1. No category

PROBLEM 6.140
m3/s and
PROBLEM 6.147
PROBLEM 6.149
Problem 7.29
As shown in Fig. P7.29, 1 kg of H2O is contained in a rigid, insulated cylindrical vessel. The
H2O is initially saturated vapor at 120°C. The vessel is fitted with a paddle wheel from which a
mass is suspended. As the mass descends a certain distance, the H2O is stirred until it attains a
final equilibrium state at a pressure of 3 bar. The only significant changes in state are
experienced by the H2O and the suspended mass. Determine, in kJ,
(a) the change in exergy of the H2O.
(b) the change in exergy of the suspended mass.
(c) the change in exergy of an isolated system of the vessel and pulley-mass assembly.
(d) the destruction of exergy within the isolated system.
Let T0 = 293 K, p0 = 1 bar.
Known:
H2O is in a rigid vessel with a paddle wheel inside that is attached to a dropping mass.
Find:
(a) the change in exergy of the H2O, (b) the change in exergy of the suspended mass, (c) the
change in exergy of an isolated system of the vessel and pulley-mass assembly, and (d) the
destruction of exergy within the isolated system.
Schematic and Known Data:
Isolated system
Q=W=0
Saturated H2O vapor
m W = 1 kg
T1 = 120 C
p2 = 3 bar
T
3 bar
2
1.985 bar
Initial
mass
120 C
20 C
1
1 bar
0
Dead State
T0 = 293 K
p0 = 1 bar
Engineering Model:
z
Final
mass
v
(1) As shown in the schematic, three systems are under consideration: the H2O, the suspended
mass, and an isolated system consisting of the vessel and pulley-mass assembly. For the
isolated system,
.
(2) The only significant changes of state are experienced by the H2O and the suspended mass.
For the H2O, there is no change in kinetic or potential energy. For the suspended mass, there
is no change in kinetic or internal energy. Elevation is the only intensive property of the
suspended mass that changes.
(3) For the environment, T0 = 293 K, p0 = 1 bar.
Analysis:
(a) The initial and final exergies of the H2O can be evaluated using Eq. 7.2. From assumption
(2), it follows that for the H2O there are no significant effects of motion or gravity, thus
the exergy at the initial state is:
[(
)
(
)
(
)]
The initial and final states of the H2O are shown on the accompanying T-v diagram. From
Tables A-2 and A-4:
Then:
#1
(
) [(
(
)(
)
)(
(
)
]
)(
( )[
)|
|
]
The final state of the H2O is fixed by p2 = 3 bar and v2 = v1. Using interpolation and
values from Table A-4:
Then:
(
) [(
)
(
(
)(
)(
( )[
)(
)
)|
|
]
]
For the H2O, the change in exergy is:
#2
(b) With assumption 2, Eq. 7.3 reduces to give the exergy change for the suspended mass:
⏟
#3
Thus, the exergy change for the suspended mass equals the change in potential energy.
The change in potential energy of the suspended mass is obtained from an energy balance
for the isolated system as follows: the change in energy of the isolated system is the sum
of the energy changes of the H2O and suspended mass. There is no heat transfer or work,
and with assumption 2 we have:
(⏟
)
(⏟
)
⏟
Solve for ΔPEm and using previously determined values for the specific internal energy
of the H2O:
(
)(
)
The exergy of the mass decreases because its elevation decreases.
(c) The change in exergy of the isolated system is the sum of the exergy changes of the water
and suspended mass. With the results from parts (a) and (b):
(d) The exergy of the isolated system decreases. With Eq. 7.9, (Ed)sys = 178.29 kJ
To summarize:
H2 O
Suspended Mass
Isolated System
Energy Change (kJ)
293.77
-293.77
0
Exergy Change (kJ)
115.48
-293.77
-178.29
For the isolated system there is no net change in energy. The increase in the internal
energy of the H2O equals the decrease in potential energy of the suspended mass.
However, the increase in exergy of the H2O is much less than the decrease in exergy of
the mass. For the isolated system, exergy decreases because stirring destroys exergy.
Comments:
1. Exergy is a measure of the departure of the state of the system from that of the
environment. At all states, E > 0. This applies when T > T0, p > p0, as at states 1 and 2,
and when T < T0, p < p0.
2. Alternatively, Eq. 7.3 can be used. This requires dead state property values only for T0
and p0. In part (a), u0, v0, and s0 are also required; so more computation is needed with the
approach shown in this problem’s solution.
3. The change in potential energy of the suspended mass cannot be determined from Eq.
2.10 (Sec. 2.1) since the mass and change in elevation are unknown. Moreover, for the
suspended mass as the system, change in potential energy of the suspended mass cannot
be obtained from an energy balance without first evaluating the work. Thus, we resort
here to an energy balance for the isolated system, which does not require such
information.
Problem 7.30
A rigid insulated tank contains 0.5 kg of carbon dioxide, initially at 150 kPa, 20°C. The carbon
dioxide is stirred by a paddle wheel until its pressure is 200 kPa. Using the ideal gas model with
cv = 0.65 kJ/kg·K, determine, in kJ, (a) the work, (b) the change in exergy of the carbon
dioxide, and (c) the amount of exergy destroyed. Ignore the effects of motion and gravity, and
let T0 = 20°C, p0 = 100 kPa.
Known:
Carbon dioxide is in a rigid tank with a paddle wheel.
Find:
Determine (a) the work, (b) the change in exergy of the carbon dioxide, and (c) the amount of
exergy destroyed.
Schematic and Known Data:
200 kPa
T
150 kPa
Carbon Dioxide
mg = 0.5 kg
T1 = 20 C
p1 = 150 kPa
p2 = 200 kPa
100 kPa
2
T2
20 C
0
1
v
Engineering Model:
(1) The carbon dioxide is the closed system and the volume remains constant throughout the
process (isometric).
(2) For the system, Q = 0 and the effects of motion and gravity can be ignored.
(3) For the environment, T0 = 20°C = 293 K, p0 = 100 kPa.
Analysis:
Using assumption (1) and the ideal gas model equation of state to obtain T2:
(a) An energy balance reduces to give,
(
)
(
)
⏟
⏟
(
)(
⏟
, therefore:
)(
)
(b) With Eq. 7.3:
⏟(
[
)
(
)]
Then, with Eq. 6.21:
[ (
)
)
(
(
) [(
)]
⏟
)
(
) [(
)(
]]
, where σ is obtained from an entropy balance:
(c)
(
)
Finally,
(
)(
(
⏟
)
(
) ((
)
)
)
Alternatively, an exergy balance reduces to
equation could also be used to solve part (c).
. This