For the specified initial state, the ideal gas law gives v1 = 0.0149 m3/mole. For the final state, the gas law gives v2 = 0.0446 m3/mole. Equation (3.5) gives w = q = 9800 J/mole

Example (b) Superheated Water Vapor

For this case, the temperature and the initial and final pressures are chosen to be the same as those in Example (a). The specific volumes and internal energies of these states are obtained from steam table A.3:

v1 = 0.825 m3/kg (0.0148 m3/mole); u = 3661.8 kJ/kg v2 = 2.475 m3/kg (0.0446 m3/mole); u2 = 3663.1 kJ/kg

The above numbers show that in this particular process, the p-v-T equation of steam very closely follows the ideal gas law. However, the slight change in the internal energy is a sign of residual nonideality. In calculating the heat absorbed using the 1st law, the ideal gas law can be used to compute the work, but Au cannot be neglected in Eq (3.2). As in Example (a), w = 9800 J/mole and u2 - u = (3663.1 - 3661.8) x 18 = 23 J/mole. From Eq (3.2), the heat absorbed is:

q = 23 + 9800 = 9823 J/mole Example (c) Solid See problem 3.10.

Figure 3.3 shows a generic constant-pressure, or isobaric, process. The means of energy input is the same as in the isochoric examples. The rigid container is replaced by a cylinder with a frictionless, massless piston, which assures equality of the system pressure and the external pressure.

In p-v coordinates, an isobaric process appears as a horizontal line as in Fig. 3.3. the work done is the rectangular area between v1 and v2. Because p is constant, Eq (3.1) reduces to w = p(v2 - v1) and Eq (3.2), yields:

q = Au + w = u2 - u1 + p(v2 - v1) = u2 - u1 + p2v2 - p1v1

Because the enthalpy is defined by h = u + pv, combining the first and third terms and the second and fourth terms simplifies the above formula to:

which provides the following useful generalization:

Heat transferred in an isobaric process in which only reversible expansion work is performed is equal to the change in the system's enthalpy.

The simplification afforded by combining u and pv for the constant-pressure case with a closed system does not in itself warrant naming the combination a new thermodynamic property h. However, as will be shown in Chap. 4, in applications to open, or flow, systems, the analog of Eq (3.6) takes on considerably greater importance.

vi v2

specific volume

Fig. 3.3 An isobaric process

Example (a) Vaporization of Water

When saturated liquid water is vaporized to saturated steam at constant temperature, the path followed is a horizontal line such as BC in the T-v projection of the EOS of water in Fig. 2.8. The enthalpy difference in converting saturated liquid to saturated vapor is the heat of vaporization. In steam-table terminology, Eq (3.6) becomes:

In many applications, the heat of vaporization is assumed to be a temperature-independent property of the substance. For water, however, hfg varies sufficiently over the range of practical interest that its temperature dependence generally must be taken into account. At 100oC, hfg = 2257 kJ/kg, but at 300oC hfg is 1405 kJ/kg.

Example (b) Heating of superheated steam

2 m3 of steam initially at 200oC are contained in a cylinder by a frictionless piston. 3500 kJ of heat are added and the gas expands at a constant pressure of 400 kPa..

From Table A.3, the pertinent properties of steam in state 1 are:

The mass of steam in the cylinder is 2/0.534 = 3.75 kg, so the heat added per unit mass is q = 3500/3.75 = 934.5 kJ/kg. Applying the First law in the form of Eq (3.6), the enthalpy in state 2 is:

This value of h is located between 600 and 700oC in the enthalpy column of the 400 kPa subtable of Table A.3. Linear interpolation yields T2 = 641oC. A variant of isobaric processes involving steam is given in problem 3.9

Example (c) Heating of a solid See problem 3.12.

Many processes do not fit the simple restraints of constant volume, temperature or pressure. These are considered in problems 3.2 and 3.4

The last of the important "iso" processes are those taking place at constant entropy. Figure 3.5 shows a simple version of this type of state change. A less evident example of an isentropic process is the blowdown of gas from a high pressure cylinder, as illustrated in Fig. 1.8.

v1 v2 specific volume

Fig. 3.4 An isentropic process

As emphasized on the diagram, such a process must be both adiabatic and reversible

No real process can be perfectly reversible, so isentropic analyses of adiabatic processes are idealizations. However, nearly-isentropic processes include the following: expansion or compression of gases (except through a valve or an orifice); liquid-vapor phase changes of water encountered in practical devices such as turbines of power plants; compression of liquids by efficient pumps. Entropy changes accompanying reversible and irreversible processes in gases and liquids are treated in problems 3.1 and 3.8.

The starting point for thermodynamic calculations involving this particular restraint is the pair of formulae relating entropy changes in a process to changes in pressure, temperature, and specific volume. These are rearranged forms of Eqs (1.15a) and (1.16a):

3.5.1 Ideal Gases

The functional dependence of s on p, v, and T can be determined from the above equations by using the following ideal-gas properties du = CVdT p/T = R/v dh = CpdT v/T = R/p

The resulting transformed equations are:

ds |
= Cv |
dT |
™ dv |

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