in \(\mathrm{lbm}\) and slugs
(c) An infrequently used unit of force, the poundal, is defined by the following expression: \[1 \text { poundal } = 1 \mathrm{~lbm} \cdot \mathrm{ft} / \mathrm{s}^{2} \nonumber \] Determine the weight of the object in poundals when it is on the surface of the earth and the surface of the moon.
A moon landing craft has a mass of \(5000 \mathrm{~kg}\) on the surface of the earth.
(a) Determine the following information when the object is on the surface of the earth:
- weight of the object in newtons
- mass of the object mass in kilograms.
(b) Determine the following information for the object when it sits on the surface of the moon if the strength of gravity on the moon is \(1 / 6\) of the value on earth:
- weight of the object in newtons
- mass of the object in kilograms.
(c) Although it is not a standard unit of measure, you will sometimes find forces expressed in terms of kilogram-force, e.g. \(1 \mathrm{~kgf}=9.81 \mathrm{~kg} \cdot\mathrm{m} / \mathrm{s}^{2}\). Determine the weight of the object in \(\mathrm{kgf}\) when it is on the surface of the earth and the surface of the moon.
Engineers often define units which make their life easier and simplify calculations. Calculate the magnitude of the physical quantities in the indicated units. Consult a good engineering handbook to find the units of terms with which you are unfamiliar. Show all of your work. [Do not just look up one single unit conversion factor in a table.] If it is impossible to make the conversions indicated, please state the reasons why you believe this to be the case.
(a) Unit of mass: \(1 \mathrm{~blob} = 1 \mathrm{~lbf} / \left(\mathrm{in} / \mathrm{s}^{2} \right)\)
\(100 \mathrm{~blobs} = \text{_______________ } \mathrm{lbm} = \text{_______________ } \mathrm{kg}\)
(b) Units of volume: \(\text{acre-foot}\)
\(10 \text{ acre-foot} = \text{_______________ } \mathrm{ft}^{3} = \text{_______________ } \mathrm{gal}\)
(c) Unit of area: \(\text{circular mils}\)
\(1000 \text{ circular mils} = \text{_______________ } \mathrm{in}^{2} = \text{_______________ } \mathrm{mm}^{2}\)
(d) Unit of electrical resistivity: \(\mathrm{microhms} \text{-} \mathrm{cm}\)
\(1000 \text{ microhms-cm} = \text{_______________ } \text{ohms-in} = \text{_______________ } \text{volt-cm} / \mathrm{amp}\)
(e) Unit of electrical inductance: \(1 \mathrm{~henry} = 1 \text{volt-s} / \mathrm{amp}\)
\(100 \mathrm{~henrys} = \text{_________________ } \mathrm{joule} / \mathrm{amp}^{2} = \text{_________________ } \text{hp-s}^{3} / \mathrm{coulomb}^{2}\)
Units of energy and power frequently occur in many different forms. Calculate the magnitude of the physical quantities in the indicated units. Show all of your work. [Do not just look up one single unit conversion factor in a table.] If it is impossible to make the conversions indicated, please state the reasons why you believe this to be the case.
(a) \(100 \mathrm{~hp} = \text{_______________ } \mathrm{kW} = \text{_______________ } \mathrm{J} / \mathrm{s}\)
(b) \(100 \text{ kW-h} = \text{_______________ } \text{hp-s} = \text{_______________ } \mathrm{J}\)
(c) \(1000 \text{ lbf-in}^{3} = \text{_______________ bar-cm}^{2} = \text{_______________ ft-lbf}\)
(d) \(1000 \mathrm{~J} = \text{_______________ N-m} = \text{_______________ Btu-h}\)
(e) \(1000 \mathrm{~Btu} = \text{_______________ J-ft} = \text{_______________ hp-h}\)