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1
Thermodynamics Vm235
Lecture 5
Chapter 5 – The First Law
Key concepts for today:
¾
Ideal gas.
¾
Examples.
¾
We now wish to apply it to an open system –
1
st
Law analysis of a control volume – Chapter 6.
A piston/cylinder contains 50 kg of water at 200 kPa with a volume of 0.1 m
3
.
Stops in the cylinder are placed to restrict the enclosed volume to a maximum
of 0.5 m
3
. The water is now heated until the piston reaches the stops. Find the
necessary heat transfer.
P
Solution:
C.V. H
2
O ;
m = constant ;
Energy Conservation (Eq.5.11):
Constant pressure example:
V
1
2
0.1
0.5
m(e
2
– e
1
) = m(u
2
– u
1
) =
1
Q
2

1
W
2
;
Process: P = constant
⇒
1
W
2
=
∫
P dV = P
1
(V
2
– V
1
)
or,
(mu
2
+ P
2
V
2
) – (mu
1
+ P
1
V
1
) = m(h
2
–h
1
) =
1
Q
2
State 1:
v
1
= 0.1/50 = 0.002 m
3
/kg
=>
2phase
as
v
1
< v
g
; (Table B.1.2)
x
1
=
v
1
– v
f
v
fg
=
0.002 – 0.001061
0.88467
= 0.001061 ;
h
1
= 504.68 + 0.001061 × 2201.96 = 507.02 kJ/kg
3
Heat transfer
from the energy equation:
1
Q
2
= m(u
2
– u
1
) +
1
W
2
= m(h
2
– h
1
)
= 50 kg × (526.92 – 507.02) kJ/kg =
995 kJ ;
[Note:
1
W
2
= P
1
(V
2
– V
1
) = 200 × (0.5 – 0.1) = 80 kJ ]
State 2:
v
2
= 0.5/50 = 0.01 m
/kg
also 2phase as
v
2
< v
g
, same P
x
2
=
v
2
– v
f
v
fg
=
0.01 – 0.001061
0.88467
= 0.01010 ;
h
2
= 504.68 + 0.01010 × 2201.96 = 526.92 kJ/kg
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.
:
gases
ideal
of
behavior
the
describing
Equation
only
gases
ideal
for
u(T)
u
RT
Pv
=
=
Internal Energy and Enthalpy of Ideal Gases
.
)
,
P
,
T
(i.e.
region
gas
ideal
in the
tables
the
checking
by
yourself
satisfy
now,
For
later.
until
proof
the
postpone
will
we
Hence,
so.
do
entropy to
need
but
this
prove
ally
mathematic
can
We
remember
fact to
important
An
or
v
↓
↑
↓
↑
ρ
remember
fact to
important
An
.
)
(
:
:
only
gases
ideal
for
T
h
h
Thus
RT
u
h
RT
Pv
and
Pv
u
h
Because
=
+
=
⇒
=
+
=
10
100
3/kg)
5000
5200
5400
)
0.1
1
Specific Volume (m
4400
4600
4800
u and h (kJ/kg
Specific Volume
0.01
1
Pressure (MPa)
4000
4200
Internal Energy
Specific Enthalpy
3
:
),
,
(
:
@
@
)
(
;
0
dv
v
u
dT
T
u
du
so
v
T
u
u
general
In
T
const
v
const
gas
ideal
for
v
u
u
general
in
v
C
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=
=
≠
=
=
±
±²
±
±³
´
´
Ideal Gases (internal energy & enthalpy)
;
),
,
(
:
)
(
&
:
.
;
:
0
0
0
0
dP
h
dT
h
dh
P
T
h
h
general
in
Again
T
C
C
dT
C
du
now
because
great
is
simplicity
This
gas
ideal
an
for
dT
du
C
Thus
v
v
v
v
⎟
⎞
⎜
⎛ ∂
+
⎟
⎞
⎜
⎛ ∂
=
=
=
=
=
.
;
)
(
&
:
0
0
0
0
@
@
)
(
;
0
gas
ideal
an
for
T
C
C
dT
C
dh
or
dT
dh
C
Thus
P
T
P
P
P
P
T
const
P
const
gas
ideal
for
P
h
h
general
in
P
C
=
=
=
⎠
⎝
∂
⎠
⎝
∂
≠
=
=
±
±
´
´
Chap 5, more examples:
A rigid
tank A
of volume 0.6 m
3
contains 3 kg water at 120
o
C and
the rigid
tank B
is 0.4 m
3
with water at 600 kPa, 200
o
C.
They are
connected to a piston cylinder initially empty with closed valves.
The pressure in the cylinder should be 800 kPa to float the piston.
Now the valves are slowly opened and heat is transferred so the
water reaches a uniform state at 250
o
C with the valves open.
C
Find the final
volume
,
pressure
,
work done
and
heat transfer
in the
process.
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This note was uploaded on 09/15/2010 for the course ME 235 taught by Professor Borgnakke during the Fall '07 term at University of Michigan.
 Fall '07
 Borgnakke

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