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Chapter 3.
Total Energy Is Not Created
or Destroyed
2012/2/6
1
3.1 CONSERVATION OF ENERGY
⎛ v2
⎞
θ = U + M ⎜⎜ + ψ ⎟⎟
⎝2
⎠
ACC = Input − Output
⎛ v2
⎞ ⎪⎫ ⎛ Rate at which energy ⎞ ⎛ Rate at which energy ⎞
d ⎪⎧
U
M
ψ
+
+
⎨
⎜
⎟⎬ = ⎜
⎟−⎜
⎟
dt ⎪⎩
⎝ 2
⎠ ⎪⎭ ⎝ enters the system ⎠ ⎝ leaves the system ⎠
(3.1-1)
Here U is the total internal energy, v2/2 is the kinetic energy
per unit mass (where v is the center of mass velocity), and Ψ
is the potential energy per unit mass.
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2
2
3
4
Q
⎛ ∧ v2
⎞
⎜ U + +ψ ⎟
2
⎝
⎠k
W = Ws + Wpv
⎛ ∧ v2
⎞
⎜ U + +ψ ⎟
2
⎝
⎠k
1
⎞
⎛ v2
U + M ⎜⎜ + ψ ⎟⎟
⎠
⎝ 2
⎛
⎞
v
⎜ U + +ψ ⎟
2
⎝
⎠k
∧
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5
Wflow
2
⎛ ∧ v2
⎞
⎜ U + +ψ ⎟
2
⎝
⎠k
3
Various energy enter/leave the system
„
(1) Energy flow accompanying mass flow
K
) u2
⎞
⎛
&
∑ M i ⎜⎜U + 2 + ψ ⎟⎟
(3.1-2)
k =1
⎠k
⎝
‰
„
)
where U is the internal energy per unit mass of the k th
flow stream, and M& i is its mass flow rate.
(2) Heat
Q& = ∑ Q& j
j
‰
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where Q& j is the heat flow at the j th heat flow port.
4
Work(3, 4)
W = Ws + Wpv
(3) Shaft work, Ws, the mechanical energy flow that occurs
without a deformation of the system boundaries. Examples,
steam turbine, internal combustion engine, pump, and
compressor.
(4) PV work, the movement of the system boundary
d ( LA )
dL
dV
&
W = −F
= − ( F / A)
= −P
dt
dt
dt
‰
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(3.1-3)
where P is the pressure exerted by the system at its boundaries.
It represents that work done on the system in compression is
positive,
5
Flow work (5) Wflow
5. Work of a flowing fluid against pressure
Figure 3.1-1 Flow through a valve
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⎛ Work done by surrounding fluid in ⎞
)
⎜
⎟
M
PV
Δ
=
pushing
fluid
element
of
mass
(
)
1 1ΔM 1
⎜
1⎟
⎜ into the valve
⎟
⎝
⎠
⎛ Work done on surrounding fluid ⎞
⎜
⎟
⎜ by movement of fluid element of ⎟
)
⎜ mass ( ΔM ) out of the valve (since ⎟ = − P V ΔM
2 2
2
2
⎜
⎟
⎜ this fluid element is pushing the ⎟
⎜
⎟
⎝ surrounding fluid
⎠
⎛
)
)
on the system due to ⎞
⎜⎜ Net work done
⎟⎟ = PV
1 1ΔM 1 − P2V2 ΔM 2
movement
of
fluid
⎝
⎠
For a more general system, with several mass flow ports,
⎛ Net work done on the system due
⎜
⎜ to the pressure forces acting on
⎜ the fluids moving into and out of
⎜
⎝ the system
⎞
⎟ K
)
⎟ = ΔM PV
k
⎟ ∑
k =1
⎟
⎠
⎛ Net rate at which work in done on
⎜
⎜ the system due to the pressure forces
⎜ acting on the fluids moving into and out
⎜
⎝ of the system
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( )k
⎞
⎟ K
)
⎟ = M& PV
k
⎟ ∑
k =1
⎟
⎠
( )k
7
Energy balance (mass basis)
⎛ v2
⎞ ⎫⎪ K & ⎛ ˆ v 2
⎞
d ⎧⎪
⎨U + M ⎜ + ψ ⎟ ⎬ = ∑ M k ⎜ U + + ψ ⎟ + Q&
2
dt ⎪⎩
⎝ 2
⎠ ⎪⎭ k =1
⎝
⎠k
( )
K
dV
& −P
+ Ws
+ ∑ M& k PVˆ
dt k =1
(3.1-4)
k
⎛ v2
⎞⎫ K & ⎛ ˆ v 2
⎞
d⎧
⎜
⎟
⎜
⎨U + M ⎜ + ψ ⎟⎬ = ∑ M k ⎜ H + + ψ ⎟⎟ + Q& + W&
2
dt ⎩
⎝2
⎠⎭ k =1 ⎝
⎠k
(3.1-4a)
where
& − P dV
W& = Ws
dt
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(3.1-4a)
No Acc.
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Energy Balance (molar basis)
⎛ v2
⎞⎫ K & ⎛
⎛ v2
⎞⎞
d⎧
⎨U + Nm⎜⎜ + ψ ⎟⎟⎬ = ∑ Nk ⎜⎜ H + m⎜⎜ + ψ ⎟⎟ ⎟⎟ + Q& + W&
dt ⎩
⎝2
⎠⎭ k =1 ⎝
⎝2
⎠ ⎠k
(3.1-4b)
Since:
M = Nm;
M& k = N& k m
H: total enthalpy
∧
H : specific enthalpy
H : molar enthalpy
∧
mH = H
) v2
) v2
⎛
⎞
⎛
⎞
&
&
M k ⎜⎜ H + + ψ ⎟⎟ = Nk m⎜⎜ H + + ψ ⎟⎟
2
2
⎝
⎠
⎝
⎠
2
2
)
⎛
⎛
⎞
⎞
⎞⎞
⎛
⎛
v
v
&
&
= N k ⎜⎜ mH + m⎜⎜ + ψ ⎟⎟ ⎟⎟ = Nk ⎜⎜ H + m⎜⎜ + ψ ⎟⎟ ⎟⎟
⎠⎠
⎠⎠
⎝2
⎝2
⎝
⎝
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Commonly Used Forms of Energy balance
Mass basis
( )
K
d
{U } = ∑ M& k Hˆ k + Q& + W&
dt
k =1
⎛ v2
⎞
M ⎜⎜ + ψ ⎟⎟ << U ;
⎝2
⎠
(3.1-5a)
v2
+ ψ << Hˆ
2
Molar basis
K
d
{U } = ∑ N& k (H )k + Q& + W&
dt
k =1
(3.1-5b)
⎛ v2
⎛ v2
⎞
⎞
Nm⎜⎜ + ψ ⎟⎟ << U ; m⎜⎜ + ψ ⎟⎟ << H
⎝2
⎠
⎝2
⎠
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To obtain its difference form from above:
Integrate from t1 to t2
⎛ v2
⎞ ⎫⎪ K & ⎛ ˆ v 2
⎞
d ⎧⎪
⎨U + M ⎜ +ψ ⎟ ⎬ = ∑ M k ⎜ H + +ψ ⎟ + Q& + W&
2
dt ⎪⎩
⎝ 2
⎠ ⎪⎭ k =1
⎝
⎠k
⎧⎪
⎛ v2
⎞ ⎫⎪ ⎧⎪
⎛ v2
⎞ ⎫⎪
⇒ ⎨U + M ⎜ +ψ ⎟ ⎬ − ⎨U + M ⎜ +ψ ⎟ ⎬
⎪⎩
⎝ 2
⎠ ⎪⎭t2 ⎪⎩
⎝ 2
⎠ ⎪⎭t1
K
t2
k =1
t1
= ∑∫
Q=∫
t2
t1
2
⎛
⎞
v
ˆ
&
M k ⎜ H + +ψ ⎟ dt + Q + W
2
⎝
⎠k
& ;
Qdt
W = Ws − ∫
V ( t2 )
V ( t1 )
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Ws = ∫
t2
t1
W&s dt ;
V ( t2 )
∫V (t )
1
(3.1-6)
t2
PdV = ∫ P
t1
dV
dt
dt
PdV
12
„
The first term on the right side of Eq. 3.1-6 is
usually the most troublesome to evaluate because
the mass flow rate and/or the thermodynamic
properties of the flowing fluid may change with time.
However, if the thermodynamic properties of the
fluids entering and leaving the system are
independent of time, we have
K
∑ ∫t
t2
k =1
1
2
2
K ⎛
⎛
⎞
⎞ t2 &
v
v
&
ˆ
ˆ
M k ⎜⎜ H + + ψ ⎟⎟ dt = ∑ ⎜⎜ H + + ψ ⎟⎟ ∫ M k dt
t
2
2
k =1 ⎝
⎝
⎠k
⎠k 1
⎞
⎛ ˆ v2
= ∑ ΔM k ⎜⎜ H + + ψ ⎟⎟
2
k =1
⎠k
⎝
K
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(3.1-7)
13
When PE, KE<<0, Ws=0, for a single
flow system, (3.1-5a) becomes:
dM
= M&
dt
dU
dV
&
&
ˆ
= MH + Q − P
dt
dt
dU dM ˆ &
dV
=
H +Q− P
dt
dt
dt
For a time interval dt
dU = Hˆ dM + Q − PdV
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(3.1-8)
(3.1-9a)
14
For a closed system
dU = Q − PdV
(3.1-9b)
Closed system
No K.E. and no P.E.
No Ws
A single flow
NOTE:
Positive Q: Added on system
Negative Q: removed from system
-PdV: PV work done on system in
compression is positive
PdV: PV work done by system in
expansion is negative
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(3.1-6)
(3.1-7)
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3.2 SEVERAL EXAMPLES OF USING
THE ENERGY BALANCE
„
„
„
„
The energy balance equations can be used for the description of
any process.
First, define the system, describe the system boundary.
The system may be selected using different boundary definitions,
such as closed or open system. This may result in different
formulation in the energy balance equation. However, processes
occurring in nature are in no way influenced by our mathematical
description of them. Therefore, if our descriptions are correct, they
must lead to the same final result for the system and its
surroundings regardless of which system choice is made.
Result (open system) = Result (closed system)
This is demonstrated in the following example, where the same
result is obtained by choosing for the system first a given mass of
material (closed) and then a specified region in space (open).
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ILLUSTRATION 3.2-1
„
„
Showing That the Final Result Should Not Depend on
the Choice of System
A compressor is operating in a continuous, steady-state
manner to produce a gas at temperature T2 and pressure P2
from one at T1 and P1. Show that for the time interval Δt
(
)
)
)
Q + Ws = H 2 − H1 ΔM
„
where ΔM is the mass of gas that has flowed into or out of
the system in the time Δt. Establish this result by
(1) first writing the balance equations for a closed system
consisting of some convenient element of mass, and then
(2) by writing the balance equations for the compressor and its
contents, which is an open system.
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3.2-1 (1) The closed-system analysis
t
t +Δt
(V2)t = (V1) t+ Δt
(1 +
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C) portion
(C +
2) portion
19
„
(1) The mass balance for the closed system
M 2 ( t + Δt ) + M c ( t + Δt ) = M 1 ( t ) + M c ( t )
The compressor is in stady-state operation, M c ( t + Δt ) = M c ( t ) . Thus,
M 2 ( t + Δt ) = M 1 ( t ) = ΔM
The energy balance for this system, neglecting the potential and kinetic energy terms, is
( M Uˆ )
2
2
t +Δt
(
+ M cUˆ c
)
t +Δt
(
) (
− M 1Uˆ1 − M cUˆ c
t
Q
)
t
ˆ
ˆ
= Ws + Q + PV
1 1 M 1 − PV
2 2 M 2 (a)
{
}
{
− ∫ PdV = − P1 ∫ dVinlet − P2 ∫ dVoutlet = − P1 (V1 )t +Δt − (V1 )t − P2 (V2 )t +Δt − (V2 )t
ˆ
ˆ
− ∫ pdV = PV
1 1 − PV
2 2 = PV
1 1 M 1 − PV
2 2M 2
L Net PV -work
The compressor is in steady-state operation,
∴ (a ) ⇒
ACC=
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(
ΔM (Uˆ
)
( M Uˆ )
c
c
t +Δt
=
( M Uˆ )
c
c
}
& − P dV
W& = Ws
dt
In (3.1-4a)
t
ˆ
ˆ
ΔM Uˆ 2 − Uˆ1 = Ws + Q + PV
1 1ΔM − PV
2 2 ΔM
2
ˆ ˆ
ˆ
+ PV
2 2 − U1 − PV
1 1
)
(
)
= ΔM Hˆ 2 − Hˆ 1 = Ws + Q
Proved
20
„
(2) The mass balance for the open system
‰
System: the gas within the compressor
„
(3.1-6)
Open steady-state process, constant energy, volume
⎧⎪
⎧⎪
⎛ v2
⎞ ⎫⎪
⎛ v2
⎞ ⎫⎪
− ⎨U + M ⎜ + ψ ⎟ ⎬
⎨U + M ⎜ + ψ ⎟ ⎬
⎪⎩
⎝ 2
⎠ ⎪⎭t +Δt ⎪⎩
⎝ 2
⎠ ⎪⎭t
2
2
⎛
⎞
⎛
⎞
t +Δt
v
v
=∫
M& 1 ⎜ Hˆ 1 + 1 + ψ 1 ⎟ dt + ∫
M& 2 ⎜ Hˆ 2 + 2 + ψ 2 ⎟ dt + Q + W
t
t
2
2
⎝
⎠
⎝
⎠
⎛ ˆ v12
⎞
⎛ ˆ
⎞
v22
ACC= 0 = ΔM1 ⎜ H1 + + ψ 1 ⎟ + ΔM 2 ⎜ H 2 + + ψ 2 ⎟ + Q + Ws
2
2
⎝
⎠
⎝
⎠
Neglecting EKE , EP ,
t +Δt
( )
( )
= ( Hˆ − Hˆ ) ΔM
0 = ΔM1 Hˆ 1 + ΔM 2 Hˆ 2 + Q + Ws
Q + Ws
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2
1
Proved
21
COMMENT
„
„
„
„
The final results are exactly the same for both closed and open
system calculations.
‰ Result (open case) = Result (closed case)
For closed case, the volume of the system changes, the PV work is
involved; for open case, the enthalpy which contain the flow work is
involved. Somehow, PV work is related to flow work.
‰ PV work (closed) = Flow work (open)
For this particular example, the choice of open case is easier
than the choice of closed case since the process is open steadystate operation.
However, in other case, the closed system may be easy to apply
such as for a piston cylinder system.
‰ Easier choice based on the nature of the operation:
„
„
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Open steady state process Î using open system
Gas within piston and cylinder Î using closed system
22
ILLUSTRATION 3.2-2
„
„
–
–
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Showing That Processes in Closed and Open
Systems Are Different
A mass M of gas is to be compressed from
temperature T1 and pressure P1 to T2 and P2 in
a one-step process in a frictionless piston and
cylinder and
a continuous process in which the mass M of
gas is part of the feed stream to the compressor
of the previous illustration(3.2-1). Compute the
sum Q + W for each process.
23
a. The piston-and-cylinder process
„
Here we take the gas within the piston and cylinder as the
system. The energy balance for this closed system is
(
)
M Uˆ 2 − Uˆ1 = Q + W = Q − ∫ PdV
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b. The flow compressor process
„
„
System: the contents of the compressor
Recalled illustration(3.2-1)
Open steady-state process 3.1-4a becomes:
(
)
0 = M Hˆ 1 − Hˆ 2 + Q + W
(
)
M Hˆ 2 − Hˆ 1 = Q + W = Q + Ws
H2
H1
dV/dt = 0
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COMMENT
„
Flow and nonflow systems
‰
For closed (nonflow) system, the changes of Q
and PV-work results in the internal energy
change of the system
(
Q − ∫ PdV = M Uˆ 2 − Uˆ1
‰
)
For flow system, the changes of Q and Ws
result in the enthalpy changes between the
inlet/outlet streams
(
)
M Hˆ 2 − Hˆ 1 = Q + W = Q + Ws
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ILLUSTRATION 3.2-3
„
„
A Joule-Thomson or Isenthalpic Expansion
A gas at pressure P1 and temperature T1 is steadily
exhausted to the atmosphere at pressure P2 through a
pressure-reducing valve. Find an expression relating
the downstream gas temperature T2 to P1, P2 and T1.
Since the gas flows through the valve rapidly one can
assume that there is no heat transfer to the gas. Also,
the potential and kinetic energy terms can be
neglected.
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SOLUTION
„
„
System: gas within (involving) the pressure-reducing valve
The flow is steady.
‰
„
No heat (Q) or work (Ws) flows
‰
„
No PV-work, constant V
Valve is not a shaft equipment (neither a compressor nor a pump)
Neglecting the kinetic and potential energy changes
Mass balance (Molar) 0 = N&1 + N& 2
Energy balance
0 = N&1 H1 + N& 2 H 2
Thus,
(2.2-1b)
H1 = H 2
H 1 (T1, P1 ) = H 2 (T2 , P2 )
The relation between T1, P1, T2 , P2 can be found.
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COMMENTS
„
„
No heat and energy flows, the
thermodynamic properties of the system
remain the same values; thus, enthalpies in
the upstream and downstream states must
be the same.
Knowing H2 = H1 and either one of T2 (or
P2), then, the outlet condition P2 (or T2) can
be found out.
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3.3 THE THERMODYNAMIC
PROPERTIES OF MATTER
„
„
The internal energy (U) and enthalpy (H) in
energy balance equation are thermodynamic
properties which depend on the state of the system.
The thermodynamic properties can be obtained from
‰ PVT data, graphical figures, or Tables (e.g.
steam tables)
‰ Equations of state calculations (f (P,V,T) = 0)
„
„
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Ideal gas law
van der Waals equation
30
An ideal gas
„
„
An ideal gas is a gas at such a low pressure that there
are no interactions among its molecules. Also, an ideal
gas occupies negligible molecule volume.
The ideal gas law can be expressed as
PV = RT
„
(3.3-1)
Since there is no intermolecular forces, the internal
energy and enthalpy of an ideal gas are function of
temperature only. U = f (T) and H = f (T)
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Constant-volume heat capacity definition
„
If a sufficiently small quantity of heat is added to the
system, it is observed that the temperature rise
produced, ΔT, is linearly related to the heat (Q)
added and inversely proportional to the number of
moles (N) in the system:
Q
= C ΔT
N
For constant V process (N =1 mol), dU = dQV
⎛ ∂U (T ,V ) ⎞
dQV ⎛ ∂U ⎞
CV =
=⎜
⎟
⎟ =⎜
dT ⎝ ∂T ⎠V ⎝
∂T
⎠V
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(3.3-2)
32
Constant-pressure heat capacity definition
Q
= C ΔT
N
At constant pressure (N = 1 mol), dH = dQP
⎛ ∂ H (T , P ) ⎞
dQP ⎛ ∂ H ⎞
=⎜
CP =
⎟
⎟ =⎜
∂T
dT ⎝ ∂T ⎠ P ⎝
⎠P
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(3.3-3)
33
Ideal gas heat capacity
CP* = a + bT + cT 2 + dT 3 + L
CP* =
H
IG
U
IG
d H d (U + RT )
=
= CV* (T ) + R
dT
dT
(T2 ) = H
IG
(T1 ) + ∫ CP* (T ) dT
T2
T1
(T2 ) = U (T1 ) + ∫T
IG
T2
CV* (T ) dT
1
Set T1 = TR = 0 K,
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H
IG
U
IG
H
IG
(0
K) = 0
constant C
→ CP* (T ) T
(T ) = ∫0 CP* (T ) dT ⎯⎯⎯⎯
T
P
constant C
→ CV* (T ) T
(T ) = ∫0 CV* (T ) dT ⎯⎯⎯⎯
T
V
34
PVT obtained from graphical figures
How to sharp see H-S, T-S, P-H diagrams?
H-S: P-solid; T-dash
T-S: P-solid; H-solid; V-dash
P-H: T-solid; V-big dash; S-dash
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P, k
Pa
Figure 3.3-1 (a) Enthalpy
Entropy of Mollier diagram
For steam (H-S diagram)
T, oC
H
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S
36
Figure 3.3-1 (b) Temperature
Entropy diagram for steam
(T-S diagram)
T
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S
37
Figure 3.3-2 Pressure-enthalpy diagram for methane (P-H)
P
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H
38
Figure 3.3-3 Pressure-enthalpy diagram for nitrogen
P
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H
39
Figure 3.3-4 Pressure-enthalpy diagram for HFC-134a
P
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H
40
Properties of a two-phase mixture: the
lever rule
θˆ = ω Iθˆ I + ω IIθˆ II = ω Iθˆ I + (1 − ω I )θˆ II
θˆ, an intensive property
(3.3-9)
ω I , the mass fraction of the system that is in phase I
θˆ I , the value of the variable in phase I
For mixtures of steam and water, the mass fraction of steam
is termed the quality and is frequently expressed as a percent %.
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Energy difference between two phases
Δ vapHˆ = Hˆ V − Hˆ L = enthalpyof vaporization per unit mass or on a molar basis
Δ vap H = H − H = molar enthalpyof vaporization
V
L
Δfus H = H − H = molar enthalpyof melting or fusion
L
S
Δsub H = H − H = molar enthalpyof sublimation
V
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S
42
Simplifications
„
Solid or liquids at low pressure
H ≈U
„
(3.3-10)
Idealized incompressible fluid or solid
⎛ ∂V ⎞
⎜ ⎟ =0
⎝ ∂P ⎠T
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(3.3-11)
43
3.4 APPLICATIONS OF THE MASS AND
ENERGY BALANCES
„
2012/2/6
Some examples as follows
44
ILLUSTRATION 3.4-1
„
„
„
„
„
Joule-Thomson Calculation Using a Mollier Diagram and
Steam Tables
Steam at 400 bar and 500 oC undergoes a Joule-Thomson
expansion to 1 bar. Determine the temperature of the steam
after the expansion using
a. Fig. 3.3-1a
b. Fig. 3.3-1b
c. The steam tables in Appendix A.III
2012/2/6
45
Solution (3.4-1a, b)
„
We start from Illustration 3.2-3, where it was shown that
Hˆ 1 = Hˆ (T1 , P1 ) = Hˆ (T2 , P2 ) = Hˆ 2
„
„
„
for a Joule-Thomson expansion. Since T1 and P1 are
known, H1 can be found from either Fig. 3.3-1 or the steam
tables. Then, since H2 = H1, and P2 are known, T2 can be
found.
a. Using Fig. 3.3-1a, we first locate the point P = 400 bar =
40 000 kPa and T = 500oC, which corresponds to H1 =
2900 kJ/kg. Following a line of constant enthalpy to P = 1
bar = 100 kPa, we find that the final temperature is about
214oC.
b. Using Fig. 3.3-1b, we locate the point P = 400 bar and T
= 500oC and follow the curved line of constant enthalpy to a
pressure of 1 bar to see that T2 = 214oC.
2012/2/6
46
Solid line
40,000 kPa
Dash line
Figure 3.3-1 (a) Enthalpy
Entropy of Mollier diagram
For steam
H1 = H2 = 2900kJ/kg
2012/2/6
100 kP
500 oC
214.0oC
47
Solution (3.4-1c)
„
c. Using the steam tables of Appendix A.III, we have that at P =
400 bar = 40 MPa and
T = 500oC, H = 2903.3 kJ/kg. At P = 1 bar = 0.1 MPa, H = 2875.3
kJ/kg at T = 200oC and H = 2974.3 kJ/kg at T = 250oC. Assuming
that the enthalpy varies linearly with temperature between 200
and 250 oC at P = 1 bar, we have by interpolation
2903.3 − 2875.3
T = 200 + (250 − 200 ) ×
= 214.1o C
2974.3 − 2875.3
2012/2/6
T, oC
P, MPa
H, kJ/kg
200
0.1
2875.3
(214.1)
0.1
2903.3
250
0.1
2974.3
48
COMMENT
„
For many problems a graphical representation of
thermodynamic data, such as Figure 3.3-1, is easiest to use,
although the answers obtained are approximate and certain
parts of the graphs may be difficult to read accurately.
„
The use of tables of thermodynamic data, such as the steam
tables, generally leads to the most accurate answers; however,
one or more interpolations may be required. For example, if
the initial conditions of the steam had been 475 bar and
530oC instead of 400 bar and 500oC, the method of solution
using Fig. 3.3-1 would be unchanged; however, using the
steam tables, we would have to interpolate with respect to
both temperature and pressure to get the initial enthalpy of the
steam, shown in next page.
2012/2/6
49
Interpolation between temperatures and
*: first cal’d; **: second cal’d
pressures
Enthalpy
kJ/kg
500oC
530oC
550oC
40 MPa
2903.3
(3050.8)*
3149.1
47.5 MPa
50 MPa
2012/2/6
[2924.0]**
2720.1
(2899.7)*
3019.5
50
ILLUSTRATION 3.4-2
„
„
Application of the Complete Energy Balance Using the
Steam Tables
An adiabatic steady-state turbine is being designed to serve
as an energy source for a small electrical generator. The inlet
to the turbine is steam at 600 oC and 10 bar, with a mass
flow rate of 2.5 kg/s through an inlet pipe that is 10 cm in
diameter. The conditions at the turbine exit are T = 400 oC
and P = 1 bar. Since the steam expands through the turbine,
the outlet pipe is 25 cm in diameter. Estimate the rate at
which work can be obtained from this turbine.
2012/2/6
51
Solution
„
„
System: the turbine and its contents
Open, steady-state flow process
Constant mass, volume (no PV work), and energy in the
system
‰
Mass Balance
dM
= 0 = M& 1 + M& 2
dt
(2.2-1b)
Energy balance
2
2
⎛
⎞
⎛
⎞ &
⎛ v2
⎞⎫
v
v
d ⎧
1
2
&
&
ˆ
ˆ
⎜
⎟
⎜
⎟⎟ + Ws
⎜
⎟
U
M
gh
=
=
M
H
+
+
M
H
+
+
+
0
⎬
⎨
1⎜
1
2⎜
2
⎟
⎜
⎟
dt ⎩
2⎠
2⎠
⎝
⎝
⎝2
⎠⎭
(3.1-4a)
2012/2/6
52
Energy balance
2
2
⎛
⎞
⎛
⎞ &
v
v
1
2
&
&
ˆ
ˆ
0 = M 1 ⎜ H1 + ⎟ + M 2 ⎜ H 2 + ⎟ + Ws
2⎠
2⎠
⎝
⎝
2
2
⎛
⎞
⎛
⎞
v
v
1
2
&
&
&
ˆ
ˆ
Ws = − M 1 ⎜ H1 + ⎟ − M 2 ⎜ H 2 + ⎟
2⎠
2⎠
⎝
⎝
2
3
3
d
π
kg
m
m
2 m
&
ˆ
v [ =]
Volumetric flow rate = MV =
[ =] [ =] m
4
s kg
s
s
kg
m3
4 ⋅ 2.5 ⋅ 0.4011
s
kg
4M& 1Vˆ1
v1 =
=
2
2
π din
3.14159 ⋅ ( 0.1m )
2012/2/6
m
m
= 127.7 ; v2 = 158.0
s
s
53
2
2
⎛
⎞
⎛
⎞
v
v
1
2
&
&
&
ˆ
ˆ
Ws = − M 1 ⎜ H1 + ⎟ − M 2 ⎜ H 2 + ⎟
2⎠
2⎠
⎝
⎝
kg ⎧ ˆ
1 2
2 ⎫ kJ
ˆ
= −2.5 ⎨ H1 − H 2 + v1 − v2 ⎬ ;
s ⎩
2
⎭ kg
(
) (
)
m2
1 J =1 kg 2
s
J
⎛
⎞
2 1
⎜
kg
kJ 1
kg 1kJ ⎟
2
2 m
⎟
= −2.5 ⎜ 419.7
+ 127.7 − 158.0
⋅ 2 ⋅
2
m 1000 J ⎟
s ⎜
kg 2
s
⎜
⎟
2
s
⎝
⎠
kg
kJ
kJ
= −2.5 ( 419.7 − 4.3)
= −1038.5
( = -1329 hp)
s
kg
s
(
2012/2/6
)
54
COMMENT
„
If we had completely neglected the kinetic
energy terms in this calculation, the error in
the work term would be 4.3 kJ/kg, or about
1%. Generally, the contribution of kinetic and
potential energy terms can be neglected
when there is a significant change in the fluid
temperature.
2012/2/6
55
ILLUSTRATION 3.4-3
„
„
„
„
Use of Mass and Energy Balances with an Ideal Gas
A compressed-air tank is to be pressurized to 40 bar by
being connected to a high-pressure line containing air at
50 bar and 20 oC. The pressurization occurs so quickly
that the process can be assumed to be adiabatic; also,
there is no heat transfer from the air to the tank.
Assuming air to be an ideal gas with CV = 21 J/(mol K).
a. If the tank initially contains air at 1 bar and 20 oC,
what will be the temperature of the air in the tank at the
end of the filling process?
b. After a sufficiently long period of time, the gas in the
tank is found to be at room temperature (20 oC) because
of heat exchange with the tank and the atmosphere.
What is the new pressure of air in the tank?
2012/2/6
56
Solution
„
System: the contents of the tank
‰
‰
‰
1. The kinetic and potential energies are small and neglected.
2. Since the tank is connected to a source of gas at constant
temperature and pressure, Hin is constant.
3. The initial process is adiabatic, so Q = 0, and the system
(the contents of the tank) is of constant volume, so ΔV=0.
mass balance:
(3.1-5a) energy balance:
2012/2/6
N2 − N1 = ΔN
N2U 2 − N1U 1 = (ΔN )H in
57
Ideal gas, U = CVT , H = CPT , N =
PV
RT
Energy balance
N 2CVT2 − N1CVT1 = ( ΔN ) CPTin = ( N 2 − N1 ) CPTin
⎛ P2V2 PV
⎞
P2V2
PV
1
1
1
1
CVT2 −
CVT1 = ⎜
−
⎟ CPTin
RT2
RT1
R
RT
T
2
1⎠
⎝
Eliminate V2 = V1 and R
⎛P
P⎞
P2CV − P1CV = ⎜ 2 − 1 ⎟ CPTin
⎝ T2 T1 ⎠
CV ( P2 − P1 ) P1 P2
+ =
CPTin
T1 T2
T2 =
P2
CV ( P2 − P1 )
CPTin
2012/2/6
+
P1
T1
P2= 40bar
Tin= 293.15K
P1= 1bar
T1= 293.15K
CV= 21 J/(molK)
CP= CV+8.314
58
„
„
„
„
„
„
„
„
„
2012/2/6
(a) T2 = 404.2 K = 132.05 oC
(b) The tank is the system: N2 =N1
T2 = 20oC, P2/P1= (N2RT2/V2) / (N1RT1/V1) =
P2 =P1 (T2 /T1)= 40(293.15/404.2)= 28.94 bar
Discussion:
(1) Note that T2 = 132.05 oC >> 20 oC, because
pressurization increases the temperature.
In other words,
due to H(20 oC) = U (20 oC) + PV(20 oC)
Î
U (20 oC) increase to final value of U (132.05 oC).
59
ILLUSTRATION 3.4-4
„
„
Example of a Thermodynamics Problem That
Cannot Be Solved with Only the Mass and Energy
Balances
A compressor is a gas pumping device that takes in gas
at low pressure and discharges it at a higher pressure.
Since this process occurs quickly compared with heat
transfer, it is usually assumed to be adiabatic: that is,
there is no heat transfer to or from the gas during its
compression. Assuming that the inlet to the compressor
is air [which we will take to be an ideal gas with CP* =
29.3 J/(mol K)] at 1 bar and 290 K. and that the
discharge is at a pressure of 10 bar, estimate the
temperature of the exit gas and the rate at which work is
done on the gas for a gas flow of 2.5 mol/s.
2012/2/6
60
Solution
„
(Since there are two unknown thermodynamic
quantities, the final temperature and the rate at which
work is being done, we can anticipate that the mass
and energy balances will not be sufficient to solve this
problem.)
„
The system will be taken to be the gas contained in the
compressor. We have used the subscript 1 to indicate the flow
stream into the compressor and 2 to indicate the flow stream out
of the compressor. Since the compressor operates continuously,
the process may be assumed to be in a steady state,
2012/2/6
61
(3.1-5a)
dN
= N& 1 + N& 2 = 0; N& 1 = − N& 2
dt
dU
= N& 1 H 1 + N& 2 H 2 + Q& + W& = 0;
dt
*
&
&
&
&
Ws = N1 H 1 − N1 H 2 = N1CP (T2 - T1 )
W s = C (T2 - T1 )
*
P
Two unknows : T2 and W s
2012/2/6
62
„
We need additional information about the
system before we can solve the problem.
„
This additional information will be
obtained using an additional (Entropy)
balance equation developed in the next
chapter (Chapter 4).
2012/2/6
63
ILLUSTRATION 3.4-5
„
„
Use of Mass and Energy Balances to Solve an Ideal Gas Problem
A gas cylinder of 1 m3 volume containing nitrogen initially at a pressure
of 40 bar and a temperature of 200 K is connected to another cylinder
of 1 m3 volume that is evacuated. A valve between the two cylinders is
opened until the pressures in the cylinders equalize. Find the final
temperature and pressure in each cylinder if there is no heat flow into or
out of the cylinders or between the gas and the cylinder. You may
assume that the gas is ideal with a constant-pressure heat capacity of
29.3 J/(mol K).
40 bar/200K
2012/2/6
0 bar
1 m3
1 m3
Cylinder 1
Cylinder 2
20 bar / ???K
20 bar / ???K
64
Solution
„
This problem is more complicated than the previous ones
because we are interested in changes that occur in two separate
cylinders. We can try to obtain a solution to this problem in two
different ways. First, we could consider each tank to be a
separate system, and so obtain two mass balance equations
and two energy balance equations, which are coupled by the fact
that the mass flow rate and enthalpy of the gas leaving the first
cylinder are equal to the like quantities entering the second
cylinder. Alternatively, we could obtain an equivalent set of
equations by choosing a composite system of the two
interconnected gas cylinders to be the first system and the
second system to be either one of the cylinders. In this way
the first (composite) system is closed and the second system is
open. We will use the composite system here: you are
encouraged to explore the first system choice independently and
to verify that the same solution is obtained.
2012/2/6
65
Mass balance: N1i = N1f + N 2f
(a)
Energy balance: N1i U 1 = N1f U 1 + N 2f U 2
i
f
f
(b)
The ideal gas equation, N = PV
RT
P1iV1i P1 f V1 f P2 f V2 f
i
f
f
=
+
N1 = N1 + N 2 ⇒
RT1i
RT1 f
RT2 f
P1i P1 f P2 f
Same values for R and V ,
= f + f
i
T1 T1 T2
(a')
QU = CV T ,EQ( b ) ⇒
⎛ P1iV1i ⎞
⎛ P1 f V1 f
i
⎜ RT i ⎟ CV T1 = ⎜ RT f
⎝ 1 ⎠
⎝
1
Q
⎞
⎛ P2 f V2 f
f
⎟ CV T1 + ⎜ T f T f
⎠
⎝ 2 2
P1i = P1 f + P2 f
⎞
f
C
T
⎟ V 2
⎠
(c)
1
= P1i = 20 bar
2
∴ EQ (a') ⇒
P1 f = P2 f
2
1
1
=
+
T1i T1 f T2 f
2012/2/6
(c')
66
the open but not S.S system: the gas initially filled LHS cylinder
dN1 &
=N
dt
d ( N1U 1 ) &
ACC =
= N H 1 which is not ZERO
dt
d ( N1U 1 )
d (U 1 )
d ( N1 )
dN
= N1
+U1
= H1 1
dt
dt
dt
dt
d (U 1 )
dN
N1
= ( H1 −U1 ) 1
dt
dt
For ideal gas,
⎛ PV ⎞
d⎜ 1 1⎟
d (T1 )
RT
PV
1 1
CV
= ⎡⎣( CP − CV )T1 ) ⎤⎦ ⎝ 1 ⎠
dt
RT1
dt
⎛P⎞
d⎜ 1⎟
d (T1 )
T
P1
CV
= (T1 ) ⎝ 1 ⎠
RT1
dt
dt
⎛P⎞
d ln ⎜ 1 ⎟
CV d lnT1
⎝ T1 ⎠
=
R dt
dt
2012/2/6
67
⎛ P1 ⎞
d ln ⎜ ⎟
CV d lnT1
⎝ T1 ⎠
=
R dt
dt
Integrate from i to f for the LHS cyclinder:
⎛ T1 f ⎞
⎜ Ti ⎟
⎝ 1 ⎠
CV
R
⎛ P1 f ⎞⎛ T1i ⎞
= ⎜ i ⎟⎜ f ⎟
⎝ P1 ⎠⎝ T1 ⎠
CP
R
⎛ T1 f ⎞
⎛ P1 f ⎞
⎜ T i ⎟ = ⎜ Pi ⎟
⎝ 1 ⎠
⎝ 1 ⎠
T1 f = ( 200 )e
⎛ ln 0.5 ⎞
⎜
⎟
⎝ 3.488 ⎠
EQ( f )
= 164.3 K;
Substituted into EQ(a'): T2 f = 255.6 K
Using ideal gas EOS:
N1f = 1.464 mol; N 2f = 0.941 mol
2012/2/6
68
ILLUSTRATION 3.4-6
„
„
„
„
Showing That the Change in State Variables
between Fixed Initial and Final States Is
Independent of the Path Followed
ΔU=f (state); ΔH=f (state)
Q, W = f (path)
One mole of a gas at a temperature of 25 oC
and a pressure of 1 bar (the initial state) is to
be heated and compressed in a frictionless
piston (reversible process) and cylinder to
300 oC and 10 bar (the final state).
2012/2/6
69
ILLUSTRATION 3.4-6 (continued)
„
Compute the heat and work required along each
of the following paths.
‰
‰
‰
2012/2/6
Path A. Isothermal (constant temperature) compression to
10 bar, and then isobaric (constant pressure) heating to
300 oC.
Path B. Isobaric heating to 300 oC followed by isothermal
compression to 10 bar.
Path C. Adiabatic compression in which PVγ = constant,
where γ = Cp/Cv followed by an isobaric cooling or heating,
if necessary, to 300 oC. For simplicity, the gas is assumed
to be ideal with CP = 38 J/(mol K).
70
2012/2/6
71
Q, W, ΔU, and ΔH for an ideal gas system
Reversible
process
Isothermal
(CONSTANT T)
Isobaric
(CONSTANT P)
Isochoric
(CONSTANT V)
Q
W
ΔU
(=CV ΔT)
- RTln(P2/P1)
RTln(P2/P1)
0
CP ΔT = ΔH
- R ΔT
CV ΔT = ΔU
0
0
CV ΔT
Adiabatic
(Q = 0)
T2 ⎛ P2 ⎞
=⎜ ⎟
T1 ⎝ P1 ⎠
(γ −1) γ
2012/2/6
C
where γ = P
CV
ΔH
(=CP ΔT)
0
72
Path A
i. Isothermal compression at 298.15K
Wi = − ∫V PdV = − ∫V
V2
V2
1
1
RT
V
P
dV = − RT ln 2 = RT ln 2
V
V1
P1
= 8.314 J/ ( mol K ) × 298.15 K × ln
10
= 5707.7 J/mol
1
ΔU i = CV* ΔT = 0 = Δ H i
Qi = ΔU i − Wi = 0 − 5707.7 = −5707.7 J/mol
ii. Isobaric heating
Wii = − ∫V P2 dV = − P2 ∫V dV = − P2 (V 3 − V 2 )
V3
V3
2
2
= − R (T3 − T2 ) = −8.314 J/ ( mol K ) × 275 = −2286.3 J/mol
Qii = Δ H i = CP* ΔT = 38 J/ ( mol K ) × 275 = 10450 J/mol
For the overall process
Q = Qi + Qii = 4742.3 J/mol
W = Wi + Wii = 3421.4 J/mol
ΔU = 8163.7 J/mol
2012/2/6
73
Path B
i. Isobaric heating at 1 bar
Qi = Δ H i = CP* ΔT =10450 J/mol
Wi =-R (T2 -T1 ) = −2286.3 J/mol; (work done by system)
ii. Isothermal compression at 300K
P2
10
Wii = RT ln = 8.314(300+273.15) ln = 10972.2 J/mol
1
P1
Qii = −10972.2 J/mol
Q = Qi + Qii = -552.2 J/mol
W = Wi + Wii = 8685.9 J/mol
ΔU = 8163.7 J/mol
2012/2/6
74
Path C
γ
i. Compression with P V = constant
γ
γ
⎛ RT1 ⎞
⎛ RT2 ⎞
γ
γ
PV
=
P
=
P
V
=
P
⎟
⎟
1 1
1⎜
2 2
2⎜
P
P
⎝ 1 ⎠
⎝ 2 ⎠
T2 ⎛ P2 ⎞
=⎜ ⎟
T1 ⎝ P1 ⎠
( γ −1) / γ
T2 = 298.15 K (10 )
0.28/1.28
=493.38 K
Wi = CV (T2 − T1 ) = 5795.6 J/mol
Qi = 0
ΔU i = Wi = 5795.6 J/mol
ii. Isobaric heating
Qii = C p (T3 − T2 ) = 3031.3 J/mol
Wii = − R (T3 − T2 ) = - 663.2 J/mol; T3 =573.15K (work done by system)
Q =3031.3 J/mol
W =5132.4 J/mol
2012/2/6
75
Overall process
2012/2/6
Path
Q(J/mol)
W(J/mol)
Q+W=ΔU
(J/mol)
A
4742.3
3421.4
8163.7
B
-522.2
8685.9
8163.7
C
3031.3
5132.4
8163.7
76
Homework assignments
„
„
Problems
3.3, 3.5, 3.13, 3.15, 3.19, 3.29
2012/2/6
77
Illistration 3-4.1 Regression of Steam Table Data at 47.5MPa + 500C Using MATHCAD
Method − ( 1 )
⎛⎜ 40
40
Mxy := ⎜
⎜ 50
⎜ 50
⎝
500 ⎞
⎟
550 ⎟
500 ⎟
⎟
550 ⎠
⎛⎜ 2903.3 ⎞⎟
3149.1 ⎟
vz := ⎜
⎜ 2720.1 ⎟
⎜ 3019.5 ⎟
⎝
⎠
v :=
⎛ 47.5 ⎞
⎜
⎟
⎝ 530 ⎠
cc := regress( Mxy , vz , 1 )
z := interp( cc , Mxy , vz , v )
z = 2.9362 × 10
3
ANSWER
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Method − ( 2 ) Since doing the first two interpolations, doing the last interpolations as follows:
d := 3050.8 + 7.5⋅ ⎛⎜
⎝
2899.7 − 3050.8⎞
10
3
d = 2.9375× 10
⎟
⎠
3
2903.3 + 0.6⋅ ( 3149.1 − 2903.3) = 3.0508× 10
3
2720.1 + 0.6⋅ ( 3019.5 − 2720.1) = 2.8997× 10
vvx :=
⎛ 500 ⎞
⎜ ⎟
⎝ 550 ⎠
vvy :=
⎛ 2903.3⎞
⎜
⎟
⎝ 3149.1⎠
vvs := regress( vvx , vvy , 1 )
dd1 := interp( vvs , vvx , vvy , 530 )
vvx :=
⎛ 500 ⎞
⎜ ⎟
⎝ 550 ⎠
vvy :=
dd1 = 3.0508 × 10
3
⎛ 2720.1 ⎞
⎜
⎟
⎝ 3019.5 ⎠
vvs := regress( vvx , vvy , 1 )
dd2 := interp( vvs , vvx , vvy , 530 )
dd2 = 2.8997 × 10
3
Using dd1 and dd2, put into the following regression (REGRESS) and interpolation (INTERP):
vvx :=
⎛ 40 ⎞
⎜ ⎟
⎝ 50 ⎠
vvy :=
⎛ dd1 ⎞
⎜ ⎟
⎝ dd2 ⎠
vvs := regress( vvx , vvy , 1 )
dd := interp( vvs , vvx , vvy , 47.5)
3
dd = 2.9375 × 10
ANSWER
Summary for Mass and Energy Balance Equations
Referred to 4th Ed., Ch 3. pp50 (or 2nd Ed.; 3rd Ed., pp25-30) of Sandler textbook
General mass and energy balance equations in general form:
Accumulation = In − Out + Generation
1. Mass balance equation in mass rate of change form, defined Table 2.2-1 (3rd Ed.):
dM K &
= ∑ Mk
dt
k
K: total # of entry ports; k: an arbitrary port
2. Conservation of energy balance equation: Defined in 2.3 (3rd); 3.1(4th)
K
v2
(1) ∑ M& k ( U +
of kth flow stream {energy come into system}
+ψ )
2
k =1
k
v2
d
(2)
( U + M ( + ψ ))
dt
2
&
(3) Q
of the system {energy accumulation}
{Heat added to the system}
K
(4) ∑ M& k ( PVˆ ) k
{Net work done on the system due to pressure forces}
k =1
(5) − P
(6) Ws
dV
dt
{Work done by the system through an expansion}
{Net shaft work done by the system through shaft work}
Differential form of energy balances (i.e. mass rate of change form) become:
(2) = (1) + (4) + (3) + (6) + (5):
2
K
⎞ K
⎛
⎞
d ⎛⎜
v2
⎟ = ∑ M& k ⎜Û + v + ψ ⎟ + ∑ M& k ( PV̂ )k + Q& + W& s + ( − P dV )
U
M
(
ψ
)
+
+
⎟ k =1 ⎜
⎟ k =1
2
2
dt ⎜⎝
dt
⎠
⎝
⎠k
⇒
2
⎞ K
⎛
⎞
d ⎛⎜
v2
⎟ = ∑ M& k ⎜ Ĥ + v + ψ ⎟ + Q& + W&
U
M
(
ψ
)
+
+
⎟ k =1
⎜
⎟
dt ⎜⎝
2
2
⎠
⎝
⎠k
where W& = W& s + ( − P dV )
dt
Table 2.3-1 (3rd); 3.1-4a (4th)
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