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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 = . 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 U = W = nC V , m ( T 2 T 1 ) (3.49) and H = U + ( PV ) (3.50) 319 NOTE: Here, PV = nRT ; We mean ( PV ) = nR T , as both volume and pressure are changing here. Previously, we had ( PV ) = P V for an isobaric process. Lets 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) U = 2.5mol12.47 J molK ( 268K 325K ) = 1780J (3.53) H = U + 2.5mol8.314 J molK ( 268K 325K ) = 2960J (3.54) 320 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 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 321 of the mass in the control volume in time unit t is m cv = m 1 m 2 (3.55) = A 1 v 1 t A 2 v 2 t (3.56) d m cv d t = A 1 v 1 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 A 1 v 1 = A 2 v 2 (3.58) m 1 = m 2 . (3.59) NOTE: The expression Av is called mass flux m (again in general a vector quantity) Units: A v = kg m 3 m 2 m s . = kg s (3.60) 322 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

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