define some units
Some Relationships for Gases
these are extracted from Van Wylen & Sonntag, Fudamentals of Classical
Thermodynamics, Third Edition to which page numbers and equation numbers apply
2006: included reference to text: Woud section 2.23 in [W n.nn]
3
kJ := 10 ⋅J
3
kmol := 10 mole
section 3.4 Equations of state for the vapor phase
of a simple compressile substance - page 41
(Woud page 20)
⎯ ⎯
p ⋅ v = R⋅T
gas at low density (experiment)
R_bar := 8.3144
R=
⎯
R
kJ
p ⋅ V = m⋅ R⋅T
or ...
p ⋅ v = R⋅ T
for R_bar above
p 1 ⋅v 1
(3.2)
(3.1)
kJ
units sometimes ...
kmol⋅ K
mw = molecular_weight
mw
⎯
R = universal_gas_constant
⎯
xxx = mole_basis
T1
=
mw =
p 2 ⋅v 2
kg_mol⋅ K
kg
kmol
[W 2.32, 2.33[
(3.5)
T2
section 4.3 Work done at moveable boundary of
simple compressible system - page 63
V
if ...
n
p ⋅ V = constant
V
⌠ 2 1
⌠ 2
⎛ V2 ⎞
dV = p 1 ⋅ V1 ⋅ln⎜
W1_2 = ⎮
p dV = p 1 ⋅ V1 ⋅ ⎮
⎟
V
⎮
⌡V
⎝ V1 ⎠
1
⌡V
n=1
(4.5)
1
section 5.6 The Constant-Volume and
Constant-Pressure Specific Heats - page 98
specific heat = increment of heat Q to change T by 1 deg
two cases: 1) constant volume
δQ = dU + p⋅δV
cp =
10/23/2006
cv =
δQ = dU + p⋅δV = δH
1 δQ
1 δH
δh
⋅
= ⋅
=
m δT
m δT
δT
cp =
1
m
= specfic
constant volume
(5.4)
δW = p⋅δV = 0
dKE = dPE = 0
1 δQ
1 δU
δu
⋅
= ⋅
=
m δT
m δT
δT
2) constant pressure
1 δQ
⋅
m δT
1 δQ
⋅
m δT
δQ = dE + δW = dU + dKE + dPE + δW
1st law ...
cv =
cv =
c=
δu
δT
as ....
δh
δT
1
constant volume
(5.14)
[W 2.36]
dH = d ( U + p ⋅ V) = dU + p ⋅ dV + V⋅ dp dp = 0
constant pressure
(5.15)
[W 2.37]
section 5.7 The Internal Energy, Enthalpy
and Specific Heats of Ideal Gases - page 100
p ⋅ v = R⋅ T
ideal gas
cv =
also ...
δu
experiment (Joule)
u = f( T)
du = cvo⋅dT
u not a function of v =>
δT
h = u + p ⋅ v = u ( T) + R ⋅ T = h ( T)
cp =
δh
δT
po
cpo = cvo + R
(
du
dT
+R
differentiate w.r.t T
(5.27)
)
section 7.10 Entropy Change of an
Ideal Gas - page 206
du = cvo⋅dT
p ⋅ v = R⋅ T
and ...
dT
dv
ds = cvo⋅
+ R⋅
T
v
(7.19)
T⋅ ds = dh − v ⋅dp
dh = cpo⋅dT
dp
T⋅ ds = du + p ⋅ dv
(7.21)
=>
T
p = R⋅
v
=>
(7.7)
p ⋅ v = R⋅T
=>
(5.29)
[W 2.38]
[W 2.18]
T
T⋅ ds = cvo⋅dT + R⋅ ⋅dv
v
⎛ T2 ⎞
⎛ v2 ⎞
s2 − s1 = cvo⋅ln⎜
⎟ + R⋅ ln⎜ ⎟ (7.24)
⎝ T1 ⎠
⎝ v1 ⎠
(7.7)
and ...
ds = cpo⋅
− R⋅
T
p
10/23/2006
=
h 2 − h 1 = cpo⋅ T2 − T1
otherwise integrate if c(T) known or tables
with constant c
dT
dT
cpo − cvo = R
or ...
(5.24)
=> constant pressure
ideal gas
dh
h = u + pv = u + R⋅ T
relation between c vo and c po ...
(5.20)
i.e. h = F(T) only
dh = cpo⋅dT
=>
=> constant volume
ideal gas
vo
cvo = constant
otherwise integrate or use
tables
[W 2.21]
=>
v = R⋅
T
=>
p
⎛ T2 ⎞
⎛ p2 ⎞
s2 − s1 = cpo⋅ln⎜
⎟ − R⋅ ln⎜ ⎟
⎝ T1 ⎠
⎝ p1 ⎠
2
T
T⋅ ds = cpo⋅dT − R⋅ ⋅dp
p
(7.23)
cpo = constant
otherwise integrate or use
tables
γ=
page 211 introduce specific heat
ratio γ
cpo
cpo − cvo = R
from above
(7.30)
cvo = cpo − R = γ⋅cvo − R
changing signs ...
cvo =
similarly ...
γ = γ ( T) [W 2.44]
cvo
cpo − cvo = R
c = c ( T)
cvo⋅ ( 1 − γ ) = −R
=>
R
(7.31)
γ−1
cpo = R + cvo = R +
cpo = R⋅
as ...
cpo
cpo⋅ ⎛⎜ 1 −
=>
γ
⎝
γ
1⎞
γ−1
⎟ = R = cpo⋅
γ⎠
γ
(7.31)
γ−1
for constant specific heat = perfect gas
T⋅ ds = du + p ⋅dv = 0
reversible, adiabatic process ...
du + p ⋅dv = cvo⋅dT + p ⋅ dv
reversible, adiabatic ...
p ⋅ v = R⋅ T
=>
dT =
1
R
⋅( p ⋅ dv + v ⋅ dp)
cvo
cvo
R 1
0 = cvo⋅dT + p ⋅ dv =
⋅(p ⋅ dv + v ⋅ dp) + p ⋅ dv =
⋅(p ⋅ dv + v ⋅ dp) + p ⋅ dv =
⋅ ⋅( p ⋅ dv + v ⋅ dp) + p ⋅ dv
R
R
γ−1 R
( p ⋅ dv + v ⋅ dp) + p ⋅ dv( γ − 1 ) = v ⋅ dp + γ⋅p⋅dv = 0
=>
ln( p ) + γ⋅ln( v ) = constant
integrating ...
p1⋅ v1
T1
γ
+ γ⋅
dv
v
γ
or ...
dividing by pv
= 0
(7.32)
p ⋅ v = constant
[W 2.49]
γ
=
p2⋅ v2
γ
v1
=>
T2
v2
γ
⎛ v 1 ⎞ ⎛ T1 p 2 ⎞ ⎛ T1 ⎞ ⎛ p 2 ⎞
=⎜ ⎟ =⎜
⋅ ⎟ =⎜
⎟ ⋅⎜ ⎟
p1
⎝ v 2 ⎠ ⎝ T2 p 1 ⎠ ⎝ T2 ⎠ ⎝ p 1 ⎠
p2
p
⎛ v 1 ⎞
= ⎜ ⎟
p1
⎝ v 2 ⎠
p2
rearranging ...
for ideal gas
dp
=>
p 2 T1
T1 p 2
⋅
=
⋅
T2 p 1
T2 p 1
=
γ
=>
⎛ p2 ⎞
⎜ ⎟
⎝ p1 ⎠
1−γ
=>
γ
⎛ T1 ⎞
=⎜
⎟
⎝ T2 ⎠
γ
⎛ T1 ⎞
=⎜
⎟
p1
⎝ T2 ⎠
p2
1−γ
γ
γ−1
⎛ T2 ⎞
=⎜
⎟
⎝ T1 ⎠
and ... for reversible adiabatic process constant specific heat (ideal gas)
⎛ p2 ⎞
=⎜ ⎟
T1
⎝ p1 ⎠
T2
10/23/2006
γ−1
γ
γ
γ−1
γ−1
⎛ v 1 ⎞
= ⎜ ⎟
⎝ v 2 ⎠
(7.35)
or ...
T⋅ p
T⋅ v
3
γ−1
1−γ
= T⋅ p
γ
= constant
= constant
[W 2.47]
[W 2.48]
to explore the effect of temperature on the coefficients of specific heat the following is
provided
to calculate enthalpies at non-standard conditions one approach is to use constant-pressure specific heats from
Table A.9 of Van Wylen and Sonntag ... The applicable temperature range for these materials is 300 - 3500 deg K
with less than approximately 0.5 % maximum error to experimental values. see also figure 5.10 on page 103.
O2
Cpo_O2 ( θ ) := 37.432 + 0.020102⋅ θ
C_O2
Cpo_C_O2 ( θ ) := −3.7357 + 30.529⋅ θ
N2
Cpo_N2 ( θ ) := 39.060 − 512.79⋅ θ
H2_O
Cpo_H2_0 ( θ ) := 143.05 − 183.54⋅θ
1.5
− 178.57⋅ θ
0.5
− 1.5
+ 236.88⋅ θ
− 4.1034⋅θ + 0.024198⋅ θ
+ 1072.7⋅ θ
0.25
− 1.5
−2
+ 82.751⋅θ
− 820.40⋅ θ
0.5
−2
θ =
kJ
T
100
kmol⋅ K
given in
kJ/kmole*K so
divide by
molecular weight
to get kJ/kg*K
kJ
2
kmol⋅ K
−3
kJ
kmol⋅ K
kJ
− 3.6989⋅θ
kmol⋅ K
air based on volumetric average as these are on mole basis ...
air
Cpo_air( θ ) := 0.21⋅ Cpo_O2 ( θ ) + 0.79⋅ Cpo_N2 ( θ )
T := 300 .. 3500
Specific Heat (Constant Pressure) Cp
Specific Heat (Constant Pressure) Cp
70
70
60
specific heat
specific heat
60
O2
CO2
N2
H2O
air
50
40
20
200
400
600
800
50
40
20
0
1000
1000
2000
Temperature deg K
Temperature deg K
relatively constant at reasonably low temperatures
10/23/2006
O2
CO2
N2
H2O
air
30
30
4
T in deg
K
3000
the next section was added Nov 2005 to organize plots for Brayton and dual (Seiliger) cycles
applications of above relationships to processes
R := 0.287
γ := 1.4
kJ
cpo := 1.0035
cvo :=
kg⋅
K
cpo
cvo = 0.717
γ
units assumed are p = bar, v = m 3 /kg, T = K, s = kJ/kg*K
100 necesary for consistency in numerical calculations without units.
isentropic adiabatic compression (expansion) _______________________________
pressure ratio known
p initial := 1
Tinitial := 400
sinitial := 1
v initial :=
R⋅ Tinitial
v initial = 1.148
p initial⋅100
γ−1
⎛ p final ⎞
Tfinal := Tinitial⋅ ⎜
⎟
⎝ pinitial ⎠
p final := 10
1
γ
⎛ pinitial ⎞
v final := v initial⋅ ⎜
⎟
⎝ p final ⎠
sfinal := sinitial
Tfinal = 772.279
γ
v final = 0.222
sfinal = 1
p - v plot parameterized on either p or v say p
δp :=
p final − p initial
to plot 20 points and accomodate increases and decreases
20
pp := p initial , p initial + δp .. p final
T - s is straight line, need 2 points
1
γ
⎛ p initial ⎞
v_plot( pp) := v initial⋅ ⎜
⎟
⎝ pp ⎠
Ts_plot :=
⎛⎜ Tinitial sinitial ⎞⎟
⎜⎝ Tfinal sfinal ⎟⎠
p-v isentropic compression (expansion)
p
T
800
5
600
400
0.99
0.4
0.6
0.8
1
1.2
0.995
1
s
v
10/23/2006
⎛ 400 1 ⎞
⎜
⎟
⎝ 772.279 1 ⎠
T-s isentropic compression (expansion)
10
0
0.2
Ts_plot =
5
1.005
isentropic adiabatic compression (expansion) _______________________________
volume ratio known
v initial := 1
Tinitial := 400
sinitial := 1
γ−1
⎛ vinitial ⎞
Tfinal := Tinitial⋅ ⎜
⎟
⎝
v final ⎠
1
v final :=
10
Tfinal = 1.005 × 10
R⋅ Tinitial
p initial :=
p initial = 1.148
v initial⋅100
γ
⎛ vinitial ⎞
p final := p initial⋅ ⎜
⎟
⎝
v final ⎠
sfinal := sinitial
3
p final = 28.836
sfinal = 1
p - v plot parameterized on either p or v say p
δp :=
p final − p initial
to plot 20 points and accomodate increases and decreases
20
pp := p initial , p initial + δp .. p final
T - s is straight line, need 2 points
1
γ
⎛ pinitial ⎞
pv_plot( pp) := v initial⋅ ⎜
⎟
⎝
pp ⎠
Ts_plot :=
⎛⎜ Tinitial sinitial ⎞⎟
⎜⎝
Tfinal sfinal ⎟⎠
p-v isentropic compression (expansion)
Ts_plot =
1⎞
⎛ 400
⎜
3 ⎟
⎝
1.005 × 10 1 ⎠
T-s isentropic compression (expansion)
40
1500
p
T
1000
20
500
0
0.99
0
0
0.2
0.4
0.6
or ....
1
s
0.8
v
δv :=
v final − v initial
to plot 20 points and accomodate increases and decreases
20
vv := v initial , v initial + δv .. v final
γ
⎛ vinitial ⎞
pv_plot( vv) := p initial⋅ ⎜
⎟
⎝
vv ⎠
p-v isentropic compression (expansion)
40
p
0.995
20
0
same plot as above
0
0.2
0.4
0.6
0.8
v
10/23/2006
6
1.005
heat transfer at constant pressure .... ________________________________________
set up to go from T initial to Tfinal at p = constant, sinitial assumed = 1
Tinitial := 298
sinitial := 1
Tfinal := 500
p constant := 3 bar
v initial = 1
final end state calculation
⎛ Tfinal ⎞
sfinal := sinitial + cpo⋅ ln⎜
⎟
⎝ Tinitial ⎠
sfinal = 1.519
v final = 0.1
inbetween states for plotting .......
p - v is a straight line
needing only end
points to plot
pv_plot :=
if desired for a simulation where interim p v points are
required, choose one parameter, e.g. T or v and
calculate v or T using pv/T = R for the other
⎛⎜ p constant vinitial ⎞⎟
⎜⎝
p constant v final ⎟⎠
pv_plot =
TT := Tinitial .. Tfinal
⎛3 1 ⎞
⎜
⎟
⎝
3 0.1 ⎠
splot(TT) := ⎛⎜ sinitial + cpo⋅ ln⎛⎜
T
⎞⎞
⎟⎟
⎝ initial ⎠ ⎠
⎝
heat transfer constant pressure
TT
heat transfer constant pressure
3.004
500
3.002
400
T
p
3
300
2.998
2.996
0
0.2
0.4
0.6
0.8
200
1
1.2
1.4
v
s
10/23/2006
7
1.6
heat transfer at constant volume .... ________________________________________
set up to go from T initial to Tfinal at v = constant, s initial assumed = 1
Tinitial := 450
sinitial := 1
Tfinal := 298
v constant := 0.3
p initial :=
R⋅ Tinitial
p initial = 4.305
v constant⋅100
final end state calculation
⎛ Tfinal ⎞
sfinal := sinitial + cvo⋅ ln⎜
⎟
⎝ Tinitial ⎠
sfinal = 0.705
p final :=
R⋅ Tfinal
p final = 2.851
v constant⋅100
inbetween states for plotting .......
p - v is a straight line
needing only end
points to plot
pv_plot :=
⎛⎜ p initial
⎜⎝
pfinal
pv_plot =
if desired for a simulation where interim p v points are
required, choose one parameter, e.g. T or v and
calculate v or T using pv/T = R for the other
v constant ⎞
⎟
⎠
TT := Tinitial .. Tfinal
v constant ⎟
⎛ 4.305 0.3 ⎞
⎜
⎟
⎝
2.851 0.3 ⎠
splot(TT) := ⎛⎜ sinitial + cvo⋅ ln⎛⎜
T
⎝
heat transfer constant volume
4.5
450
4
400
3.5
T
p
heat transfer constant volume
⎞⎞
⎟⎟
⎝ initial ⎠ ⎠
TT
350
3
300
2.5
0.2996
0.2998
0.3
0.3002
0.3004
250
0.7
v
0.8
0.9
s
10/23/2006
8