Modern Engineering Thermodynamics

4.6 Mechanical Work Modes of Energy Transport 115
Exercises
13. What type of rigid system has zero elastic work regardless of the loading? Answer: A perfectly rigid system (E = ∞).
14. If the system analyzed in Example 4.6 was a rectangular steel bar, 1.0 inch square by 12 inches long, determine the elastic
work required to stress it from 0.0 to 10. × 10 3 lbf/in 2 . Use E steel = 30. × 10 6 lbf/in 2 . Answer: ( 1 W 2 ) elastic = − 1.7 ft·lbf.
15. Ten joules of elastic work is applied to a circular brass rod 0.0100 m in diameter and 1.00 m long. Determine the
resulting stress and strain in the bar if it is initially unloaded. Use E brass = 1.05 × 10 11 Pa. Answer: σ = 164 MPa and
ε = 1.56 × 10 −3 m/m.
4.6.4 Surface Tension Work
Surface tension work is the two-dimensional analog of the elastic work just considered. Figure 4.5d shows a
soap film on a wire loop. One side of the loop has a movable wire slider that can either compress or extend the
film. As in the case of the elastic solid, the force and deflection are always in the same direction and the force is
applied to the system, so we can modify Eq. (4.36) to read
dW = − F ! .d x ! = − Fdx= −ð2σ s bÞ dx (4.41)
where σ s is the surface tension of the film, and b is the length of the moving part of the film. The factor of 2
appears because the film normally has two surfaces (top and bottom) in contact with air. Now, 2b · dx = dA =
change in the film’s surface area, so Eq. (4.41) becomes
and, for the surface tension work,
dW = − σ s dA (4.42)
Surface tension work
ð 1
W 2 Þ surface = −
tension
Z 2
1
σ s dA
(4.43)
EXAMPLE 4.7
Determine the amount of surface tension work required to inflate the soap
bubble shown in Figure 4.14 from a diameter of zero to 0.0500 m. The surface
tension of the soap film can be taken to be a constant 0.0400 N/m.
D 1 = 0
State 1
W surf. tension = ?
State 2
D 2 = 0.0500 m
Solution
Here, σ s = constant = 0.0400 N/m. Note that we are not calculating the
surface area of the bubble here from its geometric elements, but wish
only to find the change in area between states 1 and 2. Consequently,
the area integral in this instance can be treated as a point function rather
than as a path function. So Eq. (4.43) becomes
FIGURE 4.14
Example 4.7.
ð 1
W 2 Þ surface
tension
Z 2
= − σ s dA = − σ s ðA 2 − A 1 Þ
1
where A 1 = 0. Now, since a soap bubble has two surfaces (the outside and inside films),
and
A 2 = 2ð4πR 2 Þ = 2ð4πÞ 0:0500 m 2
= 0:0157 m
2
2
ð 1
W 2 Þ surface
= − ð0:0400 N/mÞð0:0157 − 0m 2 Þ
tension
= − 6:28 × 10 − 4 N.m = − 6:28 × 10 − 4 J
= − ð6:28 × 10 − 4 JÞð1 Btu/1055 JÞ = − 5:96 × 10 − 7 Btu