2
2.1
d (C V )
dM
=
=M IN -M OUT
dt
dt
a) at steady state
g
M IN =1
= M OUT = C volumetric flow rate out
min
g
10m3
1 g
g
1
= C
C=
= 0.1 3
min
min
10 m3
m
dC
m3
dC
m3
V
= M C 10
V
= M C 10
b)
dt
min
dt
min
3
V = 6 6 3 = 108 m
V = 6 6 3 = 108 m3
3
dC
g
10
3
3.11 System = gas contained in piston and cylinder (closed)
0
Energy balance: U t2 U t1 = Q + Ws PdV
(a) V = constant,
z
z
d
i
a
PdV = 0 , Q = U t2 U t1 = N U t2 U t1 = NCV T2 T1
f
From ideal gas law
N=
.
PV
114,367 Pa 0120
m3
.
=
= 5539
mol (see note f
4
4.9
System = contents of the turbine. This is a steady-state, adiabatic, constant volume system.
dM
= M
+M
or M
=0= M
(a) Mass balance
2
1
1
2
dt
Energy balance
constant
dU
H + M
H + Q adiabatic +W P dV volume
=0= M
s
1 1
2 2
dt
dt
Entropy balance
4
4.44
c
1 bar
20C
10 bar
e
d
Liquid, 85C
10 bar = 1000 kPa =1 MPa
200C
1 bar, 20C
L
1 bar,85C
= 83.96
H
1
H 3 = 355.90
S 1 = 0.2966
S 3 = 1.1343
S = 6.6940
= 2827.9
V cfw_10 bar, 200C H
2
M.B.
+M
+M
=0
M
1
2
3
2
Enthalpy and entropy values
written w