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Unformatted text preview: EXAMPLE: (Engel et al) Adiabatic expansion. n mol of an ideal gas, with heat capacity C V , m , is expanded adiabatically against a constant external pressure P ext . The initial temperature T 1 and pressure P 1 are given. QUESTION: Calculate the final temperature, work, energy and enthalpy for this process. REMINDER: adiabatic: Q = . ⇒ Δ U = W (3.45) nC V , m ( T 2 T 1 )= P ext ( V 2 V 1 ) (3.46) nC V , m T 2 nC V , m T 1 = P ext nRT 2 P 2 + P ext nRT 1 P 1 T 2 n C V , m + RP ext P 2 = T 1 n C V , m + RP ext P 1 (3.47) ⇒ T 2 = T 1 C V , m + RP ext P 1 C V , m + RP ext P 2 (3.48) • We can now find Δ U = W = nC V , m ( T 2 T 1 ) (3.49) and Δ H = Δ U + Δ ( PV ) (3.50) 3–19 NOTE: Here, PV = nRT ; We mean Δ ( PV ) = nR Δ T , as both volume and pressure are changing here. Previously, we had Δ ( PV ) = P Δ V for an isobaric process. • Let’s try this with numbers: ‹ n = 2.5mol , C V , m = 12.47 J molK › T 1 = 325K , P 1 = 2.5bar fi P ext = 1bar , P 2 = 1.25bar T 2 = 325K 12.47 J molK + 8.314 J molK 1bar 2.5bar 12.47 J molK + 8.314 J molK 1bar 1.25bar (3.51) = 268K (3.52) Δ U = 2.5mol12.47 J molK ( 268K 325K ) = 1780J (3.53) Δ H = Δ U + 2.5mol8.314 J molK ( 268K 325K ) = 2960J (3.54) 3–20 3.4 The First Law in Open Systems AIM: See how the first law is applied to open systems. MOTIVATION: Many important systems in engineering are open because they involve mass flow (pumps, turbines, generators, some kinds of reactors. . . ) QUESTION: We now have changing mass as well as moving systems . How do we introduce these effects into our reasoning about a system? ANSWER: ‹ Introduce the Continuity Equation (mass conservation) › Build this into the First Law, taking motion of the liquid into account. The Continuity Equation • We have a chamber with an input pipe of cross section area A 1 and an output pipe of cross section area A 2 . • We have a fluid of density ρ flowing in at a velocity v 1 and out a velocity of v 2 . v1 v2 A1 A2 • Because of Conservation of Mass we can write that the change 3–21 of the mass in the control volume in time unit Δ t is Δ m cv = m 1 m 2 (3.55) = ρ A 1 v 1 Δ t ρ A 2 v 2 Δ t (3.56) ⇒ d m cv d t = ρ A 1 v 1 ρ A 2 v 2 . (3.57) NOTE: In principle, velocity is a vector quantity ; however, here we are only interested in the velocity component normal to A 1,2 ⇒ we write this component as a scalar value v 1,2 . NOTE: An important special case ( steady flow ): the mass in the control volume is constant ( d m cv d t =0). In this case, we can write ρ A 1 v 1 = ρ A 2 v 2 (3.58) ˙ m 1 = ˙ m 2 . (3.59) NOTE: The expression ρ Av is called mass flux ˙ m (again in general a vector quantity) • Units: ρ A v = kg m 3 m 2 m s . = kg s (3.60) 3–22 The First Law with an Open System • We now have to know, in addition, the entry and exit pressure of the fluid; we can also have heat and work processes acting on the...
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 Winter '09
 scotty
 Energy, First Law, dQ spec spec

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