Modern Engineering Thermodynamics

12 CHAPTER 1: The Beginning
WHICH WEIGHS MORE—A POUND OF FEATHERS
OR A POUND OF GOLD?
The avoirdupois (from the French meaning “to have weight”) pound contains 7000 barleycorns and is divided into 16
ounces. It was used primarily for weighing ordinary commodities, such as wood, bricks, feathers, and so forth. The troy
pound was named after the French city Troyes and was used to weigh only precious metals (gold, silver, etc.), gems, and
drugs. The English troy pound contains only 5760 barleycorns and is subdivided into 12 ounces, as was the original
Roman pound. The English word ounce is also derived from the Latin word uncia, meaning “the twelfth part of.”
Consequently, the avoirdupois pound is considerably larger (by a factor of 7000/5760 = 1.215) than the troy pound and
the coexistence of both pound units produced considerable confusion over the years. So a pound of feathers actually does
weigh more than a pound of gold, because the weight of the feathers is measured with the avoirdupois pound, whereas
the weight of the gold is measured with the troy pound. Today, all engineering calculations done in an English units
system are done with the 16 ounce, 7000 grain, avoirdupois pound.
Table 1.3 Units of Energy and Power
System Name Energy Power
MKS (SI) N·m =kg·m 2 /s 2 = joule (J) N·m/s = kg·m 2 /s 3 = J/s = watt (W)
CGS dyn·cm = g·cm 2 /s 2 = erg dyn·cm/s = g·cm 2 /s 3 = erg/s
Absolute English foot·poundal (ft·pdl) ft·pdl/s
Technical English ft·lbf ft·lbf/s
Engineering English ft·lbf (1 Btu = 778.17 ft·lbf) ft·lbf/s (1 hp = 550 ft·lbf/s)
Note: 1 dyn = 10 –5 N and 1 erg = 10 –7 J.
EXAMPLE 1.3
In Table 1.2, the Technical English units system uses force (F), length (L), and time (t) asthe
fundamental dimensions. Then, the mass unit “slug” was defined such that k 1 and g c came
out to be unity (1) and dimensionless. Define a new units system in which the force, mass,
and time dimensions are taken to be fundamental with units of lbf, lbm, and s, and the
length unit is defined such that k 1 is unity (1) and dimensionless. Call this new length unit
the chunk and find its conversion factor into the Engineering English and SI units systems
(Figure 1.10).
1 chunk = ? feet = ? meters
FIGURE 1.10
Example 1.3.
Solution
From Eq. (1.1), we see that the length unit must be defined via Newton’s secondlaw,F = k 1 ma. Sincewewantk 1 to be
unity and dimensionless, we set
k 1 =
F ma = 1 ðdimensionlessÞ
In our new system, we arbitrarily require 1 lbf to be the force calculated from Newton’s second law when 1 lbm is accelerated
at a rate of 1 chunk/s 2 . Then, from the preceding k 1 equation, we get
1 lbf
1 lbm ð1 chunk/s 2 Þ = 1 ðdimensionlessÞ
so that
1 chunk = 1 lbf .s 2
lbm
In the Engineering English units system, 1 lbf accelerates 1 lbm at a rate of
a = F m ðg cÞ = 1 lbf
1 lbm 32:174 lbm
.ft
lbf .s 2 = 32:174 ft
s 2