Chapter 5
The First Law of Thermodynamics
Mixing Chambers and Heat Exchangers
5101C
Yes, if the mixing chamber is losing heat to the surrounding medium.
5102C
Under the conditions of no heat and work interactions between the mixing chamber and the
surrounding medium.
5103C
Under the conditions of no heat and work interactions between the heat exchanger and the
surrounding medium.
5104
A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass
flow rate of cold water is to be determined.
Assumptions
1
Steady operating conditions exist.
2
The mixing chamber is wellinsulated so that heat loss
to the surroundings is negligible.
3
Changes in the kinetic and potential energies of fluid streams are
negligible.
4
Fluid properties are constant.
5
There are no work interactions.
Properties
Noting that T < T
sat @ 250 kPa
= 127.44°C, the water in all three streams exists as a compressed
liquid, which can be approximated as a saturated liquid at the given temperature.
Thus,
h
1
≅
h
f
@ 80
°
C
=
334.91 kJ/kg
h
2
≅
h
f
@ 20
°
C
=
83.96 kJ/kg
h
3
≅
h
f
@ 42
°
C
=
175.92 kJ/kg
Analysis
We take the mixing chamber as the system, which is a control
volume. The mass and energy balances for this steadyflow system can be
expressed in the rate form as
Mass balance:
&&
&
&
mm
E
m
in
out
−=
=⎯
→
⎯+
=
Δ
system
(steady)
Ê
0
12
0
3
Energy balance:
H
2
O
(P = 250 kPa)
T
3
= 42
°
C
T
1
= 80
°
C
m
1
= 0.5 kg/s
T
2
= 20
°
C
m
2
·
·
0)
pe
ke
(since
0
3
3
2
2
1
1
energies
etc.
potential,
kinetic,
internal,
in
change
of
Rate
(steady)
0
system
mass
and
work,
heat,
by
nsfer
energy tra
net
of
Rate
≅
Δ
≅
Δ
=
=
=
+
=
=
Δ
=
−
W
Q
h
m
h
m
h
m
E
E
E
E
E
out
in
out
in
&
&
&
&
&
&
&
4
43
4
42
1
&
1
&
&
©
Combining the two relations and solving for
gives
&
m
2
( )
3
2
1
2
2
1
1
h
m
m
h
m
h
m
&
&
&
&
+
=
+
m
hh
m
2
13
32
1
=
−
−
Substituting, the mass flow rate of cold water stream is determined to be
()
kg/s
0.864
=
−
−
=
kg/s
0.5
kJ/kg
83.96
175.92
kJ/kg
175.92
334.91
2
m
&
5105
Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing
temperature, the mass flow rate of the steam is to be determined.
590
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View Full DocumentChapter 5
The First Law of Thermodynamics
Assumptions
1
This is a steadyflow process since there is no change with time.
2
Kinetic and potential
energy changes are negligible.
3
There are no work interactions.
4
The device is adiabatic and thus heat
transfer is negligible.
Properties
Noting that
T
<
T
sat @ 300 kPa
= 133.55
°
C, the cold water stream and the mixture exist as a
compressed liquid, which can be approximated as a saturated liquid at the given temperature.
Thus,
h
1
≅
h
f @ 20
°
C
=
83.96 kJ/kg
h
3
≅
h
f @ 60
°
C
= 251.13 kJ/kg
and
kJ/kg
3069.3
C
300
kPa
300
2
2
2
=
⎪
⎭
⎪
⎬
⎫
=
=
h
T
P
o
Analysis
We take the mixing chamber as the system, which is a control volume since mass crosses the
boundary. The mass and energy balances for this steadyflow system can be expressed in the rate form as
Mass balance:
&&
&
&
mm
m
m
in
out
in
out
−=
=→
=
→+
=
Δ
system
(steady)
Ê
0
12
0
3
Energy balance:
&
&
&
&
EE
E
mh
Q W
in
out
in
out
=
=
+=
≅
≅
≅
≅
Rate of net energy transfer
by heat, work, and mass
system
(steady)
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 Spring '07
 Dr.Kinne
 Thermodynamics, Energy, Heat, Heat Transfer, The First Law of Thermodynamics

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