Energy Analysis of Closed Systems

1. Energy Analysis of Closed Systems Chapter 4

2. Recall that a closed system does not include mass transfer • Heat can get in or out • Work can get in or out • Matter does not cross the system boundaries

3. Mechanical Work • There are many kinds of mechanical work • The most important for us will be moving boundary work • Wb • Sometimes called PdV work • The primary form of work involved in automobile engines is moving boundary work produced by piston cylinder devices

4. Boundary Work The mass of the substance contained within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and make it move.

6. The area under the curve is the work

7. The work depends on the process path

8. Consider some special cases • Constant volume • Constant pressure • Isothermal for an ideal gas • Polytropic

9. P V Constant Volume 1 2

10. P W V Constant Pressure 1 2

11. P V Isothermal for an ideal gas 1 2

12. P V Polytropic 1 2

13. P V n=1 equivalent to the isothermal case for an ideal gas PV1= constant PV= mRT

14. 2 1 Lets go through the polytropic case integration step-wise

15. Wb But… And…. So…

16. So far, we have not assumed an ideal gas in this derivation, if we do, then….

17. Energy Balance for a Closed System (Chapter 2) Total Energy entering the system Total Energy leaving the system The change in total energy of the system - = How can energy get in and out of a closed system? Heat and Work

18. Rate form

19. 0 0 E = U + KE + PE If the system isn’t moving

20. In a cyclic process, one where you end up back where you started: You convert heat to work, or vise versa

21. The net work is the area inside the figure If we can calculate the work done for each step in this process, we can find the net work produced or consumed by the system 1 Wnet Pressure 2 Volume

22. Calculating Properties Chapter 4b

23. We know it takes more energy to warm up some materials than others For example, it takes about ten times as much energy to warm up a pound of water, as it does to warm up the same mass of iron.

24. Specific Heats – Cp and Cv • Also called the heat capacity • Energy required to raise the temperature of a unit mass one degree • Units • kJ/(kg 0C) or kJ/(kg K) • cal/(g 0C) or cal/(g K) • Btu/(lbm0F) or Btu/(lbm0R)

25. Consider a stationary constant volume system E=U+KE +PE DE = DU First Law Q-W=ΔU dE = dU = mCvdT du = CvdT

26. Consider a stationary constant pressure system • It takes more energy to warm up a constant pressure system, because the system boundaries expand • You need to provide the energy to • increase the internal energy • do the work required to move the system boundary

27. Consider a stationary constant pressure system E=U+KE +PE DE = DU Q-W=ΔU Q-PΔV=ΔU Q=ΔU+PΔV = ΔH dH = mCpdT dh = CpdT

28. Cp is always bigger than Cv h includes the internal energy and the work required to expand the system boundaries

29. Cp and Cv are properties • Both are expressed in terms of u or h, and T, which are properties • Because they are properties, they are independent of the process!! • The constant volume or constant pressure process defines how they are measured, but they can be used in lots of applications

30. Ideal Gases Joule determined that internal energy for an ideal gas is only a function of temperature

31. Which means that h is also only a function of temperature for ideal gases!! For non ideal gases both h and u vary with the state

32. Specific Heats vary with temperature – but only with temperature –for an ideal gas Note that the Noble gases have constant specific heats Why is water on this chart?

34. If Cv and Cp are functions of temperature – how can we integrate to find DU and DH? Use an average value, and let the heat capacity be a constant

35. Crummy Approximation OK Approximation That only works, if the value of heat capacity changes linearly in the range you are interested in. Sometimes the best you can do is the room temperature value

36. What if you need a better approximation? All of these functions have been modeled using the form Cp = a + bT + cT2 + dT3 The values of the constants are in the appendix of our book – Table A-2c

37. This is a pain in the neck!! Only do it if you really need to be very accurate!! Isn’t there a better way?

38. Use the Ideal Gas Tables • Table A-17 pg 910 (air) • Both u and h are only functions of T – not pressure • Relative values have been tabulated for many ideal gases • If the gas isn’t ideal – then it’s a function of both T and P and these tables don’t work!!

39. Cp is modeled in the Appendix as a function of temperature – so you could calculate Dh, but what if you want to calculate Du? You’d need Cv There is no corresponding Cv table !! Cp = Cv + R

40. Three Ways to Calculate Du

41. Use in Chapter 7 Specific Heat Ratio k does not vary as strongly with temperature as the heat capacity k = 1.4 for diatomic gases (like air) k = 1.667 for noble gases

42. Solids and Liquids • Treat as incompressible fluids Cp = Cv = C

43. But dv is 0 if the system is incompressible 0 small

44. Summary • Boundary work • Looked at four different special cases • Constant volume • Constant pressure • Constant temperature • Polytropic

45. Summary • First Law for a closed system

46. Summary • Defined • Constant pressure heat capacity • Constant temperature heat capacity By performing a first law analysis of a closed system

47. Summary • The three ways to calculate changes in • Internal energy (u) • Enthalpy (h) • Look up properties at state one and two in the tables • Assume constant values of specific heat, then integrate • Use the curve fit equation for specific heat, then integrate