Modern Engineering Thermodynamics

110 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms
EXAMPLE 4.3 (Continued )
Solution
Let the system be the material inside the tank. The process of
heating the tank is one of constant volume (the tank is “rigid”).
Therefore, since the system volume, V, is constant, dV = 0 and
the moving boundary work is:
p 1 = 0.100 MPa
T 1 = 20°C
Heated sealed
rigid container
Work = ?
p 2 = 0.800 MPa
ð 1
W 2 Þ moving
=
boundary
Z 2
1
pdV = 0
Therefore, no moving boundary work occurs during this process.
FIGURE 4.6
Example 4.3.
State 1 State 2
Since a “rigid” container cannot change its volume, its moving boundary work is always zero regardless of the
process it undergoes.
EXAMPLE 4.4
The weather balloon in Figure 4.7 is inflated from a constant pressure, compressed gas source at 20.0 psia. Determine the
moving system boundary work as the balloon expands from a diameter of 1.00 ft to 10.0 ft.
Solution
Assume the balloon is a sphere, then V = 4 3 πR 3 = 1 6 πD 3 . The process here
is one of constant pressure, so p = constant, and
p 1 = 20.0 psia
p 2 = 20.0 psia
ð 1
W 2 Þ moving
=
boundary
Z 2
1
pdV = p
Z 2
= 20:0 lbf 144 in
2
in 2 ft 2
= 1:51 × 10 6 ft.lbf
1
dV = pðV 2
− V 1
Þ
π
6
½ð10:0 3 − 1:00 3 Þ ft 3 Š
The work is positive because the balloon does work on the atmosphere as
it expands and pushes the atmosphere out of the way.
D 1 = 1.00 ft dia
D 2 = 10.0 ft dia
State 1 State 2
FIGURE 4.7
Example 4.4.
Exercises
7. In Example 4.3, is the moving boundary work always zero for a sealed, rigid container? Are any other work modes
always zero for this type of system? Could a piston-cylinder apparatus be modeled as a sealed, rigid system? Answers:
Yes, no, no. (It is sealed and the components, the piston and the cylinder, are rigid, but the piston can move, producing
a change in the enclosed volume.)
8. Determine the moving boundary work for the balloon in Example 4.4 as it deflates from a diameter of 10. ft to a diameter
of 5.0 ft at a constant pressure of 20. psia. What does the work on the balloon? Answer:( 1 W 2 ) moving boundary = –1.3 ×
10 6 ft · lbf. The surrounding atmosphere does work on the balloon as it deflates, that is why the work is negative.
9. If the pressure inside a system depends on volume according to the relation p = K 1 + K 2 V + K 3 /V, where K 1 , K 2 , and K 3
are constants, determine the appropriate equation for the moving boundary work done as the volume changes from
V to V : Answer: ð 1 2
1W 2 Þ moving boundary = K 1 ðV − V Þ + K 2 ðV 2 − 2 1 2 V2Þ/2 + K 3 ln ðV /V Þ:
1 2 1
To carry out the integration indicated in Eq. (4.26), the exact p = pðVÞ pressure volume function must be
known. This function is usually given in the process path specification of a problem statement. For example, in
Example 4.3, the process is one of constant volume (the container is rigid), so dV = 0; and in Example 4.4, the
filling process is isobaric (p = constant), so the integral of Eq. (4.26) is very easy. In general, outside of these
two cases, the integration of Eq. (4.26) is not trivial and must be determined with great care.
As an example of a nontrivial integration of Eq. (4.26), consider a process that obeys the relation
pV n = constant (4.27)