Download State Equations The Thermodynamics of State An Isentropic ... and more Summaries Thermodynamics in PDF only on Docsity! State Equations Reading Problems 6-4 → 6-12 The Thermodynamics of State IDEAL GAS The defining equation for a ideal gas is Pv T = constant = R Knowing that v = V/m PV Tm = constant = R where R is a gas constant for a particular gas (as given in C&B Tables A-1 and A-2). An Isentropic Process for an Ideal Gas Given: • constant specific heats over a wide range of temperature • ds = 0 • du = cvdT ≡ cv = ( ∂u ∂T ) V • dh = cpdT ≡ cp = ( ∂h ∂T ) P 1 Gibb’s equation can be written as Tds = du + Pdv = cvdT + Pdv = 0 (1) where ds = 0 because we have assumed an isentropic process. The definition of enthalpy is h = u + Pv Taking the derivative yields dh = du + Pdv︸ ︷︷ ︸ ≡Tds +vdP dh = Tds + vdP ⇒ Tds = 0 = dh − vdP cpdT − vdP = 0 (2) Equating Eqs. (1) and (2) through the dT term gives dP P = −cp cv dv v (3) Integrating (3) from its initial state to a final state P1v k 1 = P2v k 2 = constant = Pvk where k = cp cv The product of P · vk remains constant for an ideal gas when: • specific heats are constant 2 A General Formulation Steady State, Steady Flow in a Flow Channel of Arbitrary Cross-section with Work and Heat Transfer dĖ = Ėfinal − Ėinitial = Ėx+dx − Ėx where Ė = ṁ(e + Pv) = ṁ(u + (v∗)2 2 + gz + Pv) From the 1st law rate of energy storage = rate of work + rate of heat transfer + net rate of energy leaving the system dECV dt = dẆ − dQ̇ − dĖ (1) 5 where dECV dt = 0 for steady state. Equation (1) becomes 0 = dẆ − dQ̇ − ṁ d [ u + Pv + (v∗)2 2 + gz ] (2) From the 2nd law rate of entropy storage = rate of entropy inflow − rate of entropy outflow + rate of entropy production dSCV dt = [ṁs]x − [ṁs]x+dx − dQ̇ TTER + ṖS where dSCV dt = 0 for steady state. 0 = −ṁds − dQ̇ TTER + ṖS or dQ̇ = TTERṖS − TTERṁds (3) Combining (2) and (3) through dQ̇ TTER ṖS − TTER ṁds = dẆ − ṁ d ( u + Pv + (v∗)2 2 + gz ) (4) Equation (4) can be used for any SS-SF process. 6 Special Cases Reversible, SS-SF Process Reversible implies ⇒ ṖS = 0 • frictionless process • heat transfer is allowed but must be across ∆T → 0 • which means TTER ≈ TCV = T Equation 4 becomes dẆ ṁ = −Tds + du + d(Pv)︸ ︷︷ ︸ =du + Pdv︸ ︷︷ ︸ =T ds +vdP︸ ︷︷ ︸ +d ( (v∗)2 2 ) + d(gz) (5) Therefore dẆ ṁ = vdP + d ( (v∗)2 2 ) + d(gz) (6) Integrating Eq. (6) between the inlet and the outlet Ẇ ṁ = ∫ out in vdP + (v∗)2 2 ∣∣∣∣∣ out in︸ ︷︷ ︸ ∆KE + gz ∣∣∣∣∣ out in︸ ︷︷ ︸ ∆P E (7) but ∆KE and ∆PE are usually negligible. If ∆KE + ∆PE = 0 Ẇ ṁ = ∫ out in vdP (8) Equation can be used for a reversible, SS-SF flow in a liquid or a gas. 7
