PHGN341: Thermal Physics
Exam II - April 12, 2013
NAME: KEY
1. (30) A nuclear power plant generates 3 GW (3 × 109 W) of electricity by operating on a Rankine cycle (steam engine)
between {T1 = 20 C,P1 = .023 bar} and {T3 = 600 C, P3 = 300 bar}. (The steam tables are attached.)
2
(a) What is the enthalpy of 1 kg of water after the turbine power step (3 → 4)?
3
Pressure
(Steam)
(Water)
1
(Water + Steam)
4
Solution: The power step is adiabatic so the entropy is conserved which then
determines the admixture of water/steam at point 4:
S3 = αSW ater + (1 − α)SSteam . Using the steam tables we
find: α = .2908. Thus, the enthalpy at point 4 is:
H4 = αHW ater + (1 − α)HSteam = 1824 kJ.
Volume
(b) Suppose the plant operates at an efficiency of 48%, what is the minimum mass of water that must pass through the
generating plant turbines per second to generate 3 GW electric power?
Solution: Since the efficiency is the work done divided by the heat delivered, we can solve for the heat delivered (per
1 dW
h
unit time): dQ
dt = dt = (3GW )/0.48 = 6.25 GW. For one kg of water the heat absorbed is: H3 − H2 = 3.360 MJ.
Thus, the amount of water that must pass through the Rankine cycle per second is: 1.86 × 103 kg/s.
(c) The waste heat is dumped into a nearby river which has a flow rate of 2 × 105 kg per second. Upstream from the
plant the temperature of the river water is 20 C. What is the average temperature of the river water downstream from
the plant? (The heat capacity of water at constant pressure is cp = 4.18 kJ/(kg-K).)
Solution: Using the efficiency above, the reactor must dump heat into the river at the rate of
GW. Therefore, the temperature rise is
∆T =
dQc
dt
=
dQh
dt
−
dW
dt
= 3.25
3.25 × 109 J/s
= 3.89 C
(4.18 × 103 J/(kg · C)(2.0 × 105 kg/s)
.
2. (20) Estimate the temperature of the melting point of ice at 200 bar pressure. (Data: Latent heat: L = 0.333 MJ/kg,
upon freezing 1000 cm3 of water (1 kg) expands to 1091 cm3 .)
∆P
L
∆T ' T (Vwater −Vice ) .
−5
Pa·273 K·9.1×10
m3 /kg
=
0.333×106 J/kg
Solution: The Clausius-Clapeyron relation gives the relation for the phase boundary:
200×10
5
Thus the
−1.49 K.
change in temperature needed to stay on the phase boundary is ∆T = −
Note the minus sign which is due to the fact that water expands when it freezes. Thus, the freezing temperature at 200
Pa is 271.5 K.
3. (20) A methane (CH4 ) fuel cell operates by combining oxygen and methane to form water (liquid) carbon dioxide
at standard pressure and temperature with the transfer of 4 electrons per molecule of water formed: CH4 + 2O2 →
2H2 O + CO2 .
Table 1: Thermodynamic properties per mole of substance.
Substance ∆f H (kJ) ∆f G (kJ) S (J/K) CP (J/K)
H2 O (l)
-285.83
-237.13
69.91
75.29
O2 (g)
0.0
0.0
205.14
29.38
CH4 (g)
-74.81
-50.72
186.26
35.31
CO2 (g)
-393.51
-394.36
213.74
37.11
a. Calculate the voltage of a methane fuel cell?
Solution: The voltage is given the magnitude of the change in the Gibbs free energy divided by the amount of charge
transferred. We are told that four electrons are transferred per mole of water produced. From the tables we calculate
1
the change in the Gibbs free energy per mole of water produced to be: ∆G = −408.95 kJ. The voltage is therefore
V = |∆G|/(4eNA ) = (4.0895 × 105 J/mole)/(4 × (1.6 × 10−19 C) × (6.022 × 1023 /mole)) = 1.06 V.
b. How much heat is delivered to the environment per mole of water produced?
Solution: The heat delivered to the environment can be calculated two ways. First, one can calculate the difference
between the change in enthalpy (i.e. Heat plus Work for this constant pressure process) and the change in the Gibbs
Free Energy (i.e. Work). Thus, Q = ∆H − ∆G = 36.23 kJ. Alternatively, one could calculate the change in entropy
and multiply by the temperature: Q = T ∆S = 36.23 kJ.
4. (30) Consider an ideal gas consisting of N indistinguishable (massless) neutrinos at temperature, T , confined to an
area, A. The neutrino’s kinetic energy is given by the ultrarelativistic relation: Ep = pc where c is the speed of light.
A
2
(Recall that the particle-in-a-box result for the density of states in 2 dimensions is d2 n = (2πh̄)
2 d p.)
(a) Calculate the partition function for a single neutrino confined to area, A. Solution:(In this problem we ignore the
spin-1/2 (Pauli exclusion) nature of the neutrinos.) Performing the indicated integral gives the partition function for
one neutrino:
Z
Z ∞
A
A
2
−βcp
d pe
=
2π
dp p e−βcp
Z1 =
(2πh̄)2
(2πh̄)2
0
Z ∞
2πA
A
=
dx x e−x =
.
(2πh̄)2 β 2 c2 0
2πh̄2 β 2 c2
(b) Calculate the average energy of one neutrino in two dimensions.
Solution:The average energy is given by
Ē
= −
∂
∂
2
log Z1 = −
[−2 log β + log{other stuff}] = = 2kT.
∂β
∂β
β
This is twice what is expected from the equipartition theorem. This is due to the fact that the energy for the ultrarelativistic neutrino is linear in the momentum, not quadratic, so the equipartition theorem doesn’t apply.
(c) What is the partition function for the N-neutrino gas in terms of the single neutrino partition function? (Use the
Sterling’s approximation result, N ! ' N N e−N .) Solution:The partition function for the N-neutrino gas is
Z1N
ZN
' N 1 −N
N!
N e
where we divide by N ! to account for the fact that the neutrinos are indistinguishable.
Z
=
(d) Use your result from part (c) to calculate the 2-dimensional analog of the “pressure” (i.e. force per unit length
along the boundary of the confining area) for the neutrino gas. Solution:The Helmholtz free energy for the N-neutrino
gas is
Z1N
Z1
= −N kT log
+1
= −kT log
N N e−N
N
A
= −N kT log
+1 .
N 2πh̄2 β 2 c2
F
The analog of the “pressure” for this 2-dimensional case is the force per unit length, λ, and is given by:
N kT
∂F
λ = −
=
.
∂A T,N
A
which is the standard ideal gas law in two-dimensions. This is expected since the pressure arises from the momentum
change at the bounding surface, not the energy.
2