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EGR 334 Thermodynamics Chapter 3: Section 9-10 Lecture 08: Specific Heat Capacity Quiz Today? Today’s main concepts: Introduce Specific Heats, cv and cP. 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, cv ≈ cp • the specific heat ratio is defined as k = cp/cv • • • • Reading Assignment: • Read Chap 3: Sections 11 Homework Assignment: 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…. You performed experiments to try and figure out how the world around you worked. … it requires 1 calorie of heat to raise the temperature of 1 gram of water by 1 deg C. Q cm T In one experiment you dropped hot aluminum pellets into a container of water and measured the change of temperature the water underwent. Using that information you found a property of aluminum called specific heat capacity. Sec 3.9 : Specific Heat 4 In Thermodynamics, two common forms of specific heat capacity are used. u cv T v Constant Volume process ----- cp and h T p Constant Pressure process 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 cv and cp are often treated as constants) u 2 u 1 c v (T 2 T1 ) and h 2 h1 c p (T 2 T1 ) Sec 3.9 : Specific Heat 5 Specific Heats isobars T2 the internal energy change or enthalpy change when heat is added at constant pressure or constant volume. T At constant volume U f T , V T1 v This is an exact differential U U U dU dT dV c V dT dV T V V T V T U where cV U T V At constant volume, dV=0; thus, dV 0 V T Finally: dU cV dT U T2 T1 cV dT Sec 3.9 : Specific Heat 6 isobars At Constant Pressure H f T , P T2 This is an exact differential T H H dH dT dP T P P T H c P dT dP P T At constant pressure, dP=0; thus, Finally, T1 v H cP where T P H dP 0 P T dH c P dT H T2 T1 c P dT Sec 3.9 : Specific Heat 7 Specific Heat Ratio: k cP cV For air (at 68oF (20oC) and 14.7 psia (1 atm)): k a ir 1 .0 1 kJ / kg K 0 .7 1 8 kJ / kg K 0 .2 4 B tu / lb m F o 0 .1 7 B tu / lb m F o 1 .4 Fig. 3.9 Shows that cv and cp for water/steam vary with temperature and pressure. In practice, specific heats will be used as constants which are looked up on tables based on standard values of temperature and pressure. Example of Table: Gases - Specific Heats and Individual Gas Constants 10 Evaluating Properties of Liquids and Solids: 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? For liquids and solids, it is acceptable practice to approximate v (T , p ) v f (T ) and u (T , p ) u f (T ) h (T , p ) h f (T ) if a super cooled table is not available. Sec 3.10.1 : 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, v f T , P v f T u f T , V hf uf T , P h f T T Evaluate liquids at the saturated state Since these properties are essentially only a function of T and not P, we call them Incompressible. 11 12 Example: What is the enthalpy for Refrigerant 22 at T = 10 deg F. and p = 40 psi. (Refer to Table A-7E) Temperature pressure T=10 deg C * spec. vol. h (T , p ) h f (T ) * o p = 40 psi spec. vol. at T=10 deg….hf=13.33 Btu/lbm Sec 3.10.2 : Incompressible Substance Model 13 Incompressible Substance: Includes any substance whose properties do not change with pressure. For liquids and solids: v f , s co n stan t Thus c V u f ,s u T h f ,s h T dU U c V dT T V But, h T , P u T PV So now what? Take partial with respect to T H U PV U 0 T P T P T P T P Thus, H dU c P cv c dT T P Sec 3.10.2 : Incompressible Substance Model 14 Use the heat capacity to calculate the change in internal energy. u u 2 u1 T2 T1 c T dT h h 2 h1 u 2 u 1 v P2 P1 T2 T1 c T dT v P2 P1 If c = constant then that means, v P2 P1 is small and can usually be dropped Therefore, h h 2 h1 T2 T1 c T dT c T 2 T1 15 Table of specific heat for incompressible materials. See the course website for the complete tables of specific heats for both compressible and incompressible materials. 16 End of Slides for Lecture 08