**2. Description of Systems**

For recovering heat with low-temperature sources, water is not a good option as a working fluid in cycles. ORCs are suitable cycles for the conversion of low-quality heat to electricity. ORCs are similar to the conventional steam Rankine cycles, but the working fluids in these cycles are organic fluids with low boiling points. Using this kind of fluid, it is possible to use heat from lower temperature heat sources. In ORC applications, organic fluids such as alkanes, aromates and siloxanes are used as working fluids instead of water. The reason to use these fluids is that they have lower boiling temperatures and higher molecular weights compared to water, which causes higher thermodynamic performances in lower temperatures. Since most of the organic fluids do not go to wet regions when they are expanded in turbines, they prevent erosion in turbine blades. Additionally, compared with the steam Rankine cycles, ORC applications work under lower working fluid pressures, which leads to lower turbine costs [23].

In ORC system designing, at least five components are needed for basic types: turbines, evaporators, condensers, pumps and working fluids. In the present study, R11 is selected as the working fluid for the analysis because of its high exergy efficiency [12]. Three different ORC systems (BORC, SRORC and DRORC) are chosen in this research. The schematic of these cycles with T-S (temperature-entropy) diagrams are shown respectively in Figures 1–3. In the basic ORC, first, the fluid is heated by the evaporator, and its temperature is increased. Then, the working fluid (R11) expands in the turbine from high pressure to low pressure to produce power. The outlet flow of the turbine goes to the condenser for cooling, and the saturated liquid enters the pump. The pump increases the flow pressure to reach the evaporator pressure.

**Figure 1.** Schematic diagram of the cycle and T-S thermodynamic diagram of the basic organic Rankine cycle (BORC) system.

**Figure 2.** Schematic diagram of the cycle and T-S thermodynamic diagram of the regeneration ORC (RORC) system.

As it is seen in Figures 2 and 3, in regenerative cycles, the turbine outlet flow is split into two parts in single, and three parts in double, systems. These parts are in the vapor phase, which enters feed-water heaters to preheat the working fluid before going to the evaporator. By this method, more potential of the cycle energy is used, so a higher efficiency, compared with the basic type, is gained. In order to simulate the cycle performance in this study, the following assumptions are employed:


Exiting fluid from the condenser is a saturated liquid.

**Figure 3.** Schematic diagram of the cycle and T-S thermodynamic diagram of the double-regeneration ORC (DRORC) system.

#### **3. Energy and Exergy Analysis**

#### *3.1. Energy and Conventional Exergy Analysis*

The purpose of the energy analysis is to design thermal systems. However, the aim of the exergy analysis is to determine the thermodynamic efficiency of the cycles and compare them to find out the best system performance. There are two important basic laws in thermodynamics: the first and the second law. The first law analyzes the conservation of energy in processes, while the second law is used to discuss the quality of energy and materials. The main equations for energy and mass balance used in the first law analysis for a controlled volume of each component are written below:

Mass balance:

$$
\sum \dot{m}\_{\text{i}} = \sum \dot{m}\_{\text{c}} \tag{1}
$$

.

.

Energy balance:

$$
\sum \dot{Q} - \sum \dot{W} + \sum \dot{m}\_i h\_i - \sum \dot{m}\_e h\_e = 0 \tag{2}
$$

Exergy analysis has two advantages compared with the conventional energy analysis method. It provides a more precise measurement of the real ine fficiencies in a system, as well as their exact places. The main goal of the second law is to recognize the place, type and quantity of energy loss resources in system processes and the factors a ffecting them. The exergy balance equation for a control volume of each component can be written as [24]:

$$
\dot{I} = \dot{E}\_{in} - \dot{E}\_{out} + \dot{E}\_Q - \dot{E}\_W \tag{3}
$$

$$
\dots \quad \dots
$$

.

where *I* is the total exergy rate of a destroyed system, and *EQ* and *Ew* are the exergy transfer rates for heat and work, respectively. The equivalent work of any form of energy is defined as its exergy, so work is equivalent to exergy in every respect.

For a heat transfer rate ( *Q*r) and system temperature where heat transfer happens (Tr), the maximum rate of conversion from thermal energy to work is:

$$
\dot{w}\_{\text{max}} = \dot{\bar{E}}^{\text{Q}} = \mathcal{Q}^{\text{r}} \,\text{\text{\textdegree}}\tag{4}
$$

$$
\pi = 1 - \mathcal{T}\_0 / \mathcal{T}\_r \tag{5}
$$

τ is called the dimensionless exergetic temperature and is equal to the Carnot e fficiency for special cases when the environment is at temperature T0.

Exergy of each stream in a system, . *E*, is divided into four di fferent parts:

.

.

$$
\dot{E} = \dot{E}\_{\text{k}} + \dot{E}\_{\text{P}} + \dot{E}\_{\text{Ph}} + \dot{E}\text{ch} \tag{6}
$$

.

where *E*k is kinetic exergy, *E*p is potential exergy, . *<sup>E</sup>*p<sup>h</sup> is physical exergy and *E*ch is chemical exergy. . *E*k and . *E*p are associated with high-grade energies and . *<sup>E</sup>*p<sup>h</sup> and . *E*0 with low-grade energies. In this research, kinetic and potential exergy is neglected.

Physical exergy of each stream can be calculated from the following Equation (0 refers to environmental state):

$$
\sigma\_{\rm plr} = (h - h\_0) - T\_0 (\mathbb{S} - \mathbb{S}\_0) \tag{7}
$$

and the chemical exergy of the mixture for ideal gas can be expressed as below:

$$x\_{ch} = \sum\_{i=0}^{n} x\_i \mathbf{x}\_{chi} + \text{RT}\_0 \sum\_{i=0}^{n} \mathbf{x}\_i \ln(\mathbf{x}\_i) \tag{8}$$

Here, *xi* is the mole fraction of the ith component in the mixture, and *exchi* in this relation is the chemical exergy of each component. As it is seen, exergy loss depends on the amount of heat that is transferred to ambient. So, convectional exergy analysis is only influenced by the entire condition of a system and does not deal with the e ffect of system components on other components' exergy destructions.

The basic equations used for the kth component in the conventional exergy analysis are shown in Equations (9)–(12).

$$E\_{D,k} = E\_{F,k} - E\_{P,k} \tag{9}$$

$$
\kappa\_k = \frac{\dot{E}\_{P,k}}{\dot{E}\_{D,k}} = 1 - \frac{\dot{E}\_{D,k}}{\dot{E}\_{F,k}} \tag{10}
$$

$$y\_k = \frac{\dot{E}\_{D,k}}{\dot{E}\_{F,k}} \tag{11}$$

.

.

.

$$y\_k^\* = \frac{\dot{E}\_{D,k}}{\dot{E}\_{D,\text{tot}}} \tag{12}$$

In these equations, *ED*.*k*, *EF*.*<sup>k</sup>* and *EP*.*<sup>k</sup>* represent the exergy destruction rate, exergy of fuel and exergy of product, respectively. Additionally, ε*k*, *yk* and *y*∗*k* are the exergy efficiency, the exergy loss ratio and the exergy of fuel with the total exergy destruction in the kth component, respectively. The exergy equations for components of the three cycles are expressed in Table 1.

**Table 1.** Balance of exergy for any components of the systems. BORC: basic organic Rankine cycle, SRORC: single-regeneration ORC and DRORC: double-regeneration ORC.


In order to observe the system performance for the research, the first and the second law efficiencies are calculated from the equations below [12]:

$$
\eta\_t = \frac{\mathcal{W}\_{net}}{\mathcal{Q}} \tag{13}
$$

$$\eta\_{\ell} = \mathcal{W}\_{\text{net}} / \left[ Q \left( 1 - \frac{T\_0}{T\_m} \right) \right] \tag{14}$$

where T0 is environmental temperature, and Tm is source of heat; mean temperature can be calculated by the following equation:

$$T\_m = (T\_{in} - T\_{out}) / \ln(T\_{in} / T\_{out}) \tag{15}$$

By the uses of Equations (13) and (14), the energy efficiency and exergy efficiency for the basic cycle can be obtained:

$$
\eta\_t = \mathcal{W}\_{\text{net}} / Q \tag{16}
$$

$$
\eta\_c = \frac{\mathcal{W}\_{net}}{\mathcal{Q}} \left( 1 - \frac{T\_0}{T\_m} \right) \tag{17}
$$

For SRORC, the first and the second law efficiencies are presented as:

$$\eta\_t = \frac{\mathcal{W}\_{\text{net}}}{Q} = \frac{\mathcal{W}\_t - \mathcal{W}\_{p1} - \mathcal{W}\_{p2}}{Q} \tag{18}$$

$$\eta\_c = \frac{\mathcal{W}\_{\text{net}}}{Q \left(1 - \frac{T\_0}{T\_m}\right)} = \frac{\mathcal{W}\_t - \mathcal{W}\_{p1} - \mathcal{W}\_{p2}}{\left[Q \left(1 - \frac{T\_0}{T\_m}\right)\right]} \tag{19}$$

For DRORC, efficiencies are calculated from the equations below:

$$
\eta\_l = \frac{\mathcal{W}\_{\text{net}}}{Q} = \frac{\mathcal{W}\_l - \mathcal{W}\_{p1} - \mathcal{W}\_{p2} - \mathcal{W}\_{p3}}{Q} \tag{20}
$$

$$\eta\_{l} = \frac{\mathcal{W}\_{\text{net}}}{Q \left(1 - \frac{T\_0}{T\_m}\right)} = \frac{\mathcal{W}\_l - \mathcal{W}\_{p1} - \mathcal{W}\_{p2} - \mathcal{W}\_{p3}}{Q(1 - T\_0/T\_m)}\tag{21}$$
