Equations of State Ideal Gas, Lecture notes of Thermodynamics

Download Equations of State Ideal Gas and more Lecture notes Thermodynamics in PDF only on Docsity! Equations of State Reading Problems 3-6, 3-7, 4-3→ 4-5 3-54, 3-64, 3-80, 4-25, 4-39, 4-42 6-6 7-3, 7-4, 7-7→ 7-10 7-43, 7-116, 7-166, 7-193 Ideal Gas • When is the ideal gas assumption viable? – for a low density gas where: ∗ the gas particles take up negligible volume ∗ the intermolecular potential energy between particles is small ∗ particles act independent of one another – Under what conditions can it be used? ∗ low density ∗ high temperatures - considerably in excess of the saturation region ∗ at very low pressures • for any gas whose equation of state is exactly Pv = RT the specific internal energy depends only on temperature u = u(T ) • the specific enthalpy is given by h = u + Pv where h(T ) = u(T ) + RT Note: Since u = u(T ), and R is a constant, enthalpy is only a function of temperature. 1 • for a ideal gas cv = du dT ⇒ cv = cv(T ) only cp = dh dT ⇒ cp = cp(T ) only From the equation for enthalpy, RT = h(T )− u(T ) If we differentiate with respect to T R = dh dT − du dT R = cp − cv Is Water Vapor an Ideal Gas? • Figure 3-49 can be used to determine the regions where water vapor behaves as an ideal gas – YES - at pressures below 10 kPa regardless of temperature – NO - at high pressure – what about at atmospheric pressure and temperature (Patm ≈ 100 kPa and Tatm < 50 ◦C)? Figure 3-49 would indicate YES ∗ look at the T − s diagram for water: h ≈ constant for atmospheric conditions ∗ for an ideal gas: h = h(T ) - since h only varies with respect to T , it must behave like an ideal gas In Summary For an ideal gas with constant cp and cv Pv = RT u2 − u1 = cv(T2 − T1) h2 − h1 = cp(T2 − T1) 2 Relative Pressure and Relative Specific Volume • typically we assume specific heat to be constant with respect to temperature • but when temperature swings are significant, this assumption can lead to inaccuracies, i.e. T (K) cp (kJ/kg ·K) % difference 300 1.0057 1000 1.1417 13.5 2500 1.688 67.8 • the relative pressure and relative volume tables (C&B Table A-17), provide an accurate way of including the temperature effects on specific heat for ideal gases during isentropic processes • note: the specific heat ratio term given by k = cp/cv will also be influenced by temperature • Procedure: – given T1, P1 and P2 for an isentropic process – determine Pr1 at T1 from Table A-17 – calculate Pr2, where( P2 P1 ) s=const = Pr2 Pr1 – read T2 from Table A-17 for the calculated value of Pr2 • use a similar procedure if volume is known instead of pressure, where ( v2 v1 ) s=const = vr2 vr1 5 Sign Convention There are many potential sign conventions that can be used. Cengel Approach Heat Transfer: heat transfer to a system is positive and heat transfer from a system is negative. Work Transfer: work done by a system is positive and work done on a system is negative. Culham Approach Anything directed into the system is positive, anything directed out of the system is negative. 6 Incompressible Liquids • a substance whose volume cannot be changed • no substance is truly incompressible, but this model is good for most liquids and solids State Postulate • the number of independent intensive thermodynamic properties is equal to the number of relevant reversible work modes plus one. • the “plus one” is for the independent control on energy through heat transfer • we know that for a simple (has only one work mode), compressible (the work mode is Pdv work) substance – 2 thermodynamic properties will fix the rest ∗ list of intensive properties includes, T, u, P, v, s ∗ state postulate says two will fix the rest, i.e. if u and v are known, the equations of state are T = T (u, v) P = P (u, v) s = s(u, v) Hence, if the substance is assumed to be incompressible, then its internal energy, for example, cannot be varied independently by work transfer −→ but it can be varied by heat transfer at constant volume, i.e. Pdv = 0 since dv = constant. There are no reversible work modes. In summary u2 − u1 = c (T2 − T1) h2 − h1 = (u2 − u1) + v(P2 − P1) s2 − s1 = c ln(T2/T1) cp = cv = c 7