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# Mass and Energy Analysis of Control Volumes – Part 2

Mass and Energy Analysis of Control Volumes – Part 2. Chapter 5b. So far…. We’ve developed a general energy balance We’ve developed a general material balance We’ve only actually looked at systems that are steady state Now we are going to do more complicated problems

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## Mass and Energy Analysis of Control Volumes – Part 2

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1. So far… • We’ve developed a general energy balance • We’ve developed a general material balance • We’ve only actually looked at systems that are steady state • Now we are going to do more complicated problems • Ones that change with time

2. For example…..

3. We need to make some assumptions • Uniform flow • The system can change with time, but the inlet conditions are constant • Everything in the system is in the same state • Fluid exiting the system is at the same state as the system

4. We need to look at our balances again We aren’t using the rate form of the balances here. Why?

5. Usually, they both equal 0 This is the kinetic energy of the system This is the potential energy of the system

6. Inlet and Exit conditions, assuming only a single inlet and a single exit stream Time 1 and time 2 Usually both the kinetic energy and potential energy of thefluidare zero too

7. Consider a bottle filling problem • What happens to the temperature when you fill an empty tank with air?

8. Consider a bottle filling problem • What happens to the temperature when you fill an empty tank with air? • The air gets hot • Why? • It takes energy to push the air into the tank (flow work). That energy is converted into internal energy

9. For our bottle filling problem… hi is bigger than ui, so… u2 is bigger than ui That means the temperature in the tank is higher than the inlet temperature

10. Try an example • Example 5-12 page 248 • Charging of a rigid tank • Note that the inlet conditions stay constant throughout the process • There is no exit term

11. Try a harder example • Example 5-13 page 250 • Cooking with a pressure cooker • Because the system is two phase and constant pressure, the exiting steam is always at the same conditions • There is no inlet term

12. What if… • How would we handle inlet or exit conditions that change with time? • The best we can do at this point is to take the average • If we knew more, we could integrate over time

13. For example… • What happens to the temperature of the container when you use a bottle of canned air? • The bottle gets cold. • Why? • It takes energy to push the air out of the can (flow work) • That energy comes from the energy of the air that remains in the can

14. But…. • The air coming out of the can gets colder with time • That means the exit conditions are not constant What conditions should you use for he?

15. SummaryControl Volumes • A control volume differs from a closed system in that it involves mass transfer. Mass carries energy with it, and thus the mass and energy content of a system change when mass enters or leaves.

16. SummaryMass and Energy Balance • The mass and energy balances for any system undergoing any process can be expressed as :

17. SummaryRate form of the mass and energy balance • The mass and energy balances for any systemundergoing any processcan be expressed in the rate form as

18. SummaryMass flow rates • Mass flow through a cross section per unit time is called the mass flow rate and is denoted m. It is expressed as .

19. SummaryVolumetric flow rates • The mass and volume flow rates are related by

20. SummaryTypes of Control Volume problems • Thermodynamic processes involving control volumes can be considered in two groups: • steady-flow processes and • unsteady-flow processes.

21. SummarySteady Flow Equations • Taking heat transfer to the system and work done by the system to be positive quantities, the conservation of mass and energy equations for steady-flow processes are expressed as where the subscript i stands for inlet and e for exit. These are the most general forms of the equations for steady-flow processes. for each exit for each inlet

22. SummarySteady Flow equations for a single inlet and single exit system • For single-stream (one-inlet--one-exit) systems such as nozzles, diffusers, turbines, compressors, and pumps, the steady flow equations simplify to:

23. SummaryUniform Flow Process • During a uniform-flow process, • the state of the control volume may change with time, but it may do so uniformly. • Also, the fluid properties at the inlets are assumed to remain constant during the entire process. • The fluid properties of the exit streams are the same as the fluid properties of the system

24. SummarySimplified Energy Balance for the Uniform Flow Case • When the kinetic and potential energy changes associated with the control volume and the fluid streams are negligible, the conservation of energy equation for a uniform-flow process simplifies to

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