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Unformatted text preview: rs constant and observing the change in the function. The variation of z(x, y) with x when y is held constant is called the partial derivative of z with respect to x, and it is expressed as
0z ab 0x y ¢z lim a b ¢x S 0 ¢x y
¢x S 0 y x lim z 1x ¢ x, y 2 ¢x z 1 x, y 2 (12–2) FIGURE 12–3 Geometric representation of partial derivative ( z/ x)y. This is illustrated in Fig. 12–3. The symbol represents differential changes, just like the symbol d. They differ in that the symbol d represents the total differential change of a function and reflects the influence of all variables, whereas represents the partial differential change due to the variation of a single variable. Note that the changes indicated by d and are identical for independent variables, but not for dependent variables. For example, ( x)y dx but ( z)y dz. [In our case, dz ( z)x ( z)y.] Also note that the value of the partial derivative ( z / x)y, in general, is different at different y values. To obtain a relation for the total differential change in z(x, y) for simultaneous changes in x and y, consider a small portion of the surface z(x, y) shown in Fig. 12–4. When the independent variables x and y change by x and y, respectively, the dependent variable z changes by z, which can be expressed as
¢z z 1x z 1x ¢ x, y z 1 x, y ¢y 2 z 1 x, y 2 z 1 x, y z 1 x, y z z(x, y) z(x + ∆ x, y + ∆y) x x + ∆ x, y y x, y + ∆y x + ∆ x, y + ∆y Adding and subtracting z(x, y
¢z ¢ x, y ¢y 2 ¢y 2 y), we get ¢y 2 ¢y 2 ¢y 2 z 1 x, y 2 z 1 x, y 2 or
¢z FIGURE 12–4 Geometric representation of total derivative dz for a function z(x, y).
¢y z 1x ¢ x, y ¢x z 1 x, y ¢y 2 ¢x ¢y Taking the limits as x → 0 and y → 0 and using the definitions of partial derivatives, we obtain
dz a 0z b dx 0x y a 0z b dy 0y x
(12–3) Equation 12–3 is the fundamental relation for the total differential of a dependent variable in terms of its partial derivatives with respect to the independent variables. This relation can easily be extended to include more independent variables. cen84959_ch12.qxd 4/5/05 3:58 PM Page 654 654 | Thermodynamics
EXAMPLE 12–2 Total Differential versus Partial Differential Consider air at 300 K and 0.86 m3/kg. The state of air changes to 302 K and 0.87 m3/kg as a result of some disturbance. Using Eq. 12–3, estimate the change in the pressure of air. Solution The temperature and specific volume of air changes slightly during a process. The resulting change in pressure is to be determined. Assumptions Air is an ideal gas. Analysis Strictly speaking, Eq. 12–3 is valid for differential changes in variables. However, it can also be used with reasonable accuracy if these changes are small. The changes in T and v, respectively, can be expressed as dT
and ¢T 1 302 300 2 K 2K 0.01 m3> kg dv ¢v 1 0.87 0.86 2 m3> kg RT v An ideal gas obeys the relation Pv RT. Solving for P yields P
Note that R is a constant and P average values for T and v, P(T, v). Applying Eq. 12–3...
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This note was uploaded on 03/09/2009 for the course ME 430 taught by Professor Y during the Spring '09 term at CUNY City.
- Spring '09