VIEWS: 6 PAGES: 37 POSTED ON: 8/14/2012
Mathematical Thermo Thermo and Math • Thermodynamic properties can be treated as pure mathematical functions and this can help one find ways to compute changes in them. – One mathematical fact is that thermodynamic properties depend on exactly two other properties. They are said to be exact functions of two others. What this means to the engineer is that within a single phase, there is one and only one value of U (internal energy), or H, or S, or V at a given T and P. Expressing Property Changes Mathematically • “Exactness” means it is possible to mathematically express a change in one of the properties in terms of the changes in two others. For example one important change is mathematically written: ~ ~ ~ ~ ~ ~ U U ~ U U (T , V ) dU dT ~ dV T ~ V V T How would you mathematically express the change in the enthalpy in terms of the change in temperature and pressure? ~ ~ ~ ~ ~ H H H H (T , P) dH dT T dP P P T How would you mathematically express the change in the entropy in terms of the change in temperature and pressure? ~ ~ ~ ~ S S (T , P ) ~ S S dS dT dP T P P T You would “say” these equations: “The change in the enthalpy of a substance is the sum of the change in the enthalpy with respect to temperature at constant pressure times the change in temperature and the change in the enthalpy with respect to pressure at constant temperature times the change in pressure”. If two things determine a property, then the change in the property is a result of changes in each of the two things. The Secret The secret to thermodynamic calculations is to understand how the partial differentials (the coefficients) depend on things one can measure. If the coefficients were known then one could always mathematically determine a finite change in a property by integrating each term on the RHS. ~ ~ ~ H H dH T dT P dP P T ~ ~ ~ ~ 2 ~ H T2 P2 H H 2 H 1 dH T dT P dP 1 T 1 P P1 T Heat Capacity • The “constant pressure” heat capacities of fluids have been experimentally measured by noting how much a joule of electrical work addition raised the temperature. ~ This is measured ~ H Cp and tabulated T P The pressure used is low (at or below 1 atm). This is called the ideal gas heat capacity. Tabulated Cp’s This means B and C are really small numbers! Heat capacity • The heat capacity of a substance behaving as an ideal gas may change with temperature but does not change with pressure. • Either the heat capacity is tabulated or it can be computed from group contribution models. (See Perry’s Manual) C po A BT CT 2 The Trick • The trick is to manipulate the mathematical equations using thermodynamic definitions of the property changes. – This requires one to express and use some relationships called the Maxwell Equations – This requires one to know how to prepare partial differentials from total differentials. Thermodynamic Definitions ~ ~ ~ dU Td S PdV Most basic definition ~ ~ ~ ~ ~ ~ ~ H U PV dH dU PdV VdP ~ ~ ~ dH Td S VdP ~ ~ ~ ~ ~ ~ ~ G H TS dG dH TdS S dT ~ ~ ~ dG S dT VdP ~ ~ ~ ~ ~ ~ ~ A U TS dA dU Td S S dT ~ ~ ~ dA S dT PdV Exactness Property These “thermodynamic” equations help one simplify the mathematical equations: Since U, H, G, and A are exact functions “the derivative of the first coefficient with respect to the second variable holding the first variable constant is equal to the derivative of the second coefficient with respect to the first variable holding the second variable constant”. variables ~ ~ ~ dU Td S PdV Coefficients The Maxwell Equations ~ ~ ~ ~ ~ ~ dU Td S PdV dH Td S VdP ~ T P T V ~ ~ ~ V S ~ S V~ P S S P ~ ~ ~ ~ ~ ~ ~ dG S dT VdP dA S dT PdV ~ ~ S ~ V S P ~ P T V T P T T V~ Differentiation Rule ~ ~ ~ dH Td S VdP ~ ~ H S ~ P P T P V P T T T ~ ~ H S ~ T V P P T T Using Maxwell Relationship with ~ Differential Rule ~ H S ~ T V P P T T ~ ~ H V ~ P T T V T P Note: All these terms depend on P,V, and T and can be found if the EOS is known Differentiation Rules ~ ~ ~ dU Td S PdV ~ ~ U S V ~ T ~ P ~ V V V T T T ~ U S ~ T ~ P V V T T Including Maxwells ~ U S ~ T ~ P V V T T U P ~ T P V T T V~ Note: All these terms depend on P,V, and T and can be found if the EOS is known Joule Thompson Coef Show how the J-T coefficient depends on Cp and PVT behavior T H P H ~ Joule Thompson Coef Show how the J-T coefficient depends on Cp and PVT behavior T H P H ~ ~ ~ ~ ~ H H ~ ~ H dH T dT P dP dH C p dT P dP P T T ~ ~ H ~ T H P ~ Cp T 1 H T ~ P T ~ T H ~ P H ~ T H P H ~ Cp T ~ T 1 V ~ Note: All these terms depend on ~ T T V P,V, and T and can be found if P H C p ~ P the EOS is known JT coefficient Find the J-T coefficient for an ideal gas and for a “Clausius Gas” Clausius RT Gas P ~ V b Explain why a Clausius gas would always rise in temperature as it flowed through a value. J-T Coefficient Show that if V is a constant (Like a liquid) that the temperature will go up when the pressure drops through a valve. In a valve, the pressure drops but the enthalpy stays constant. The question is what the temperature will do. ~ V T 1 ~ ~ T T V P H ~ Cp P Heat capacity relationship Show how the difference in Cp and Cv depends on PVT relationships ~ ~ ~ H ~ U Cp T CV ~ T ~ P V ~ ~ ~ ~ H H U V ~ P V ~ ~ dH C p dT dP P P T ~ T ~ T ~ T V T V V V ~ ~ ~ H ~ T H P T ~ C p T ~ P T ~ ~ V ~ V V T V H ~ ~ P T ~ C p T T V T ~ V P V ~ ~ ~ ~ ~ ~ dH dU dPV dU PdV VdP Heat capacity relationship ~ V ~ H ~ ~ P T ~ C p T T V T ~ V P V ~ ~ ~ H U V ~ P P V T ~ T ~ T ~ T V V V V ~ ~ H ~ P CV V ~ T ~ ~ T V V ~ ~ ~ ~ V P C p CV T ~ T T ~ P V Show how Cp and Cv are related for an ideal gas: ~ ~ ~ V P C p CV T ~ T T ~ P V Show how Cp and Cv are related for a constant volume fluid: ~ ~ ~ V P C p CV T ~ T T ~ P V Ideal gases ~ V ~ ~ ~ dH C p dT T V dP T P ~ ~ P ~ dU CV dT T ~ P dV T V~ Show that dH = CpdT for an ideal gas Show that dU = CvdT for an ideal gas Show that dS = CpdT/T – RdP/P for an ideal gas Liquids ~ V ~ ~ ~ dH C p dT T V dP T P ~ ~ P ~ dU CV dT T ~ P dV T V~ Show that dH = CpdT + VdP for a constant specific volume fluid (approx by a liquid) Show that dU = CvdT for a constant volume fluid Show that dS = CpdT/T for a constant volume fluid van der Waals gas For a vdW gas, it is possible to calculate the difference in the Internal energy between two temperatures and pressures fairly easily. Problem 6 For a gas obeying the van der Waals equation: RT a P ~ ~2 Find an expression for the V b V difference U1-U2. ~ ~ ~ T1 V1 P ~ U 1 (T1 ,V1 ) U 2 (T2 ,V2 ) CV dT T T ~ P dV V 2 V ~ T2 3 Step Process • Step 1 – Calculate the difference between the internal energy at starting condition and the internal energy when the volume is large (ideal gas behavior) at the same temperature. ~ ~ ~ V1 P ~ a a U 1 U 1 * T P dV ~ V T V ~ 1 Evaluated at T1 3 Step Process • Step # 2 – Calculate the difference in internal energy between T1 and T2 at constant V. (V = infinity) T1 ~ ~ ~ U 1 * (T1 , ) U 2 * (T2 , ) CV * dT T2 3 Step Process • Calculate the difference between the internal energy at T2 and large volume (ideal gas) and at T2 and the final volume ~ ~ P ~ a a U 2 * U 2 V 2T P dV ~ ~ T V~ V 2 Evaluated at T2 Problem 7. A heat exchanger is used to cool 2 kg/s of 50 bar HFC 134a from 140 C to 50 C using 25 C air. Use van der Waals equation and Cp = 100 J/mole-K to calculate the rate of heat release to the surroundings. Steps: 1. Energy balance gives heat release = m(Hin- Hout) 2. Hin – Hout = Uin – Uout + PinVin-PoutVout 3. Calculate Uin – Uout 4. Solve vdW for Vin and Vout and then calculate PinVin -PoutVout Summary • Thermodynamic properties have both mathematical and thermodynamic definitions • The mathematical relationships can be manipulated so that property changes can be related to heat capacity data and PVT relationships. • The heat capacity and EOS are used to solve for the way H,U, and S are related to changes in T,P, and V Addendum Example calculations using vdW equation Problem 1. Rework Problem 6 of worksheet 7 step by step to find an expression for the difference in internal energy between condition 1 U1(T1,V1) and condition 2 U2(T2,V2) for a van der Waals gas that depends on (only) the ideal gas constant volume heat capacity (Cv) and the van der Waals parameters a and b. Problem 2. Use the van der Waals equation to compute the specific volumes of methane at 50 bar and 330 K (condition 1) and 50 bar, 200 K (condition 2). (Note: Find the values of a and b using either CRC or, failing that, the van der Waals equations for a and b given in the notes) Problem 3. Use the van der Waals equation and an ideal gas constant pressure heat capacity (Cp) value (35 J/mole-K) to compute the internal energy difference for methane between 50 bar, 330 K (condition 1) and 50 bar, 200 K (Condition 2). (How is Cp related to Cv in the ideal gas state?) Problem 4 Use the results from Problem 2 and 3 to compute the enthalpy difference for methane between 50 bar, 330 K (condition 1) and 50 bar, 200 K (Condition 2). (What is the relationship between U and H?)