Modern Engineering Thermodynamics

104 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms
EXAMPLE 4.1 (Continued )
and
U 2 = mu 2 = ð3:00 lbm
Þð950:0 Btu/lbmÞ = 2850 Btu
so
E T = ð2850 − 3216:6Þ Btu −
3:00 lbm
2
2
5280 ft/mile
ð25,000: mile/hÞ
3600 s/h
1 Btu
×
778:16 ft.lbf
32:174 lbm − 3:00 lbmð32:174 ft/s2 Þ
.ft
lbf .s 2 32:174 lbm ð200: milesÞð5280 ft/mileÞ
.ft
lbf .s 2
= − 366:6 − 80,550 − 4071 = − 85,000 Btu ðto three significant figuresÞ
1 Btu
778:16 ft.lbf
Therefore, 85,000 Btu of energy must be transferred out of the water (E T is negative here) by some mechanism. This can be
done, for example, by having the spaceship (and the water) do work on the atmosphere by aerodynamic drag as it lands.
Exercises
1. What would be the value of u 2 in Example 4.1 if E T were zero? Answer: u 2 = 29,300 Btu/lbm. (What is the physical state
of the water now?)
2. Which causes the larger change in E G :
a. A velocity increase from 0 to 1 ft/s or an increase in height from 0 to 1 ft?
b. A velocity increase from 0 to 100 ft/s or a height increase from 0 to 100 ft?
Answers: (a) height, (b) velocity.
3. Determine the value of E T that must occur when you stop a 1300. kg automobile traveling at 100. km/h on a level road
with no change in internal energy. Answer: E T = 502 kJ.
In nonequilibrium systems, we use the energy rate balance equation with _E G defined as
_E G = d
U + m V 2 + mg
dt 2g c
g
Z = _E T (4.9)
c
Equation (4.9) can become quite complicated for open systems whose total mass is rapidly changing (such as
with rockets), because it expands as follows (using U = mu):
_E G = m _u +
g V ð _V Þ + g
c g
ð _Z Þ + u + V 2
+ gZ
c 2g c
g
_m = _E T (4.10)
c
Notice that, in this equation, _V = dV/dt is the magnitude of the instantaneous acceleration, and _Z is the
magnitude of the instantaneous vertical velocity.
The equilibrium thermodynamics energy balance and the nonequilibrium energy rate balance are fairly simple
concepts; however, their implementation can be quite complex. Each of the gain, transport, and production
terms may expand into many separate terms, all of which must be evaluated in an analysis. Next, we investigate
the structure of the energy transport and energy transport rate terms.
system
4.4 ENERGY TRANSPORT MECHANISMS
There are three energy transport mechanisms, any or all of which may be operating in any given system: (1) heat,
(2) work, 4 and (3) mass flow. These three mechanisms and their sign conventions are illustrated in Figure 4.3.
Note that the sign conventions for heat and work shown in Figure 4.3 are not the same. Heat transfer into
a system is taken as positive, whereas work must be produced by or come out of a system to be positive. This is
the conventional mechanical engineering sign convention and reflects the traditional view that heat coming out
4 The types of work transports of energy included here are only those due to dissipative or nonconservative forces. For example, the
work associated with gravitational or electrostatic forces is not considered a work mode because it is conservative (i.e., it is
representable by the gradient of a scalar quantity) and is consequently nondissipative. Energy transports resulting from the actions of
conservative forces have their own individual terms in the energy balance equation (such as mgZ/g c for the gravitational potential
energy).