Industrial waste heat recovery and cogeneration involving organic Rankine cycles

Abstract

This paper proposes a systematic approach for energy integration involving waste heat recovery through an organic Rankine cycle (ORC). The proposed approach is based on a two-stage procedure. In the first stage, heating and cooling targets are determined through heat integration. This enables the identification of the excess process heat available for use in the ORC. The optimization of the operating conditions and design of the cogeneration system are carried out in the second stage using genetic algorithms. A modular sequential simulation approach is proposed including several correlations to determine the properties for the streams in the ORC. The proposed approach is applied to a case study which addresses the tradeoffs among the different forms of energy and associated costs. The results show that the optimal selection of the operating conditions and working fluid is very important to reduce the costs associated to the process.

Appendix 1: Correlations for the thermodynamic properties

In this section, representative correlations for the properties of the refrigerant R245fa are described. Additional relationships for other refrigerants are presented in the Electronic Supplementary Material. The correlation coefficients for these relationships are greater than 0.99 with respect to REFPROP (Lemmon et al. 2014), which is software dedicated to determine refrigerant properties. The streams involved are identified in Fig. 2.

The saturation temperature, T _Sat1 (°C), for stream 1 at the output of the boiler is a function of the pressure P ₁ (bar) as follows.

For pressures from 0.6 to 5 bars:

$$\begin{aligned} T_{\text{Sat1}} & = 0.066568672192 \times P_{1}^{5} - 1.108775860189 \times P_{1}^{4} + 7.363209229763 \times P_{1}^{3} \\ & - 25.624800698713 \times P_{1}^{2} + 58.471405699693 \times P_{1} - 24.406058535636. \\ \end{aligned}$$

(12)

For pressures from 5 to 10 bars:

$$T_{\text{Sat1}} = \,0.022355866356 \times P_{1}^{3} - 0.778937062934 \times P_{1}^{2} + 13.169856254828 \times P_{1} + 13.590459207549.$$

(12)

For pressures from 10 to 36 bars:

$$T_{\text{Sat1}} = \,0.001083730652 \times P_{1}^{3} - 0.119432214832 \times P_{1}^{2} + 6.034814366651 \times P_{1} + 40.332775394524.$$

(14)

The enthalpy for stream 1, H ₁ (kJ/kg), is a function of the inlet pressure to the turbine, and for pressures between 1 and 10 bars, this is given as follows:

$$H_{ 1} = \, - 0.011040 \times P_{1}^{4} + 0.309896 \times P_{1}^{3} - 3.404426 \times P_{1}^{2} + 21.106078 \times P_{1} + 397.735358.$$

(15)

For pressures between 10 and 30 bars:

$$\begin{aligned} H_{ 1} = & - 8.69711 \times 10^{ - 5} \times P_{1}^{4} + 0.0072280959 \times P_{1}^{3} - 0.2875158903 \times P_{1}^{2} + 6.577807603 \times P_{1} . \\ & + 424.6546520126. \\ \end{aligned}$$

(16)

For pressures between 30 and 36 bars:

$$\begin{aligned} H_{ 1} = & - 0.0047916667 \times P_{1}^{5} + 0.7708522735 \times P_{1}^{4} - 49.5997917182 \times P_{1}^{3} + 1595.3600584406 \times P_{1}^{2} . \\ & - 25648.7329419049 \times P_{1} + 165367.4014206060. \\ \end{aligned}$$

(17)

The entropy for stream 1 (S ₁ (kJ/kg °C)) is a function of the inlet pressure to the turbine. For pressures between 1 and 25 bars, this is given as follows:

$$\begin{aligned} S_{1} & = - 1.464 \times 10^{ - 6} \times P_{1}^{5} + 0.0002145718 \times P_{1}^{4} - 0.0125588231 \times P_{1}^{3} + 0.3666710619 \times P_{1}^{2} \\ & - 5.3386278379 \times P_{1} + 32.8066383969 \\ \end{aligned}$$

(18)

For pressures between 25 and 36 bars:

$$\begin{aligned} S_{ 1} = & - 1.464 \times 10^{ - 6} \times P_{1}^{5} + 0.0002145718 \times P_{1}^{4} - 0.0125588231 \times P_{1}^{3} + 0.3666710619 \times P_{1}^{2} \\ & \quad- 5.3386278379 \times P_{1} + 32.8066383969. \\ \end{aligned}$$

(19)

The isentropic temperature at the exit of the turbine for stream 2 (T _2S (°C)) is calculated in terms of the inlet entropy S ₁ (kJ/kg °C) as follows:

$$T_{{ 2 {\text{S}}}} = \, 3 6 2. 0 0 2 6 7 9\times S_{ 1} - 6 2 1. 3 1 4 2 8 3.$$

(20)

The isentropic enthalpy at the exit of the turbine (H _2S (kJ/kg)) is given in terms of the outlet temperature (T _2S (°C)) as follows:

$$H_{{ 2 {\text{S}}}} = \, 0. 9 8 9 9 4 1\times T_{{ 2 {\text{S}}}} { + 398} . 5 4 9 9 6 0.$$

(21)

The enthalpy at the exit of the turbine (H ₂ (kJ/kg)) is given in terms of the efficiency of the turbine (η _turbine), the isentropic enthalpy H _2S (kJ/kg), and the inlet enthalpy H ₁ (kJ/kg) as follows:

$$H_{ 2} = H_{ 1} - \eta_{\text{turbine}} \left( {H_{ 1} - H_{{ 2 {\text{S}}}} } \right).$$

(22)

The temperature at the exit of the turbine (T ₂ (°C)) is obtained in terms of the enthalpy H ₂ (kJ/kg) at a pressure P ₂ (bar) as follows:

$$T_{ 2} = 1. 0 0 9 4 7 2\times H_{ 2} - 4 0 2. 2 6 5 8 4 7.$$

(23)

The entropy at the exit of the turbine (S ₂ (kJ/kg °C)) is given in terms of the enthalpy H ₂ (kJ/kg) as follows:

$$S_{ 2} = - 4 \times 10^{ - 6} \times H_{ 2}^{2} + 0.006602 \times H_{ 2} - 0.314007.$$

(24)

The saturation temperature T _Sat3 (°C) at the outlet of the condenser (stream 3) is the function of the pressure P ₂ (bar) as follows. For pressures from 0.6 to 5 bars, the correlation is

$$\begin{aligned} T_{\text{Sat3}} = & 0.066568672192 \times P_{2}^{5} - 1.108775860189 \times P_{2}^{4} + 7.363209229763 \times P_{2}^{3} \\ & - 25.624800698713 \times P_{2}^{2} + 58.471405699693 \times P_{2} - 24.406058535636. \\ \end{aligned}$$

(25)

For pressures from 5 to 10 bars:

$$T_{\text{Sat3}} = 0.022355866356 \times P_{2}^{3} - 0.778937062934 \times P_{2}^{2} + 13.169856254828 \times P_{2} + 13.590459207549.$$

(26)

For pressures from 10 to 36 bars:

$$T_{\text{Sat3}} = 0.001083730652 \times P_{2}^{3} - 0.119432214832 \times P_{2}^{2} + 6.034814366651 \times P_{2} + 40.332775394524.$$

(27)

In addition, the enthalpy H ₃ (kJ/kg) for the stream 3 at the inlet to the pump is function of the saturation temperature T _Sat3 (°C):

$$H_{ 3} = 0.000031 \times T_{\text{Sat3}}^{3} - 0.004074 \times T_{\text{Sat3}}^{2} + 1.493974 \times T_{\text{Sat3}} + 198.439968,$$

(28)

while the entropy S ₃ (kJ/kg) is given in terms of the saturation temperature T _Sat3 (°C), which is as follows,

$$S_{ 3} = 0. 0 0 4 3 7 7\times T_{\text{Sat3}} { + 1} . 0 0 0 2 6 6.$$

(29)

The power consumed by the pump is obtained through the following equation:

$$W_{P} = \frac{{v\left( {P_{1} - P_{2} } \right)}}{{\eta_{\text{pump}} }},$$

(30)

where $\eta_{\text{pump}}$ is the efficiency for the pump, and $v$ (m³/kg) is the volume of the stream inlet to the pump, which is determined by the inverse of the density of this fluid that is calculated as follows:

$${\text{Density}} = - 0.006593 \times T_{\text{Sat3}}^{2} - 2.388728 \times T_{\text{Sat3}} + 1403.123984.$$

(31)

The enthalpy H ₄ (kJ/kg) for the stream at the outlet of the pump is calculated with the following relationship:

$$H_{ 4} = H_{ 3} + W_{\text{P}} .$$

(32)

The entropy for the stream 4 at the exit of the pump (S ₄ (kJ/kg °C)) is given in terms of the enthalpy H ₄ (kJ/kg) at a pressure P ₁ (bar) as follows:

$$S_{ 4} = - 4.2844 \times 10^{ - 6} \times H_{ 4}^{2} + 0.0053489173 \times H_{ 4} + 0.1012742328.$$

(33)

The temperature for stream 4 at the exit of the pump (T ₄ (°C)) is obtained in terms of the enthalpy H ₄ (kJ/kg) at a pressure P ₁ (bar) as follows:

$$T_{ 4} = - 4.698168 \times 10^{ - 4} \times H_{ 4}^{2} + 0.9749078129 \times H_{ 4} - 176.2592280050.$$

(34)

The isentropic enthalpy for stream 4 at the exit of the pump (H _4S (kJ/kg)) is given in terms of the following expression:

$$H_{{ 4 {\text{S}}}} = \eta_{\text{pump}} \left( {H_{ 4} - H_{ 3} } \right) + H_{ 3} .$$

(35)

The isentropic entropy for stream 4 at the exit of the pump (S _4S (kJ/kg °C)) is given in terms of the enthalpy H _4S (kJ/kg) at a pressure P ₁ (bar) as follows:

$$S_{{ 4 {\text{S}}}} = - 4.2844 \times 10^{ - 6} \times H_{{ 4 {\text{S}}}}^{2} + 0.0053489173 \times H_{{ 4 {\text{S}}}} + 0.1012742328.$$

(36)

The isentropic temperature for stream 4 at the exit of the pump (T _4S (°C)) is obtained in terms of the isentropic enthalpy H _4S (kJ/kg) at a pressure P ₁ (bar) as follows:

$$T_{{ 4 {\text{S}}}} = - 4.698168 \times 10^{ - 4} \times H_{{ 4 {\text{S}}}}^{2} + 0.9749078129 \times H_{{ 4 {\text{S}}}} - 176.2592280050.$$

(37)

The produced power in the turbine is calculated using the following expression:

$$W_{\text{T}} = \left( {H_{ 1} - H_{{ 2 {\text{S}}}} } \right)\eta_{\text{turbine}} .$$

(38)

The needed cooling in the ORC is calculated from the following relationship:

$$F_{\text{refrigerant}} = \frac{{Q^{\text{ORC}} }}{{\left( {H_{ 1} - H_{ 4} } \right)}},$$

(39)

where Q ^ORC is the heat required in the boiler (kJ/s) and F _refrigerant is the flow rate for the refrigerant (kg/s) in the ORC.

The proposed correlations to determine the thermodynamic properties for refrigerants R123 and n-butane are shown in Tables 5 and 6, respectively. These tables are in the Electronic Supplementary Material available in the WEB.