Availability
From chapter 8 VW&S
control volume in uniform state, uniform flow process USUF, irreversible Q c.v. and Wc.v.
what if process were reversible, how much work would have been done if the process had been reversible
<= reversible heat
engine
Wrev = Wc_v_rev +
Wc
reversible heat transfer takes place through reversible heat engine
with output Wc
(8.1)
I = irreversibility = Wrev −
Wcv
first law for uniform state, uniform process ... from first_law.mcd
uniform state, uniform flow process (USUF)
(5.54) = (8.3)
Qc_v_rev +
∑
n
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2
⎞⎟⎤⎥
⎡⎢ ⎛⎜
Vi
⎢min⋅ ⎜ hi + 2 + g⋅ zi⎟⎥ =
⎣ ⎝
⎠⎦
2
⎞⎟⎤⎥
⎡⎢
⎛⎜
Ve
⎢men⋅ ⎜ he + 2 + g⋅ ze⎟⎥ ...
⎣
⎝
⎠⎦
∑
n
2
2
⎛⎜
⎛⎜
⎞⎟
⎞⎟
V2
V1
+ m2 ⋅ ⎜ u 2 +
+ g ⋅ z2⎟ − m1 ⋅ ⎜ u 1 +
+ g ⋅ z1⎟ + Wc_v_rev
2
2
⎝
⎠
⎝
⎠
1
Wc = Qo − Qc_v_rev
for the reversible heat engine ...
ΔS
from second law
for Qo and Qc_v_rev are the same, Qo at constant temperature
t
ΔS = ΔSc_v_rev
⌠ Q_dot
c_v_rev
⎮
dt
ΔSc_v_rev =
⎮
T
⌡
Qo = To ⋅ ΔS
that is express as integral of rate
uniform state => T constant in c.v.
0
t
⌠ Q_dot
c_v_rev
⎮
=
dt
To ⎮
T
⌡
Qo
t
and
0
⌠ Q_dot
c_v_rev
⎮
dt
Qo = To ⋅
⎮
T
⌡
0
t
⌠ Q_dot
c_v_rev
⎮
dt − Qc_v_rev
Wc = Qo − Qc_v_rev = To ⋅
⎮
T
⌡
substituting into ...
(8.4)
0
t
m2 ⋅ s2 − m1 ⋅ s1 +
also for USUF
from second law
∑ (
me⋅se)
− ∑ (
)
n
substituting (8.5) into (8.4) ...
⌠ Q_dot
c_v
⎮
dt
mi ⋅ si =
⎮
T
⌡
n
0
Wc = Qo − Qc_v_rev = To ⋅ ⎡m2 ⋅ s2 − m1 ⋅ s1 +
⎢
⎣
so ... the bottom line, substitute (8.3) rearranged and (8.6) into (8.1)
∑ (
me⋅se)
− ∑ (
mi⋅si)⎤⎥ − Qc_v_rev
n
⎦
n
Wrev = Wc_v_rev + Wc
2
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥ ⎡ ⎡⎢
⎛⎜
⎞⎟⎤⎥⎤
Vi
Ve
⎢
⎥
Wrev = Qc_v_rev +
⎢min⋅ ⎜ h i + 2 + g ⋅ zi⎟⎥ − ⎢ ⎢men⋅ ⎜ he + 2 + g⋅ ze⎟⎥⎥ ...
⎣ ⎝
⎠⎦
⎣
⎝
⎠⎦
n
⎣n
⎦
2
2
⎡⎢ ⎛⎜
⎞⎟
⎛⎜
⎞⎟⎤⎥
V1
V2
+ −⎢m2 ⋅ ⎜ u 2 +
+ g ⋅ z2⎟ − m1 ⋅ ⎜ u 1 +
+ g ⋅ z1⎟⎥ ...
2
2
⎣ ⎝
⎠
⎝
⎠⎦
+ To ⋅ ⎡m2 ⋅ s2 − m1 ⋅ s1 +
me⋅ se)
−
mi⋅ si)⎤ − Qc_v_rev
(
(
⎢
⎥
n
n
⎣
⎦
∑
(7.56) = when
reversible (8.5)
∑
∑
∑
} (8.3)
} (8.6)
Qc_v_rev cancels and rearranging (moving T o and s terms into mass flow terms) ...
reversible work (maximum) of a control volume that exchanges heat with the
surroundings at To
2
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥ ⎡ ⎡⎢
⎛⎜
⎞⎟⎤⎥⎤
Vi
Ve
⎢
⎥
Wrev =
⎢min⋅ ⎜ hi − To ⋅ si + 2 + g⋅ zi⎟⎥ − ⎢ ⎢men⋅ ⎜ h e − To⋅ se + 2 + g ⋅ ze⎟⎥⎥ ...
⎣ ⎝
⎠⎦
⎣
⎝
⎠⎦
n
⎣n
⎦
2
2
⎞
⎞
⎤
⎡⎢ ⎛⎜
⎛
V
V2
⎟
⎜
⎟⎥
1
+ −⎢m2 ⋅ ⎜ u 2 − To ⋅ s2 +
+ g ⋅ z2⎟ − m1 ⋅ ⎜ u 1 − To ⋅ s1 +
+ g ⋅ z1⎟⎥
2
2
⎣ ⎝
⎠
⎝
⎠⎦
∑
∑
(8.7)
latter [..] is total for
c.v.
two special cases: a system (fixed mass) and steady-state, steady flow process for a control volume
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2
system (fixed mass)
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥
Vi
⎢min⋅ ⎜ h i − To⋅si + 2 + g ⋅ zi⎟⎥ = 0
⎣ ⎝
⎠⎦
2
⎡⎢
⎛⎜
⎞⎟⎤⎥
Ve
⎢men⋅ ⎜ h e − To⋅se + 2 + g ⋅ ze⎟⎥ = 0
⎣
⎝
⎠⎦
∑
∑
n
n
m1 = m2 = m
system (fixed mass
Wrev_1_2
m
2
⎛⎜
⎞⎟
V1
= wrev_1_2 = ⎜ u 1 − To ⋅s1 +
+ g ⋅ z1 ⎟ −
2
⎝
⎠
2
⎛⎜
⎞⎟
V2
⎜ u 2 − To ⋅s2 + 2 + g⋅ z2⎟
⎝
⎠
(8.8)
steady-state, steady flow process
2
2
⎛⎜
⎞⎟
⎛⎜
⎞⎟
V2
V1
m2 ⋅ ⎜ u 2 − To ⋅s2 +
+ g ⋅ z2⎟ − m1 ⋅ ⎜ u 1 − To ⋅ s1 +
+ g ⋅ z1 ⎟ = 0
2
2
⎝
⎠
⎝
⎠
steady-state, steady flow process - rate form
W_dot rev =
2
⎡⎢ ⎛⎜
⎞⎟⎤⎥ ⎡
Vi
⎢
⎢min⋅ ⎜ hi − To ⋅si + 2 + g⋅ zi⎟⎥ − ⎢
⎣ ⎝
⎠⎦
⎣n
∑
n
∑
2
⎡⎢
⎛⎜
⎞⎟⎤⎥⎤
Ve
⎥
⎢men⋅ ⎜ h e − To⋅se + 2 + g ⋅ ze⎟⎥⎥
⎣
⎝
⎠⎦
⎦
(8.9)
single flow of fluid
2
⎛⎜
⎞⎟
Ve
= wrev = h i − To ⋅si +
+ g ⋅ zi − ⎜ h e − To ⋅se +
+ g ⋅ z e⎟
2
2
m_dot
⎝
⎠
W_dot rev
Vi
2
(8.10)
above represents maximum work for given change of state of a system
what is maximum work that can be done by system in a given state???
answer: when system is in equilibrium with the environment, no spontaneous change of state can occur, and is
incapable of doing work. therefore if system in a given state undergoes a completely reversible process until it
is in equilibrium with the environment, the maximum reversible work will have been done by the system
steady state, steady flow
process ...(e.g. single flow)
2
⎛⎜
⎞⎟
Ve
wrev = h i − To ⋅si +
+ g ⋅ zi − ⎜ h e − To ⋅se +
+ g ⋅ z e⎟
2
2
⎝
⎠
Vi
2
(8.10)
maximum when mass leaving c.v. is in equilibrium with environment. define ψ = availability (per unit mass flow)
steady state, steady flow process ...(e.g. single flow ...availability (per unit mass flow)
2
⎛⎜
⎞⎟
Vo
ψ = h − To ⋅s +
+ g ⋅ z − ⎜ h o − To ⋅so +
+ g ⋅ zo ⎟
2
2
⎝
⎠
2
V
(8.16)
reversible work between any two states = decrease in availablity between them
(
)
(
)
wrev = ψ i − ψ e = h 1 − To ⋅s1 − h 2 + To ⋅s2 = h 1 − To ⋅s1 − h 2 + To ⋅s2 = h 1 − h 2 − To ⋅ s1 − s2 (8.17) extended
extension
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3
W − dotrev =
can be written for more than one
flow ...
∑ ⎛⎝ min⋅ψ in⎞⎠ − ∑ ⎛⎝ men⋅ψ en⎞⎠
n
n
(8.18)
for a system (no flow across the control surface)
. need to account for work done by system against the surroundings ...
... assume kinetic and potential energy changes negligible ...
Wrev_1_2
m
2
2
⎛⎜
⎞⎟ ⎛⎜
⎞⎟
V1
V2
= wrev_1_2 = ⎜ u 1 − To ⋅ s1 +
+ g ⋅ z1⎟ − ⎜ u 2 − To ⋅s2 +
+ g ⋅
z2⎟
2
2
⎝
⎠ ⎝
⎠
(
) (
wrev_1_2 = u 1 − To ⋅s1 − u 2 − To ⋅s2
becomes ...
(
) (
wrev_max = u − To ⋅s − u o − To ⋅so
(8.8)
)
)
(8.19)
availability per unit mass is then ... this maximum work - that done against the surroundings
(
)
(
Wsurr = p o ⋅ Vo − V = m⋅ p o ⋅ v o − v
)
(8.20)
(
) (
)
(
(
)
φ = availability_w_o_KE_PE = wrev_max − wsurr = u − To ⋅s − u o − To ⋅so + p o ⋅ v − v o
)
availability w/o KE and PE per unit mass of system
(
) (
)
(
)
φ = u + p o ⋅v − To ⋅s − u o + p o ⋅v o − To ⋅so = u − u o + p o ⋅ v − v o − To ⋅ s − so
(8.21)
and reversible work maximum between states 1 and 2 is ...
2
(
)
wrev_1_2 = φ1 − φ2 − p o ⋅ v 1 − v 2 +
V1 − V2
2
2
(
+ g ⋅ z1 − z2
)
(8.22)
check ...
(
)
(
φ1 := u 1 − u o + p o ⋅ v 1 − v o − To ⋅ s1 − so
2
(
)
wrev_1_2 := φ1 − φ2 − p o ⋅ v 1 − v 2 +
V1 − V2
2
)
(
)
(
φ2 := u 2 − u o + p o ⋅ v 2 − v o − To ⋅ s2 − so
)
2
(
+ g ⋅ z1 − z2
)
1
2 1
2
wrev_1_2 simplify → u 1 − To ⋅ s1 − u 2 + To ⋅s2 + ⋅V1 − ⋅ V2 + g ⋅ z1 − g ⋅ z2
2
2
matches ...
Wrev_1_2
m
9/27/2006
2
2
⎞⎟ ⎛⎜
⎞⎟
⎛⎜
V1
V2
= wrev_1_2 = ⎜ u 1 − To ⋅s1 +
+ g ⋅ z1⎟ − ⎜ u 2 − To ⋅s2 +
+ g ⋅ z2 ⎟
2
2
⎝
⎠ ⎝
⎠
4
(8.8) from above
geothermal well example
3
3
kPa := 10 Pa
define some units ...
kJ := 10 J
example ... geothermal well
water as saturated liquid issues from a process at 200 deg C. What is maximum power if the environment is at 10^5 N/m^2 at 30 deg C
T1_C := 200
2
⎛⎜
⎞⎟
Vo
ψ = h − To ⋅s +
+ g ⋅ z − ⎜ h o − To ⋅so +
+ g ⋅ zo ⎟
2
2
⎝
⎠
2
V
1 saturated
(
)
T1 := 273 + T1_C ⋅K
environment
(dead state)
p 0 := 10
5 N
2
(8.16)
T1 = 473 K p 1 := 1.5538MPa
(
T0_C := 30
h 1 := 852.45
kJ
kg
kJ
s1 := 2.3309
kg⋅
K
)
T0 := 273.16 + T0_C ⋅K T0 = 303.16 K
p 0 = 100 kPa
m
water at this state is "compressed liquid" as pressure exceeds saturation pressure at 30 deg C
p sat_30 := 4.246kPa
ref: water saturated liquid at 30 deg C
v f_30 := 1.004 × 10
3
−3m
h f_30 := 125.79
kg
kJ
kJ
sf_30 := 0.437
kg⋅ K
kg
limited values for compressed liquid are in Table A.1.4 well beyond this pressure
water is ~ incompessible
values for u, v and s can be estimated to be the saturation values at the T so... (see example validation below)
v 0 := v f_30
s0 := sf_30
but work must be done to compress to higher pressure than saturated
estimate from definition of enthalpy:
2
2
2
2
⌠
⌠
⌠
⌠
h = u + p⋅ v
dh = du + p ⋅ dv + v ⋅ dp
⎮ 1 dh = h 2 − h 1 = ⎮ 1 du + ⎮ p dv
+ ⎮ v dp
⌡
⌡
⌡
⌡
1
2
⌠
⎮ 1 du = cv ⋅ T2 − T1
⌡
we'll see later
(
)
1
dv ~ 0 <<<1 => v = constant so ...
1
2
⌠
h 2 − h 1 = v ⎮ 1 dp = v f( T)⋅ p 2 − p 1
⌡
(
)
1
(
h 0 := h f_30 + v f_30⋅ p 0 − p sat_30
(
ψ := h 1 − T0 ⋅s1 − h 0 − T0 ⋅s0
9/27/2006
)
)
ψ = 152.409
1
h 0 = 125.886
kJ
kg
5
kJ
kg
1
to show example of estimates u, v, and s of compressed liquid = saturation, but not h consider value in Table vs
stauration at T
i := 1 .. 4
Table A.1.1.1
saturation
t=40 deg C
Table A.1.1.4
p=10MPa
T=40 deg C
difference
(col 2 - col 3)
-------------------col2
⎛⎜ p 10⋅ 106 7.384 ⋅ 103 ⎞⎟
⎜ v 0.0010034 0.001008 ⎟
⎟
data := ⎜ u 166.35
167.56 ⎟
⎜
167.57 ⎟
⎜ h 176.38
⎜⎝
s 0.5686
0.5725 ⎟⎠
data
i, 1
− data
data
i, 2
⋅100 =
i, 1
-0.5
-0.7
differences all < 1% for 106 pressure
difference
h differs by 5%
5
-0.7
using estimate from definition of enthalpy
h := data
kJ
3 , 2 kg
3
+ data
m
1 , 2 kg
and difference is ...
(
⋅ data
data
0, 1
Pa − data
kJ
3 , 1 kg
data
9/27/2006
kJ
6
Pa
)
−h
3 , 1 kg
geothermal well example
0, 2
⋅100 = −0.716
h = 177.643
kJ
kg
difference now < 1%