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Lec 12: Closed system, open system

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• For next time:– Read: § 5.4

• Outline:– Conservation of energy equations– Relationships for open systems– Example problem

• Important points:– Memorize the general conservation of mass

and energy equations– Know how to translate the problem statement

into simplifications in the mass and energy conservation equations.

– Remember the conversion factor between m2/s2 and J/kg.

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Remember the difference between closed and open systems

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Property variations

• Closed system--properties at any location in the system are the same (though in a transient problem they may change with time).

• Open system--properties vary with location in a control volume--for example between the entrance to an air compressor and the exit.

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Steady-flow assumption

Extensive and intensive properties within the control volume don’t change with time, though they may vary with location.

Thus mCV, ECV, and VCV are constant.

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• With VCV constant and VCV=0, there is no boundary work.

Steady-flow assumption

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Steady-flow assumption

• With mCV and ECV constant,

0dt

dEcv 0dt

dmcv

• This allows the properties to vary from point-to-point but not with time.

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Steady-flow assumption

• However, material can still flow in and out of the control volume.

• The flow rate terms are not zero.m

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Consider a simple two-port system (one inlet/one outlet)

Control Volume

m cv, Ecv

Q W

m 2

m 1

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Assumptions

• No generation of mass or energy in the control volume

• No creation of mass or energy in the control volume

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Conservation of mass

c.v1 2

dMm m

dt

C.V.

INTO RATE

FLOW MASS

C.V.

OF OUT RATE

FLOW MASS

C.V.

THE IN MASS OF

CHANGE OF RATE

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Conservation of mass

• If we assume steady-flow,

dt

dmmm c.v

21 0

21 mm

outin mm

• and

• or

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• This can be extended to multiple inlets and outlets:

Conservation of mass

outin mm

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Apply energy conservation equation

C.V.

INTO ENERGY

OFRATE

C.V.

THE IN ENERGY OF

CHANGE OF RATE

C.V.OF

OUT ENERGY

OFRATE

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How does energy enter the control volume?

Heat Energy

Work Energy

Movement of fluid

inQ

inW

ininθm

[Could be if there is more than one source]

inQ

[Could be if there is more than one source]

inW

(this also could be a summed term)

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Conservation of energy

• Remember

kepehθ

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Thus, energy input is:

C.V.

INTO ENERGY

OFRATE

inQ inW ininθm

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We can develop a similar expression for rate of energy leaving the control volume

C.V.OF

OUT ENERGY

OFRATE

outQ outW outoutθm

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We’ve seen the transient term before in the closed system

analysis

C.V.

THE IN ENERGY OF

CHANGE OF RATE

dt

dEcv

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Let’s look at the heat transfer terms first:

We want to combine them into a single term to give us the net heat transfer

outQnetQ inQ

For simplicity, we’ll drop the “net” subscript

Q netQ

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We’ll do the same thing with work

Work becomes

netWW outin WW

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Energy equation

dt

dEmθmθ)W-W(Q-Q c.v

outoutininoutinoutin

Q Wdt

dEmθmθ c.v

outoutinin

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We’ll now write the energy equation as:

dt

dEmθmθW-Q c.v

outoutinin

dt

dEmθmθW-Q c.v

eeii

dt

dEmθmθW-Q c.v

2211

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Comment on work

• Work includes, in its most general case, shaft work, such as that done by moving turbine blades or a pump impeller; the work due to movement of the CV surface (usually the surface does not move and this is zero); the work due to magnetic fields, surface tension, etc., if we wished to include them (usually we do not); and the work to move material in and out of the CV. However, we have already included this last pv term in the enthalpy.

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e

2e

eei

2i

iiCV gz

2

Vhmgz

2

VhmWQ

dt

dE

and we finally have a useful expression for conservation of energy for an open system:

eeiiCV θmθmWQ

dt

dE

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More on the work term

The work term does not include boundary work (=0 because the control volume does not change size) and it does not include flow work.

Shaftcv WWW

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For multiple inlets and outlets, the first law will look like:

2

1 2

ki

Shaft i i ii

Q W h gz mV

2.

1 2

lj c v

j j jj

dEh gz m

dt

V

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Two port devices with steady state steady flow (SSSF) assumption

Conservation of mass:

1 2m m m

Conservation of energy:

21

1 12ShaftQ W h gz mV

2

22 2 0

2h gz m

V

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The energy equation can be simplified even more…..

Divide through by the mass flow:

m

Qq

Heat transfer per unit mass

m

Ww shaft

shaft

Shaft work per unit mass

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We get the following for the energy equation

2 2

2 12 1 2 1( - )

2 2Shaftq w h h g z zV V

Or in short hand notation:

pekehwq shaft

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TEAMPLAYTEAMPLAY

On a per unit mass basis, the conservation of energy for a closed system is

pekehwq shaft

pekeu wq Conservation of energy for an open system was just derived

Explain the difference of the meaning of each term between the open and closed system expressions.

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Let’s Review - for two port system

Conservation of Mass

dt

dmmm c.v

21

Conservation of energy

2

1 2

ki

Shaft i i ii

Q W h gz mV

2.

1 2

lj c v

j j jj

dEh gz m

dt

V

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Example ProblemExample Problem

Steam enters a two-port device at 1000 psia and 1000F with a velocity of 21.0 ft/s and leaves as a dry saturated vapor at 2 psia. The inlet area is 1 ft2 and the outlet area is 140 ft2.

A) What is the mass flow (lb/hr)?

B) What is the exit velocity (ft/s)?

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Draw DiagramDraw Diagram

1

2

Turbine

P1 = 1000 psia

T1 = 1000F

V1 = 21.0 ft/s

A1=1 ft2

P2 = 2 psia

x2 = 1.0

A1=140 ft2

STATE 1 STATE 2

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State assumptionsState assumptions

• Steady state (dm/dt = 0)• One inlet/one outlet• Uniform properties at inlet and outlet

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Basic EquationsBasic Equations

Conservation of mass for 2 port steady state:

21 )A

()A

(VV

m

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Get property data from steam tables:

31 m0.831ft / lb

32 m173.75 ft / lb

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Calculate mass flow:Calculate mass flow:

1)A

(V

m

)3600(831.0

)1(0.21

3

2

hr

s

hrft

ftsft

m

hr/lb975,90m m

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Exit VelocityExit Velocity

2)A

(V

m

A

mV

2

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Exit Velocity - page 2Exit Velocity - page 2

)s3600

hr(

ft140

)hr

lb975,90(

lb

ft75.173

V2

m

m

3

2

sftV /4.312

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TEAMPLAYTEAMPLAY

Water at 80 ºC and 7 MPa enters a boiler tube of constant inside diameter of 2.0 cm at a mass flow rate of 0.76 kg/s. The water leaves the boiler tube at 350 ºC with a velocity of 102.15 m/s. Determine (a) the velocity at the tube inlet (m/s) and (b) the pressure of the water at the tube exit (MPa).