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Unformatted text preview: wo properties. Mathematically speaking,
f(x) z z 1 x, y 2 (x +∆x) ∆f f(x) ∆x Slope x x+∆x x where x and y are the two independent properties that fix the state and z represents any other property. Most basic thermodynamic relations involve differentials. Therefore, we start by reviewing the derivatives and various relations among derivatives to the extent necessary in this chapter. Consider a function f that depends on a single variable x, that is, f f (x). Figure 12–1 shows such a function that starts out flat but gets rather steep as x increases. The steepness of the curve is a measure of the degree of dependence of f on x. In our case, the function f depends on x more strongly at larger x values. The steepness of a curve at a point is measured by the slope of a line tangent to the curve at that point, and it is equivalent to the derivative of the function at that point defined as
¢x S 0 FIGURE 12–1 The derivative of a function at a specified point represents the slope of the function at that point. lim ¢f ¢x ¢x S 0 lim f 1x ¢x 2 ¢x f 1x 2 (12–1) Therefore, the derivative of a function f(x) with respect to x represents the rate of change of f with x. EXAMPLE 12–1 Approximating Differential Quantities by Differences The cp of ideal gases depends on temperature only, and it is expressed as cp(T ) dh(T )/dT. Determine the cp of air at 300 K, using the enthalpy data from Table A–17, and compare it to the value listed in Table A–2b.
h (T ), kJ/kg Slope = cp(T ) 305.22 Solution The cp value of air at a specified temperature is to be determined using enthalpy data. Analysis The cp value of air at 300 K is listed in Table A–2b to be 1.005 kJ/kg · K. This value could also be determined by differentiating the function h(T ) with respect to T and evaluating the result at T 300 K. However, the function h(T ) is not available. But, we can still determine the cp value approximately by replacing the differentials in the cp(T ) relation by differences in the neighborhood of the specified point (Fig. 12–2):
cp 1 300 K 2 c dh 1 T 2 dT d c ¢h 1T 2 ¢T d h 1 305 K 2 1 305 295 2 K h 1 295 K 2 295.17 1 305.22
295 300 305 T, K T 1 305 295.17 2 kJ> kg 295 2 K 300 K T 300 K 1.005 kJ/kg # K FIGURE 12–2 Schematic for Example 12–1. Discussion Note that the calculated cp value is identical to the listed value. Therefore, differential quantities can be viewed as differences. They can cen84959_ch12.qxd 4/5/05 3:58 PM Page 653 Chapter 12
even be replaced by differences, whenever necessary, to obtain approximate results. The widely used finite difference numerical method is based on this simple principle.
z ∂ ( ––xz )y ∂ | 653 Partial Differentials
Now consider a function that depends on two (or more) variables, such as z z(x, y). This time the value of z depends on both x and y. It is sometimes desirable to examine the dependence of z on only one of the variables. This is done by allowing one variable to change while holding the othe...
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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