# PHGN341: Thermal Physics Exam I - February 18, 2011 NAME:

```NAME:
PHGN341: Thermal Physics
Exam I - February 18, 2011
1. (30) One of the major problems facing the large scale deployment of renewable energy technologies like wind and solar
is their inherent variability. For example, the Bonneville Power Authority (BPA) has approximately 3 GW of installed
wind generating capacity, yet their published wind generation data shows that it rarely generates that much power
and shows large power fluctuations on the time scale of hours to days. Fortunately, BPA also has a large hydroelectric
capacity that can quickly fill the gaps. If BPA were to rely on wind alone, it would need a backup capacity roughly
equal to the desired total generating capacity, an impractical situation. One proposed solution is energy storage, that
is, install an excess renewable capacity and store the excess energy for use when the renewable resource drops. One
proposed method for large scale energy storage is Compressed Air Energy Storage (CAES) whereby air is compressed
and stored underground. The energy is recovered by letting the compressed air expand in a generator turbine.
(a) (10) Consider a CAES system that compresses air rapidly so the process may be treated as adiabatic. Calculate
the energy stored when 1 m3 of air at standard conditions (P0 = 105 Pa, T0 = 300 K) is compressed to 20 MPa.
R
Solution: The energy stored is equal to the work done in compressing the gas: U = P dV . For adiabatic processes we
have P V γ = constant, where for air γ = 1 + 52 = 75 . Thus, P = P0 ( VV0 )γ and
Z Vf
dV
P0 V0
Vf (1−γ)
Pf (1−1/γ)
P0 V 0
γ
=−
∆U = W = −P0 V0
( )
( )
(1)
−1 =
−1
γ
(1 − γ)
V0
(γ − 1)
P0
V0 V
where the adiabatic condition for the volume ratio has been used. Plugging in the numbers gives U = 8.86 &times; 105 J.
Alternatively, one can use the equipartition theorem for the change in the internal energy:
5
5
5
5
Pf Vf
∆U = N k∆T = ∆(P V ) = (Pf Vf − P0 V0 ) = P0 V0
(2)
−1 .
2
2
2
2
P0 V0
1/γ
V
The adiabatic condition yields Vf0 = PPf0
, which then gives the same expression above.
(b) (5) After compression, what is the stored energy density (J/m3 )? (For comparison a lead-acid battery stores about
360 J/m3 .)
Solution: The final volume is found from the adiabatic condition to be: Vf = V0
P0
Pf
γ1
= 0.227 m3 . Thus, the energy
density is W/Vf = 3.90 &times; 107 J/m3.
(c) (5) What volume of compressed air is needed to store 3 GW-days of energy?
Solution: 3 GW-days is equal to 2.59 &times; 1014 J. Thus, using the result from part (b), the volume needed is 6.65 &times; 106
m3 .
(d) (5) What is the temperature of the compressed gas?
Solution: Use the adiabatic condition and the ideal gas law to find: Tf = T0
P0 (1/γ−1)
Pf
= 1363.2 K.
(e) (5) Based on your results, discuss the practical issues associated with adiabatic CAES.
Solution: From the answers above, the practical issues are (1) the energy density appears to be very favorable (at least
compared to a lead-acid battery), (2) the volume required to store utility-scale energies is very large; so one would need
to build gigantic storage tanks or use large underground caverns. It would be expensive to do extensive underground
excavation for CAES purposes only; so in practice application of CAES technology would be limited to regions that
have geological or mined underground caverns already in place. This limits the locations where CAES can be used.
(3) the high temperature of the compressed air would cause additional problems with material degradation and energy
dissipation. Additional comments: Despite these problems, CAES continues to be one of the best hopes for large scale
energy storage. There is a 110 MW CAES plant in operation in the U.S. today at McIntosh, Alabama. It stores the
compressed air underground. The compressed air cools, but is partially reheated by turbine exhaust before being fully
reheated by natural gas before entering the turbines. Sadly, as currently configured, CAES requires natural gas (or
some other heat source) to reheat the gas upon expansion. Nevertheless, the net efficiency of the energy stored is about
0.5 making it commercially competitive.
1
2. (40) Consider two different Einstein solids initially separated and thermally isolated. System A consists of four oscillators
with quantum of energy, h̄ωA . System B consists of three oscillators with a quantum of energy twice that of System A,
h̄ωB = 2h̄ωA . Let qA be the number of h̄ωA energy quanta in system A, and qB be the number of h̄ωB energy quanta
in system B. Suppose the total energy is fixed at 6 h̄ωA (or equivalently, 3 h̄ωB ).
(a) (20) It takes two energy quanta from system A to excite one quantum of energy in system B; therefore system A
can only exchange even multiples of its energy quanta with system B; so qA can only take on the values {0, 2, 4, 6}
corresponding to qB having the values {3, 2, 1, 0}. Complete the multiplicity tables below for the two systems.
Solution: Use the Einstein solid multiplicity formula to fill out the table:
qA
ΩA (4, qA )
qB
ΩA (4, qA ) &middot; ΩB (3, qB )
ΩB (3, qB )
5!
3!2!
0
Ω(4, 0) =
3!
0!3!
=1
2
Ω(4, 2) =
5!
2!3!
= 10
2
Ω(3, 2) =
4!
2!2!
=6
60
3
Ω(4, 4) =
7!
4!3!
= 35
1
Ω(3, 1) =
3!
1!2!
=3
105
6
Ω(4, 6) =
9!
6!3!
= 84
0
Ω(3, 0) =
2!
0!1!
=1
84
3
Ω(3, 3) =
= 10
10
Total:
259
(b) (10) Initially, suppose that system A has qA = 0 and system B has qB = 3. Now let the two systems be in thermal
contact, but are otherwise isolated so that the net energy in h̄ωA units is fixed at 6 = qA + 2qB . Use your multiplicity
tables to find the probability that qA = 4 (qB = 1). Explain your reasoning and show all work.
Solution: Since all accessible microstates are equally likely, using the table one finds that the probability that qA = 4
is P (qA = 4) = 105
259 = 0.405.
(c) (10) What is the entropy of the combined system at equilibrium (in units of Boltzmann’s constant)?
Solution: The equilibrium state is the one with the largest multiplicity. From the table one finds the greatest multiplicity
for qA = 4. The entropy (in units of Boltzmann’s constant) is therefore: S/k = ln 105 = 4.654.
2
3. (30) The Sackur-Tetrode expression for the entropy of N atoms of mass, m, confined to a two-dimensional area, A, is
A 2πmU
S2d (N, A, U ) = k ln Ω2d (N, A, U ) = kN ln
+2 ,
N h2 N
where U is the total energy and h is Planck’s constant.
(a) (10) From the entropic definition of temperature, derive an expression for the total energy as a function of the
∂S
. Separating out the energy dependence from the other
Solution: The entropic definition of temperature is: T1 = ∂U
parameters in the expression for the entropy gives: S2d = kN (ln U + {other factors not containing U}). Taking the
necessary derivative gives: T1 = kN
U or, putting this in standard form: U = N kT .
Comment: This is another example of the equipartition theorem, U = f2 N kT , where in this case the number of degrees
of freedom is two.
(b) (10) From the entropic definition of ”pressure” in two dimensions (i.e. force per unit length, instead of force per
unit area), derive an expression for the force per unit length, λ, as a function of the temperature. Comment on your
result.
Solution: In analogy to the 3-dimensional definition of pressure, the 2-dimensional entropic definition of force per unit
length on the boundary is
∂S2d
λ
.
=
T
∂A N,U
As in the previous part, separating out the area dependence from the other parameters in the expression for the
entropy gives: S2d = kN (ln A + {other factors not containing A}).Taking the indicated derivative gives: Tλ = kN
A , or,
λA = N kT .
Comment: This is the ideal gas law in two dimensions comparable t P V = N kT in three dimensions. Note that λA
has units of energy just as P V .
(c) (10) Find an expression for the two-dimensional entropy as a function of temperature, T , and the force per unit
length, λ.
Solution: To get the entropy in terms of T and λ, one needs to eliminate U and A in terms of T and λ. From part (a)
we have U = N kT and from part (b) we have A = N kT /λ. Substituting in the expression for entropy gives:
2πmk 2 T 2
+
2
.
S(N, λ, T ) = kN ln
λh2
3
```