Lec 12: Closed system, open system

For next time: Outline: Important points: Read: § 5.4 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.

Remember the difference between closed and open systems

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.

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.

Steady-flow assumption With VCV constant and VCV=0, there is no boundary work.

Steady-flow assumption With mCV and ECV constant, This allows the properties to vary from point-to-point but not with time.

Steady-flow assumption However, material can still flow in and out of the control volume. The flow rate terms are not zero.

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

Assumptions No generation of mass or energy in the control volume No creation of mass or energy in the control volume

Conservation of mass

Conservation of mass If we assume steady-flow, and or

Conservation of mass This can be extended to multiple inlets and outlets:

Apply energy conservation equation

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

Conservation of energy Remember

Thus, energy input is:

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

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

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

We’ll do the same thing with work Work becomes

Energy equation

We’ll now write the energy equation as:

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.

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

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.

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

Two port devices with steady state steady flow (SSSF) assumption Conservation of mass: Conservation of energy:

The energy equation can be simplified even more….. Divide through by the mass flow: Heat transfer per unit mass Shaft work per unit mass

We get the following for the energy equation Or in short hand notation:

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.

Let’s Review - for two port system Conservation of Mass Conservation of energy

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)?

Draw Diagram STATE 1 STATE 2 P1 = 1000 psia T1 = 1000F V1 = 21.0 ft/s A1=1 ft2 P2 = 2 psia x2 = 1.0 A1=140 ft2

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

Basic Equations Conservation of mass for 2 port steady state:

Get property data from steam tables:

Calculate mass flow:

Exit Velocity

Exit Velocity - page 2

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).