Lec 12: Closed system, open system

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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 dont change with time, though
they may vary with location. Thus mCV, ECV, and
VCV are constant.
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Steady-flow assumption

• With VCV constant and ?VCV0, there is no
  boundary work.

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Steady-flow assumption

• With mCV and ECV constant,

• 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.

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

• If we assume steady-flow,

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• and

• or

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Conservation of mass

• This can be extended to multiple inlets and
  outlets

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Apply energy conservation equation
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How does energy enter the control volume?
Could be if there is more than one
source
Heat Energy
Could be if there is more than one
source
Work Energy
(this also could be a summed term)
Movement of fluid
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Conservation of energy

• Remember

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Thus, energy input is
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We can develop a similar expression for rate of
energy leaving the control volume
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Weve seen the transient term before in the
closed system analysis
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Lets look at the heat transfer terms first
We want to combine them into a single term to
give us the net heat transfer
For simplicity, well drop the net subscript
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Well do the same thing with work
Work becomes
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Energy equation
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Well now write the energy equation as
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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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and we finally have a useful expression for
conservation of energy for an open system
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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.
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For multiple inlets and outlets, the first law
will look like
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Two port devices with steady state steady flow
(SSSF) assumption
Conservation of mass
Conservation of energy
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The energy equation can be simplified even more..
Divide through by the mass flow
Heat transfer per unit mass
Shaft work per unit mass
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We get the following for the energy equation
Or in short hand notation
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TEAMPLAY
On a per unit mass basis, the conservation of
energy for a closed system is
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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Lets Review - for two port system
Conservation of Mass
Conservation of energy
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Example Problem
Steam enters a two-port device at 1000 psia and
1000?F 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 Diagram
STATE 2
STATE 1
P1 1000 psia T1 1000?F V1 21.0 ft/s A11 ft2
P2 2 psia x2 1.0 A1140 ft2
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State assumptions

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

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Basic Equations
Conservation of mass for 2 port steady state
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Get property data from steam tables
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Calculate mass flow
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Exit Velocity
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Exit Velocity - page 2
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TEAMPLAY
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).