List of Symbols
Chapter 1 Introduction
1.7 ENERGY
Equation (1.4) expresses the work done by a finite compression or expansion process.6 Figure 1.3 shows a path for compression of a gas from point 1, initial volume V1t at pressure P1, to point 2, volume V2t at pressure P2. This path relates the pressure at any point of the process to the volume. The work required is given by Eq. (1.4) and is proportional to the area under the curve of Fig. 1.3.
1.7 ENERGY
The general principle of conservation of energy was established about 1850. The germ of this principle as it applies to mechanics was implicit in the work of Galileo (1564–1642) and Isaac Newton (1642–1726). Indeed, it follows directly from Newton’s second law of motion once work is defined as the product of force and displacement.
Kinetic Energy
When a body of mass m, acted upon by a force F, is displaced a distance dl during a differen- tial interval of time dt, the work done is given by Eq. (1.2). In combination with Newton’s second law this equation becomes:
dW = ma dl
By definition the acceleration is a ≡ du/dt, where u is the velocity of the body. Thus,
dW = m ___du
dt dl = m __dl dt du
Because the definition of velocity is u ≡ dl/dt, this expression for work reduces to:
dW = mu du
6However, as explained in Sec. 2.6, its use is subject to important limitations.
Figure 1.3: Diagram showing a P vs. Vt path.
P2
P1 P
Vt 0
1 2
V2t V1t
Integration for a finite change in velocity from u1 to u2 gives:
W = m
∫ u 1u 2
u du = m (
u 22
___2 − u 12
___2 ) or
W = _____mu 22
2 − mu 12
____2 = Δ ( _mu 2
2 ) (1.5)
Each of the quantities _12 mu 2 in Eq. (1.5) is a kinetic energy, a term introduced by Lord Kelvin7 in 1856. Thus, by definition,
E K ≡ __1
2 mu 2 (1.6)
Equation (1.5) shows that the work done on a body in accelerating it from an initial velocity u1 to a final velocity u2 is equal to the change in kinetic energy of the body. Conversely, if a moving body is decelerated by the action of a resisting force, the work done by the body is equal to its change in kinetic energy. With mass in kilograms and velocity in meters/second, kinetic energy EK is in joules, where 1 J = 1 kg⋅m2⋅s−2 = 1 N⋅m. In accord with Eq. (1.5), this is the unit of work.
Potential Energy
When a body of mass m is raised from an initial elevation z1 to a final elevation z2, an upward force at least equal to the weight of the body is exerted on it, and this force moves through the distance z2 − z1. Because the weight of the body is the force of gravity on it, the minimum force required is given by Newton’s law:
F = ma = mg
where g is the local acceleration of gravity. The minimum work required to raise the body is the product of this force and the change in elevation:
W = F( z 2 − z 1 ) = mg( z 2 − z 1 ) or
W = mz 2 g − mz 1 g = mgΔz (1.7)
We see from Eq. (1.7) that work done on a body in raising it is equal to the change in the quan- tity mzg. Conversely, if a body is lowered against a resisting force equal to its weight, the work done by the body is equal to the change in the quantity mzg. Each of the quantities mzg in Eq. (1.7) is a potential energy.8 Thus, by definition,
E P = mzg (1.8)
7Lord Kelvin, or William Thomson (1824–1907), was an English physicist who, along with the German physicist Rudolf Clausius (1822–1888), laid the foundations for the modern science of thermodynamics. See http://en . wikipedia.org/wiki/William_Thomson,_1st_Baron_Kelvin. See also http://en.wikipedia.org/wiki/Rudolf_Clausius.
8This term was proposed in 1853 by the Scottish engineer William Rankine (1820–1872). See http://en.wikipedia .org/wiki/William_John_Macquorn_Rankine.
1.7. Energy 13 With mass in kg, elevation in m, and the acceleration of gravity in m·s−2, EP is in joules, where 1 J = 1 kg⋅m2⋅s−2 = 1 N⋅m. In accord with Eq. (1.7), this is the unit of work.
Energy Conservation
The utility of the energy-conservation principle was alluded to in Sec. 1.1. The definitions of kinetic energy and gravitational potential energy of the preceding section provide for limited quantitative applications. Equation (1.5) shows that the work done on an accelerating body produces a change in its kinetic energy:
W = Δ E K = Δ ( _mu 2 2 )
Similarly, Eq. (1.7) shows that the work done on a body in elevating it produces a change in its potential energy:
W = E P = Δ ( mzg)
One simple consequence of these definitions is that an elevated body, allowed to fall freely (i.e., without friction or other resistance), gains in kinetic energy what it loses in poten- tial energy. Mathematically,
Δ E K + Δ E P = 0 or
mu 2 2
____2 − mu 12
____2 + mz 2 g − mz 1 g = 0
The validity of this equation has been confirmed by countless experiments. Thus the develop- ment of the concept of energy led logically to the principle of its conservation for all purely mechanical processes, that is, processes without friction or heat transfer.
Other forms of mechanical energy are recognized. Among the most obvious is potential energy of configuration. When a spring is compressed, work is done by an external force.
Because the spring can later perform this work against a resisting force, it possesses potential energy of configuration. Energy of the same form exists in a stretched rubber band or in a bar of metal deformed in the elastic region.
The generality of the principle of conservation of energy in mechanics is increased if we look upon work itself as a form of energy. This is clearly permissible because both kinetic- and potential-energy changes are equal to the work done in producing them [Eqs. (1.5) and (1.7)].
However, work is energy in transit and is never regarded as residing in a body. When work is done and does not appear simultaneously as work elsewhere, it is converted into another form of energy.
With the body or assemblage on which attention is focused as the system and all else as the surroundings, work represents energy transferred from the surroundings to the system, or the reverse. It is only during this transfer that the form of energy known as work exists. In contrast, kinetic and potential energy reside with the system. Their values, however, are measured with reference to the surroundings; that is, kinetic energy depends on velocity with respect to the surroundings, and gravitational potential energy depends on elevation with respect to a datum level. Changes in kinetic and potential energy do not depend on these reference conditions, provided they are fixed.
Example 1.4
An elevator with a mass of 2500 kg rests at a level 10 m above the base of an elevator shaft. It is raised to 100 m above the base of the shaft, where the cable holding it breaks. The elevator falls freely to the base of the shaft and strikes a strong spring. The spring is designed to bring the elevator to rest and, by means of a catch arrangement, to hold the elevator at the position of maximum spring compression. Assuming the entire process to be frictionless, and taking g = 9.8 m⋅s−2, calculate:
(a) The potential energy of the elevator in its initial position relative to its base.
(b) The work done in raising the elevator.
(c) The potential energy of the elevator in its highest position.
(d) The velocity and kinetic energy of the elevator just before it strikes the spring.
(e) The potential energy of the compressed spring.
(f) The energy of the system consisting of the elevator and spring (1) at the start of the process, (2) when the elevator reaches its maximum height, (3) just before the elevator strikes the spring, and (4) after the elevator has come to rest.
Solution 1.4
Let subscript 1 denote the initial state; subscript 2, the state when the elevator is at its greatest elevation; and subscript 3, the state just before the elevator strikes the spring, as indicated in the figure.
10 m
100 m
State 1
State 2
State 3
(a) Potential energy is defined by Eq. (1.8):
E P 1 = mz 1 g = 2500 kg × 10 m × 9.8 m·s −2
= 245,000 kg·m 2 ⋅ s −2 = 245,000 J
1.7. Energy 15 (b) Work is computed by Eq. (1.7). Units are as in the preceding calculation:
W = mg( z 2 − z 1 ) = ( 2500 ) ( 9.8 ) ( 100 − 10 ) = 2,205,000 J (c) Again by Eq. (1.8),
E P 2 = mz 2 g = ( 2500 ) ( 100 ) ( 9.8 ) = 2,450,000 J Note that W = EP2 − E P1 .
(d) The sum of the kinetic- and potential-energy changes during the process from state 2 to state 3 is zero; that is,
Δ E K 2→3 + Δ E P 2→3 = 0 or E K 3 − E K 2 + E P 3 − E P 2 = 0 However, E K 2 and E P 3 are zero; hence E K 3 = E P 2 = 2,450,000 J.
With E K 3 = _12 m u 32
u 32 = ____2 E K 3
m = _____________2 × 2,450,000 J
2500 kg = 2 × 2,450,000 kg·m _____________________2 ⋅ s −2 2500 kg = 1960 m 2 · s −2 and u 3 = 44.272 m·s −1
(e) The changes in the potential energy of the spring and the kinetic energy of the elevator must sum to zero:
Δ E P ( spring ) + Δ E K ( elevator ) = 0
The initial potential energy of the spring and the final kinetic energy of the eleva- tor are zero; therefore, the final potential energy of the spring equals the kinetic energy of the elevator just before it strikes the spring. Thus the final potential energy of the spring is 2,450,000 J.
(f) With the elevator and spring as the system, the initial energy is the potential energy of the elevator, or 245,000 J. The only energy change of the system occurs when work is done in raising the elevator. This amounts to 2,205,000 J, and the energy of the system when the elevator is at maximum height is 245,000 + 2,205,000 = 2,450,000 J. Subsequent changes occur entirely within the system, without interaction with the surroundings, and the total energy of the system remains constant at 2,450,000 J. It merely changes from potential energy of posi- tion (elevation) of the elevator to kinetic energy of the elevator to potential energy of configuration of the spring.
This example illustrates the conservation of mechanical energy. However, the entire process is assumed to occur without friction, and the results obtained are exact only for such an idealized process.
Example 1.5
A team from Engineers Without Borders constructs a system to supply water to a mountainside village located 1800 m above sea level from a spring in the valley below at 1500 m above sea level.
(a) When the pipe from the spring to the village is full of water, but no water is flow- ing, what is the pressure difference between the end of the pipe at the spring and the end of the pipe in the village?
(b) What is the change in gravitational potential energy of a liter of water when it is pumped from the spring to the village?
(c) What is the minimum amount of work required to pump a liter of water from the spring to the village?
Solution 1.5
(a) Take the density of water as 1000 kg⋅m−3 and the acceleration of gravity as 9.8 m⋅s−2. By Eq. (1.1):
P = hρg = 300 m × 1000 kg·m −3 × 9.8 m·s −2 = 29.4 × 10 5 kg·m −1 ⋅ s −2 Thus, P = 29.4 bar or 2940 kPa
(b) The mass of a liter of water is approximately 1 kg, and its potential-energy change is:
Δ E P = Δ(mzg) = mgΔz = 1 kg × 9.8 m·s −2 × 300 m = 2940 N·m = 2940 J (c) The minimum amount of work required to lift each liter of water through an ele- vation change of 300 m equals the potential-energy change of the water. It is a mini- mum value because it takes no account of fluid friction that results from finite-velocity pipe flow.