2 For next time: Outline: Important points: Read: § 5.4 Conservation of energy equationsRelationships for open systemsExample problemImportant points:Memorize the general conservation of mass and energy equationsKnow 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.
3 Remember the difference between closed and open systems
4 Property variationsClosed 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.
5 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.
6 Steady-flow assumption With VCV constant and VCV=0, there is no boundary work.
7 Steady-flow assumption With mCV and ECV constant,This allows the properties to vary from point-to-point but not with time.
8 Steady-flow assumption However, material can still flow in and out of the control volume.The flow rate terms are not zero.
9 Consider a simple two-port system (one inlet/one outlet)
10 Assumptions No generation of mass or energy in the control volume No creation of mass or energy in the control volume
15 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
24 Comment on workWork 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.
25 and we finally have a useful expression for conservation of energy for an open system:
26 More on the work termThe work term does not include boundary work (=0 because the control volume does not change size) and it does not include flow work.
27 For multiple inlets and outlets, the first law will look like:
28 Two port devices with steady state steady flow (SSSF) assumption Conservation of mass:Conservation of energy:
29 The energy equation can be simplified even more….. Divide through by the mass flow:Heat transfer per unit massShaft work per unit mass
30 We get the following for the energy equation Or in short hand notation:
31 TEAMPLAYOn a per unit mass basis, the conservation of energy for a closed system isConservation of energy for an open system was just derivedExplain the difference of the meaning of each term between the open and closed system expressions.
32 Let’s Review - for two port system Conservation of MassConservation of energy
33 Example ProblemSteam 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)?
41 TEAMPLAYWater 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 m/s. Determine (a) the velocity at the tube inlet (m/s) and (b) the pressure of the water at the tube exit (MPa).