Thermodynamics lecture 10

1. BITS
Pilani
BITS
Pilani
Pilani Campus
Lecture10
FirstLawforControlMass
Lecture
10
–
First
Law
for
Control
Mass

2. Problem 1 A piston/cylinder contains 50 kg of wate
r
at 200 kPa with a
volume of 0.1 m
3
. Stops in the
c
y
linde
r
restricts the enclosed
y
volume to 0.5 m
3
,asshownin
fig. The water is now heated to
200
o
C
Find
the
final
pressure
200
o
C
.
Find
the
final
pressure
,
volume and the work done by
thewater.
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3. Example 2
A piston-cylinder arrangement has a linear
sprin
g
and the outside atmosphere actin
g
g
g
on the piston. It contains water at 3 MPa and 400
o
Cwithavolumeof0.1m
3
.Ifthe
piston
is
at
the
bottom
the
spring
exerts
a
piston
is
at
the
bottom
,
the
spring
exerts
a
force such that a pressure of 200 kPa insideisrequiredtobalancetheforces.The
t
l
til
th
sys
t
em now coo
lsun
til
th
e pressure
reaches 1 MPa. Find the heat transfer for
theprocess.
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4. Example 2:Solution Initial state determined by P and v, it is superheated. Final
state fixed by P (given) and v again. How do we find v?
P= 200 kPa+ k
s
V/A
2
, where k
s
is the spring constant, and A
the area of cross-section. This is since the pressure to just
balancewhenpistonisatbottomisgivenas200
kPa
Since
balance

when

piston

is

at

bottom

is

given

as

200

kPa
.
Since

P
1
= 3 MPaand V
1
= 0.1 m
3
given, k
s
/A
2
= 2.8x10
4
kJ/m
3
, so
V
2
= 0.02857 m
3
, v
2
= 0.02840 m
3
/kg, saturated mixture with
x
2
= 0.14107
State P (kPa) T (C) V (m
3
)v (m
3
/kg) Mass M (kg) x u (kJ/kg) U(kJ)
1 3000 400 0.1 0.09936 1.006‐2950
2 1000 143.6 0.02857 0.02840 1.006 0.14107 1019
W
=∫
PdV
=200(
∆
V)+28x10
4
(V
2
V
2
)/2=
143kJ
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1
W
2
=

∫
PdV
=

200

(
∆
V)

+

2
.
8x10
4
(V
2
2
–
V
1
2
)/2

=
-
143

kJ
1
Q
2
= ∆U +
1
W
2
= -1925 -143 = -2068 kJ

5. Problem 3
An insulated piston cylinder device contains
5
L
of
saturated
liquid
contains
5
L
of
saturated
liquid
water at a constant pressure of
175kPa.Waterisstirredbyapeddle
wheel while current of 8
A
flows fo
r
45 min through a resistor place in the wate
r
.If50%ofli
q
uid
(
b
y
mass
)
q
(y
)
is evaporated during this constant pressure process and the peddle work
amounts
to
300
kJ
determine
work
amounts
to
300
kJ
,
determine
thevoltageofthesource.Alsoshow theonP-vdiagram.
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6. Fi 42 E l f k i th b d f t
BITS Pilani, Pilani Campus
Fi
g
4
.
2

E
xamp
le o
f
wor
k
cross
ing
th
e
b
oun
d
ary o
f
a sys
t
em
because of electric current flow across the system boundary

7. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by m ass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram. Solution:Assumptions 1The cylinde
r
is stationary and thus the kinetic and potential
energychangesarezero.
2
The
cylinder
is
well
-
insulated
and
thus
heat
transfer
is
2
The
cylinder
is
well
insulated
and
thus
heat
transfer
is
negligible.
3Thecompressionorexpansionprocessisquasi-equilibrium.
BITS Pilani, Pilani Campus

8. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by m ass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram. •Wetakethecontentsofthecylinderasthesystem. •
This
is
a
closed
system
since
no
mass
enters
or
leaves
•
This
is
a
closed
system
since
no
mass
enters
or
leaves
.
•The energy balance for this stationary closed system can be
expressedas
)
(
)
(
2
1
2
2
Z
Z
V Vm
U
U
W
Q
+
−
+
)
(
2
)
(
1 2
1
2
1 2 2 1 2 1
Z
Z
m
g
U
U
W
Q
−
+
+
−
= −
1 2 2 1
U U
W
−
=
−
BITS Pilani, Pilani Campus

9. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by m ass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram.
U
W W W
=
−
+
out
b,
in
p
w
,
ine
,
  
∆
)
h
h
(
m
W
)
t
I
(
)h h(m W W
1
2
1 2
−
=
+
− = +
i
p
in pw, ine,
,
p,
,
∆
V
)
h
h
(
m
W
)
t
I
(
1
2
+
i
n
p
w,
∆
V
/
kJ/kg 486.97  kPa 175
3
kPa 175
=
=
⎬⎫ =
@f 1 1
h h P
  kPa,  175
/
kg m 0.001057  liquid sat.
3
kPa 175
= =
=
=
⎭⎬
2 2
@f 1
5.0 x P
v
v
(
)
k
47304
m 0.005
kJ/kg 1593.755
3
=
×
+
= + =
1
fg 2 f 2
57. 2213 5.0 97.
4
86 hx h h
V
BITS Pilani, Pilani Campus
k
g
4
.
7304
/kg m 0.001057 
3
=
= =
11
m
v

10. An insulated piston cylinder device contains 5 L of saturated liquid water at a constant
pressure of 175kPa. Water is stirred by a peddle wheel while current of 8 A flows for 45 min
through a resistor place in the water. If 50% of liquid (by m ass) is evaporated during this
constant pressure process and the peddle work amounts to 300kJ, determine the voltage of
the source. Also show the on P-v diagram.
SUBSTITUTING
=
−
=
+
kJ 
kg 486.97)kJ/ 55 kg)(1593.7 (4.7304 kJ) (300
V
V
536. 4935 tI
t
I
∆
∆
V 228.497 =
⎟
⎟
⎠⎞
⎜
⎜
⎝
⎛
×
=
kJ/s  1
VA  1000
s) 60 A)(45  (8
kJ  4935.536
V
P
12
BITS Pilani, Pilani Campus
v

11. First law as a rate equation
W Q PE KE U
δ
δ
−
=
∆
+
∆+ ∆
t
W
tQ
t
PE
t
KE
tU
δδ
δδ
δ
δ
δ
− =
∆
+
∆
+
∆
dt
dU
tU
t
=
∆
δ
δ
0
lim
dt
t
δ
KE d KE) ( ) (
lim
=
∆
dt t
t
lim
0
δ
δ
PE d PE) ( ) (
lim
=
∆
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dt t
t
lim
0
=
δ
δ

12. First law as a rate equation
Q
Q
&
δ
lim
Q
tQ
t
=
δ
δ
0
lim
W
δ
W
t
W
t
&
=
δδ
δ
0
lim
W Q
d
t
PEd
d
t
KE d
d
t
dU
& &
− = + +
) ( ) (
W
Q
dE
& &
=
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W
Q
dt
−
=

13. BITS
Pilani
BITS
Pilani
Pilani Campus
FIRST
LAWANALYSISFORA
FIRST
-
LAW
ANALYSIS
FOR
A

CONTROL VOLUME

14. FIRST-LAW ANALYSIS FOR A CONTROLVOLUME CONTROL
VOLUME
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15. Conservation of mass and ControlVolume Contro
l
volumes:
Control
Volume
Masscancrosstheboundaries,andsowemustkeeptrack
of
the
amount
of
mass
entering
and
leaving
the
control
of
the
amount
of
mass
entering
and
leaving
the
control
volume.
ConservationofMassPrinciple
Netmasstransfertoorfromasystemduringaprocessis
equal to the net change in total mass of the system
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duringthatprocess.

16. Conservation of mass and ControlVolume Control
Volume
∑
∑
V
C
dm
& &
∑
∑
−
=
e i
V
C
m m
dt
..
Total mass inside the control volume
∫
∫
+ + + = = =....... )/1(
..C B A VC
m m m dV v dV m
ρ
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17. Conservation of mass and ControlVolume Control
Volume
Th l fl t i Th
e

vo
l
ume
fl
ow

ra
t
e
i
s
∫
∨= ∨=dA A V
local
&
The mass flow rate becomes
∫
The mass flow rate becomes
A
dA
V
V
m
local
∨
=
∨
=
=
=
∫
)
(
&
& &
ρ
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v
dA
v v
V
m
avg
=
=
=
=
∫
)
(
ρ

18. FIRST-LAW ANALYSIS FOR A CONTROLVOLUME CONTROL
VOLUME
For a Fixed mass
1
2
E E−
For a Fixed mass
=
2
1
Q
−
2
1W
1
2
2
1
Q
2
1
The instantaneous rate equation is The instantaneous rate equation is
dE
M
C
Q
&
W
&
dt
M
C
..
=
Q
−
W
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