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Lec 14: Heat exchangers

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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 can take a variety of shapes

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Condenser/evaporator for heat pump

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Cooling towers are a type of heat exchanger.

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Something a little closer to home..

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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

• Another type is parallel flow, where the fluids flow in the same direction:

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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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With heat exchangers, we have to deal with multiple inlets and

outlets

1m

2m

4m

3m

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• At steady flow, what is the relationship between

?mand,m,m,m 4321

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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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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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Work problem 5-100E

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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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The mixer may look more like a tank

1m

2m 3m

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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

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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

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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

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Work problem 5-96