1 Lec 14: Heat exchangers. 2 Heat Exchangers and mixing devices Heat exchangers are devices which...
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Transcript of 1 Lec 14: Heat exchangers. 2 Heat Exchangers and mixing devices Heat exchangers are devices which...
2
Heat Exchangers and mixing devices
• Heat exchangers are devices which transfer heat between different fluids
• Mixing devices (also called open heat exchangers) combine two or more fluids to achieve a desired output, such as fluid temperature or quality
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Heat exchangers are used in a variety of industries
• Automotive - radiator• Refrigeration - evaporators/condensers• Power production - boilers/condensers• Power electronics - heat sinks• Chemical/petroleum industry- mixing
processes
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Heat Exchangers
• In a closed heat exchanger, the fluids do not mix. This is a shell-and-tube heat exchanger.
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Heat Exchangers
• Your book has very simple examples of heat exchangers. One is counterflow where the fluids flow in opposite directions in the heat exchanger:
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Heat Exchangers
• Yet another type is cross-flow, shown below. These are common in air conditioning and refrigeration systems.
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What assumptions to make?
Ask yourself:• See any devices producing/using shaft
work?• What about potential energy effects?• What about kinetic energy changes?• Can we neglect heat transfer?
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Apply conservation of mass on both streams...
dt
dmCV 0
A21 mmm
If we have steady flow, then:
And1m
2m
4m
3m
B43 mmm
Fluid A
Fluid B
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Conservation of energy can be a little more complicated...
1m
2m
4m
3m
I’ve drawn the control volume around the whole heat exchanger.
Implications:
•No heat transfer from the control volume.
Fluid A
Fluid B
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Conservation of energy looks pretty complicated:
A21 mmm
B43 mmm
2 2• • • •1 31 31 1 3 32 2CV
V VQ W m h gz m h gz
2 2
• •2 42 42 2 4 4 0
2 2V V
m h gz m h gz
We know from conservation of mass:
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Conservation of energy equation for the heat exchanger
2 2• • • 1 2
1 2 1 22 2ACV
A
V VQ W m h h g z z
2 2
• 3 43 4 3 4 0
2 2B
B
V Vm h h g z z
Apply what we know about the mass flow relationships:
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Heat Exchangers
• Generally, there is no heat transfer from or to the heat exchanger, except for that leaving or entering through the inlets and exits.
• So,• And, because the device does no work,
• Also, potential and sometimes kinetic energy changes are negligible.
0Q
0WCV
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Heat Exchangers - apply assumptions
2 2• • • 1 2
1 2 1 22 2ACV
A
V VQ W m h h g z z
2 2
• 3 43 4 3 4 0
2 2B
B
V Vm h h g z z
0 0 0 0
0 0
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Heat Exchangers
)hh(m)hh(m 34B21A
After throwing away a bunch of terms, we’re left with:
The energy change of fluid A is equal to the negative of the energy change in fluid B.
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TEAMPLAYTEAMPLAY
How would the energy equation differ if we drew the boundary of the control volume around each of the fluids?
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Heat Exchangers
• Now if we want the energy lost or gained by either fluid we must let that fluid be the control volume, indicated by the red.
A1 mm
2m
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Heat Exchangers
0hhmQ 21AA,CV
12A hhq
The energy equation for one side:
Or dividing through by the mass flow:
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Example Problem
Refrigerant 134a with a mass flow rate of 5 kg/min enters a heat exchanger at 1.2 MPa and 50C and leaves at 1.2 MPa and 44C. Air enters the other side of the heat exchanger at 34 C and 1 atmosphere and leaves at 42 C and 1 atmosphere. Calculate:
a) the heat transfer from the refrigerant in (kJ/min)
b) the mass flow rate of the air (kg/min)
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Draw diagram
R134am 5 kg/ min
?mair
R134a INLET
T1=50C
P1 = 1.2 MPa
R134a OUTLET
T2=44C
P2 = 1.2 MPa
AIR INLET
T3=34C
P3 = 101 kPa
AIR OUTLET
T3=42C
P3 = 101 kPa
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State assumptions
• Steady state, steady flow• No work• Air is ideal gas• Kinetic energy change is zero• Potential energy change is zero
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Start analysis with R134a
2 2• • • 2 1ReRe 2 1 2 1[( ) ( )]
2ff
V VQ W m h h g z z
Apply assumptions
0 0 0
)h(hmQ 12RefRef
We can get h1 and h2 from tables. The refrigerant mass flow is given.
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From R134a tables
h1 = 275.52 kJ/kg h2 = 112.22 kJ/kg
Plugging back into energy equation:
kg
kJ)52.27522.112(
min
kg5QRef
min
kJ5.816QRef
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On to part (b) of the problem.
We want to get the mass flow of the air...
Start by writing the energy equation for the air side:
2 2• • • 4 34 3 4 3[( ) ( )]
2airair
V VQ W m h h g z z
Simplify
0 0 0
)h(hmQ 34airair
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Sample Problem, Con’t
If air is an ideal gas, then we can rewrite the enthalpy difference as:
)T(TcmQ 34pairair
Rearrange to solve for mass flow:
)T(Tc
Qm
34p
airair
How do we get the heat transfer rate to/from the air?
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Almost there!!!!!
refair QQ
We can write:
somin
kJ5.816Qair
Get specific heat from table:
kgK
kJ006.1cp
Plug in numbers from here: )K34)(42
kgKkJ
(1.006
minkJ
816.5mair
min
kg4.101mair
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Open heat exchangers (mixers)
• In an open heat exchanger, the fluids mix.
1,airm
2,airm
3,airm
3,air2,air1,air mmm
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Apply conservation equations for steady state, steady flow
321 mmm Mass
Energy2 2• • • •1 2
1 21 1 2 22 2V V
Q W m h gz m h gz
2• 3
3 3 3 02V
m h gz
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Look at some typical assumptions
• Steady state, steady flow• No work• No heat transfer - not always true• Kinetic energy change is zero - usually• Potential energy change is zero - usually
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Look at the impact on the energy equation
332211 hmhmhm
Physically, what does this equation tell you?
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Sample Problem
Water is heated in an insulated tank by mixing it with steam. The water enters at a rate of 200 lbm/min at 65°F and 50 psia. The steam enters at 600°F and 50 psia. The mixture leaves the tank at 200°F and 48 psia. How much steam is needed?
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Draw system with knowns:
lb/min200m1
?m2 ?m3
T1 = 65°F
P1 = 50 psia
Water
Steam
T2 = 600°F
P2 = 50 psia
T3 = 100°F
P3 = 50 psia
Outlet
42
Assumptions
• Steady state, steady flow• No work• No heat transfer • Kinetic energy change is zero• Potential energy change is zero
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Applying assumptions gives us:
321 mmm Mass
Energy
332211 hmhmhm
We have two unknowns (m2 and m3) and two equations.
)h-(h
)h-(hmm
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3112
44
Get some properties and complete the solution
h1 = 33.1 btu/lbm
h2 = 1332.8 btu/lbm
h1 = 33.1 btu/lbm
lbmbtu
)8.1332-(68.1
lbmbtu
)1.68-(33.1
min
lbm002m2
min
lbm5.5m2