Modern Engineering Thermodynamics

370 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.6 (Continued )
b. The combined energy and entropy balance for this material is
so that
du = Tds+ Fdl
∂u
∂l T
= T ∂s
+ F
∂l T
From the third Maxwell equation for this substance listed in part a and the given equation of state, we have
and
Therefore,
T
∂s
∂l
∂s
∂l T
T
= − ∂F
∂T l
= −KðL/L o − 1Þ 2
= −KTðL/L o − 1Þ 2 = −F
∂u
= −F + F = 0
∂l T
If we now set u = u(T, l) and differentiate it, we get
du =
∂u
dT +
∂u
dl
∂T l ∂l T
= c l dT + ∂u
dl
∂l T
where c l is the constant length specific heat. Now since (∂u/∂l) T = 0 here, then this equation reduces to du = c l dT, sou
is only a function of T.
c. A closed system energy balance applied to this material for an isothermal process with (u 2 – u 1 ) T = 0 gives
1Q 2 = 1 W 2 = −
= −KT
Z L
Lo
Z L
Lo
FdL
ðL/L o − 1Þ 2 dL
= −KTL o ðL/L o − 1Þ 3 /3
0:200 3 .
= −ð0:150 N/KÞð293 KÞð0:0700 mÞ
0:0700 − 1 3
1Q 2 = − 6:57 N.m
Consequently, there is a heat transfer out of the system equal in magnitude to the work input.
Exercises
16. Determine the heat transfer and work required to stretch the rubber band in Example 11.6 if the elastic constant of the
rubber is increased from 0.150 N/K to 10.0 N/K. Answer: 1 Q 2 = 1 W 2 = –438 N · m.
17. If the temperature of the rubber band in Example 11.6 is increased from 20.0°C to60.0°C, determine the heat transfer and
work required to stretch the rubber band assuming all the other variables remain unchanged. Answer: 1 Q 2 = 1 W 2 = –7.47 N·m.
18. How much heat transfer and work is required to stretch the rubber band in Example 11.6 twice as far, to L = 0.400 m
instead of 0.200 m, if everything else remains constant? Answer: 1 Q 2 = 1 W 2 = –107 N·m.
11.5 THE CLAPEYRON EQUATION
Benoit Pierre Emile Clapeyron (1799–1864) was a French mining engineer and a contemporary of Carnot who,
in the, 1830s, took an interest in studying the physical behavior of gases and vapors. He was able to derive a
relation for the enthalpy change of the liquid to vapor phase transition (h fg ) in terms of pressure, temperature,
and specific volume, thus providing one of the first equations for calculating a property that is not directly
measurable in terms of properties that are directly measurable. Today, this relation is most easily derived from
one of the Maxwell equations, Eq. (11.15). For an isothermal phase change from a saturated liquid to a saturated
vapor, the pressure and temperature are independent of volume. Then, Eq. (11.15) becomes
∂p
∂T
v
= dp
= s g − s f
dT
sat
v g − v f
= s fg /v fg