Mathematical Theorems
If there exists a relation among x , y & z ; −→ f (x , y, z ) = 0
Thermodynamics Relations
Dr. M. Zahurul Haq
Professor
Department of Mechanical Engineering
Bangladesh University of Engineering & Technology (BUET)
Dhaka-1000, Bangladesh
∂y
∂x
z
∂x
∂y
∂x
∂y
=
z
z
∂y
∂z
z = z (x , y) → dz =
ME 6101: Classical Thermodynamics
For continuous functions,
Thermodynamics Relations
ME 6101 (2011)
1 / 22
Maxwell Relationships
Enthalpy, h ≡ u + Pv ⇒ dh = du + Pdv + vdP = Tds + vdP
Helmholtz Energy, f ≡ u − Ts ⇒ df = −Pdv − sdT
Gibbs Energy, g ≡ h − Ts ⇒ dg = dh − Tds − sdT = vdP − sdT
∂2 z
∂N
∂M
∂2 z
For continuous functions, ∂x
=
=
=⇒
∂y
∂y∂x
∂y
∂x y
x
du = +Tds − Pdv =⇒
2
3
4
∂T
x
∂z
∂y
= −1
y
=
x
∂z
∂x
c Dr. M. Zahurul Haq (BUET)
∂z
∂x
∂2 z
∂x ∂y
dx +
y
=
∂2 z
∂y∂x
∂z
∂y
y
∂x
∂z
=1
y
=⇒
dy = Mdx + Ndy
x
∂M
∂y
Thermodynamics Relations
=
x
∂N
∂x
ME 6101 (2011)
y
2 / 22
=+
∂v
∂s P
(B)
=+
∂P
∂T v
(C)
Ideal gas: Pv = RT
∂2 P
R
∂P
&
=
= 0 7−→ Cv 6= f (v )
=⇒ ∂T
2
v
∂T
v
=−
∂v
(D)
=⇒
dh = +Tds + vdP =⇒
∂T
∂P s
df = −sdT − Pdv =⇒
∂s
∂v T
dg = −sdT + vdP =⇒
∂s
∂P T
If P-v-T data or mathematical relationship is available, it is possible to
evaluate ∂2 P/∂T 2 , and then (∂Cv /∂v )T .
(A)
=−
∂v
∂P
du = Tds − Pdv =⇒ ∂u
∂s v = T
∂s ∂u
∂s
Cv ≡ ∂T
= ∂u
∂s v ∂T v = T ∂T v
v
2 ∂s ∂Cv
∂
∂s
∂2 s
∂ P
∂
∂v T = ∂v T ∂T v = T ∂v ∂T = T ∂T ∂v T = T ∂T 2 v
∂P
∂s
= + ∂T
using Maxwell’s relation: ∂v
T
v
v
s
c Dr. M. Zahurul Haq (BUET)
Cv & Thermodynamics Relationships
Internal Energy, du = ∂q + ∂w = Tds − Pdv (for rev. process)
1
x
∂z
∂x
f (x , y, z ) = 0 7−→ z = z (x , y), y = y(z , x ), x = x (y, z )
zahurul@me.buet.ac.bd
http://teacher.buet.ac.bd/zahurul/
c Dr. M. Zahurul Haq (BUET)
∂y
∂z
∂s
∂T P
Thermodynamics Relations
ME 6101 (2011)
van
3 / 22
v
RT
der Wall’s gas: P = v −b − va2
2 ∂ P
∂P
R
&
=
∂T v
v −b
∂T 2 = 0 7−→
c Dr. M. Zahurul Haq (BUET)
Cv 6= f (v )
Thermodynamics Relations
ME 6101 (2011)
4 / 22
CP & Thermodynamics Relationships
Tds Relationships
dh = Tds + vdP =⇒ ∂h
∂s P = T
∂s
∂h
∂s
∂h
=
=
T
CP = ∂T
∂s
∂T
∂T
P
P
P
P
2 ∂s ∂CP
∂
∂s
∂
∂ v
∂P T = ∂P T ∂T P = T ∂T ∂P T = −T ∂T 2 P
∂v
∂s
= − ∂T
using Maxwell’s relation: ∂P
T
P
Ideal gas: =⇒
∂v
∂T P
=
R
P
&
∂v
∂2 v
∂T 2
P
= 0 7−→ CP 6= f (P)
van der Wall’s gas: =⇒ ∂T P = P − a R1− 2b
( v)
v2
2 2 R P2
R 2 2a3 − 6ab
v
v4
∂ v
∂ v
−
7
→
C
=
f
(P)
=
−T
=
−
dP
P
3
P1
∂T 2 P
∂T 2 P
2b
a
P − 2 (1− v )
v
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
5 / 22
Internal Energy, u
∂u
∂T v
∂u
∂v
RT
∂u
∂v T
dT +
du = Tds − Pdv = Cv dT + T
∂P
∂T v
=T
T
dv = Cv dT +
∂u
∂v T
dv
dv − Pdv (←- 1st Tds Eq.)
∂P
∂T
Thermodynamics Relations
v
a
v2
−→ du = Cv dT +
Thermodynamics Relations
a
dv
v2
∂h
∂T P
∂h
∂P
∂h
∂P T
dT +
dh = Tds + dP = CP dT − T
ME 6101 (2011)
6 / 22
∂v
∂T P
=v −T
T
dP = Cp dT +
dP + vdP (←
∂v
∂T
∂h
∂P T
2nd Tds
dP
Eqn.)
P
∂h
= 0 −→ h 6= f (P)
= v − TR
Ideal gas: ∂P
T P
∂v
∂h
∂h
∂h
=
0
=
=
− Pv ∂h
∂P T
∂v T ∂P T
∂v T → ∂v T = 0
∂h
∂h
∂P T = 0 −→ h 6= f (P) :
∂v T = 0 → h 6= f (v )
−→ du = Cv dT = f (T )
van der Waals’ gas:
∂u
RT
∂v T = v −b − P =
h = h(P, T ) → dh =
−P
Ideal gas: ∂v T = v − P = 0
∂P ∂u
∂u
P ∂u
∂u
∂v T = 0 = ∂P T ∂v T = − v ∂P T → ∂P T = 0
∂u
∂u
∂v T = 0 −→ u 6= f (v ) :
∂P T = 0 → u 6= f (P)
c Dr. M. Zahurul Haq (BUET)
c Dr. M. Zahurul Haq (BUET)
Enthalpy h
u = u (v , T ) −→ du =
∂u
∂s
∂s
s = s(v , T ) −→ ds = ∂T
dT
+
dv
∂v
v
T
∂s
∂P
∂s
&
=
+
,
Maxwell’s
Relation (C)
Cv = T ∂T
∂v T
∂T v
v
∂P
dv
· · · 1st Tds
Tds = Cv dT + T
∂T v
∂s
∂s
s = s(P, T ) −→ ds = ∂T
dT + ∂P
dP
P
T
∂s
∂v
∂s
&
=
−
,
Maxwell’s
Relation (D)
CP = T ∂T
∂P T
∂T P
P
∂v
· · · 2nd Tds
Tds = CP dT − T
dP
∂T P
∂s
∂s
dP + ∂v
dv
s = s(P, v ) −→ ds = ∂P
v
P
∂s
∂s
Cv = T ∂T v & CP = T ∂T P
∂T
∂T
Tds = Cv
dP + CP
dv
· · · 3rd Tds
∂P v
∂v P
−→ dh = CP dT = f (T )
van der Waals’ gas: =⇒ h = f (T , P)
= f (T , v )
ME 6101 (2011)
7 / 22
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
8 / 22
CP − Cv
∂s
∂s
s = s(v , T ) → ds = ∂T
dT + ∂v
dv
v
h T CP
∂s
∂v
∂s
∂s
∂s
∂s
,
=
+
←=
=
∂T P
∂T v
∂v T ∂T P
∂T P
∂vT ∂v T
∂s
∂v
∂P
−→ CP − CV = T ∂v T ∂T P = T ∂T v ∂T P
∂y ∂x ∂z
f (x , y, x ) = 0 → ∂x
= −1
y ∂z x ∂y z
∂T
∂v
∂v
∂P
∂P
−→ ∂P
∂v T ∂P v ∂T P = 1 −→ ∂v T ∂T P = − ∂T v
∂v ∂v 2
∂P
= −T ∂P
CP − Cv = T ∂T
∂v T ∂T P
v ∂T P
for isobar:
⇒ dhP = CP dTP
for isotherm:
⇒ dhT = v − T
∂v
∂T P
dPT
1
T011
h2 − h1
= (h2 − h2∗ ) + (h2∗ − h1∗ ) + (h1∗ − h1 )
RP RT
∂v
= P02 v − T ∂T
dP + T12 CPo dTP
P T2
RP ∂v
− P01 v − T ∂T
dP
P T
2
3
1
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
Isothermal compressibility, kT ≡ − v1
∂v
Volume expansivity, β ≡ v1 ∂T
P
CP − Cv = −T ∂P
∂v T
Ideal gas: Pv = RT −→ kT =
1
P,
ME 6101 (2011)
9 / 22
β=
1
T
i
∂v
→ 0 ⇒ CP ≈ Cv ≈ C
For liquids & solids, ∂T
P
2
∂P
∂v
∂T P is +ve & ∂v T is -ve for all known substances, CP ≥ Cv
as T → 0, Cp → Cv , at T = 0, CP = Cv
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
10 / 22
Example: Liquid Water 1 atm & 20o C
∂v
∂P T
∂v 2
∂T P
Cv
T
=
105 N2
(206.6×10−6 K1 )2
1
m
(293K
)
1
1bar
45.95×10−6 bar 998.21 kg2
m
J
J
, ∵ CP = 4188 kg.K
7→ CP w Cv
CP − Cv = 27.29 kg.K
β
∂P
1
P = P(v , T ) → dP = ∂T
dT + ∂P
∂v T dv = kT dT − kT v dv
v
β2
kT vT
CP − Cv =
=⇒ Cp − Cv = R
for liquid water at 1 bar:
β2
kT vT
=
dP =
1
β
dv
dT −
kT
kT v
If liquid water temperature is raised form 19.5 to 20.5o C at
constant volume: dP = kβT dT 7→ ∆P = kβT ∆T
∆P =
T012
β = 0.0@4o C for water.
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
206.6×10−6 K1
bar
1
45.95×10−6 bar
2
(1K ) = 450 kPa
10 kPa
ME 6101 (2011)
11 / 22
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
12 / 22
Example: 15 cm3 Hg @0o C & 1 bar → 1000 bar
Applications of Tds Relationships
at 0o C
2nd Tds
: ···
Tds = CP dT − T
∂v
∂T P
dP
Reversible isothermal change in pressure:
∂v
−→ Tds = −T ∂T
dP
R
R ∂v P
⇒ q = −T
∂T P dP = −T βvdP = −T v β∆P
R P2
R
R ∂v v kT
2
2
PdP
=
⇒ w = − Pdv = − ∂P
P1 vkT PdP = 2 (P2 − P1 )
T
for a liquid & solid, v , β & kT are insensitive to change in P
Thermodynamics Relations
ME 6101 (2011)
= 1.5 × 10−5 m 3
= 178 × 10−6 K −1
CP
kT
= 28.6J /K
= 3.88 × 10−6 bar −1
isothermal compression: P1 = 1 bar, P2 = 1000 bar.
⇒ Q = mq = −T (mv )β∆P = −T V β∆P = −78.2 J
⇒ W = mw = V 2kT (P22 − P12 ) = 2.91 J
⇒ ∆U = Q + W = −75.29 J
78.2 J heat is liberated but only 2.91 J work is done. The extra
amount of heat comes from the store of the internal energy.
For a substance with a negative expansivity, heat is absorbed and
the internal energy is increased.
Reversible adiabatic change in pressure:
Tvβ
β
∂v
−→ dT = CTP ∂T
dP = Tv
CP dP =⇒ ∆T = CP ∆P
P
Experiments show that CP hardly changes for a solid & liquid
even for an increase of 10,000 bar.
c Dr. M. Zahurul Haq (BUET)
V
β
isentropic compression: P1 = 1 bar, P2 = 1000 bar.
=⇒ ∆T = TCvPβ ∆P = 2.55 K
13 / 22
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
14 / 22
Low Temperature Refrigeration
Clausius-Clapeyron Equation (for Phase Change)
Tc & Pc of Common Substances
s2 −s1
∂s
∂P
∂T v = ∂v T = v2 −v1 : phase change → T = Tsat = const.
∂P
dP
∂T v = dT , for mixture of 2 phases, P = f (T )
s2 −s1
h2 −h1
h12
dP
dP
dT = v2 −v1 = T (v2 −v1 ) ⇒ dT sat = Tv12 Clapeyron Eqn.
Example: using only P-v-T data, estimate hfg of R-134a at 20o C
o
vfg = (vg − vf )sat ,@20
m3 kg
C = 0.035153
Psat ,@24o C −Psat ,@16o C
dP
∆P
= 17.70 kPa/K
dT sat ,@20o C =
240 C −16o C
∆T sat ,@20o C =
dP
=
(293.15)(0.035153)(17.70)
= 182.40 kJ/kg
hfg = Tvfg dT
sat ,@20o C
o
Tabulated value of hfg @20 C is 182.27 kJ/kg.
hfg
d ln P
⇒
v2 >> v1 → v2 = RT
P
d (1/T ) sat = − R Clausius-Clapeyron Eq.
P2
∆h
1
1
=
−
−→ ln P
R
T
T
1
1
2
sat
ln Psat
B
+ C ln T + DT , widely used vapour-pressure Eq.
=A+
T
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
15 / 22
e789
If the temperature and pressure of a gas can be brought into the region
between the saturated liquid and saturated vapour lines then the gas
will become ’wet’ and this ’wetness’ will condense giving a liquid.
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
16 / 22
Liquefaction by Cooling
Liquefaction by Cooling
Liquefaction by Cooling
Liquefaction by Expansion
This method is satisfactory if the liquefaction process does not require
very low temperatures. Example butane, propane, Examples of these
are the hydrocarbons butane and propane, which can both exist as
liquids at room temperature if they are contained at elevated pressures.
Mixtures of hydrocarbons can also be obtained as liquids and these
include liquefied petroleum gas (LPG) and liquefied natural gas (LNG).
e792
1
2
3
e790
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
17 / 22
Compress isentropically to 2, where P2 > Pc
As T2 > Ta , cool it to Ta using ambient sources, and further cool
to T3 using available cold sources.
Expand isentropically form 3 to 4 ⇒ liquid formation.
c Dr. M. Zahurul Haq (BUET)
Liquefaction by Cooling
Thermodynamics Relations
ME 6101 (2011)
18 / 22
Liquefaction by Cooling
Gas Expansion & Joule-Thomson coefficient
The temperature behaviour of a fluid during a throttling process is
described by Joule-Thomson coefficient, µJT .
e772
e773
µJT

 −ve
∂T
=
≡
0
∂P h 
+ve
c Dr. M. Zahurul Haq (BUET)
: temperature increase
: temperature same
: temperature drop
Thermodynamics Relations
ME 6101 (2011)
e769
A cooling effect cannot be achieved by throttling unless the fluid is
below its maximum inversion temperature. For hydrogen its value
is -68o C and hydrogen must be cooled below this temperature if
further cooling is to be achieved.
19 / 22
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
20 / 22
Liquefaction by Cooling
Liquefaction by Cooling
Simplified Linde Liquefaction Plant
e793
P=
RT
v −b
−
a
v2
=⇒ µJT =
1
CP
"
RT
P + a2 −(v −b) 2a3
v
v
−v
#
e797
Maximum inversion temperature = 6.75Tc
Minimum inversion temperature = 0.75Tc
I If air is compressed to a pressure of 200 bar and a temperature of
52o C, after the throttling to 1 bar it will be cooled to 23o C. In case of
helium, throttling from the came condition will result in 64o C.
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
21 / 22
e796
Two performance parameters:
Yield, z : mass of liquid produced per unit mass of gas compressed.
Sp. work required, wz : work per unit mas of liquid produced.
y
h7 − h2
Win
z =
=
wz =
m
h7 − h5
z
c Dr. M. Zahurul Haq (BUET)
Thermodynamics Relations
ME 6101 (2011)
22 / 22