25
2.2. EXERGY
Control
Volume
h + (1/2) v . v+ g z
ho + gz o
q cv = - q H,Carnot
Carnot
Engine
w
Carnot
q L,Carnot
To
Figure 2.2: Sketch of control volume balance for exergy discussion.
Now take the i
45
2.5. BRAYTON
x 3: combustion chamber,
3 4: expansion in turbine, and
4 y : turbine exit gas goes through regenerator (heat exchanger).
A schematic for the Brayton cycle is shown in Figure 2.11.
237
7.1. ISOTHERMAL, ISOCHORIC KINETICS
Now in the Dalton mixture model, all species share the same T and V . So the mixture
temperature and volume are the same for each species Vi = V , Ti = T . But
47
2.5. BRAYTON
If we have a CPIG, then
reg =
Tx T2
.
Tx T2
(2.277)
Example 2.7
(extension for Moran and Shapiro, Example 9.6) How would the addition of a regenerator aect
the thermal eciency of the i
239
7.1. ISOTHERMAL, ISOCHORIC KINETICS
We notice that e cancels. This so-called third body will in fact never aect the equilibM
rium state. It will however inuence the dynamics. Removing e and slight
241
7.1. ISOTHERMAL, ISOCHORIC KINETICS
Near the rst non-physical root at e = 0.003, a positive perturbation from equiO
librium induces f < 0, which induces dO /dt < 0, so O returns to its equilibriu
51
2.5. BRAYTON
Solving for the thrust force, neglecting the small dierences in pressure, one gets,
F = m (v 5 v 1 ) .
(2.294)
The work done by the thrust force is the product of this force and the ai
243
7.1. ISOTHERMAL, ISOCHORIC KINETICS
7.1.1.2
Single reversible reaction
The two irreversible reactions studied in the previous section are of a class that is common
in combustion modeling. However,
245
7.1. ISOTHERMAL, ISOCHORIC KINETICS
Thus, we once again nd
O + 2O2 = O + 2O2 = constant.
(7.117)
As before, we can say
O
O2
=
=
O
1
+
1
O2
2
=D
b
=
O .
(7.118)
=
This gives the dependent variables
55
2.6. RECIPROCATING ENGINE POWER CYCLES
B
S
2Rcrank
Figure 2.15: Piston-cylinder congurations illustrating geometric denitions.
For these engines the volumes are closed in expansion and compression,
249
7.1. ISOTHERMAL, ISOCHORIC KINETICS
Again, A is an arbitrary constant. Obviously the equilibrium is stable. Moreover the time
constant of relaxation to equilibrium is
=
1
= 4.77156 108 s.
2.09575
43
2.5. BRAYTON
Now state three has T3 = 1400 K . Recall that 2 3 involves heat addition in a combustion chamber.
Now calculate T4s for the ideal turbine. Recall that the expansion is to the same pres
233
7.1. ISOTHERMAL, ISOCHORIC KINETICS
exp(- E j /(R T)
1
T
Ej/ R
Figure 7.1: Plot of exp(E j /R/T ) versus T ; transition occurs at T E j /R.
for small T , the modulation is extreme, and the reacti
219
6.3. CONCISE REACTION RATE LAW FORMULATIONS
pliers. Continue to nd
N
i=1
N
exp
i=1
RT
ln i
Po
N
RT
Po
ij
RT
ln i
Po
i=1
PN
i=1
=
Go
j
RT
,
ij
= exp
ij
RT
i
Po
= exp
ij N
i=1
i ij = exp
N
i
ij
P
27
2.2. EXERGY
So we get
= cP (T To ) To cP ln
T
To
R ln
1
+ v v + g (z zo ).
2
P
Po
(2.94)
For T To P Po , we can use Taylor series to simplify this somewhat. First, we recall the general
Taylor se
221
6.4. ADIABATIC, ISOCHORIC KINETICS
Note nally that it can be shown that D, of dimension N (N L), and , of dimension
N J , share the same column space, which is of dimension (N L); consequently, bo
29
2.3. RANKINE
T
3
2
1
4
s
Figure 2.4: T s plane for Rankine cycle.
thermal eciency
back work ratio
mass ow rate of steam
rate of heat transfer Qin to the uid in the boiler
rate of heat transfer
223
6.4. ADIABATIC, ISOCHORIC KINETICS
This problem requires a detailed numerical solution. Such a solution was performed by solving
Eq. (6.14) along with the associated calorically imperfect species
31
2.3. RANKINE
The necessary heat transfer rate in the boiler is then
Qin
= m(h3 h2 ),
kg
=
104.697
s
= 269706 kW ,
=
(2.123)
2758
kJ
kg
181.94
kJ
kg
,
(2.124)
(2.125)
269.7 M W.
(2.126)
In the cond
33
2.3. RANKINE
Relative to the pump, the boiler has added much more exergy to the uid. After the turbine, just
before the condenser, we have
4
= (h4 To s4 ) (ho To so ) ,
kJ
kJ
(298.15 K ) 5.7432
=
Chapter 7
Kinetics in some more detail
Here we give further details of kinetics. These notes are also used to introduce a separate
combustion course and have some overlap with previous chapters.
Let u
37
2.3. RANKINE
Now get the turbine work
= h5 h6 + 1
wt
=
3213.6
kJ
kg
m6
m5
kJ
kg
kJ
2685.6
kg
(2.179)
2685.6
+ (1 0.165451)
= 979.908
(h6 h7 ),
2144.1
kJ
kg
,
(2.180)
kJ
.
kg
(2.181)
Now get the
229
7.1. ISOTHERMAL, ISOCHORIC KINETICS
The standard model for chemical reaction, which will be generalized and discussed in
more detail later, induces the following two ordinary dierential equations
1.
V_1 : 0.2 m3 P_1 : 200 Kpa
V_2 : 0.02 m3
a) P is constant,
P_2 : 200 Kpa
V_2
W= P dv=P( V 2 - V 1 ) =200 Kpa* ( 0.02-0.2 ) =-36 KJ
V_1
b) T is constant
Its mean PV=k=40 (KJ)
W=
P=k/V=40/V
V_2
V_2
EML 3100 - Homework 2
Due 9/25/17
Problem 1
Can you carry 0.5 m3 of liquid water?
A: 0.5 m3 is 500 kg. So, We cant carry it.
Problem 2
What is the saturation pressure for water at a temperature of 325