1 / 16

EGR 334 Thermodynamics Chapter 4: Section 1-3

EGR 334 Thermodynamics Chapter 4: Section 1-3. Lecture 12: Control Volumes and Conservation of Mass. Quiz Today?. Today’s main concepts:. Be able to explain what a control volume is Be able to write mass balance and mass rate balance equations for a control volume.

erin-larson
Télécharger la présentation

EGR 334 Thermodynamics Chapter 4: Section 1-3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. EGR 334 ThermodynamicsChapter 4: Section 1-3 Lecture 12: Control Volumes and Conservation of Mass Quiz Today?

  2. Today’s main concepts: • Be able to explain what a control volume is • Be able to write mass balance and mass rate balance equations for a control volume. • Be able to explain the continuity of mass flow equation. • Explain the difference between mass flow rate and volumetric flow rate. • Be able to set up problems involving mass balance Reading Assignment: • Read Chap 4: Sections 4-5 Homework Assignment: From Chap 4: 1, 6, 11, 22

  3. Sec 4.1: Conservation of Mass Q and W Now move to open system, mass can also pass the boundary. add the mass balance. Mass So far we have looked at closed systems Only Q and W pass the boundary Energy Balance From Chapter 2: Energy Balance • [ ] E within the system net Qinput • net W • output • [ ] • [ ] + = MassBalance • [ ] m within the system • [ ] • [ ] net minput • net m output + =

  4. Sec 4.1: Conservation of Mass MassBalance • [ ] m within the system • [ ] • [ ] net minput • net m output + = CV = control volume i = inlet, e = exit General form for multiple inlets/exits

  5. Sec 4.1.2: Mass Flow Rate Continuity Principle: mass flow is steady and continuous where ρ = density Vn = normal velocity component A = cross sectional area V = volume Mass flow rate: for constant Area for variable Area MassFlux

  6. Sec 4.2 Mass Rate Balance One-Dimensional Flow (Continuity Equation) • Flow is normal to the boundary • The fluid is homogeneous (intensive properties are uniform with position). where: = mass flow rate ρ = density = volumetric flow rate V = velocity v = specific volume A = area General form for multiple inlets/exits At steady State, and

  7. Sec 4.3 : Applications of the Mass Rate Balance Example: (4.16) Ammonia enters a control volume operating at steady state at pA= 14 bar, TA= 28oC, with a mass flow rate of 0.5 kg/s. Saturated vapor at pB = 4 bar leaves through one exit, with a volumetric flow rate of 1.036 m3/min and saturated liquid at pC=4 bar leaves through a second exit. Determine (a) the minimum diameter of the inlet pipe, in cm, so the ammonia velocity does not exceed 20 m/s (b) the volumetric flow rate of the second exit stream in m3/min. • Saturated vapor • pB= 4 bar, • 1.036 m3/min • pA= 14 bar, TA= 28oC • 0.5 kg/s • pC= 4 bar • Saturated liquid

  8. Sec 4.3 : Applications of the Mass Rate Balance • pA= 14 bar • TA= 28oC • mA=0.5 kg/s • Saturated vapor • pB= 4 bar, • 1.036 m3/min • pC= 4 bar • Saturated liquid Ammonia What else can you determine about the states? State 1: from table A-14 (state A is compressed liquid… let vA≈ vf @28oC = 1.6714x10-3 m3/kg) State 2: from table A-14 ( vB = vg @4bar) = 0.3094 m3/kg and TB=Tsat = -1.9oC) State 3: from table A-14 ( vC = vf @4bar)= 0.0015597 m3/kg and TC=Tsat = -1.9oC)

  9. Sec 4.3 : Applications of the Mass Rate Balance • Ammonia • Saturated vapor • Liquid • Saturated • liquid Consider mass flows: State A: already known… State B: can be found from volumetric flow rate State C: can be found from mass rate balance

  10. Sec 4.3 : Applications of the Mass Rate Balance • Ammonia • Saturated vapor • Liquid • Saturated • liquid What can we learn from the continuity equation: State A: State B: (already known) State C:

  11. Sec 4.3 : Applications of the Mass Rate Balance • Ammonia • Liquid • Saturated • vapor • Saturated • liquid Determine the size of the inlet pipe so that the velocity does not exceed VA = 20 m/s from the continuity equation: Area of a circular cross section: therefore:

  12. Sec 4.3 : Applications of the Mass Rate Balance Example 2: ( fromProb4.16) Liquid water at 70 oF enters a pump through an inlet pipe having a diameter of 6 in. The pump operates at steady state and supplies water to two exit pipes having diameters of 3 in and 4 in. The velocity of the 3 in pipe is 1.31 ft/s. At the exit of the 4 in pipe the velocity is 0.74 ft/s. The temperature of the water in each exit is 72 deg F. Determine a) the mass flow rate in lb/s in the inlet and each of the exit pipes. b) the volumetric flow rate at the inlet in ft3/min. Exit B Exit C Inlet A Identify what you know:

  13. Sec 4.3 : Applications of the Mass Rate Balance Use continuity to find volumetric flow rates Volumetric Flow Rate: Area: Exit B Exit C

  14. Sec 4.3 : Applications of the Mass Rate Balance Next find the mass flow rates Specific Volumes may be found on Table A-2E: vA= vf@ 70 oF =0.01605 ft3/lbm vB= vC = vf@ 72 oF = 0.01606 ft3/lbm Exit B Exit C Inlet A: (apply mass balance)

  15. Sec 4.3 : Applications of the Mass Rate Balance Finally, the volumetric flow rate of the inlet may be found: Summary:

  16. End of slides for Lecture 12

More Related