Modern Engineering Thermodynamics

7.8 Numerical Values for Entropy 223
DO ALL ELASTIC MATERIALS HAVE ENTROPIC ELASTICITY?
No, most elastic solids do not have entropic elasticity, but the elasticity present in rubber and polymers is largely entropic. When
work is done adiabatically on a material with entropic elasticity, the temperature of the material increases. You can demonstrate
this by stretching a rubber band rapidly then immediately touching it to your lips (which are very sensitive to temperature). The
rubber band is warmer than it was before it was stretched. Then, if you hold the stretched rubber band long enough for it to
return to room temperature and suddenly release it and touch it to your lips, it is colder than it was before it was stretched.
When an elastic deformation produces a decrease in the specific entropy of a material, it is said to have entropic
elasticity. In the case of an ideal gas, Eq. (7.36) shows that an isothermal (T 2 = T 1 ) compression (v 2 < v 1 )
produces a decrease in the specific entropy of the gas. Consequently, ideal gases have entropic elasticity.
Any process in which entropy remains constant is called an isentropic process. The term isentropic comes from the
Greek words for “constant entropy.”
If an ideal gas with constant specific heats undergoes an isentropic process, then s 2 − s 1 and Eqs. (7.36) and
(7.37) give
or
ln T 2
T 1
= − R c v
ln v 2
v 1
= R c p
ln p 2
p 1
For an isentropic process with an ideal gas,
T 2
= v 1−k
2
= p ðk−1Þ/k
2
(7.38)
T 1 v 1 p 1
and
p 1 v1 k = p 2v2 k = constant (7.39)
where
k = c p /c v and R = c p − c v (7.40)
Consequently, from Eq. (4.27), we see that, in the case of an ideal gas with constant specific heats, an isentropic
process is the same as a polytropic process with n = k.
These equations for the specific entropy of an incompressible substance and an ideal gas are the only such formulae
to be introduced at this point. Specific entropy equations for more complex substances are introduced
later in the text as they are needed.
EXAMPLE 7.4
An insulated apparatus contains 1.5 kg of saturated liquid water at 20.°C. Determine the change in specific entropy of the
water as it is pressurized from 0.10 MPa to 10. MPa. Assume the liquid water is an incompressible material.
Solution
First, draw a sketch of the system (Figure 7.12).
The unknown is the change in specific entropy, s 2 − s 1 , for the system. The material is liquid water.
An energy balance for this process gives
1Q 2 − 1 W 2 = mu ð 2 − u 1 Þ
Z
and since the apparatus is insulated 1 Q 2 = 0. The only possible work mode here is moving boundary work, so 1 W 2 = pdV.
But the water is to taken as incompressible, so V = constant and dV = 0. Also, Eq. (3.33) gives the specific internal energy
change of an incompressible material as u 2 − u 1 = cT ð 2 − T 1 Þ,wherec is the specific heat of the material. Then, the energy
balance equation gives for this process
0 − 0 = mcðT 2 − T 1 Þ
(Continued )