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Unformatted text preview: and using dP a 0P b dT 0T v a 0P b dv 0v T R dT v RT dv v2 1 0.287 kPa 0.664 kPa # m3> kg # K2 c 2K 0.865 m3> kg 1 301 K 2 1 0.01 m3> kg 2 1 0.865 m3> kg 2 2 d 1.155 kPa 0.491 kPa
Therefore, the pressure will decrease by 0.491 kPa as a result of this disturbance. Notice that if the temperature had remained constant (dT 0), the pressure would decrease by 1.155 kPa as a result of the 0.01 m3/kg increase in specific volume. However, if the specific volume had remained constant (dv 0), the pressure would increase by 0.664 kPa as a result of the 2K rise in temperature (Fig. 12–5). That is, P, kPa (∂P)v = 0.664 dP = – 0.491 (∂P)T = –1.155 a a
and 0P b dT 0T v 0P b dv 0v T 1 0P 2 T 1 0P 2 v 1 0P 2 T 0.664 kPa 1.155 kPa 0.86 0.87 300 302 T, K v, m3/kg dP 1 0P 2 v 0.664 1.155 0.491 kPa FIGURE 12–5 Geometric representation of the disturbance discussed in Example 12–2. Discussion Of course, we could have solved this problem easily (and exactly) by evaluating the pressure from the idealgas relation P RT/v at the final state (302 K and 0.87 m3/kg) and the initial state (300 K and 0.86 m3/kg) and taking their difference. This yields 0.491 kPa, which is exactly the value obtained above. Thus the small finite quantities (2 K, 0.01 m3/kg) can be approximated as differential quantities with reasonable accuracy. cen84959_ch12.qxd 4/5/05 3:58 PM Page 655 Chapter 12  655 Partial Differential Relations
Now let us rewrite Eq. 12–3 as
dz M dx N dy a 0z b 0y x
(12–4) where
M a 0z b 0x y and N Taking the partial derivative of M with respect to y and of N with respect to x yields
a 0M b 0y x 0 2z 0x 0y and a 0N b 0x y 0 2z 0y 0x The order of differentiation is immaterial for properties since they are continuous point functions and have exact differentials. Therefore, the two relations above are identical:
a 0M b 0y x a 0N b 0x y
(12–5) This is an important relation for partial derivatives, and it is used in calculus to test whether a differential dz is exact or inexact. In thermodynamics, this relation forms the basis for the development of the Maxwell relations discussed in the next section. Finally, we develop two important relations for partial derivatives—the reciprocity and the cyclic relations. The function z z(x, y) can also be expressed as x x(y, z) if y and z are taken to be the independent variables. Then the total differential of x becomes, from Eq. 12–3,
dx a 0x b dy 0y z a a 0x b dz 0z y a 0x 0z b a b dz 0z y 0x y
(12–6) Eliminating dx by combining Eqs. 12–3 and 12–6, we have
dz ca 0z 0x ba b 0x y 0y z 0z b d dy 0y x Rearranging,
ca 0z 0x ba b 0x y 0y z a 0z b d dy 0y x c1 a 0x 0z b a b d dz 0z y 0x y
(12–7) The variables y and z are independent of each other and thus can be varied independently. For example, y can be held constant (dy 0), and z can be varied over a range of values (dz 0). Therefore, for this equation to be valid at all times, the terms in the brackets must equal zero, regardless of the values of y and z. Setting the terms in each bracket equal to zero gives
a a 0z 0x ba b 0z y 0x y a 1...
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