Department of Mechanical Engineering ME 322 Mechanical Engineering
Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics Lecture 5 Thermodynamic Properties Why Study Properties of Substances? In thermal systems, the goal is energy conversion. A working fluid usually transports the energy from one component to another. In order to analyze and design the components of this system, the properties of the working fluid need to be known. 2 2 Thermodynamic Properties We have referred to the following properties several times p Pressure
T Temperature Density We have also previously introduced the specific volume, V 1 v specific volume m For a single phase pure substance, the State Postulate says that the thermodynamic state can be fixed by two independent intensive properties. 3 P-v-T Behavior of Pure Substances Substance that expands upon freezing (water) 4
Substance that contracts upon freezing (everything else) Thermodynamic Properties Using the chain rule from calculus, we can write v v p, T v v dv dp dT T p p T In mathematical expansions like this, the partial derivatives are often properties or related to properties. For example, 1 v isobaric coefficient of volume expansion v T p
1 v isothermal coefficient of compressibility v p T 5 dv dT dp v and for Various Substances In most engineering tables, the scientific notation in the column heading implies that the number in the table had to be multiplied by 10n to be recorded in the table as shown.
Therefore, the value of for benzene from Table 3.2 is, 0.689 10 6 1/R 1.24 10 6 1/K Notice the sign in the exponent of 10 6 Significance Using the chain rule and introducing and , we found, dv dT dp v This equation can be integrated between any two states to determine the volume change of a substance! This equation allows us to determine the volume change of a substance as it undergoes changes in its thermodynamic state from p1,T1 to p2,T2. Study Example 3.2 in your textbook!
7 Why is P-v-T Behavior Important? p p v, T Equation of State All thermodynamic properties can be determined from the equation of state. They are hidden and need to be extracted with mathematics. For example 1 v v T p 8 1 v
v p T Example Consider the Ideal Gas Equation of State (EOS) pv RT Lets find an expression for for an ideal gas. RT v p 1 v v T p R v T p p
1 v 1 R p v T p v p RT R 1 p T We just discovered that the isobaric coefficient of volume expansion for a substance that can be modeled as an ideal gas is 1/T ! A related derivation for kappa is on the website. 9 Properties from P-v-T Behavior Research in the University of Idaho Center for Applied Thermodynamic Studies (CATS)
sion s e r Reg lysis Ana Mathematical model based on experimental data Results published in high-quality scientific journals Applications Experimental data Computer Software Tables & Charts in Books 10 Theres an
app for that! Internal Energy V V z 35, 000 ft above sea level To an observer on the ground, the water in the glass has a total energy content made up of, E U KE PE U = the internal energy of the water. This is the energy resulting from molecular motion (translation, vibration, rotation, etc.) consistent with its thermodynamic state (e.g. quantified by its temperature and pressure). 11 Internal Energy The total energy (extensive form) of any system is given by, E U KE PE U, KE, and PE are all properties of the system. The specific energy (intensive form) of any system is given by,
E U KE PE e m m m m e u ke pe u, ke, and pe are all specific properties of the system. Question to think about: I know how to find ke and pe. But how can I find u? 12 Isochoric Specific Heat For a single phase pure substance, u can be specified by any two independent intensive properties. Therefore consider, u u du dT
dv T v v T u u T , v The isochoric specific heat is defined as, u cv T v Therefore, 13 u u T , v
u du cv dT dv v T Enthalpy and Isobaric Specific Heat A convenient grouping of other properties H U pV h u pv Specific enthalpy Total enthalpy h h dh dT dp T p p T
h h T , p The isobaric specific heat is defined as, h c p T p Therefore, 14 h h T , p h dh c p dT dp p T
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