Thermodynamic Properties

Abstract

The equilibrium thermodynamic properties of nanofluids are important variables that are used in all engineering applications. Since the nanofluids are mixtures of liquids and solid particles, the determination of their thermodynamic properties may be accomplished using the theory of heterogeneous mixtures and the corresponding properties of the constituent materials (The Collective Works of J. Willard Gibbs, Longmans, New York, NY, 1878). In this chapter we derive the functional form of several equilibrium thermodynamic properties of these heterogeneous mixtures. Pressure and temperature are intensive properties and do not depend on the mass or the volume of the system. Of the other properties, the extensive properties are proportional to the mass of the system. For every extensive property, one may define the relevant specific property, which is equal to the ratio of the corresponding extensive property and the mass of the system. The density of the system, which is the inverse of the specific volume, and its derivative properties are treated separately.

Intrinsic chemical reactions or other energy releases that may take place between the constituents of the nanofluid are taken into account in the derivation of the energy-related properties. Because certain inaccurate “alternative models” for well-defined properties have appeared in the nanofluids literature, the derivation of the thermodynamic properties of nanofluids is accomplished in detail using the methodology of thermodynamics and the applicable conservation principles. Two implicit assumptions in all the derivations of this chapter—as well as in the engineering uses of the equilibrium thermodynamic properties—are the following:

• (a) The system under observation is in local thermodynamic equilibrium.

• (b) The volume of the system is large enough to be treated as a continuum.

Appendix: Specific Heat Differences for Liquids and Solids

Consider the Gibbs equations for internal energy and enthalpy :

$$ \mathrm{d}u=T\mathrm{d}s-P\mathrm{d}v $$

(3.58)

and

$$ \mathrm{d}h=T\mathrm{d}s-v\mathrm{d}P. $$

(3.59)

One may use these differentials to obtain two expressions for the specific heats in terms of the entropy function as follows:

$$ {c}_v={\left(\frac{\partial u}{\partial T}\right)}_v=T{\left(\frac{\partial s}{\partial T}\right)}_v $$

(3.60)

and

$$ {c}_P={\left(\frac{\partial P}{\partial T}\right)}_P=T{\left(\frac{\partial s}{\partial T}\right)}_P, $$

(3.61)

Because entropy is a thermodynamic property, for homogeneous substances it may be expressed in terms of any two independent variables. Using the functional forms for entropy, s = s(T, v) and s = s(T, P), we obtain the following two expressions for the differential ds:

$$ \mathrm{d}s={\left(\frac{\partial s}{\partial T}\right)}_v\mathrm{d}T+{\left(\frac{\partial s}{\partial v}\right)}_T\mathrm{d}v=\frac{c_v}{T}\mathrm{d}T+{\left(\frac{\partial s}{\partial v}\right)}_T\mathrm{d}v $$

(3.62)

and

$$ \mathrm{d}s={\left(\frac{\partial s}{\partial T}\right)}_P\mathrm{d}T+{\left(\frac{\partial s}{\partial P}\right)}_T\mathrm{d}P=\frac{c_P}{T}\mathrm{d}T+{\left(\frac{\partial s}{\partial P}\right)}_T\mathrm{d}P. $$

(3.63)

Since the l.h.s. of Eqs. (3.62) and (3.63) are identical, we may use the equality of the r.h.s. of these equations to obtain the following expression for the difference of the two specific heats:

$$ \left({c}_P-{c}_v\right)\mathrm{d}T=T{\left(\frac{\partial s}{\partial v}\right)}_T\mathrm{d}v-T{\left(\frac{\partial s}{\partial P}\right)}_T\mathrm{d}P. $$

(3.64)

Now we use the functional form of the equation of state P = (T, v) to obtain the following expression for the differential dP:

$$ \mathrm{d}P={\left(\frac{\mathrm{d}P}{\mathrm{d}T}\right)}_v\mathrm{d}T+{\left(\frac{\mathrm{d}P}{\mathrm{d}v}\right)}_T\mathrm{d}v. $$

(3.65)

Substitution of the dP term in Eq. (3.64) from Eq. (3.65) yields

$$ \left({c}_P-{c}_v\right)\mathrm{d}T=T{\left(\frac{\partial s}{\partial v}\right)}_T\mathrm{d}v-T{\left(\frac{\partial s}{\partial P}\right)}_T\left[{\left(\frac{\mathrm{d}P}{\mathrm{d}T}\right)}_v\mathrm{d}T+{\left(\frac{\mathrm{d}P}{\mathrm{d}v}\right)}_T\mathrm{d}v\right] $$

(3.66)

or

$$ \left[\left({c}_P-{c}_v\right)+T{\left(\frac{\partial s}{\partial P}\right)}_T{\left(\frac{\mathrm{d}P}{\mathrm{d}T}\right)}_v\right]\mathrm{d}T=T\left[{\left(\frac{\partial s}{\partial v}\right)}_T-{\left(\frac{\partial s}{\partial P}\right)}_T{\left(\frac{\mathrm{d}P}{\mathrm{d}v}\right)}_T\right]\mathrm{d}v. $$

(3.67)

The partial derivatives of the entropy function may be expressed in terms of derivatives involving the variables, P, v, and T, using the two Maxwell’s equations:

$$ {\left(\frac{\partial s}{\partial P}\right)}_T=-{\left(\frac{\partial v}{\partial T}\right)}_P\kern0.5em \mathrm{and}\kern0.5em {\left(\frac{\partial s}{\partial v}\right)}_T={\left(\frac{\partial P}{\partial T}\right)}_v. $$

(3.68)

Substituting the expression from Eq. (3.68) into Eq. (3.67) yields

$$ \left[\left({c}_P-{c}_v\right)-T{\left(\frac{\partial v}{\partial T}\right)}_P{\left(\frac{\mathrm{d}P}{\mathrm{d}T}\right)}_v\right]\mathrm{d}T=T\left[{\left(\frac{\partial P}{\partial T}\right)}_v+{\left(\frac{\partial v}{\partial T}\right)}_P{\left(\frac{\mathrm{d}P}{\mathrm{d}v}\right)}_T\right]\mathrm{d}v. $$

(3.69)

The variables T and v are independent variables in all substances—homogeneous and nonhomogeneous—and may vary independently. Therefore, the terms in square brackets must vanish for the equation to hold. This yields the following expression for the difference of the two specific heats:

$$ \left({c}_P-{c}_v\right)=T{\left(\frac{\partial v}{\partial T}\right)}_P{\left(\frac{\mathrm{d}P}{\mathrm{d}T}\right)}_v. $$

(3.70)

The last equation expresses the difference of the two specific heats in terms of the measurable P, v, and T variables. It is apparent that for incompressible substances (v = constant) the difference of the two specific heats vanishes, that is,

$$ {c}_P={c}_v=c. $$

(3.71)

Liquid water and other commonly used base fluids are not exactly incompressible substances: their specific volume is a weak function of pressure and temperature. The effect of the pressure and temperature on the specific heats of all the materials may be calculated using the two expansion coefficients, β and κ _T , which are defined as

$$ {\beta}_P\equiv \frac{1}{v}{\left(\frac{\partial v}{\partial T}\right)}_P\kern0.5em \mathrm{and}\kern0.5em {\kappa}_T\equiv -\frac{1}{v}{\left(\frac{\partial v}{\partial P}\right)}_T. $$

(3.72)

From the functional form of the equation of state for any substance, P = (T, v), we obtain the identity

$$ {\left(\frac{\partial P}{\partial T}\right)}_v{\left(\frac{\partial T}{\partial v}\right)}_P{\left(\frac{\partial v}{\partial P}\right)}_T\equiv -1\kern0.5em \mathrm{and}\kern0.5em {\left(\frac{\partial P}{\partial T}\right)}_v\equiv -{\left(\frac{\partial v}{\partial T}\right)}_P{\left(\frac{\partial P}{\partial v}\right)}_T. $$

(3.73)

Equation (3.70) may now be transformed into the following form in terms of the variables P, v, and T and the two expansion coefficients:

$$ \left({c}_P-{c}_v\right)=-\frac{T{\left(\frac{\partial v}{\partial T}\right)}_P^2}{{\left(\frac{\mathrm{d}v}{\mathrm{d}P}\right)}_T}\Rightarrow \left({c}_P-{c}_v\right)= Tv\frac{\beta_P^2}{\kappa_T}. $$

(3.74)

For liquid water in the range of 0–100 °C (273–373 K), v ≈ 0.001 m³/kg, β _P  ≈ 3 × 10⁻⁷ K⁻¹, and κ _T  ≈ 4 × 10⁻¹² Pa⁻¹. Then the last equation gives for the difference of the two specific heats as c _P  − c _v  ≈ 0.007 J/kg K. Given that the value of c _P for water is approximately 4,180 J/kg K, this difference is negligible and the approximation of Eq. (3.71) is accurate.