Like length and time, mass is a fundamental quantity. Mass is the amount of material an object, or body, contains. Mass makes a body have weight in a gravitational field and creates inertia in the body. Inertia makes a stationary body tend to remain stationary or, if the body is moving, makes the body continue to move in a straight line until acted upon by a directional force.
Thus, any force applied to start a body moving or to change the direction of a moving body has to overcome the body’s inertia. Derived quantities related to mass are weight and force, density and specific gravity, work and power, and pressure.
Weight and Force
Under most circumstances, weight and mass are the same thing. And, sometimes weight and force are considered to be the same. For example, in rotary drilling operations, weight, or force, must be placed on the bit to enable it to drill. So, a common expression is weight on the bit. In SI units, weight on the bit is expressed in decanewtons, which are units of force. In conventional units, weight on the bit is expressed in pounds-force. This pound-force is, however, usually stated as simply pounds. So, in Canada, for instance, a driller may say that the force on the bit is 8,896 decanewtons, while in the U.S., a driller may say that the weight on the bit is 20,000 pounds.
In the United States, pound-mass is the unit of mass and is symbolized as lbm to distinguish it from the pound-force, lbf. Interestingly, the metric system’s
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102 SOME PHYSICAL QUANTITIES AND THEIR MEASUREMENT
unit of mass, the kilogram, defines a pound-mass. One pound-mass (1 lbm) equals 0.45359237 kilogram. The conventional pound-force is defined as the force necessary to accelerate a mass of 1 pound at 32.15 feet per second squared (32.15 ft/s2). Table 4.8 lists the most-used units of mass, weight, and force in the conventional and SI metric systems.
Example Problem: A spring is acting against a heavy door with a force of 30 decanewtons (daN). What is the force in pounds-force (lbf)?
Solution: A newton is equal to 0.2247 pounds-force (from table 4.8), and a dec-anewton is 10 newtons, or
10 × 0.2247 = 2.247 lbf. Then, 30 daN is 30 × 2.247 = 67.41 lbf.
Example Problem: On a rotary drilling rig, the driller reports to the toolpusher that the weight on the bit is 35,800 pounds. Because the toolpusher has spent a great deal of time in Canada, he asks the driller to give him the force on the bit in decanewtons.
Solution: 1 lbf is equal to 4.45 N (from table 4.8). Since a daN is 10 N, then 10 ÷ 4.45 = 0.445 daN.
Then, 0.445 daN is 35,800 × 0.445 = 15,931 daN.
TABLE 4.8
Units of Weight, Mass, and Force Measurement
Conventional Conventional Metric
Unit Equivalents Equivalents
ounce (oz) 0.0625 lbm 28.35 g
pound-mass (lbm) 16 oz 0.4536 kg
453.6 g
ton (tn) 2,000 lbm 0.9072 t
907.185 kg pound-force (lbf) lbm × 32.15 ft/s2 4.45 N
Metric Conventional
Metric Unit Equivalents Equivalents
gram (g) 0.002 kg 0.03527 oz
kilogram (kg) 1,000 g 2.2046 lbm
tonne (t) 1,000 kg 1.10231 tn
newton (N) kg•m/s2 0.2247 lbf
Density and Specific Gravity Density
Density is the weight of a substance per unit volume. A unit volume can be a cubic inch, a cubic foot, a gallon, a barrel, a cubic yard, or any of several measurement units. In conventional units, pounds and cubic feet are the most commonly paired units, although pounds per gallon is used to measure density in some instances, particularly when dealing with liquids.
In the metric system, grams per cubic centimetre (g/cm3) is in common use, although in SI only kilograms per cubic metre (kg/m3) is recognized as a density measure. A cubic centimetre is (10–2m)3, or 10–6 m3, while a gram is 10–3 kg; so,
g/cm3 = 10–3 ÷ 10–6 m3 = 103 kg/m3 = 1,000 kg/m3.
One gram per cubic centimetre and 1,000 kilograms per cubic metre are equal quantities as far as density is concerned and represent the density of pure water at a temperature of 4° Celsius.
The density of a material indicates the weight of a product made from the material. Weight is an important consideration when engineers design a product.
For example, they favor aluminum alloy over steel as a material for aircraft because aluminum is less dense than steel for a given strength. Thus, they can design an aircraft of adequate strength but of less weight.
The weight of a large body may be calculated if the density of the material and the volume of the body are known. Table 4.9 gives the densities of several
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TABLE 4.9
Densities of Some Solids, Liquids, and Gases
Material lb/ft3 kg/m3 g/cm3 (sg)
Solids
Gold 1,206.2 19,300 19.3
Mercury 846.0 13,500 13.5
Lead 712.5 11,400 11.4
Iron 485.0* 7,700* 7.7*
Aluminum 165.6 2,600 2.6
Wood 50.0* 800* 0.8*
Ice 56.9 900 0.9
Liquids
Sulfuric acid 125.0 2,000 2.0
Seawater 64.3 1,030 1.03
Fresh (pure) water 62.5 1,000 1.00
Kerosene 50.0 800 0.80
Gasoline 46.8 750 0.75
Gases
Air 0.075 1.20 0.0012
Oxygen 0.084 1.34 0.00134
Nitrogen 0.0737 1.18 0.0018
Carbon monoxide 0.0734 1.17 0.0017 Hydrogen 0.0053 0.085 0.000085
*Wood and iron vary in density. The values shown are approximate.
104 SOME PHYSICAL QUANTITIES AND THEIR MEASUREMENT
solids, liquids, and gases. The density of each is given in pounds per cubic foot (lb/
ft3), kilograms per cubic metre (kg/m3), and grams per cubic centimetre (g/cm3).
Example Problem: The volume of a gasoline storage tank is 5,360 cubic feet. When the tank is full, what is the weight of the gasoline?
Solution: From table 4.9, find the density of gasoline in pounds per cubic foot.
Then set up the proportion, 46.8 pounds is to 1 cubic foot as x pounds is to 5,360 cubic feet:
46.8 x
—— = ——1 5,360
x = 46.8 × 5,360= 250,848.
The gasoline in the tank weighs 250,848 pounds.
Specific Gravity
For liquids and solids, the specific gravity (sp gr) is the ratio between the weight of a given volume of a substance and the weight of an equal volume of pure water at 39°F (water is at its densest at 39.2°F or 4°C):
weight of a volume of substance sp gr = ———————————————— weight of an equal volume of pure water
or density of substance sp gr = ———–————— density of pure water
For gases, specific gravity is the ratio between the weight of a given volume of a gas to an equal volume of air or hydrogen under prescribed conditions of temperature and pressure. (The density of a gas varies with the temperature and pressure. Thus, a specific temperature and pressure must be stated—for example, 60°F and 14.7 psi.)
density of gas
sp gr = ———–––––————— (at a specific pressure and temperature). density of air or hydrogen Like density, weight per volume can be expressed in lb/ft3, kg/m3, or g/cm3. When the density is expressed in grams per cubic centimetre, the specific grav-ity of a substance is the same as its densgrav-ity because the densgrav-ity of pure water is 1 gram per cubic centimetre, and any number divided by 1 equals the same number. Thus, the densities given under the column headed g/cm3 in table 4.9 are also the specific gravities for those substances. Incidentally, the SI system calls specific gravity “relative density.”
Example Problem: What is the specific gravity of a drilling mud that weighs 78.54 lb/ft3?
Solution: From table 4.9, the density of pure water is 62.5 lb/ft3. sp gr = 78.54 ÷ 62.5 = 1.257.
The same equation may be used to find the weight or density of a substance by transposing it to read:
density of substance = density of pure water × sp gr
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or
wt per vol of substance = wt per vol of pure water × sp gr.
As mentioned earlier, the specific gravities of gases are frequently expressed as a ratio between the weight of a given volume of the gas and the weight of an equal volume of air or hydrogen at a given temperature and pressure. Specific gravity of gases may also be referred to as vapor density.
Table 4.10 shows specific gravities based on air for several gases.
Baumé Scales
Several scales based on specific gravity have been developed for specialized use.
In the early 1800s, the French chemist Antoine Baumé developed two scales for measuring liquid density, or gravity. Today, two Baumé scales are used: one for measuring the gravity of heavy liquids and another for measuring the gravity of liquids lighter than fresh water. The increments in gravity are called degrees Baumé (°Baumé). Two equations express the two scales:
1. For heavy liquids: °Baumé = 145 – 145 ÷ . 2. For light liquids: °Baumé = 140 ÷ – 130.
Note that in equation 1 if the liquid has a specific gravity of 1, the degrees Baumé is 0, but in equation 2 it is 10. In equation 2, the degrees Baumé increase as the specific gravity decreases. In equation 1, the degrees Baumé increase as the specific gravity increases.
Example Problem: A heavy syrup has a specific gravity of 1.153. What is its density in degrees Baumé?
Solution: Solve the problem by using the Baumé scale for heavy liquids:
°Baumé = 145 – 145 ÷ 1.153 = 19.2.
Example Problem: A liquid has a specific gravity of 0.7006. What is its gravity in degrees Baumé?
Solution: Solve the problem by using the Baumé scale for light liquids:
°Baumé = 140 ÷ 0.7006 – 130 = 69.8.
°API Gravity Scale
During the early 1920s, the American Petroleum Institute (API), the Bureau of Mines, and the National Bureau of Standards adopted a special gravity scale for measuring crude petroleum and some of its products. It is designated °API and bears a resemblance to the Baumé scale for light liquids:
141.5
°API = ––––– – 131.5.
Example Problem: A sample of crude oil has a specific gravity of 0.8692. What is its API gravity?
Solution: Use the formula for determining API gravity.
141.5
106 SOME PHYSICAL QUANTITIES AND THEIR MEASUREMENT
= 162.8 – 131.5
°API = 31.3.
Using the Baumé and API scales expands changes in specific gravity, which are expressed in very small increments. For example, a change from 28°API to 30°API is equivalent to a change from 0.8348 to 0.8251 in specific gravity, which is a change of only 0.0097. The API scale makes the differences in crude oil densities much more evident. Notice, too, that the higher the API gravity, the lower is the specific gravity. For example, a liquid with an API gravity of 30°
has a specific gravity of 0.8251, whereas a liquid with an API gravity of 28° has a specific gravity of 0.8348.
To solve for specific gravity when the °API gravity is known, the equation is:
141.5
= ––––––––––– . °API + 131.5
Example Problem: What is the specific gravity of a 43°API gravity oil?
Solution:
= 141.5 ÷ (43 + 131.5) = 0.810882
= 0.8109.
Pressure
One definition of pressure is that it is a force per unit area. The force can be pounds-force or newtons. The area can be square inches, square centimetres, square metres, and so on. Pounds per square inch (psi) is the most often used unit of pressure in the English, or conventional, system. SI stipulates that pressure be measured in newtons per square metre (N/m2), which it terms a pascal (Pa)—that is, 1 N/m2 = 1 Pa. One psi equals 6,895 pascals or 6.895 kilopascals (kPa).
A measure of pressure that falls outside the definition of force per unit area is kilograms per square centimetre (kg/cm2). A kilogram is the unit of mass in SI, and SI convention does not allow it to be used as a unit of force. However, kg/
cm2 is still encountered where people employ the old metric system. A kilogram-force per square centimetre is about 98 kPa, or about 14.2 psi.
Another form of pressure measurement outside the force-per-unit-area definition is the height of a liquid column, typically a column of mercury or water. In this instance, mercury or water is placed in a length of glass tubing that is bent into a U shape. Each end of the U is open and each leg is of equal height. Mercury or water is put into the tube and, under atmospheric pressure, the liquid rises in both legs of the U to an equal height. However, if pressure above atmospheric is put on one leg of the tube, the water or mercury rises in the other side. By marking the tube in inches or millimetres, the height of the rise can be read and the pressure given in inches or millimetres of mercury or water. Such equipment is called a mercury or water manometer. Manometers are often used to measure small changes in pressure, such as in the atmosphere.
In this case, the manometer is often called a barometer.
As is the case with temperature, an absolute scale of pressure also exists. The air that surrounds the earth exerts a pressure of about 14.7 pounds per square inch at sea level. The pressure value of 14.7 psi is above absolute zero pressure, and it
is designated as pounds per square inch, absolute, or psia. In SI, the value is about 101.35 kilopascals (kPa), and for mercury columns it is about 760 millimetres or 29.92 inches—that is, the pressure of the atmosphere on one side of a column of mercury in a U-tube forces the mercury to rise to a level of 760 millimetres or 29.92 inches on the other side.
A typical pressure gauge, such as a tire-pressure gauge, shows gauge pressure (psig). Gauge pressure means that the atmospheric pressure of 14.7 corresponds to zero on the pressure gauge. Put another way, before the gauge is attached to measure the pressure of a vessel, the gauge reads zero, although the atmosphere is placing 14.7 psi on it.
Some gauges show readings below atmospheric pressure. On such gauges, the values may be expressed in inches of mercury, vacuum. In this case, vacuum means a pressure below that of the atmosphere. For example, if a vacuum exists in a vessel, then a mercury manometer used to measure the vacuum reads lower than 29.92 inches or 760 millimetres.
In converting gauge pressure to absolute pressure, the formula is psia = psig + 14.7.
Example Problem: A gauge attached to a vessel containing a fluid under pressure reads 25 psig. What is the absolute pressure of the fluid in this vessel?
Solution:
psia = 25 + 14.7 = 39.7.
Physical laws of gases state that the absolute pressure of a gas varies directly with its absolute temperature as
P2 T2
— = — P1 T1
where P and T signify pressure and temperature before (P1,T1) and after (P2,T2) a change. This formula can be applied only where the volume of gas does not change.
Example Problem: Suppose the pressure on a 10 ft3 tank of air was 30 psig at 70°F. After the sun had shone on it all morning, the temperature rose to 110°F.
How much did the pressure increase?
Solution: In this case, P1, T1, and T2 are known but P2 is not. First, convert both temperatures to absolute values:
70° + 460° = 530°.
110° + 460° = 570°.
Then, convert the pressure to absolute value:
30 psig + 14.7 = 44.7 psia.
Then
P2 ÷ 44.7= 570 ÷ 530
P2 = 44.7 × 570 ÷ 530 = 48.07 = 48.1 psia.
The pressure increase is
48.1 – 44.7 = 3.4 psi.
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108 SOME PHYSICAL QUANTITIES AND THEIR MEASUREMENT
Hydrostatic pressure is the force exerted by a body of fluid at rest. Hydro-static pressure increases directly with the density and the depth of the fluid. The hydrostatic pressure of fresh water is 0.433 psi per foot of depth, or 9.792 kPa per metre of depth.
Example Problem: What is the hydrostatic pressure in psi at the bottom of a 150-foot (46-metre) column of water? In kPa?
Solution:
150 × 0.433 = 65 psi.
46 × 9.792 = 450 kPa.