*2.2. Exergy Analysis*

The analysis of the spray drying system was performed by using the engineering equation solver (EES) software for the formulation of mass, energy, and exergy balances for each component. In their general form, they are, respectively:

$$
\sum\_{in} \dot{m}\_{in} - \sum\_{out} \dot{m}\_{out} = 0 \tag{1}
$$

$$\sum\_{in} h\_{in} \dot{m}\_{in} - \sum\_{out} h\_{out} \dot{m}\_{out} + \dot{W}\_k + \dot{Q}\_k = 0 \tag{2}$$

$$
\sum\_{k} \dot{E}\_{q,k} + \dot{\mathcal{W}}\_{k} + \sum\_{in} \dot{E}\_{in} - \sum\_{out} \dot{E}\_{out} - \dot{E}\_{D,k} = 0 \tag{3}
$$

The exergy rate, specific exergy, physical exergy, kinetics exergy, and potential exergy were calculated using Equations (4)–(8). Table 1 shows the expressions of both fuel and product exergy of each component.

$$
\dot{E} = \dot{m} \ast e\tag{4}
$$

$$e = e^{PH} + e^{CH} + e^{KN} + e^{PT} \tag{5}$$

$$e^{PH} = (h - h\_0) - T\_0(s - s\_0) \tag{6}$$

$$\mathbf{e}^{PT} = \mathbf{g}\mathbf{z} \tag{7}$$

$$
\varepsilon^{\text{KN}} = \frac{v^2}{2} \tag{8}
$$

#### **Table 1.** Composition of the different states.


The velocities of different streams were estimated by the Bernoulli relationship, Equation (9), where γ is the specific heat ratio and ρ is the density of the stream.

$$\frac{\Delta v^2}{2} + \left(\frac{\gamma}{\gamma - 1}\right) \* \frac{P}{\rho} = \left(\frac{\gamma}{\gamma - 1}\right) \* \frac{P\_0}{\rho\_0} \tag{9}$$

For the streams that had soluble coffee solids as part of their compositions, Equations (10) and (11) were used to determine the thermodynamic properties such as entropy and enthalpy. The *cp* value was

obtained from Burmester et al. [25]. The dead state conditions have been taken as *T*0 = 27.5 ◦C and *P*0 = 101.13 kPa.

$$h - h\_0 = c\_p (T - T\_0) \tag{10}$$

$$\mathbf{s} - \mathbf{s}\_0 = \mathbf{c}\_p \ln\left(\frac{T}{T\_0}\right) - R \ln\left(\frac{P}{P\_0}\right) \tag{11}$$

The composition for the different states of the system is shown in Table 1. This information was used to calculate the different thermodynamic properties.

For the calculation of chemical exergy of each state point that has soluble coffee solids and water, Equation (12) [17] was used. The concentration of water and coffee in equilibrium with the environment (*xei*) was chosen as the dead state of reference. Those values were obtained from previous studies on Arabica coffee by Yao et al. [26]. For the calculation of the chemical exergy of each state point that has soluble coffee solids, water, and air, Equation (13) [17] was used, where *xi* is the mole fraction of the different substances.

$$
\sigma\_{CE}^{CH} = -RT\_0 \sum \mathbf{x}\_i \ln \left(\frac{\mathbf{x}\_i^{\mathcal{A}}}{\mathbf{x}\_i}\right) \tag{12}
$$

$$
\varepsilon\_{\rm mix}^{CH} = \sum \mathbf{x}\_i \mathbf{e}\_i^{ch} + RT\_0 \sum \mathbf{x}\_i \ln(\mathbf{x}\_i) \tag{13}
$$

The chemical exergy of air for the different moisture content in air was calculated using an expression from Wepfer et al. [27], according to Equation (14), where *wo* and *w* are mole fraction of water vapor at environmental conditions and operational conditions, respectively.

$$e\_{\rm air}^{CH} = 0.2857 c\_{p, \rm air} T\_o \ln\left[ \left[ \frac{1 + 1.6078 w\_o}{1 + 1.6078 w} \right]^{(1 + 1.6078 w)} \left[ \frac{w}{w\_o} \right]^{1.6078 w} \right] \tag{14}$$

The exergy balance can also be formulated as Equation (15).

$$
\dot{E}\_{F,k} - \dot{E}\_{P,k} = \dot{E}\_{D,k} - \dot{E}\_{L,k} \tag{15}
$$

where . *EF*,*<sup>k</sup>* corresponds to the fuel exergy, . *EP*,*<sup>k</sup>* is the product exergy, . *ED*,*<sup>k</sup>* is the destroyed, exergy and . *EL*,*<sup>k</sup>* is the exergy loss. The exergy of the fuel and the exergy of the product for each single component were formulated following Lazzareto and Tsatsaronis rules [28] and they are shown in Table 2.

For the total system the exergetic efficiency was calculated as the sum of the product exergy rates divided by the sum of the fuel exergy rates.

Other interesting parameters involved in an exergy analysis were the relative exergy destruction (*y*<sup>∗</sup>*D*,*<sup>k</sup>*), which represents the relationship between the destroyed exergy of a component and the total destroyed exergy of the system, as shown in Equation (16) [17]. The exergy destruction ratio (*yD*,*<sup>k</sup>*), which relates the destroyed exergy of a component with the total fuel exergy of the system, is shown in Equation (17). The exergetic efficiency (*nex*,*<sup>k</sup>*), which represents the amount of exergy that is useful in relation to the fuel exergy in the component, is shown in Equation (18).

.

$$y\_{D,k}^\* = \frac{E\_{D,k}}{\dot{E}\_{D, \text{tot}}} \tag{16}$$

$$y\_{D,k} = \frac{\dot{E}\_{D,k}}{\dot{E}\_{F, \text{tot}}} \tag{17}$$

$$m\_{\rm ex\,k} = \frac{\dot{E}\_{P,k}}{\dot{E}\_{F,k}}\tag{18}$$


**Table 2.** Definitions of fuel and product exergy for each component.
