Units and Standards

As with all disciplines, a set of standards has evolved over the years to ensure consistency and avoid confusion. The units of measurement fall into two distinct systems; first, the English system and second, the International system, SI (Systéme International D’Unités) based on the metric system, but there are some differences. The English system has been the standard used in the United States, but the SI system is slowly making inroads, so that students need to be aware of both systems of units and be able to convert units from one system to the other. Confusion can arise over some units such as pound mass and pound weight. The unit for pound mass is the slug (no longer in common use), which is the equivalent of the kilogram in the SI system of units whereas pound weight is a force similar to the newton, which is the unit of force in the SI system. The conversion factor of 1 lb 0.454 kg, which is used to convert mass (weight) between the two systems, is in effect equating 1-lb force to 0.454-kg mass; this being the mass that will produce a force of 4.448 N or a force of 1 lb. Care must be taken not to mix units of the two systems. For consistency some units may have to be converted before they can be used in an equation.Table 1.1 gives a list of the base units used in instrumentation and measurement in the English and SI systems and also the conversion factors, other units are derived from these base units.Example 1.2 How many meters are there in 110 yard?

110 yard 330 ft (330 0.305) m 100.65 m

Example 1.3 What is the equivalent length in inches of 2.5 m?2.5 m (2.5/0.305) ft 8.2 ft 98.4 in

Example 1.4 The weight of an object is 2.5 lb. What is the equivalent force and mass

in the SI system of units?2.5 lb (2.5 4.448) N 11.12 N2.5 lb (2.5 0.454) kg 1.135 kg

Table 1.2 gives a list of some commonly used units in the English and SI systems, conversion between units, and also their relation to the base units. As explained above the lb is used as both the unit of mass and the unit of force.

TABLE 1.1 Basic Units

TABLE 1.2 Units in Common Use in the English and SI System

Hence, the unit for the lb in energy and power is mass, whereas the unit for the lb in pressure is force, where the lb (force) lb (mass) g (force due to gravity).

Example 1.5 What is the pressure equivalent of 18 psi in SI units?1 psi 6.897 kPa18 psi (18 6.897) kPa 124 kPa

Standard prefixes are commonly used for multiple and submultiple quantities to cover the wide range of values used in measurement units. These are given in Table 1.3

TABLE 1.3 Standard Prefixes

Konversi1 gallon (gal) = 231.0 cubic inches (in3) = 4 quarts (qt) = 8 pints (pt) = 128 fuid

ounces (fl. oz.) = 3.7854 liters (l)

Contoh:

1. 40 gallons converted into fluid ounces:

2. v5.5 pints converted into cubic inches:

3. 1170 liters converted into quarts:

4. 5.5 pints converted into cubic feet (our first attempt! ):

Unfortunately, this will not give us the result we seek. Even though in is a

valid unity fraction, it does not completely cancel out the unit of inches. What we

need is a unity fraction relating cubic feet to cubic inches. We can get this, though,

simply by cubing the in unity fraction:

5.5 pints converted into cubic feet (our second attempt! ):

Distributing the third power to the interior terms of the last unity fraction:

Calculating the values of 13 and 123 inside the last unity fraction, then canceling units

and solving:

Conversion formulae for temperature

Conversion factors for distance

1 inch (in) = 2.540000 centimeter (cm)

1 foot (ft) = 12 inches (in)

1 yard (yd) = 3 feet (ft)

1 mile (mi) = 5280 feet (ft)

Conversion factors for volume

1 gallon (gal) = 231.0 cubic inches (in3) = 4 quarts (qt) = 8 pints (pt) = 128 fuid

ounces (fl. oz.) = 3.7854 liters (l)

1 milliliter (ml) = 1 cubic centimeter (cm3)

Conversion factors for velocity

1 mile per hour (mi/h) = 88 feet per minute (ft/m) = 1.46667 feet per second (ft/s) =

1.60934 kilometer per hour (km/h) = 0.44704 meter per second (m/s) = 0.868976

knot (knot - international)

Conversion factors for mass

1 pound (lbm) = 0.45359 kilogram (kg) = 0.031081 slugs

Conversion factors for force

1 pound-force (lbf) = 4.44822 newton (N)

Conversion factors for area

1 acre = 43560 square feet (ft2) = 4840 square yards (yd2) = 4046.86 square meters

(m2)

Conversion factors for pressure (either all gauge or all absolute)

1 pound per square inch (PSI) = 2.03603 inches of mercury (in. Hg) = 27.6807

inches of water (in. W.C.) = 6.894757 kilo-pascals (kPa)

Conversion factors for pressure (absolute pressure units only)

1 atmosphere (Atm) = 14.7 pounds per square inch absolute (PSIA) = 760

millimeters of mercury absolute (mmHgA) = 760 torr (torr) = 1.01325 bar (bar)

Conversion factors for energy or work

1 British thermal unit (Btu – “International Table") = 251.996 calories (cal –

“International Table") = 1055.06 joules (J) = 1055.06 watt-seconds (W-s) =

0.293071 watt-hour (W-hr) = 1.05506 x 1010 ergs (erg) = 778.169 foot-pound-force

(ft-lbf)

Conversion factors for power

1 horsepower (hp - 550 ft-lbf/s) = 745.7 watts (W) = 2544.43 British thermal units

per hour (Btu/hr) = 0.0760181 boiler horsepower (hp - boiler)

Terrestrial constants

Acceleration of gravity at sea level = 9.806650 meters per second per second (m/s2)

= 32.1740 feet per second per second (ft/s2)

Atmospheric pressure = 14.7 pounds per square inch absolute (PSIA) = 760

millimeters of mercury absolute (mmHgA) = 760 torr (torr) = 1.01325 bar (bar)

Properties of water

Freezing point at sea level = 32oF = 0oC

Boiling point at sea level = 212oF = 100oC

Density of water at 4oC = 1000 kg/m3 = 1 g/cm3 = 1 kg/liter = 62.428 lb/ft3 = 1.951

slugs/ft3

Specific heat of water at 14oC = 1.00002 calories/g.oC = 1 BTU/lb.oF = 4.1869

joules/g.oC

Specific heat of ice≈ 0.5 calories/g¢oC

Specific heat of steam ≈ 0.48 calories/g.oC

Absolute viscosity of water at 20oC = 1.0019 centipoise (cp) = 0.0010019 Pascal-

seconds (Pa.s)

Surface tension of water (in contact with air) at 18oC = 73.05 dynes/cm

pH of pure water at 25oC = 7.0 (pH scale = 0 to 14 )

Weight densities of common materials

All density figures approximate for samples at standard temperature and pressure.

Liquids:

Gasoline: = 41 lb/ft3 to 43 lb/ft3

Naphtha, petroleum: = 41.5 lb/ft3

Acetone: = 49.4 lb/ft3

Ethanol (ethyl alcohol): = 49.4 lb/ft3

Methanol (methyl alcohol): = 50.5 lb/ft3

Kerosene: = 51.2 lb/ft3

Toluene: = 54.1 lb/ft3

Benzene: = 56.1 lb/ft3

Olive oil: = 57.3 lb/ft3

Coconut oil: = 57.7 lb/ft3

Linseed oil (boiled):= 58.8 lb/ft3

Castor oil: = 60.5 lb/ft3

Sea water: = 63.99 lb/ft3

Milk: = 64.2 lb/ft3 to 64.6 lb/ft3

Ethylene glycol (ethanediol): = 69.22 lb/ft3

Glycerin: = 78.6 lb/ft3

Mercury: = 849 lb/ft3

Dimensional analysis

An interesting parallel to the “unity fraction" unit conversion technique is something

referred to in physics as dimensional analysis. Performing dimensional analysis on a

physics formula means to set it up with units of measurement in place of variables, to

see how units cancel and combine to form the appropriate unit(s) of measurement for

the result.

For example, let's take the familiar power formula used to calculate power in a

simple DC electric circuit:

P = IV

[Watts] = [Amperes] x [Volts] or [W] = [A][V]

Where,

P = Power (watts)

I = Current (amperes)

V = Voltage (volts)

Pressure Measurement

There are six terms applied to pressure measurements. They are as follows:

Total vacuum–which is zero pressure or lack of pressure, as would be experienced

in outer space.

Vacuum is a pressure measurement made between total vacuum and normal

atmospheric pressure (14.7 psi).

Atmospheric pressure is the pressure on the earth’s surface due to the weight of the

gases in the earth’s atmosphere and is normally expressed at sea level as 14.7 psi or

101.36 kPa. It is however, dependant on atmospheric conditions.

The pressure decreases above sea level and at an elevation of 5000 ft drops to about

12.2 psi (84.122 kPa).

Absolute pressure is the pressure measured with respect to a vacuum and is

expressed in pounds per square inch absolute (psia).

Gauge pressure is the pressure measured with respect to atmospheric pressure and is

normally expressed in pounds per square inch gauge (psig). Figure 5.2a shows

graphically the relation between atmospheric, gauge, and absolute pressures.

Differential pressure is the pressure measured with respect to another pressure and

is expressed as the difference between the two values. This would represent two

points in a pressure or flow system and is referred to as the delta p

PressureIt is very useful to quantify force applied to a fuid in terms of force per unit area,

since the force applied to a °uid becomes evenly dispersed in all directions to the

surface containing it. This is the defnition of pressure (P): how much force (F) is

distributed across how much area (A).

P= FA

In the metric system, the standard unit of pressure is the Pascal (Pa), defined as one

Newton (N) of force per square meter (m2) of area. In the English system of

measurement, the standard unit of pressure is the PSI : pounds (lb) of force per

square inch (in2) of area. Pressure is often expressed in units of kilo-pascals (kPa)

when metric units are used because one pascal is a rather low pressure in most

engineering applications.

Figure 5.2 Illustration of (a) gauge pressure versus absolute pressure and (b) delta or differential pressure.

TABLE 5.2 Pressure Conversions

or ∆p. Figure 5.2b shows two situations, where differential pressure exists across a

barrier and between two points in a flow system.

Example 5.2 The atmospheric pressure is 14.5 psi. If a pressure gauge reads 1200

psf, what is the absolute pressure?

A number of measurement units are used for pressure. They are as follows:

1. Pounds per square foot (psf) or pounds per square inch (psi)

2. Atmospheres (atm)

3. Pascals (N/m2) or kilopascal (1000Pa)*

4. Torr = 1 mm mercury

5. Bar (1.013 atm) = 100 kPa

Table 5.2 gives a table of conversions between various pressure measurement

units.

Example 5.3 What pressure in pascals corresponds to 15 psi?

p = 15 psi (6.895 kPa/psi) = 102.9 kPa

The mathematical relationship between vertical liquid height and hydrostatic

pressure is quite simple, and may be expressed by either of the following formulae:

P = gh

P = h

[ lbft2 ]=[ lb

ft3 ][ ft1 ]

Where,

P = Hydrostatic pressure in units of weight per square area unit: Pascals (N/m2) or

lb/ft2

= Mass density of liquid in kilograms per cubic meter (metric) or slugs per cubic

foot (British)

g = Acceleration of gravity (9.8 meters per second squared or 32 feet per second

squared)

= Weight density of liquid in newtons per cubic meter (metric) or pounds per cubic

foot (British)

h = Vertical height of liquid column

Applying this to a realistic problem, consider the case of a tank fllled with 8 feet

(vertical) of castor oil, having a weight density of 60.5 pounds per cubic foot. This is

how we would set up the formula to calculate for hydrostatic pressure at the bottom

of the tank:

P=( 60,5lbft3 ) (8 ft )

P=484 lbft2

If we wished to convert this result into a more common unit such as PSI (pounds per

square inch), we could do so using an appropriate fraction of conversion units:

P=( 484 lbft2 )( 1 ft2

144 ¿2 )

p=3,36 lb¿2 =3,36PSI

Fluid density expressions

Fluid density is commonly expressed as a ratio in comparison to pure water at

standard temperature. This ratio is known as specific gravity. For example, the

specific gravity of glycerin may be determined by dividing the density of glycerin by

the density of water:

Specific gravity of any liquid=Dliquid

Dwater

Specific gravity of glycerin=Dglycerin

Dwater=

78,6 l bft3

62,4 lbft3

=1,26

Systems of pressure measurement

Pressure measurement is often a relative thing. What we mean when we say there is

35 PSI of air pressure in an in°ated car tire is that the pressure inside the tire is 35

pounds per square inch greater than the surrounding, ambient air pressure. It is a fact

that we live and breathe in a pressurized environment. Just as a vertical column of

liquid generates a hydrostatic pressure, so does a vertical column of gas. If the

column of gas is very tall, the pressure generated by it will be substantial enough to

measure. Such is the case with Earth's atmosphere, the pressure at sea level caused

by the weight of the atmosphere is approximately 14.7 PSI.

You and I do not perceive this constant air pressure around us because the pressure

inside our bodies is equal to the pressure outside our bodies. Thus our skin, which

serves as a differential pressure-sensing diaphragm, detects no difference of pressure

between the inside and outside of our bodies. The only time the Earth's air pressure

becomes perceptible to us is if we rapidly ascend or descend in a vehicle, where the

pressure inside our bodies does not have time to equalize with the pressure outside,

and we feel the force of that di®erential pressure on our eardrums.

If we wish to speak of a fluid pressure in terms of how it compares to a perfect

vacuum (absolute zero pressure), we specify it in terms of absolute units. For

example, when I said earlier that the atmospheric pressure at sea level was 14.7 PSI,

what I really meant is that it is 14.7 PSIA (pounds per square inch absolute),

meaning 14.7 pounds per square inch greater than a perfect vacuum.

When I said earlier that the air pressure inside an inflated car tire was 35 PSI, what I

really meant is that it was 35 PSIG (pounds per square inch gauge), meaning 35

pounds per square inch greater than ambient air pressure. When units of pressure

measurement are specified without a “G" or “A" suffix, it is usually (but not

always!) assumed that gauge pressure (relative to ambient pressure) is meant.

This offset of 14.7 PSI between absolute and gauge pressures can be confusing if we

must convert between different pressure units. Suppose we wished to express the tire

pressure of 35 PSIG in units of inches of water column ("W.C.). If we stay in the

gauge-pressure scale, all we have to do is multiply by 27.68:

35 PSI1

×27,68 W.C} over {1 PSI} =968.8W .C ¿

If, however, we wished to express the car's tire pressure in terms of inches of water

column absolute (in reference to a perfect vacuum), we would have to include the

14.7 PSI offset in our calculation, and do the conversion in two steps:

35 PSIG + 14:7 PSI = 49:7 PSIA

49,7 PSIA1

×27,68 W.C.A} over {1 PSIA} =1375,7 W . C . A ¿

There are some pressure units that are always in absolute terms. One is the unit of

atmospheres, 1 atmosphere being 14.7 PSIA. There is no such thing as \atmospheres

gauge" pressure. For example, if we were given a pressure as being 4.5 atmospheres

and we wanted to convert that into pounds per square inch gauge (PSIG), the

conversion would be a two-step process:

4,5 atm1

× 14,7 PSIA1 atm

=66,15 PSIA

66,15 PSIA – 14,7 PSI = 51,45 PSIG

Another unit of pressure measurement that is always absolute is the torr, equal to 1

millimeter of mercury column absolute (mmHgA). 0 torr is absolute zero, equal to 0

atmospheres, 0 PSIA, or -14.7 PSIG. Atmospheric pressure at sea level is 760 torr,

equal to 1 atmosphere, 14.7 PSIA, or 0 PSIG.

If we wished to convert the car tire's pressure of 35 PSIG into torr, we would once

again have to offset the initial value to get everything into absolute terms.

35 PSIG + 14,7 PSI = 49,7 PSIA

( 49,7 PSIA1 ) x 760 torr

14,7 PSIA=2569,5 torr

Relating 4 to 20 mA signals to instrument variablesCalculating the equivalent milliamp value for any given percentage of signal range is

quite easy. Given the linear relationship between signal percentage and milliamps,

the equation takes the form of the standard slope-intercept line equation y = mx + b.

Here, y is the equivalent current in milliamps, x is the desired percentage of signal,

m is the span of the 4-20 mA range (16 mA), and b is the offset value, or the \live

zero" of 4 mA:

Current = (16 ma )( x100 %)+ (4 mA )

This equation form is identical to the one used to calculate pneumatic instrument

signal pressures (the 3 to 15 PSI standard):

pressure=(12 PSI )( x100 % )+(3PSI )

The same mathematical relationship holds for any linear measurement range. Given

a percentage of range x, the measured variable is equal to:

measured variable=(span )( x100 % )+ ( LRV )

Some practical examples of calculations between milliamp current values and

process variable values follow:

1. Example calculation: controller output to valve

An electronic loop controller outputs a signal of 8.55 mA to a direct-responding

control valve (wherem4 mA is shut and 20 mA is wide open). How far open should

the control valve be at this MV signal level?

We must convert the milliamp signal value into a percentage of valve travel. This

means determining the percentage value of the 8.55 mA signal on the 4-20 mA

range. First, we need to manipulate the percentage-milliamp formula to solve for

percentage (x):

(16 mA )( x100 % )+(4 mA )=current

(16 mA )( x100 % )=current− (4mA )

x100 %

=current−(4 mA )

16 mA

x=( current−(4mA )16 mA )× 100 %

Next, we plug in the 8.55 mA signal value and solve for x:

x=( 8,55mA−( 4mA )16 mA )100%

x=28,4 %

Therefore, the control valve should be 28.4 % open when the MV signal is at a value

of 8.55 mA.

2. Example calculation: flow transmitter

A flow transmitter is ranged 0 to 350 gallons per minute, 4-20 mA output, direct-

responding.

Calculate the current signal value at a flow rate of 204 GPM.

First, we convert the flow value of 204 GPM into a percentage of range. This is a

simple matter of division, since the °ow measurement range is zero-based:

204 GPM350 GPM

=0,583=68,3 %

Next, we take this percentage value and translate it into a milliamp value using the

formula previously shown:

(16 ma )( x100 %)+ (4mA )=current

(16 mA )( 58,3 %100 % )+(4 mA )=13,3 mA

Therefore, the transmitter should output a PV signal of 13.3 mA at a °ow rate of 204

GPM.

3. Example calculation: temperature transmitter

A pneumatic temperature transmitter is ranged 50 to 140 degrees Fahrenheit and has

a 3-15 PSI output signal. Calculate the pneumatic output pressure if the temperature

is 79 degrees Fahrenheit.

First, we convert the temperature value of 79 degrees into a percentage of range

based on the knowledge of the temperature range span (140 degrees ¡ 50 degrees =

90 degrees) and lower-range value (LRV = 50 degrees). We may do so by

manipulating the general formula for any linear measurement to solve for x:

measured variable=(Span )( x100 % )+( LRV )

measured variable−( LRV )= (Span )( x100 % )

measured variable−( LRV )Span

= x100 %

x=( measured variable−( LRV )Span )100%

x=(79O F−500 F900 F )100 %

x=32,2 %

Next, we take this percentage value and translate it into a pneumatic pressure value

using the formula previously shown:

(12 PSI )( x100 % )+ (3 PSI )=pressure

(12 PSI )( 32,2 %100 % )+(3 psi )=6,87 PSI

Therefore, the transmitter should output a PV signal of 6.87 PSI at a temperature of

79o F.

4. Example calculation: pH transmitter

A pH transmitter has a calibrated range of 4 pH to 10 pH, with a 4-20 mA output

signal. Calculate the pH sensed by the transmitter if its output signal is 11.3 mA.

First, we must convert the milliamp value into a percentage. Following the same

technique we used for the control valve problem:

( current−(4 mA )16 mA )100 %=persenof range

( 11,3mA−(4 mA )(16 mA ) )100 %=0,456=45,6 %

Next, we take this percentage value and translate it into a pH value, given the

transmitter'smeasurement span of 6 pH (10 pH - 4 pH)and offset of 4 pH:

(10 pH )( x100 % )+( 4 mA )=pH value

(10 pH )( 45,6 %100 % )+(4 pH )=8,56 pH

Therefore, the transmitter's 11.3 mA output signal re°ects a measured pH value of

8.56 pH.

5. Example calculation: reverse-acting I/P transducer signal

A current-to-pressure transducer is used to convert a 4-20 mA electronic signal into a

3-15 PSI pneumatic signal. This particular transducer is configured for reverse action

instead of direct, meaning that its pressure output at 4 mA should be 15 PSI and its

pressure output at 20 mA should be 3 PSI. Calculate the necessary current signal

value to produce an output pressure of 12.7 PSI.

Reverse-acting instruments are still linear, and therefore still follow the slope-

intercept line formula y = mx + b. The only differences are a negative slope and a

different intercept value.

Instead of y = 16x + 4 as is the case for direct-acting instruments, this reverse-acting

instrument follows the linear equation y = -16x + 20:

(−16 mA )( x100 % )+(20 mA )=current

First, we need to to convert the pressure signal value of 12.7 PSI into a percentage of

3-15 PSI range. We will manipulate the percentage-pressure formula to solve for x:

(12 PSI )( x100 % )+ (3 PSI )=pressure

(12 PSI )( x100 % )=preeure−(3 PSI )

( x100 % )= pressure−(3 PSI )

(12 PSI )

x=( pressure−(3 PSI )(12 PSI ) )100 %

x=(12,7 PSI−(3 PSI )(12 PSI ) )100 %

x=80.8 %

This tells us that 12.7 PSI represents 80.8 % of the 3-15 PSI signal range. Plugging

this percentage value into our modified (negative-slope) percentage-current formula

will tell us how much current is necessary to generate this 12.7 PSI pneumatic

output:

(−16 mA )( x100 % )+(20 MA )=current

(−16 mA )( 80,8 %100 % )+(20 mA )=7,07 mA

Therefore, a current signal of 7.07 mA is necessary to drive the output of this

reverse-acting I/P transducer to a pressure of 12.7 PSI.

6. Graphical interpretation of signal ranges

A helpful illustration for students in understanding analog signal ranges is to

consider the signal range to be expressed as a length on a number line. For example,

the common 4-20 mA analog current signal range would appear as such:

If one were to ask the percentage corresponding to a 14.4 mA signal on a 4-20 mA

range, it would be as simple as determining the length of a line segment stretching

from the 4 mA mark to the 14.4 mA mark:

As a percentage, this thick line is 10.4 mA long (the distance between 14.4 mA and 4

mA) over a total (possible) length of 16 mA (the total span between 20 mA and 4

mA). Thus:

Percentage=( 14,4 mA−4 MA40 mA−4 mA )100 %

Percentage=65 %

This same “number line" approach may be used to visualize any conversion from

one analog scale to another. Consider the case of an electronic pressure transmitter

calibrated to a pressure range of -5 to +25 PSI, having an (obsolete) current signal

output range of 10 to 50 mA. The appropriate current signal value for an applied

pressure of +12 PSI would be represented on the number line as such:

Finding the \length" of this line segment in units of milliamps is as simple as setting

up a proportion between the length of the line in units of PSI over the total (span) in

PSI, to the length of the line in units of mA over the total (span) in mA:

17 PSI30 PSI

= ? mA40 mA

Solving for the unknown (?) current by cross-multiplication and division yields a

value of 22.67 mA. Of course, this value of 22.67 mA only tells us the length of the

line segment on the number line; it does not directly tell us the current signal value.

To ¯nd that, we must add the \live zero"offset of 10 mA, for a final result of 32.67

mA.

Thus, an applied pressure of +12 PSI to this transmitter should result in a 32.67 mA

output signal.

Pneumatic instrumentation

While electricity is commonly used as a medium for transferring energy across long

distances, it is also used in instrumentation to transfer information. A simple 4-20

mA current \loop" uses direct current to represent a process measurement in

percentage of span, such as in this example: