Today it is very important to be able to understand measurements in both the English system of units and the SI (Systems International) system of units. The SI system of units (often referred to as the metric system) is the official system for measurements in nearly all countries of the world, but not in the U.S. As our society and economy become more global in scope, it is increasingly important to fully comprehend both the English and the SI systems of units and to accurately and conveniently convert from one system to another. World-wide communications, commerce, finance and many other functions depend upon everyone involved in the processes to be knowledgeable of both systems of measurements. There have been some very costly failures associated with insufficient attention being given to the units of measurements involved in calculations and communications.
A methodical approach is needed to avoid errors encountered when converting from one system of units to another. Unit conversion multipliers are presented in this fact sheet along with several examples to describe the use of these multipliers. The conversion factors presented are for general measurements as well as those primarily associated with energy calculations; additional conversion multipliers can be found in the the references. This fact sheet is intended to serve as a reference for converting many different types of measurements quickly and accurately from one system of units to another. The fact sheet is intended for use in the classroom as well as for extension outreach programs.
Table 1 presents the conversion multiplication factors to convert from English units to SI units (and vice versa) for basic measurements including length, weight, area, volume, bulk density, mass flow rate, volume flow rate, pressure and power. Additional conversion factors are presented for energy, energy density, and energy yields. SI prefixes indicating orders of magnitude are provided in Table 2. Use of these prefixes avoids the need for powers of 10 notations.
Table 1. Conversion Multipliers
|ounce||pound||kgram||ton (short)||ton (long)||tonne (metric)|
|sq in.||sq ft||acre||sq m||hectare|
|cu in.||cu ft||gallon||bushel||liter||cu m|
|pound/cu in.||pound/cu ft||pound/cu yd||pound/gallon||g/cu cm||kg/cu m|
|cu ft/sec||cu ft/min (cfm)||gallons/hr||cu meters/sec||liters/sec|
|psi||in. water||in. Hg||mm Hg||atmospheres||Pascals|
|BTU/sq ft||BTU/ac||kJ/sq m||KJ/ha|
Table 2. SI Prefixes
No single conversion factor can be used to convert temperatures in the English system (degrees Fahrenheit or degrees Rankine) to temperatures in the SI system (degrees Celsius and degrees Kelvin) or the other way around. The equations for converting temperatures and temperature difference from one system to another are presented in Table 3.
Table 3. Equations for Temperature Conversions
F = (9/5) x C + 32
C = (5/9) x (F-32)
R = F + 460
K = C + 273
Δt (F) = Δt (C) x 1.8
Δt (C) = Δ (F) x 0.556
F = degrees Fahrenheit; C = degrees Celcius; R = degrees Rankine; K = degrees Kelvin; Δt = temperature difference
Examples Illustrating the Use of the Conversion Factors
A field is 160 acres and you want to express that area in hectares. Find the Area category in Table 1. Since the known unit is "Acres", find the numeral "1" under the acre column. Follow that line to the right to find the corresponding multiplication factor of 0.4047 under the hectare column. The area of the 160 acre field in hectares is 64.8 hectares (160 x 0.4047). For another exercise involving area, calculate how many sq ft (square feet) are in a plot that is 0.6 hectare. Find the numeral "1" under hectare column and read the multiplier of 107,639 on the same line to the left under sq ft column. Then calculate the number of square feet as 64,583 (0.60 x 107,639).
Energy content of buckwheat is 15.5 MJ/kg (Mega Joules per kilogram). What is the energy density in units of BTU/lb? Go to the category of Energy Density in Table 1 and find the column for MJ/kg. Locate the numeral "1" under that column and then read to the left to get the multiplication factor of 429.922 under the BTU/pound column. The energy density of the buckwheat as 6,664 BTU/lb (15.5 x 429.922).
An electric resistance heating system rated at 25 kW is to be replaced with a propane furnace. What BTU/Hr rating of propane furnace is needed for the same heating capacity? Under the Power category in Table 1, locate the numeral "1" under the kW column. Then under the BTU/Hr column, find the multiplier of 3,412. The BTU/Hr rating of the propane furnace to be equivalent to the electric heating system is 85,300 BTU/Hr (25 x 3,412).
A utility company decides to promote distributed generation by building 1,500 ministations with each powered by a 1,200 hp diesel engine. What is the total capacity of the 1,500 stations? First, convert the 1,200 hp to kW using the factor of 0.7457 (from Power category), yielding 895 kW (1,200 x 0.7457). The power that can be generated by 1,500 plants is 1,340,000 kW (1,500 x 895). To avoid use of trailing 0's or powers of 10, use the SI prefixes in Table 2. The answer can be expressed as 1,340 MW or 1.34 GW.
The temperature on a very cold day in Moscow in February is -27 C. Calculate the corresponding temperature in degrees F. Using Table 3, find the equation to change from degrees C to degrees F. So a temperature of -27 C is -16.6 F (-27 C x 9/5 + 32), or rounded off to -17 F.
On a winter day in State College, the outside air temperature is 20 F and the temperature inside a building is 70 F. So the temperature difference is 50 F. Calculate the temperature difference in degrees C. First of all, the temperature difference is not 10 C [5/9 x (50 - 32)]. Actually, this is a very common mistake. Referring again to Table 3, temperature difference in degrees Celsius is calculated to be 27.8 C (50 F x 5/9). You can verify this by converting the inside temperature of 70 F to 21.1 C and the outside temperature of 20 F to -6.7 C. The difference is 27.8 C.
Practical calculations often involve more than one step. Consider the situation of a material imported from France with a bulk density of 1.33 g/cu cm (grams per cubic centimeter). What is the equivalent bulk density in tons/cu yard (tons per cubic yard)? In the category of Bulk Density in Table 1, find the numeral "1" under the g/cu cm column. Reading to the left on that line, get the multiplication factor of 1685.564 under the pounds/cu yd column. So a bulk density of 1.33 g/cu cm is equivalent to 2,241.8 pounds/cu yd (1.33 x 1685.564). But we want to know how many tons per cubic yard. So now proceed to the Weight category and find the numeral "1" under the pound column and read to the right to either ton (short) or ton (long). If we want the answer in short tons, the multiplication factor is 5.000E-04. The final answer is that a bulk density of 1.33 g/cu cm is equivalent to 1.12 short tons/cu yd (2241.8 x 5.000E-04).
Careful attention to dimensions cannot be over-emphasized. The consistent use of the conversion factors and equations presented in this fact sheet should practically eliminate any errors associated with converting from the English system of units to the SI system and vice versa. Contemporary global systems for communications, commerce, and finance dictate that all participants are capable of dealing with both systems of units and accurately converting from one system to another.
- ASHRAE Handbook of Fundamentals, Atlanta, GA, 2005.
- Marks' Standard Handbook for Mechanical Engineers, 2006.
Prepared by Dennis E. Buffington, Professor, Agricultural and Biological Engineering