Conversion of Units

Mistakes are the usual bridge between inexperience and wisdom. Phyllis Theroux, Night Lights (Viking Penguin)

As an example of a serious unit conversion error, in 1999 the Mars Climate Orbiter was lost because engineers failed to make a simple conversion from English units to SI, an embarrassing lapse that sent the $125 million craft fatally close to the Martian surface, destroying it and causing the mission to be a failure.

As an engineer you must be able to handle all kinds of units and be able to convert a given set of units to another with ease. As you probably already know, the procedure for converting one set of units to another is simply to multiply any number and its associated units by ratios termed conversion factors to arrive at the desired answer with its associated units. Conversion factors are statements of

equivalent values of different units in the form of ratios. Note that because conversion factors are composed of equivalents between units, multiplying a quantity by one or more conversion factors does not actually change the basic quantity, only its numerical value and its units. Table 2.4

demonstrates how several equivalents between units can be converted into ratios that are deemed conversion factors.

Table 2.4. Examples of Conversion Factors

On the inside of the front cover of this book you will find tables listing commonly used conversion factors. To obtain unit conversion factors from these tables, use the following procedure: (1) Locate the name of the current unit in the row on the left-hand side of the table. (2) Locate the name of the desired unit in the column at the top of the table. (3) The number in the box that is at the intersection of the row and column of the table is the value of the conversion factor, that is, is equal to the value of the current unit divided by the desired unit. For example, look at the first table on the inside cover: “Volume Equivalents.” To find the conversion factor to convert U.S. gallons to liters (L), locate “liters” on the top row and “U.S. gal” in the left column. At the

intersection of these two units is a box that contains the number 3.785, which means that the conversion factor is U.S. gal/L = 3.785. What is the conversion factor to convert liters to U.S. gallons? Is the numerical value of 0.2642 correct? You will find engineering handbooks as well as certain Web sites to be good sources of tables of conversion factors. We recommend that you memorize a few of the most common conversion factors to avoid looking them up in tables.

In this book, to help you follow the calculations and emphasize the use of units, we frequently make use of a special format in the calculations, as shown below, as a substitute for parentheses, times signs, and so on.

Consider the following problem:

If a plane travels at the speed of sound (assume that the speed of sound is 1100 ft/s), how fast is it going in miles per hour? First, let’s convert feet (ft) to miles (mi). From the table “Linear Measure Equivalents,” 1 mile is equal to 5280 ft. Therefore, using this conversion factor to convert feet to miles yields

Note that when the conversion factor is applied correctly, the units of feet are eliminated from the problem, which is indicated by the slashes through “ft.” Note the format of these calculations. We have set up the calculations with vertical lines separating each ratio. We will use this formulation frequently in this text to enable you to see clearly how units are handled in each case.

Next, let’s convert from seconds to hours to obtain an answer with the desired units. You know that there are 60 s in a minute and 60 min in an hour; therefore, 1 hr is equal to 3600 s. Thus

This unit conversion problem can be more efficiently implemented by applying both unit conversions in one step:

We recommend that you always write down the units next to the associated numerical value to ensure accurate unit conversions. By striking out units that cancel at any stage in the conversion, you can determine the consolidated net units and see what conversions are still required. In this manner, you can reliably perform very complex unit conversions by checking to ensure that all the units have been correctly converted.

Consistent use of units in your calculations throughout your professional career will assist you in avoiding silly mistakes such as converting 10 cm to inches by multiplying by 2.54:

By three methods we may learn wisdom: First, by reflection, which is noblest; second, by imitation, which is easiest; and third, by experience, which is the bitterest.

Confucius

on the CD in the back of this book you can insert almost any units you want in order to retrieve property values. Nevertheless, being able to make conversions yourself is important.

Now let’s look at some examples of using conversion factors. Example 2.1. Use of Conversion Factors

Change 400 in3/day to cm3/min. Solution

In this example note that not only are the numbers raised to a power, but the units also are raised to the same power.

Example 2.2. Nanotechnology

Nanosize materials have become the subject of intensive investigation in the last decade because of their potential use in semiconductors, drugs, protein detectors, and electron transport.

Nanotechnology is the generic term that refers to the synthesis and application of such small particles. An example of a semiconductor is ZnS with a particle diameter of 1.8 nm. Convert this value to (a) decimeters (dm) and (b) inches (in.).

Solution

a.

b.

Example 2.3. Conversion of Units Associated with Biological Materials

In biological systems, enzymes are used to accelerate the rates of certain biological reactions. Glucoamylase is an enzyme that aids in the conversion of starch to glucose (a sugar that cells use for energy). Experiments show that 1 μg mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 μg mol/(mL)(min). Determine the production rate of glucose for this system in units of lb mol/(ft3)(day).

Solution

The production rate of glucose is stated in the problem as 0.6 μg mol/(mL)(min). Therefore, to solve this problem, you just have to convert this quantity into the specified units:

In the AE system the conversion of terms involving pound mass and pound force deserve special attention. Let us start the discussion with Newton’s law, which states that force (F) is proportional to the product of mass (m) and acceleration (a), that is,

where C is a constant whose numerical values and units depend on the units selected for F, m, and a. In the SI system, the unit of force is defined to be the newton (N), which corresponds to 1 kg

accelerated at 1 m/s2. Therefore, the conversion factor C = 1 N/(kg)(m)/s2 results so that the force is expressed in newtons (N):

Note that in this case C has the numerical value of 1; hence the conversion factor seems simple, even nonexistent, and the units are usually ignored.

In the American Engineering system an analogous conversion factor is required. In the AE system, one pound force (1 lb_f) corresponds to the action of the Earth’s gravitational field on one pound mass (1 lb_m):

where g is the acceleration of gravity, which at 45° latitude and sea level has the following value:

A numerical value of 1/32.174 has been chosen for the value of C_AE to maintain the numerical

correspondence between lb_f and lb_m on the surface of the Earth. The acceleration caused by gravity, you may recall, varies by a few tenths of 1% from place to place on the surface of the Earth but, of course, is quite different on the surface of the moon.

The inverse of this conversion factor C_AE is given the special symbol g_c:

a conversion factor that you will see included in some texts to remind you that the numerical value of the conversion factor in the AE system is not unity. To avoid confusion, we will not place the symbol g_c in the equations in this book because we will be using both SI and AE units.

To sum up, you can see that the American Engineering system has the convenience that the numerical value of a pound mass is also that of a pound force if the numerical value of the ratio g/g_c is equal to 1, as it is approximately on the surface of the Earth. No one should get confused by the fact that a person 6 feet tall has only two feet. In this book, we will not subscript the symbol lb with m (for mass) unless it becomes essential to do so to avoid confusion. We will always mean by the unit lb without a subscript the quantity pound mass. But never forget that the pound (mass) and pound (force) are not the same units in the American Engineering system even though we speak of pounds to express force, weight, or mass.

What is the difference between mass and weight? Is weight ever expressed in grams or kilograms? The weight of a mass is the value of an external force required to maintain the mass at rest in its frame of reference, the Earth usually. People often refer to astronauts in a space station as weightless. They are with respect to the frame of reference of the space station because the centrifugal force present essentially is equivalent to the gravitational force. With respect to the Earth as the frame of reference, they are far from weightless (if they were, how would they get home?). Thus, when an engineer says a drum (on the Earth’s surface) weighs 100 lb or 39.4 kg, you can interpret this

statement to mean the drum has a mass of 100 lb_m or 39.4 kg. That is, a weight of 39.4 kg would exert a downward force of 915 N (i.e., 39.4 kg × 9.8 m/s) at the surface of the Earth.

Example 2.4. A Conversion Involving Both lb_m and lb_f

What is the potential energy in (ft)(lb_f) of a 100 lb drum hanging 10 ft above the surface of the Earth with reference to the surface of the Earth?

Solution

The first thing to do is read the problem carefully. What are the unknown quantities? The potential energy (PE) is unknown. What are the known quantities? The mass and the height of the drum are known. How are they related? You have to look up the relation unless you recall it from physics:

Potential energy = PE = mgh

The 100 lb means 100 lb mass; let g = acceleration of gravity = 32.2 ft/s2. Figure E2.4 is a sketch of the system.

Figure E2.4

Now substitute the numerical values of the variables into the equation and perform the necessary unit conversions.

are essentially equal. A good many engineers would solve the problem by saying that 100 lb × 10 ft = 1000 (ft)(lb) without realizing that in effect they are canceling out the numbers in the g/g_c ratio, and that the lb in the solution means lb_f.

Some Trivia Concerning Unit Conversions

A U.S. frequent flier mile is not the same as a U.S. mile—the former is a nautical mile (1.85 km) whereas the latter is 1.61 km. In the AE system 1 m = 39.37 in., whereas for U.S. land survey applications it is 2 × 10–6 inches shorter.

Self-Assessment Test

Questions

1. What is g_c?