4
4.14
System: contents of valve (steady-state, adiabatic, constant volume system)
Mass balance 0 = N 1 + N 2
Energy balance
0 = N 1 H 1 + N 2 H 2 + Q 0
+ Ws
0
P
dV 0
dt
H1 = H 2
Q 0
0 = N 1 S 1 + N 2 S 2 + Sgen +
T
Sgen
S = S 2 S 1 =
N
(a) Using the Mo
4
4.13
dT
dV
+R
T
V
dS = C
S =
eqn. (4.4-1)
z LNM
(a R) + bT + cT 2 + dT 3 +
OP
Q
z
e dT
dV
+R
V
T2 T
so that
a
f a
f
a
f c
h
V
d
e
+ cT T h cT T h + R ln
3
2
V
T2
c
+ b T2 T1 + T22 T12
2
T1
S T2 , V 2 S T1, V 1 = (a R) ln
3
2
2
2
3
1
2
1
2
1
Now using
PV
4
4.10 Since compression is isentropic, and gas is ideal with constant heat capacity, we have
FG T IJ = FG P IJ
HT K H PK
2
2
1
1
FG 3 10 IJ
H 2 10 K
to Problem 3.31, that W = NC aT T f
So that T2 = T1
FG P IJ
HPK
2
R CP
R CP
6 8.314 36.8
= 29815
.
6
= 32
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6
6.7
(a) Ideal gas
PV = NRT
N=
c
(50 bar ) 100 m3
(27315
. + 150)K 8.314 10
2
h
bar m3 kmol K
= 1421
. kmol
Energy balance, closed nonflow system
z
U = Q PdV = Q + W .
However, for ideal gas U = 0 since T is constant (isothermal). Thus
z
W = Q = PdV =
z
Review
RECALL:
.Ws
I. FLOW Process
Ia. System = equipment
Heat
Exchanger
.N
.
1
moles/sec
P1 T V1
1
U1 H1
Ai
Compressor
moles/sec
Ao N2
P2 T2 V2
U2 H2
.Q
. = (H - H )
.Q + Ws
2
1
Note: The flow process is open to the flow of mass.
ENERGY BALANCE ON THE OP
1
1.2
(a) Water is inappropriate as a thermometric fluid between 0C and 10C, since
the volume is not a unique function of temperature in this range, i.e., two
temperatures will correspond to the same specific volume,
V (T = 1 C) ~ V (T = 7 C); V (T = 2 C)
3
3.15 System = contents of both chambers (closed, adiabatic system of constant volume. Also Ws = 0 ).
af af
af af af af
af af
Energy balance: U t2 U t1 = 0 or U t2 = U t1
(a) For the ideal gas u is a function of
U t2 = U t1 T t2 = T t1 = 500 K . From ide
.
I. FLOW Process
Ws
Ia. System = equipment
Heat
Exchanger
.N
.
1
moles/sec
P1 T V1
1
U1 H1
Ai
Compressor
.Q
moles/sec
Ao N2
P2 T2 V2
U2 H2
At what rate is energy gained by/lost from the equipment (the system)?
In other words, at what rate is energy gaine
I. State Properties and Path Independence
P, bar
S
SC
L
Path A
2
10
V
1
G
Path C
Path B
1
25
300
T, C
Consider a change in state for 1 mole of an ideal gas from state 1 (P = 1 bar, T = 25C)
to state 2 (P = 10 bar, T = 300 C) by three different paths (A, B
4
4.6
1 bar
10 bar
(a) Entropy change per mole of gas
290 K
575 K
T
P
S = CP ln 2 R ln 2 eqn. (4.4-3)
T1
P1
J
575
J
10
J
Thus S = 29.3
ln
8.314
ln = 0.9118
mol K 290
mol K 1
mol K
(b) System = contents of turbine (steady-state system)
dN
= 0 = N 1 + N 2
4
4.21 System = contents of the compressor (steady-state, constant volume). Also, gas is
ideal.
0 = N1 + N 2 N 2 = N1
(a) Mass balance
0 = N1 H 1 + N 2 H 2 +
Energy balance
0
Q
+Ws P
adiabatic
dV 0
dt
Entropy balance
reversible
Q 0
0 = N1 S 1 + N 2 S 2 +