Fig. 3.1 Isochoric process specific volume

The First law for the isochoric process is:

Fig. 3.1 Isochoric process

The final state is determined by v and u2. To translate this information to T2 and p2, both p-v-T and thermal equations of state for the substance involved are needed.

The diagram in Fig. 3.1 shows a rigid vessel with an internal electrical resistor inside to heat the contents. The 1st Law (or energy conservation) for this system is:

Eel = I2^.t = Au = q Where I is the electric current, ^_is the resistance and t is time.

Example (a) 1 mole of air (an ideal gas) initially at pi = 5 MPa and Ti = 60oC heated in a rigid vessel until p? = 15 MPa

The vessel volume is obtained from the ideal-gas law: v = RT1/p1:

5 x106 Pa

The final temperature is T? = p?v/R = 15x106x5.5x10-4/8.314 = 999 K = 726oC.

For an ideal gas, u is independent of pressure. Assuming a temperature-independent heat capacity, the increase in internal energy is related to the increase in temperature by:

q = u2 - u1 = Cv(T2 -T1) = y2 x 8.314(726 - 333) = 8170J/mole (3.4)

Problem 3.1a explores a slightly modified version of this type of process.

Example (b) Compressed liquid later - same process as in example (a) For pi = 5 MPa, Ti = 60oC, steam table A.4 gives v = 0.001015 m3/kg and ui = 250.2 kJ/kg. For the same specific volume and a final-state pressure of 15 MPa, interpolation in Table A.4 gives T2 = 67.7oC and u2 = 280.3 kJ/kg. Finally, substitution of u1 and u2 into Eq (3.3) gives q = 280.3 - 250.2 = 30.1 kJ/kg or 542 J/mole

Note that compressing liquid water requires a much smaller heat addition than the same process with an ideal gas as the substance acted upon.

Example (c) Simple Solid - same process as in example (a); properties from Table 2.1 By a simple solid is meant one:

i) for which the effect of pressure on internal energy is negligible ii) that obeys the equation of state given by Eq (2.18b).

Since v2 = v1 for an isochoric process, the EOS gives:

This solid obeys Eq (3.3) (with CV = 3R), so that q = u2 - u1 = Cy(T2 - T1) = 3 x8.314x 10 = 250 J/mole

Comparison of the examples

Table 3.1 Property changes due constant-volume compression of 1 mole from 5 to

15 MPa

Substance v, m3/mole AT, oC Au, J/mole

Air 5.5x10-4 666 8170

water 1.8x10"5 7.7 542

Simple solid ~ 10-5 10 250

The ordering of the heat inputs (or Au) is the inverse of the coefficients of compressibilities (R) of the three substances. The gas is highly compressible and so requires significant heat addition to be pressurized over the 10 MPa range. The solid, on the other hand, is quite incompressible, so achieving the same pressure increase requires a much smaller heat addition.

The temperature increases in the table are inversely proportional to the specific heats of the substances. For gas:water:solid, the relative values of CV are 1:6:2. Water has a high heat capacity, which is why its AT for the isochoric pressurization process is the smallest of the three substances. For the same reason water is an effective coolant in innumerable industrial processes.

Other examples of heating of a system in a rigid container are given in problems 3.7, 3.8, and 3.11.

3.3 Isothermal Process

Figure 3.3 shows the path of a process during which the temperature is maintained constant. The detailed shape of the curve depends on the equation of state of the substance. The area under the curve between v1 and v2 is the work performed by the system (Eq (3.1)).

Fig. 3.2 An isothermal process

vi specific volume v2

Fig. 3.2 An isothermal process

While work is done on the system by compressing the piston with a spring, heat is removed by the thermal reservoir, which maintains the system at constant temperature. For most substances, constant temperature means constant internal energy so the 1st law reduces to:

Example (a) 1 mole of an ideal gas at a constant temperature of 800oC compressed from 0.2 MPa to 0.6 MPa For this case, the curve in Fig. 3.2 is hyperbolic, or p <x 1/v. The work per mole is obtained by using p = RT/v in Eq (3.1) and integrating from v1 to v2:

Getting Started With Solar

Getting Started With Solar

Do we really want the one thing that gives us its resources unconditionally to suffer even more than it is suffering now? Nature, is a part of our being from the earliest human days. We respect Nature and it gives us its bounty, but in the recent past greedy money hungry corporations have made us all so destructive, so wasteful.

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