CHAPTER12

128 they are extremely valuable in thermodynamics

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Unformatted text preview: we obtain a a 0T b 0v s 0T b 0P s a a 0P b 0s v (12–16) 0v b 0s P (12–17) P T ( ∂∂–– )s = – (∂–– )v v ∂s T ∂v ( ∂∂–– )s = ( ∂–– )P P s ∂P ∂s ( ∂–– )T = ( ∂–– )v v T ∂s v –– ( ∂–– )T = – ( ∂∂T )P P 0s ab 0v T a 0s b 0P T 0P ab 0T v a 0v b 0T P (12–18) (12–19) These are called the Maxwell relations (Fig. 12–8). They are extremely valuable in thermodynamics because they provide a means of determining the change in entropy, which cannot be measured directly, by simply measuring the changes in properties P, v, and T. Note that the Maxwell relations given above are limited to simple compressible systems. However, other similar relations can be written just as easily for nonsimple systems such as those involving electrical, magnetic, and other effects. FIGURE 12–8 Maxwell relations are extremely valuable in thermodynamic analysis. cen84959_ch12.qxd 4/5/05 3:58 PM Page 658 658 | Thermodynamics EXAMPLE 12–4 Verification of the Maxwell Relations Verify the validity of the last Maxwell relation (Eq. 12–19) for steam at 250°C and 300 kPa. Solution The validity of the last Maxwell relation is to be verified for steam at a specified state. Analysis The last Maxwell relation states that for a simple compressible substance, the change in entropy with pressure at constant temperature is equal to the negative of the change in specific volume with temperature at constant pressure. If we had explicit analytical relations for the entropy and specific volume of steam in terms of other properties, we could easily verify this by performing the indicated derivations. However, all we have for steam are tables of properties listed at certain intervals. Therefore, the only course we can take to solve this problem is to replace the differential quantities in Eq. 12–19 with corresponding finite quantities, using property values from the tables (Table A–6 in this case) at or about the specified state. s P s P s400 kPa s200 kPa (400 200) kPa (7.3804 (400 ? T ? a a c 0v b 0T P 0v b 0T P T 250°C ? 300 kPa T 250°C ? v300°C v200°C d (300 200)°C P 300 kPa 7.7100) kJ kg # K 200) kPa (0.87535 0.71643) m3 kg (300 200)°C 0.00159 m3/kg K 0.00165 m3/kg K since kJ kPa · m3 and K °C for temperature differences. The two values are within 4 percent of each other. This difference is due to replacing the differential quantities by relatively large finite quantities. Based on the close agreement between the two values, the steam seems to satisfy Eq. 12–19 at the specified state. Discussion This example shows that the entropy change of a simple compressible system during an isothermal process can be determined from a knowledge of the easily measurable properties P, v, and T alone. 12–3 ■ THE CLAPEYRON EQUATION The Maxwell relations have far-reaching implications in thermodynamics and are frequently used to derive useful thermodynamic relations. The Clapeyron equation is one such relation, and it enables us to determine the enthalpy change...
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