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EGR 334 Thermodynamics Chapter 3: Section 9-10 Lecture 08: Specific Heat Capacity Quiz Today?

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Presentation on theme: "EGR 334 Thermodynamics Chapter 3: Section 9-10 Lecture 08: Specific Heat Capacity Quiz Today?"— Presentation transcript:

1 EGR 334 Thermodynamics Chapter 3: Section 9-10 Lecture 08: Specific Heat Capacity Quiz Today?

2 Todays main concepts: Introduce Specific Heats, c v and c P. Understand when it is acceptable to apply specific heat. Calculate changes of energy using specific heats For liquids and solids, the saturated properties values may be used to approximate properties values for supercooled property values, if the model can be treated as incompressible. For incompressible liquids and solids, c v c p the specific heat ratio is defined as k = c p /c v Reading Assignment: Homework Assignment: Read Chap 3: Sections 11 From Chap 3: 49, 55,68, 78

3 Think back to those simple, happy, carefree days of your youth, back when you were in Physics lab…. 3 In one experiment you dropped hot aluminum pellets into a container of water and measured the change of temperature the water underwent. You performed experiments to try and figure out how the world around you worked. Using that information you found a property of aluminum called specific heat capacity. … it requires 1 calorie of heat to raise the temperature of 1 gram of water by 1 deg C.

4 In Thermodynamics, two common forms of specific heat capacity are used. 4 Sec 3.9 : Specific Heat -- Used to calculate changes of energy between states -- Defined for pure, simple compressible substances -- May be used only under certain special conditions -- Usually applied to ideal gas model When it can be applied: (if c v and c p are often treated as constants) and Constant Volume process Constant Pressure process and

5 Specific Heats 5 the internal energy change or enthalpy change when heat is added at constant pressure or constant volume. Sec 3.9 : Specific Heat At constant volume This is an exact differential where At constant volume, dV=0; thus, Finally: T v isobars T1T1 T2T2

6 6 Sec 3.9 : Specific Heat At Constant Pressure This is an exact differential where At constant pressure, dP=0; thus, Finally, isobars T v T1T1 T2T2

7 7 Sec 3.9 : Specific Heat Specific Heat Ratio: For air (at 68 o F (20 o C) and 14.7 psia (1 atm)):

8 Fig. 3.9 Shows that c v and c p for water/steam vary with temperature and pressure.

9 In practice, specific heats will be used as constants which are looked up on tables based on standard values of temperature and pressure. Gases - Specific Heats and Individual Gas Constants Example of Table:

10 Evaluating Properties of Liquids and Solids: 10 For liquids and solids, it is acceptable practice to approximate and While Appendix A does have a table for super-cooled water, for many other liquids, a super-cooled table is not available. What to do? if a super cooled table is not available.

11 11 Sec : Approximations for Liquids using Saturated Data For a liquid, there is little change in v, u, h, s at different pressure and fixed T. Therefore, Evaluate liquids at the saturated state Since these properties are essentially only a function of T and not P, we call them Incompressible.

12 Example: What is the enthalpy for Refrigerant 22 at T = 10 deg F. and p = 40 psi. (Refer to Table A-7E) 12 p = 40 psi Temperature spec. vol. * pressure spec. vol. * T=10 deg C o at T=10 deg….h f =13.33 Btu/lb m

13 13 Sec : Incompressible Substance Model Thus Incompressible Substance: Includes any substance whose properties do not change with pressure. For liquids and solids: But, So now what? Take partial with respect to T Thus,

14 14 Sec : Incompressible Substance Model If c = constant then that means, Use the heat capacity to calculate the change in internal energy. is small and can usually be dropped Therefore,

15 Table of specific heat for incompressible materials. 15 See the course website for the complete tables of specific heats for both compressible and incompressible materials.

16 End of Slides for Lecture 08 16

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