1.1 – Unit Conversions

1.1.0 – Learning Objectives

By the end of this section you should be able to:

Understand the importance of unit conversion.
Understand the concept of a conversion factor.
Convert a unit from one to the other.

1.1.1 – Introduction

Knowing how to convert units is vital to engineers. Improperly converted units could mean the difference between life and death. With Canada's proximity to the USA, it is very important to learn how to convert from imperial to metric and vise versa. It is also important to know how to convert from one unit to another in the same system such as $m^{3}$ to $l$ .

1.1.2 – Conversion Factors

Conversion factors are ratios between two expressions of the same quantity. For an example:

$\frac{1 c m}{10 m m} (1 centimeter per 10 millimeters)$

or

$\frac{1 k g^{2}}{1, 000, 000 g^{2}} (1 kilogram squared per 1,000,000 grams squared)$

or

$\frac{1 m^{3}}{1, 000, 000, 000 m m^{3}} (1 cubic meter per 1,000,000,000 cubic millimeters)$

1.1.3 – Converting Units

To convert a quantity from one unit to its equivalent of another unit, we will use conversion factors. For example, if we want to convert 7.2 km to meters, we would write:

$7.2 k m \times \frac{1, 000 m}{1 k m} = 7, 200 m$

As you can see, the original unit is km, the final unit is m, and the conversion factor is $\frac{1, 000 m}{1 k m}$ .

1.1.4 – Dimensional Homogeneity

Consider the equation

$u = u_{0} + g t$

where the corresponding units are

$[m / s] = [m / s] + [m / s^{2}] \times [s]$

Now, lets input values for each of the variables in the first equation.

$45 \frac{m}{s} \neq 12 \frac{m}{s} + 9.81 \frac{m}{s^{2}} \cdot 3.36 m i n$

You may ask why $45 \neq 45$ and that is because the dimensions are not homogenous. The units are not consistent. The velocity and acceleration are measured in seconds, yet the time we used is minutes. To correct the formula, we must change the minutes to seconds.

$45 \frac{m}{s} \neq 12 \frac{m}{s} + 9.81 \frac{m}{s^{2}} \cdot 3.36 s$

1.1.5 – Problem Statement

Problem 1

Question

Suppose the average house in Vancouver uses 342 gallons of water per day. How many $c m^{3}$ of water per second is used on average? What are your conversion factors?

Answer

The conversion factors are: $\frac{86400 s}{1 d a y}$

$\frac{3.78541 l i t r e}{1 g a l l o n}$

$\frac{0.001 m^{3}}{1 l i t r e}$

$\frac{1, 000, 000 c m^{3}}{1 m^{3}}$

Written out, this would look like:

$\frac{342 g a l l o n s}{1 d a y} \cdot \frac{1 d a y}{86400 s} \cdot \frac{3.78541 l i t r e}{1 g a l l o n} \cdot \frac{0.001 m^{3}}{1 l i t r e} \cdot \frac{1, 000, 000 c m^{3}}{1 m^{3}} = 14.983914583 \frac{c m^{3}}{s}$

Problem 2

Question

Despite the odds, the Kinder-Morgan pipeline is being built to no environmental impact, which will soon supply BC and Washington state with 40,001 barrels of oil biweekly (including weekends). As an employee of Dusty Oil in Vancouver, you keep in contact with your colleagues in the United States and update them on the amount of oil being supplied. Since they use metric and you use the imperial system, what is the amount of oil transported, in metric?

Note: Use 1 oil barrel = 35 imperial gallons. Not US gallons.

Answer

According to the back of our text,

$1000 L = 220.83 imperial gallons$

Let's find out how many litres are in an imperial gallon first:

$1000 L \div 220.83 imperial gallons = 4.5283 \frac{L}{imperial gallons}$

Now we know that biweekly, the total amount of oil transported is:

$40, 001 b a r r e l s \times 35 35 \frac{imperial gallons}{b a r r e l s} \times 4.5283 \frac{L}{imperial gallons} = 6339778 l i t r e s$

Let's find how many litres are transported in seconds. Since I don't know how many seconds are in a fortnight off the top of my head:

$14 d a y s \times 24 \frac{h o u r s}{d a y} \times 3600 \frac{s e c o n d s}{h o u r} = 1209600 s e c o n d s$

Finally, you can report that the total amount of oil transported, in metric units, is:

$6339778 l i t r e s \div 1209600 s e c o n d s = 5.241 \frac{L}{s}$