Chapter 5
The First Law of Thermodynamics
5221
Two rigid tanks that contain water at different states are connected by a valve. The valve is opened
and the two tanks come to the same state at the temperature of the surroundings. The final pressure and the
amount of heat transfer are to be determined.
Assumptions
1
The tanks are stationary and thus the kinetic and potential energy changes are zero.
2
The
tank is insulated and thus heat transfer is negligible.
3
There are no work interactions.
Analysis
We take the entire contents of the tank as the system. This is a closed system since no mass
enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the
energy balance for this stationary closed system can be expressed as
]
)
(
)
(
[
]
[
0)
=
PE
=
KE
(since
)
(
)
(
1
1
1
1
2
total
2,
,
1
,
1
,
2
out
out
energies
etc.
potential,
kinetic,
internal,
in
Change
system
mass
and
work,
heat,
by
nsfer
energy tra
Net
out
in
B
A
B
A
B
A
B
A
u
m
u
m
u
m
U
U
U
Q
W
U
U
U
Q
E
E
E
−
−
−
=
−
−
−
=
=
Δ
+
Δ
=
Δ
=
−
Δ
=
−
+
43
42
1
1
H
2
O
400 kPa
H
2
O
200 kPa
×
B
A
Q
The properties of water in each tank are
(Tables A4 through A6)
Tank A:
kJ/kg
1949.3
,
31
.
604
/kg
m
0.4625
,
001084
.
0
80
.
0
kPa
400
3
1
1
=
=
=
=
⎭
⎬
⎫
=
=
fg
f
g
f
u
u
v
v
x
P
( )
[ ]
()
kJ/kg
2163.75
3
.
1949
8
.
0
31
.
604
/kg
m
0.3702
001084
.
0
4625
.
0
8
.
0
001084
.
0
1
,
1
3
1
,
1
=
×
+
=
+
=
=
−
×
+
=
+
=
fg
f
A
fg
f
A
u
x
u
u
v
x
v
v
Tank B:
kJ/kg
2304.9
,
88
.
104
/kg
m
43.36
,
001003
.
0
/kg
m
0.731
C
25
/kg
m
0.731
kg
0.957
m
0.7
kg
0.957
417
.
0
540
.
0
kg
0.417
/kg
m
1.1988
m
0.5
kg
0.540
/kg
m
0.3702
m
0.2
kJ/kg
2731.2
/kg
m
1.1988
C
250
kPa
200
3
3
2
2
3
3
2
,
1
,
1
3
3
,
1
,
1
3
3
,
1
,
1
,
1
3
,
1
1
1
=
=
=
=
⎪
⎭
⎪
⎬
⎫
=
=
=
=
=
=
+
=
+
=
=
=
=
=
=
=
=
=
⎭
⎬
⎫
=
=
fg
f
g
f
t
t
B
A
t
B
B
B
A
A
A
B
B
u
u
v
v
v
T
m
V
v
m
m
m
v
V
m
v
V
m
u
v
T
P
o
o
Thus at the final state the system will be a saturated liquidvapor mixture since
v
f
<
v
2
<
v
g
. Then the final
pressure must be
P
2
= P
sat @ 25
°
C
=
3.169 kPa
Also,
kJ/kg
143.60
2304.9
0.0168
104.88
0168
.
0
001
.
0
36
.
43
001
.
0
731
.
0
2
2
2
2
=
×
+
=
+
=
=
−
−
=
−
=
fg
f
fg
f
u
x
u
u
v
v
v
x
Substituting,
Q
out
= [(0.957)(143.6)  (0.540)(2163.75)  (0.417)(2731.2)] =
2170 kJ
5208
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentChapter 5
The First Law of Thermodynamics
5222
Problem 5221 is reconsidered. The effect of the environment temperature on the final
pressure and the heat transfer as the environment temperature varies from 0°C to 50°C is to be
investigated. The final results are to be plotted against the environment temperature.
"Knowns"
Vol_A=0.2
"[m^3]"
P_A[1]=400
"[kPa]"
x_A[1]=0.8
T_B[1]=250
"[C]"
P_B[1]=200
"[kPa]"
Vol_B=0.5
"[m^3]"
T_final=25
"[C]"
"T_final = T_surroundings.
To do the parametric study
or to solve the problem when Q_out = 0, place this statement in {}."
{Q_out=0"[kJ]"}
"To determine the surroundings temperature that
makes Q_out = 0, remove the {} and resolve the problem."
"Solution"
"Conservation of Energy for the combined tanks:"
E_inE_out=DELTAE
E_in=0
"[kJ]"
E_out=Q_out
"[kJ]"
DELTAE=m_A*(u_A[2]u_A[1])+m_B*(u_B[2]u_B[1])
"[kJ]"
m_A=Vol_A/v_A[1]
"[kg]"
m_B=Vol_B/v_B[1]
"[kg]"
u_A[1]=INTENERGY(Steam,P=P_A[1], x=x_A[1])
"[kJ/kg]"
v_A[1]=volume(Steam,P=P_A[1], x=x_A[1])
"[m^3/kg]"
T_A[1]=temperature(Steam,P=P_A[1], x=x_A[1])
"[C]"
u_B[1]=INTENERGY(Steam,P=P_B[1],T=T_B[1])
"[kJ/kg]"
v_B[1]=volume(Steam,P=P_B[1],T=T_B[1])
"[m^3/kg]"
"At the final state the steam has uniform properties through out the entire system."
u_B[2]=u_final
"[kJ/kg]"
u_A[2]=u_final
"[kJ/kg]"
m_final=m_A+m_B
"[kg]"
Vol_final=Vol_A+Vol_B
"[m^3]"
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Dr.Kinne
 Thermodynamics, Energy, net energy transfer, The First Law of Thermodynamics

Click to edit the document details