Solution to Homework 1
MAE 216: Statistical Thermodynamics
University of California, Davis, Winter Quarter 2014
1
A simple refrigerator (5 points)
Consider a refrigeration cycle in steady-state with performance coefficient β = 4.5 and work input
at the rate Ẇ = 0.8 kW. Energy is rejected from the refrigerator to the surrounds, which are at
T = 293◦ K.
1.a
Determine the rate energy is rejected, in kW.
We note Q̇in the rate at which energy is extracted from the inside of the refrigerator, and Q̇out the
rate at which energy is rejected in the surrounds. The definition of the performance coefficient of
a refrigerator provides
β=
Q̇in
Ẇin
⇒
4.5 =
Q̇in
0.8kW
⇒
Q̇in = 3.6kW .
(1)
Since the device should be in the same macrostate at the end of each cycle, the first law of
thermodynamics requires that the rate at which thermal energy enters the device is equal to the
one at which it exits it. Hence, energy is rejected at the rate
Q̇out = Ẇin + Q̇in = 3.6kW + 0.8kW = 4.4kW .
1.b
(2)
Determine the lowest theoretical temperature inside the refrigerator.
Assuming that the process is internal reversible, Carnot’s theorem provides
β=
TC
TH − TC
⇒
β TH − β TC = TC
⇒
TC =
β TH
1+β
.
(3)
Since TH = 293◦ K, the lowest theoretical temperature inside the refrigerator is
TC =
2
4.5 (293◦ K)
≈ 239.73◦ K .
5.5
(4)
Composite engines (10 points)
A power cycle drives a heat pump. At steady-state, the power cycle receives Q̇s by heat transfer at
Ts from the high-temperature source and delivers Q̇1 to a dwelling at Td . The heat pump receives
Q̇0 from the outdoors at T0 , and delivers Q̇2 to the dwelling.
1
2.a
Obtain an expression for the maximum theoretical value of the performance
parameter (Q̇1 + Q̇2 )/Q̇s in terms of the temperature ratios Ts /Td and T0 /Td .
Method I:
As an extreme case of the second law.
The rate at which heat enters the composite engine must balance the rate at which it exits it
⇒
Q̇0 + Q̇s = Q̇1 + Q̇2
Q̇0 = Q̇1 + Q̇2 − Q̇s
.
(5)
The maximum theoretical value for the performance is obtained for a reversible engine, i.e., when
the entropy remains unchanged
0=
Q̇s Q̇0 Q̇1 Q̇2
+
−
−
Ts
T0
Td
Td
⇒
Q̇1 + Q̇2
Q̇s Q̇0
Q̇s Q̇1 + Q̇2 − Q̇s
=
+
=
+
Td
Ts
T0
Ts
T0
Rearranging the terms gives
1
1
1
1
Q̇1 + Q̇2
= Q̇s
−
−
Td T0
Ts T0
"
#−1 "
#
Q̇1 + Q̇2
Ts T0 1
1
Ts T0 1
1
=
−
−
Td
Td T0
Td
Ts T0
Q̇s
Q̇1 + Q̇2
Ts /Td − T0 /Td
=
Ts /Td 1 − T0 /Td
Q̇s
.
.
(6)
(7a)
(7b)
(7c)
Equation (7c) is the sought relationship.
Method II:
As a power cycle driving a heat pump.
For a power cycle, the maximal theoretical efficiency is
ηmax =
TH − TC
Ẇout
=
TH
Q̇in
.
(8)
Noting Ẇ the work output rate from the power cycle (which is also the work input rate of the
heat pump), we obtain
Ts − Td
Ẇ
=
Ts
Q̇s
⇒
Ẇ =
Q̇s (Ts − Td )
Ts
.
(9)
For a heat pump, the maximal theoretical coefficient of performance is
γmax =
Q̇out
TH
=
TH − TC
Ẇin
⇒
Q̇2
Td
=
Td − T0
Ẇ
.
(10)
Conservation of energy in the heat pump provides Q̇2 = Ẇ + Q̇0 while conservation of energy
in the whole composite engine provides Q̇0 + Q̇s = Q̇1 + Q̇2 . Combining these relations gives
Q̇2 = Ẇ + Q̇1 + Q̇2 − Q̇s , which we use together with equation (9) in equation (10) to obtain
!
Td
Ẇ + Q̇1 + Q̇2 − Q̇s
Q̇1 + Q̇2 − Q̇s
Ts
Q̇2
Q̇1 + Q̇2
=
=
=1+
=1+
−1
.
Td − T0
Ts − Td
Ẇ
Ẇ
Q̇s (Ts − Td )/Ts
Q̇s
(11)
2
Isolating the sought ratio gives
Isolating the sought ratio gives
◆
✓ Td
(T
Q̇
+
Q̇
T
−
T
s−T
d )) T
d − ((T
d−T
0 )) +
1
2
s
d
(T
T
T
T
T
+T
Tss(T
(Tdd − T
T00))
Q̇1 + Q̇2 = Ts Td
Td − 1 + 1 = s
d
d
d
0
= Ts
1 +1=
T
T
Q̇
Ts
Tdd − T
T00
Tss(T
(Tdd − T
T00))
Q̇ss
2
2
Q̇
T
T
Q̇11 +
+ Q̇
Q̇22 = TssT
T00 − T
TddT
T00 +
+T
TssT
Tdd − TssT
T00 = T
TssT
Tdd/T
/Tdd2 − T
TddT
T00/T
/Tdd2
=
= Ts (Td − T0 )/T 22
T
Q̇
Tss(T
(Tdd − T
T00))
Ts (Td T0 )/Tdd
Q̇ss
Q̇
T
Q̇11 +
+ Q̇
Q̇22 =
Tss/T
/Tdd − T
T00/T
/Tdd =
T
Q̇
Tss/T
/Tdd 11 − T
T00/T
/Tdd
Q̇ss
(12a)
(12a)
(12b)
(12b)
(12c)
(12c)
which is in agreement with equation (7c).
2.b
Plot the results of 2.a versus Ts /Td ranging from 2 to 4 for T0 /Td = 0.85, 0.9,
and 0.95.
Performance parameter
15
T0 /Td = 0.85
T0 /Td = 0.90
10
T0 /Td = 0.95
5
0
2
2.5
3
3.5
4
Ts /Td
3
Ideal Gaz and Maxwell’s Relations (5+5 points)
3.a
Starting from dU = T dS P dV , show that the equation of state P V = mRT in
fact implies that U can only depend on T .
In general, U (T, V, P ). However, due to the equation of state P V = mRT , only two of the
three variables T , V and P are independent. Without loss of generality, we choose U (T, V ) =
U (T, V, mRT
). All other thermodynamic functions thus also depend on T and V alone, e.g., S(T, V ).
V
We now calculate the total derivative of S(T, V )
dS =
@S
@S
dT +
dV
@T V
@V T
(13)
which we then use to calculate the total derivative of U (T, V )
dU = T dS
@S
@S
P dV = T
dT + T
dV
@T V
@V T
✓
@S
@S
P dV = T
dT + T
@T V
@V
3
3
P
T
◆
dV
. (14)