3. Transiting thermo-mechanical exergy and its link to the thermal and mechanical components

Following the equations proposed by Brodyansky et al. [3], the specific transiting thermomechanical exergy etr of an analyzed system is defined as the minimum exergy value that can be assigned to a material stream, considering the pressure P and temperature T at the inlet and outlet, as well as the ambient temperature T0. With this definition, there are three possible combinations of Pin, Tin, Pout, Tout and T0 that determine the value of etr:

$$\text{If } (\mathbf{T}\_{\text{in}} > \mathbf{T}\_0 \text{ and } \mathbf{T}\_{\text{out}} > \mathbf{T}\_0): \dot{\mathbf{E}}\_{\text{tr}} = \dot{\mathbf{m}}. \mathbf{e}\_{\text{tr}}(\mathbf{P}\_{\text{min}}, \mathbf{T}\_{\text{min}}) \tag{9a}$$

$$\text{If } (\mathbf{T}\_{\text{in}} < \mathbf{T}\_0 \text{ and } \mathbf{T}\_{\text{out}} < \mathbf{T}\_0): \dot{\mathbf{E}}\_{\text{tr}} = \dot{\mathbf{m}} \cdot \mathbf{e}\_{\text{tr}}(\mathbf{P}\_{\text{min}}, \mathbf{T}\_{\text{max}}) \tag{9b}$$

$$\text{If } (\mathbf{T}\_{\text{in}} > \mathbf{T}\_0 \text{ and } \mathbf{T}\_{\text{out}} < \mathbf{T}\_0) \text{OR} \left(\mathbf{T}\_{\text{in}} < \mathbf{T}\_0 \text{ and } \mathbf{T}\_{\text{out}} > \mathbf{T}\_0\right): \ \dot{\mathbf{E}}\_{\text{tr}} = \dot{\mathbf{m}} \cdot \mathbf{e}\_{\text{tr}} (\mathbf{P}\_{\text{min}}, \mathbf{T}\_0) \tag{9c}$$

Inspection of these equations shows that for all three cases etr is determined by using the lowest pressure Pmin among the inlet and outlet values. The situation is different for temperature, where the minimum exergy value is determined using the lowest temperature Tmin for processes above ambient, the highest temperature Tmax for sub-ambient processes, and by using T0 for the case of processes operating across ambient temperature. In order to understand the physical meaning of the transiting exergy, let us analyze the throttling process of a real gas taking place under these three different temperature conditions.

#### 3.1. Adiabatic throttling process

The case of the throttling process operating above T0 is presented on an e-h diagram (see Figure 3a). According to Eq. (9a), the value of etr is:

$$\mathbf{e}\_{\rm tr} = \mathbf{e}(\mathbf{P}\_{\rm out}, \mathbf{T}\_{\rm out}) \tag{10}$$

Now, let us analyze the case of throttling at sub-ambient conditions presented in Figure 3b.

Figure 3. Transiting thermo-mechanical exergy presentation for a throttling process of a real gas operating above (a),

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature

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67

The term (ΔeT)Pout in Eq. (14) is the increase of the specific thermal exergy due to an isobaric temperature drop under sub-ambient conditions at constant pressure Pout. The term (∇eP)Tin in Eq. (15) is the decrease of the specific mechanical exergy due to an isothermal pressure drop

<sup>Δ</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � ð Þ¼ eout � etr <sup>m</sup>\_ � e Pout ð Þ� ; Tout e Pout <sup>½</sup> ð Þ ; Tin � ¼ <sup>m</sup>\_ � ð Þ <sup>Δ</sup>eT Pout (14)

<sup>∇</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � ð Þ¼ ein � etr <sup>m</sup>\_ � e Pð Þ� in; Tin e Pout <sup>½</sup> ð Þ ; Tin � ¼ <sup>m</sup>\_ � ð Þ <sup>∇</sup>eP Tin (15)

etr ¼ e Pout ð Þ ; Tin (13)

According to Eq. (9b):

Thus ΔE and \_ ∇E are calculated as: \_

below (b), and across (c) ambient temperature (T0).

This value coincides with the value e2 as illustrated in Figure 3a. The specific exergy losses (d) are also presented on the diagram. Following Eq. (8), the values ΔE and \_ ∇E are: \_

$$
\Delta \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot (\mathbf{e}\_{\text{out}} - \mathbf{e}\_{\text{tr}}) = \mathbf{0} \tag{11}
$$

$$\nabla \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot (\mathbf{e}\_{\mathrm{in}} - \mathbf{e}\_{\mathrm{tr}}) = \dot{\mathbf{m}} \cdot \mathbf{d} \tag{12}$$

where m is the gas mass flow rate. \_

As a result, the efficiency ηtr = 0, meaning that the exergy consumed is completely lost during the process and there is no produced exergy. It should be mentioned that the input-output efficiency calculated according to Eq. (6) has a negative value in this particular case and has no physical meaning. This is due to the fact that E\_ out < 0, because the throttling ended at the vacuum conditions. For this particular case the products-fuel efficiency according to Eq. (7) gives the same value as ηtr.

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature http://dx.doi.org/10.5772/intechopen.74041 67

Figure 3. Transiting thermo-mechanical exergy presentation for a throttling process of a real gas operating above (a), below (b), and across (c) ambient temperature (T0).

Now, let us analyze the case of throttling at sub-ambient conditions presented in Figure 3b. According to Eq. (9b):

$$\mathbf{e}\_{\rm tr} = \mathbf{e}(\mathbf{P}\_{\rm out}, \mathbf{T}\_{\rm in}) \tag{13}$$

Thus ΔE and \_ ∇E are calculated as: \_

3. Transiting thermo-mechanical exergy and its link to the thermal and

combinations of Pin, Tin, Pout, Tout and T0 that determine the value of etr:

If Tð Þ in <sup>&</sup>gt; T0 and Tout <sup>&</sup>gt; T0 : <sup>E</sup>\_

If Tð Þ in <sup>&</sup>lt; T0 and Tout <sup>&</sup>lt; T0 : <sup>E</sup>\_

If Tð Þ in <sup>&</sup>gt; T0 and Tout <sup>&</sup>lt; T0 OR Tð Þ in <sup>&</sup>lt; T0 and Tout <sup>&</sup>gt; T0 : <sup>E</sup>\_

real gas taking place under these three different temperature conditions.

Following the equations proposed by Brodyansky et al. [3], the specific transiting thermomechanical exergy etr of an analyzed system is defined as the minimum exergy value that can be assigned to a material stream, considering the pressure P and temperature T at the inlet and outlet, as well as the ambient temperature T0. With this definition, there are three possible

Inspection of these equations shows that for all three cases etr is determined by using the lowest pressure Pmin among the inlet and outlet values. The situation is different for temperature, where the minimum exergy value is determined using the lowest temperature Tmin for processes above ambient, the highest temperature Tmax for sub-ambient processes, and by using T0 for the case of processes operating across ambient temperature. In order to understand the physical meaning of the transiting exergy, let us analyze the throttling process of a

The case of the throttling process operating above T0 is presented on an e-h diagram (see

This value coincides with the value e2 as illustrated in Figure 3a. The specific exergy losses (d)

As a result, the efficiency ηtr = 0, meaning that the exergy consumed is completely lost during the process and there is no produced exergy. It should be mentioned that the input-output efficiency calculated according to Eq. (6) has a negative value in this particular case and has no

vacuum conditions. For this particular case the products-fuel efficiency according to Eq. (7)

are also presented on the diagram. Following Eq. (8), the values ΔE and \_ ∇E are:

tr ¼ m\_ :etrð Þ Pmin; Tmin (9a)

tr ¼ m\_ � etrð Þ Pmin;Tmax (9b)

etr ¼ e Pout ð Þ ; Tout (10)

<sup>Δ</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � ð Þ¼ eout � etr <sup>0</sup> (11)

<sup>∇</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � ð Þ¼ ein � etr <sup>m</sup>\_ � <sup>d</sup> (12)

\_

out < 0, because the throttling ended at the

tr ¼ m\_ � etrð Þ Pmin;T0 (9c)

mechanical components

66 Energy Systems and Environment

3.1. Adiabatic throttling process

where m is the gas mass flow rate. \_

gives the same value as ηtr.

Figure 3a). According to Eq. (9a), the value of etr is:

physical meaning. This is due to the fact that E\_

$$
\Delta \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot (\mathbf{e}\_{\text{out}} - \mathbf{e}\_{\text{tr}}) = \dot{\mathbf{m}} \cdot \left[ \mathbf{e} (\mathbf{P}\_{\text{out}}, \mathbf{T}\_{\text{out}}) - \mathbf{e} (\mathbf{P}\_{\text{out}}, \mathbf{T}\_{\text{in}}) \right] = \dot{\mathbf{m}} \cdot (\Delta \mathbf{e}\_{\text{T}})\_{\text{P}\_{\text{out}}} \tag{14}
$$

$$\nabla \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot (\mathbf{e}\_{\text{in}} - \mathbf{e}\_{\text{tr}}) = \dot{\mathbf{m}} \cdot \left[ \mathbf{e} (\mathbf{P}\_{\text{in}}, \mathbf{T}\_{\text{in}}) - \mathbf{e} (\mathbf{P}\_{\text{out}}, \mathbf{T}\_{\text{in}}) \right] = \dot{\mathbf{m}} \cdot (\nabla \mathbf{e}\_{\text{P}})\_{\text{T}\_{\text{in}}} \tag{15}$$

The term (ΔeT)Pout in Eq. (14) is the increase of the specific thermal exergy due to an isobaric temperature drop under sub-ambient conditions at constant pressure Pout. The term (∇eP)Tin in Eq. (15) is the decrease of the specific mechanical exergy due to an isothermal pressure drop at constant temperature Tin. Finally, the case presented in Figure 3c illustrates a throttling process started above ambient and ended at sub-ambient conditions. According to Eq. (9c):

$$\mathbf{e}\_{\rm tr} = \mathbf{e}(\mathbf{P}\_{\rm out}, \mathbf{T}\_0) \tag{16}$$

the decreasing outlet pressure P2. Less obvious is that the exergy produced ΔE rises with \_ decreasing P2, reflecting the production of a more important cooling effect. It can also be

ΔE\_ (kW)

1.0 118.6 101.8 30.8 71.0 245.9 30.3 0.9 117.3 110.8 32.4 78.4 236.9 29.2 0.7 114.7 132.0 35.6 96.4 215.7 26.9 0.5 111.9 160.1 38.8 121.3 187.6 24.2 0.3 109.1 202.3 42.1 160.2 145.4 20.8 0.1 106.0 292.3 45.5 246.8 55.5 15.6

Table 1. Variation in temperature and exergy metrics with the outlet pressure for a throttling process of air.

The expansion valve of a refrigeration mechanical vapor compression cycle is supplied with the subcooled working fluid R152a at the rate m = 0.15 kg/s at P \_ <sup>1</sup> = 615.1 kPa. The fluid is expanded to a pressure of P2 = 142.9 kPa. The ambient temperature T0 = 278 K. Calculate the

A vapor compression cycle is presented on a Ts-diagram in Figure 4. The subcooling process is represented by the line 3f-3. Given that the expansion of R152a takes place across ambient

The transiting exergy does not change with the subcooling, because it is the function of constant parameters T0 and P2, meanwhile the exergy produced increases and exergy consumed decreases. The new result is that ηtr is rising with the subcooling. It should be mentioned that increasing the amount of subcooling is well documented as a way to increase the COP (coefficient of performance) of vapor compression cycles [4]. Thus, the rise in ηtr of an expansion device guarantees the COP improvement of the overall cycle, a conclusion that may

lead to practical recommendations for optimization of refrigeration cycles.

3.2. Expansion in low temperature systems with work production and heat transfer

The primary purpose of expansion processes in the sub-ambient region is the production of cooling effect. The power that may be produced can be considered as a useful by-product. This type of expansion takes place in cryo-expanders. There are two types of these devices: adiabatic and non-adiabatic gas expansion machines. The energy and exergy balances around a non-adiabatic expander are presented in Figure 5. It should be emphasized that the directions

tr decreases with decreasing outlet pressure P2.

D\_ (kW)

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature

E\_ tr (kW)

http://dx.doi.org/10.5772/intechopen.74041

ηtr (%) 69

tr, ΔE and \_ ∇E. The results are shown in \_

tr, ΔE and \_ ∇E and \_ ηtr as a function of the subcooling ΔTsubC in the range 275–

noticed that the transiting exergy E\_

T2 (K) ∇E\_ (kW)

temperature, Eqs. (16)–(18) are used to evaluate E\_

Example 2

P2 (MPa)

variation of E\_

281 K.

Solution

Table 2.

The values ΔE and \_ ∇E are calculated as: \_

$$
\Delta \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot (\mathbf{e}\_{\text{out}} - \mathbf{e}\_{\text{tr}}) = \dot{\mathbf{m}} \cdot \left[ \mathbf{e} (\mathbf{P}\_{\text{out}}, \mathbf{T}\_{\text{out}}) - \mathbf{e} (\mathbf{P}\_{\text{out}}, \mathbf{T}\_0) \right] = \dot{\mathbf{m}} \cdot (\Delta \mathbf{e}\_{\text{T}})\_{\text{P}\_{\text{out}}} \tag{17}
$$

$$\nabla \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot (\mathbf{e}\_{\text{in}} - \mathbf{e}\_{\text{tr}}) = \dot{\mathbf{m}} \cdot \left[ \mathbf{e}(\mathbf{P}\_{\text{in}}, \mathbf{T}\_{\text{in}}) - \mathbf{e}(\mathbf{P}\_{\text{out}}, \mathbf{T}\_0) \right] = \dot{\mathbf{m}} \cdot (\nabla \mathbf{e}\_{\text{P}, \text{T}}) \tag{18}$$

Again, the input-output efficiency is not suitable for the evaluation of the processes presented in Figure 3b and c, given that the outlet exergy e2 = e(Pout, Tout) is negative. Another difficulty is linked to the application of the products-fuel efficiency for these cases. The exergy transfer of the throttling process at sub-ambient conditions consists in the partial transformation of mechanical exergy ("fuel") into thermal exergy ("product"). The problem stems from the fact that there are multiple possibilities to define "fuel" and "product" in this case; as a result multiple values of ηpr-f may be formulated, leading to the ambiguity in the products-fuel efficiency application. Indeed, the different increments of thermal exergy may be considered as a "product" for the case in Figure 3b, for example, the increase in thermal exergy following the isobar P1 or the isobar P2. In the same way, different decrements of mechanical exergy may be considered as a "fuel" in the same figure, for example, the decrease of mechanical exergy following the isotherms T1 or T2.

Contrary to "products-fuel", the transiting exergy approach does not attempt to individually compute the thermal and mechanical exergy component variations. It relies, rather, on the unaffected part of the thermo-mechanical exergy entering and leaving the system.

As illustrated in Figure 3a and c, the transiting exergy may be considered as the introduction of a new reference state to evaluate exergy consumed and produced. Instead of the reference point e = 0 (the intersection of the isobar P0 and the isotherm T0), the new reference point is presented by etr: the intersection of the isobar P2 and the isotherm T2 for the case 3a; of the isobar P2 and the isotherm T1 for the case 3b; and of the isobar P2 and the isotherm T0 for the case 3c. Finally, the transiting exergy approach provides the foundation for the non-ambiguous definition of the terms ΔE and \_ ∇E, and thus of \_ ηtr.

#### Example 1

The initial parameters of air at the inlet of a throttling valve are: m = 1 kg/s, P \_ <sup>1</sup> = 3 MPa, T1 = 140 K. The ambient temperature T0 = 283 K. Calculate the variation of E\_ tr, ΔE and \_ ∇E and \_ ηtr as a function of the outlet pressure P2 in the range 0.1–1 MPa.

#### Solution

The outlet temperature of the air is calculated by using the software Engineering Equation Solver (EES) [11]. Given that the expansion of air takes place at sub-ambient conditions, Eqs. (13)–(15) are used to evaluate E\_ tr, ΔE and \_ ∇E. The results are presented in \_ Table 1. One observation is obvious, that the rise in exergy losses as well as the decrease in ηtr go along with Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature http://dx.doi.org/10.5772/intechopen.74041 69


Table 1. Variation in temperature and exergy metrics with the outlet pressure for a throttling process of air.

the decreasing outlet pressure P2. Less obvious is that the exergy produced ΔE rises with \_ decreasing P2, reflecting the production of a more important cooling effect. It can also be noticed that the transiting exergy E\_ tr decreases with decreasing outlet pressure P2.

#### Example 2

at constant temperature Tin. Finally, the case presented in Figure 3c illustrates a throttling process started above ambient and ended at sub-ambient conditions. According to Eq. (9c):

Again, the input-output efficiency is not suitable for the evaluation of the processes presented in Figure 3b and c, given that the outlet exergy e2 = e(Pout, Tout) is negative. Another difficulty is linked to the application of the products-fuel efficiency for these cases. The exergy transfer of the throttling process at sub-ambient conditions consists in the partial transformation of mechanical exergy ("fuel") into thermal exergy ("product"). The problem stems from the fact that there are multiple possibilities to define "fuel" and "product" in this case; as a result multiple values of ηpr-f may be formulated, leading to the ambiguity in the products-fuel efficiency application. Indeed, the different increments of thermal exergy may be considered as a "product" for the case in Figure 3b, for example, the increase in thermal exergy following the isobar P1 or the isobar P2. In the same way, different decrements of mechanical exergy may be considered as a "fuel" in the same figure, for example, the decrease of mechanical exergy

Contrary to "products-fuel", the transiting exergy approach does not attempt to individually compute the thermal and mechanical exergy component variations. It relies, rather, on the

As illustrated in Figure 3a and c, the transiting exergy may be considered as the introduction of a new reference state to evaluate exergy consumed and produced. Instead of the reference point e = 0 (the intersection of the isobar P0 and the isotherm T0), the new reference point is presented by etr: the intersection of the isobar P2 and the isotherm T2 for the case 3a; of the isobar P2 and the isotherm T1 for the case 3b; and of the isobar P2 and the isotherm T0 for the case 3c. Finally, the transiting exergy approach provides the foundation for the non-ambiguous

The initial parameters of air at the inlet of a throttling valve are: m = 1 kg/s, P \_ <sup>1</sup> = 3 MPa,

The outlet temperature of the air is calculated by using the software Engineering Equation Solver (EES) [11]. Given that the expansion of air takes place at sub-ambient conditions,

observation is obvious, that the rise in exergy losses as well as the decrease in ηtr go along with

tr, ΔE and \_ ∇E. The results are presented in \_ Table 1. One

tr, ΔE and \_ ∇E and \_

unaffected part of the thermo-mechanical exergy entering and leaving the system.

\_ ηtr.

T1 = 140 K. The ambient temperature T0 = 283 K. Calculate the variation of E\_

ηtr as a function of the outlet pressure P2 in the range 0.1–1 MPa.

<sup>Δ</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � ð Þ¼ eout � etr <sup>m</sup>\_ � e Pout ð Þ� ; Tout e Pout <sup>½</sup> ð Þ ; T0 � ¼ <sup>m</sup>\_ � ð Þ <sup>Δ</sup>eT Pout (17)

<sup>∇</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � ð Þ¼ ein � etr <sup>m</sup>\_ � e Pð Þ� in; Tin e Pout <sup>½</sup> ð Þ ; T0 � ¼ <sup>m</sup>\_ � ð Þ <sup>∇</sup>eP,<sup>T</sup> (18)

The values ΔE and \_ ∇E are calculated as: \_

68 Energy Systems and Environment

following the isotherms T1 or T2.

Example 1

Solution

definition of the terms ΔE and \_ ∇E, and thus of

Eqs. (13)–(15) are used to evaluate E\_

etr ¼ e Pout ð Þ ; T0 (16)

The expansion valve of a refrigeration mechanical vapor compression cycle is supplied with the subcooled working fluid R152a at the rate m = 0.15 kg/s at P \_ <sup>1</sup> = 615.1 kPa. The fluid is expanded to a pressure of P2 = 142.9 kPa. The ambient temperature T0 = 278 K. Calculate the variation of E\_ tr, ΔE and \_ ∇E and \_ ηtr as a function of the subcooling ΔTsubC in the range 275– 281 K.

#### Solution

A vapor compression cycle is presented on a Ts-diagram in Figure 4. The subcooling process is represented by the line 3f-3. Given that the expansion of R152a takes place across ambient temperature, Eqs. (16)–(18) are used to evaluate E\_ tr, ΔE and \_ ∇E. The results are shown in \_ Table 2.

The transiting exergy does not change with the subcooling, because it is the function of constant parameters T0 and P2, meanwhile the exergy produced increases and exergy consumed decreases. The new result is that ηtr is rising with the subcooling. It should be mentioned that increasing the amount of subcooling is well documented as a way to increase the COP (coefficient of performance) of vapor compression cycles [4]. Thus, the rise in ηtr of an expansion device guarantees the COP improvement of the overall cycle, a conclusion that may lead to practical recommendations for optimization of refrigeration cycles.

#### 3.2. Expansion in low temperature systems with work production and heat transfer

The primary purpose of expansion processes in the sub-ambient region is the production of cooling effect. The power that may be produced can be considered as a useful by-product. This type of expansion takes place in cryo-expanders. There are two types of these devices: adiabatic and non-adiabatic gas expansion machines. The energy and exergy balances around a non-adiabatic expander are presented in Figure 5. It should be emphasized that the directions

The process of gas expansion in a non-adiabatic cryo-expander is presented on an e-h diagram (see Figure 6). Similar to the case of adiabatic throttling (Figure 3b), the transiting exergy in the gas flow is defined according to Eq. (9b). As a result, the exergy efficiency is calculated as:

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature

<sup>η</sup>tr <sup>¼</sup> <sup>m</sup>\_ � ð Þ <sup>Δ</sup>eT Pout <sup>þ</sup> <sup>Q</sup>\_ � <sup>Θ</sup> <sup>þ</sup> <sup>W</sup>\_ <sup>T</sup> m\_ � ð Þ ∇eP Tin

In the case of an adiabatic cryo-expander ηtr is calculated according to Eq. (19), but with the

An adiabatic turbine (η<sup>T</sup> = 0.80) is supplied with air at the rate m = 1 kg/s at P \_ <sup>1</sup> = 6 MPa,

Given that the expansion of air takes place across ambient temperature, Eqs. (16)–(18) are used

reduction, and as a result ΔE and \_ ∇E and \_ W rise, but the increase in \_ ΔE and \_ W is offset by the \_

to the right in Table 3, is because the stream at state 2 is under vacuum conditions. It should be

tr, ΔE and \_ ∇E. \_ ηtr is calculated according to Eq. (19), but with the term Q\_ :Θin

T1 = 320 K. The ambient temperature T0 = 283 K. Calculate the variation of E\_

equals to zero. The results are shown in Table 3. It is illustrated that E\_

greater increase in ∇E causing \_ ηtr to decrease. The negative value of E\_

Figure 6. Gas expansion in a non-adiabatic cryo-expander on an exergy-enthalpy diagram.

ηtr as a function of the outlet pressure P2 in the range 0.1–3 MPa.

term Q\_ :Θin equals to zero.

Example 3

Solution

to evaluate E\_

(19)

71

tr, ΔE and \_ ∇E and \_

http://dx.doi.org/10.5772/intechopen.74041

tr decreases with P2

tr in the second last row

Figure 4. A vapor compression cycle presentation on a Ts-diagram.


Table 2. Variation in exergy metrics with subcooling for the expansion process of R152a.

Figure 5. Energy (a) and exergy (b) balances around a non-adiabatic cryo-expander.

of heat flow Q and exergy \_ E\_ <sup>Q</sup> are opposite. This is due to the fact that heat transfer from a cooled object to the expanding fluid occurs at T < T0. As a result E\_ <sup>Q</sup> calculated according to Eq. (1) is presented as the outlet flow in Figure 5b.

The process of gas expansion in a non-adiabatic cryo-expander is presented on an e-h diagram (see Figure 6). Similar to the case of adiabatic throttling (Figure 3b), the transiting exergy in the gas flow is defined according to Eq. (9b). As a result, the exergy efficiency is calculated as:

$$\eta\_{\rm tr} = \frac{\dot{\mathbf{m}} \cdot (\Delta \mathbf{e}\_{\rm T})\_{\rm P\_{\rm out}} + \dot{\mathbf{Q}} \cdot \Theta\_{\rm } + \dot{\mathbf{W}}\_{\rm T}}{\dot{\mathbf{m}} \cdot (\nabla \mathbf{e}\_{\rm P})\_{\rm T\_{\rm in}}} \tag{19}$$

In the case of an adiabatic cryo-expander ηtr is calculated according to Eq. (19), but with the term Q\_ :Θin equals to zero.

#### Example 3

An adiabatic turbine (η<sup>T</sup> = 0.80) is supplied with air at the rate m = 1 kg/s at P \_ <sup>1</sup> = 6 MPa, T1 = 320 K. The ambient temperature T0 = 283 K. Calculate the variation of E\_ tr, ΔE and \_ ∇E and \_ ηtr as a function of the outlet pressure P2 in the range 0.1–3 MPa.

#### Solution

of heat flow Q and exergy \_ E\_ <sup>Q</sup> are opposite. This is due to the fact that heat transfer from a cooled object to the expanding fluid occurs at T < T0. As a result E\_ <sup>Q</sup> calculated according to

Eq. (1) is presented as the outlet flow in Figure 5b.

Figure 4. A vapor compression cycle presentation on a Ts-diagram.

ΔE\_ (kW)

Table 2. Variation in exergy metrics with subcooling for the expansion process of R152a.

Figure 5. Energy (a) and exergy (b) balances around a non-adiabatic cryo-expander.

 4.083 3.208 0.875 1.744 78.6 4.066 3.230 0.836 1.744 79.4 4.034 3.274 0.760 1.744 81.2 4.020 3.295 0.725 1.744 82.0 3.994 3.339 0.655 1.744 83.6

D\_ (kW) E\_ tr (kW) ηtr (%)

∇E\_ (kW)

70 Energy Systems and Environment

ΔTsubC (K)

Given that the expansion of air takes place across ambient temperature, Eqs. (16)–(18) are used to evaluate E\_ tr, ΔE and \_ ∇E. \_ ηtr is calculated according to Eq. (19), but with the term Q\_ :Θin equals to zero. The results are shown in Table 3. It is illustrated that E\_ tr decreases with P2 reduction, and as a result ΔE and \_ ∇E and \_ W rise, but the increase in \_ ΔE and \_ W is offset by the \_ greater increase in ∇E causing \_ ηtr to decrease. The negative value of E\_ tr in the second last row to the right in Table 3, is because the stream at state 2 is under vacuum conditions. It should be

Figure 6. Gas expansion in a non-adiabatic cryo-expander on an exergy-enthalpy diagram.


Table 3. Variation in exergy metrics with the outlet pressure for a turbine expansion process.

emphasized that the increase in the ΔE metric reflects the deeper refrigeration of air with \_ increasing pressure drop in the turbine.

#### 3.3. Expansion in a vortex tube

Figure 7a illustrates a counter flow vortex tube [12]. High pressure gas enters the tube through a tangential nozzle (point 1). Colder low-pressure gas leaves via an orifice near the centerline adjacent to the plane of the nozzle (point 2), and warmer low-pressure gas leaves near the periphery at the end of the tube opposite to the nozzle (point 3). The vortex tube requires no work or heat interaction with the surroundings to operate. The cold mass fraction is μ; the hot gas mass fraction is (1 � μ). The exergy balance around the vortex tube is:

$$\mathbf{e}\_1 = \; \mu \cdot \mathbf{e}\_2 + (1 - \mu) \cdot \mathbf{e}\_3 + \mathbf{d} \tag{20}$$

The expansion processes taking place within a vortex tube are presented on an e-h diagram (Figure 7b). The cold stream expands across T0, the hot expands at T > T0. By applying Eqs. (9a) and (9c) the transiting exergies may be determined for each mass stream, cold (1–2) and hot (1–3).

$$(\mathbf{e}\_{\rm tr})\_{\rm cold} = \; \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_0) \tag{21}$$

Thus ΔE\_

<sup>C</sup> and ΔE\_

<sup>C</sup> + ΔE\_

<sup>H</sup><sup>Þ</sup> / (∇E\_

<sup>C</sup> + ∇E\_

for each stream. ∇E\_

The ratio (ΔE\_

the variation of E\_

Example 4

0.2–0.9.

<sup>H</sup> represent the increase in the thermal exergy component due to the cooling

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73

<sup>C</sup> represents the decrease in thermo-mechanical exergy of the cold stream

tr, ΔE and \_ ∇E and \_ ηtr as a function of the cold mass fraction μ in the range

<sup>H</sup> is the

of the cold stream and the heating for the hot stream, under conditions of the outlet pressures

due to the partial thermal exergy destruction because of the temperature drop from T1 to T0,

decrease of mechanical exergy of the hot stream at conditions of constant inlet temperature.

H) gives the value of ηtr.

An adiabatic vortex tube is supplied with air as ideal gas at the rate m = 1 kg/s and \_ at P1 = 0.8 MPa, T1 = 308 K. The air expands at the cold end to pressure P2 = 0.1 MPa and at the hot end to the pressure P3 = 0.15 MPa. The ambient temperature T0 = 298 K. Calculate

and the decrease of mechanical exergy because of pressure drop from P1 to P3. ∇E\_

Figure 7. A vortex tube (a) and the presentation of the expansion process on an exergy-enthalpy diagram (b).

$$(\mathbf{e}\_{\rm tr})\_{\rm hot} = \; \mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_1) \tag{22}$$

As a result, the exergy produced and consumed within the cold and hot streams are:

$$\begin{aligned} \Delta \dot{\mathbf{E}}\_{\complement} &= \dot{\mathbf{m}} \cdot [(\mu) \cdot (\mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_2) - \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_0))] = \dot{\mathbf{m}} \cdot (\mu) \cdot (\Delta \mathbf{e}\_{\Gamma})\_{\text{P}\_2} \\ \Delta \dot{\mathbf{E}}\_{\text{H}} &= \dot{\mathbf{m}} \cdot [(1 - \mu) \cdot (\mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_3) - \mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_1))] = \dot{\mathbf{m}} \cdot (1 - \mu) \cdot (\Delta \mathbf{e}\_{\Gamma})\_{\text{P}\_3} \end{aligned} \tag{23}$$
 
$$\begin{aligned} \nabla \dot{\mathbf{E}}\_{\complement} &= \dot{\mathbf{m}} \cdot [(\mu) \cdot (\mathbf{e}(\mathbf{P}\_1, \mathbf{T}\_1) - \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_0))] = \dot{\mathbf{m}} \cdot (\mu) \cdot (\nabla \mathbf{e}\_{\Gamma, \text{T}}) \\ \nabla \dot{\mathbf{E}}\_{\text{H}} &= \dot{\mathbf{m}} \cdot [(1 - \mu) \cdot (\mathbf{e}(\mathbf{P}\_1, \mathbf{T}\_1) - \mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_1))] = \dot{\mathbf{m}} \cdot (1 - \mu) \cdot (\nabla \mathbf{e}\_{\Gamma})\_{\text{T}\_1} \end{aligned} \tag{24}$$

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature http://dx.doi.org/10.5772/intechopen.74041 73

Figure 7. A vortex tube (a) and the presentation of the expansion process on an exergy-enthalpy diagram (b).

Thus ΔE\_ <sup>C</sup> and ΔE\_ <sup>H</sup> represent the increase in the thermal exergy component due to the cooling of the cold stream and the heating for the hot stream, under conditions of the outlet pressures for each stream. ∇E\_ <sup>C</sup> represents the decrease in thermo-mechanical exergy of the cold stream due to the partial thermal exergy destruction because of the temperature drop from T1 to T0, and the decrease of mechanical exergy because of pressure drop from P1 to P3. ∇E\_ <sup>H</sup> is the decrease of mechanical exergy of the hot stream at conditions of constant inlet temperature. The ratio (ΔE\_ <sup>C</sup> + ΔE\_ <sup>H</sup><sup>Þ</sup> / (∇E\_ <sup>C</sup> + ∇E\_ H) gives the value of ηtr.

#### Example 4

emphasized that the increase in the ΔE metric reflects the deeper refrigeration of air with \_

Figure 7a illustrates a counter flow vortex tube [12]. High pressure gas enters the tube through a tangential nozzle (point 1). Colder low-pressure gas leaves via an orifice near the centerline adjacent to the plane of the nozzle (point 2), and warmer low-pressure gas leaves near the periphery at the end of the tube opposite to the nozzle (point 3). The vortex tube requires no work or heat interaction with the surroundings to operate. The cold mass fraction is μ; the hot

The expansion processes taking place within a vortex tube are presented on an e-h diagram (Figure 7b). The cold stream expands across T0, the hot expands at T > T0. By applying Eqs. (9a) and (9c) the transiting exergies may be determined for each mass stream, cold (1–2) and hot (1–3).

As a result, the exergy produced and consumed within the cold and hot streams are:

<sup>C</sup> ¼ m\_ � ½ð Þ� μ ð Þ e Pð Þ� <sup>2</sup>; T2 eðP2; T0Þ � ¼ m\_ � ð Þ� μ ð Þ ΔeT P2

<sup>C</sup> ¼ m\_ � ½ð Þ� μ ð Þ e Pð Þ� <sup>1</sup>; T1 eðP2; T0Þ � ¼ m\_ � ð Þ� μ ð Þ ∇eP,<sup>T</sup>

<sup>H</sup> ¼ m\_ � ½ð Þ� 1 � μ ð Þ e Pð Þ� <sup>3</sup>; T3 eðP3; T1Þ � ¼ m\_ � ð Þ� 1 � μ ð Þ ΔeT P3

<sup>H</sup> ¼ m\_ � ½ð Þ� 1 � μ ð Þ e Pð Þ� <sup>1</sup>; T1 eðP3; T1Þ � ¼ m\_ � ð Þ� 1 � μ ð Þ ∇eP T1

e1 ¼ μ � e2 þ ð Þ� 1 � μ e3 þ d (20)

ð Þ etr cold ¼ e Pð Þ <sup>2</sup>; T0 (21)

ð Þ etr hot ¼ e Pð Þ <sup>3</sup>; T1 (22)

(23)

(24)

gas mass fraction is (1 � μ). The exergy balance around the vortex tube is:

Table 3. Variation in exergy metrics with the outlet pressure for a turbine expansion process.

increasing pressure drop in the turbine.

3.3. Expansion in a vortex tube

P2 (MPa) T2 (K)

72 Energy Systems and Environment

∇E\_ (kW) ΔE\_ (kW) W\_ <sup>T</sup> (kW)

3.00 271.8 57.9 0.2 45.6 12.1 274.2 79.1 2.55 263.8 68.3 0.7 53.0 14.6 263.9 78.7 2.50 255.1 80.1 1.5 61.2 17.4 252.0 78.2 1.85 245.4 94.1 2.9 70.3 21.0 238.0 77.7 1.50 234.3 111.1 4.9 80.7 25.5 221.1 77.1 1.15 221.2 132.6 8.2 93.1 31.3 199.6 76.4 0.50 204.9 162.0 13.6 108.4 40.0 170.1 75.3 0.45 182.4 208.9 24.1 129.7 55.1 123.2 73.6 0.10 138.9 333.2 57.7 171.6 103.9 �1.05 68.8

D\_ (kW) E\_ tr (kW) ηtr (kW)

ΔE\_

ΔE\_

∇E\_

∇E\_

An adiabatic vortex tube is supplied with air as ideal gas at the rate m = 1 kg/s and \_ at P1 = 0.8 MPa, T1 = 308 K. The air expands at the cold end to pressure P2 = 0.1 MPa and at the hot end to the pressure P3 = 0.15 MPa. The ambient temperature T0 = 298 K. Calculate the variation of E\_ tr, ΔE and \_ ∇E and \_ ηtr as a function of the cold mass fraction μ in the range 0.2–0.9.

#### Solution

The results are shown in Table 4. It is illustrated that E\_ tr, H decreases with the cold mass fraction increase. The E\_ tr,<sup>C</sup> is zero, because it is defined by ambient conditions P2 = P0 and T2 = T0. As a result, for the cold stream ΔE\_ <sup>C</sup> decreases but ∇E\_ <sup>C</sup> increases strongly with μ. An opposite effect is observed for the hot stream, where ΔE\_ <sup>H</sup> increases and ∇E\_ <sup>H</sup> decreases. As a result, ηtr increases, despite the rise in the exergy losses with the increasing cold mass fraction. This can be explained by the fact that the rise in exergy produced in the hot stream surpasses the increase in exergy losses. The exergy efficiency of the vortex tube is relatively low.

#### 3.4. Compression across ambient temperature

In most refrigeration plants and heat pumps compression starts at T < T0 and ends at T > T0. The process is presented on an e-h diagram (see Figure 8). According to Eq. (9c), transiting exergy is:

$$\mathbf{e}\_{\rm tr} = \;\mathbf{e}(\mathbf{P}\_1, \mathbf{T}\_0) \tag{25}$$

Solution

η<sup>C</sup> (–)

W\_ <sup>C</sup> (kW)

result ηtr increases.

3.5. Compression and expansion in a one phase ejector

The results are shown in Table 5. The transiting exergy does not change with the isentropic efficiency ηC, because is a function of constant parameters T0 and P1. The exergy consumed does not change either. The produced exergy decreases. This drop in ΔE is explained by the reduction \_ in "compression reheat". The decrease in ΔE is offset by the greater decrease in \_ W\_ C, and as a

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A combination of the processes of vapor expansion and compression takes place within a onephase ejector presented in Figure 9a. A primary (pr) stream at high pressure P1 and temperature

Figure 8. Compression process across ambient temperature (T0) on an exergy-enthalpy diagram.

Table 5. Variation in exergy metrics with the isentropic efficiency for a compression process.

ΔE\_ (kW)

0.75 66.0 2.3 53.3 15.0 12.6 78.0 0.80 61.9 2.3 52.9 11.3 12.6 82.4 0.85 58.3 2.3 52.6 8.0 12.6 86.8 0.90 55.0 2.3 52.3 5.0 12.6 91.3

D\_ (kW) E\_ tr (kW) ηtr (kW)

∇E\_ (kW)

The produced and consumed exergies are:

$$
\Delta \dot{\mathbf{E}} = \dot{\mathbf{m}} \cdot \left[ \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_2) - \mathbf{e}(\mathbf{P}\_1, \mathbf{T}\_0) \right] = \dot{\mathbf{m}} \cdot (\Delta \mathbf{e}\_{\mathbf{P}, \mathbf{T}}) \tag{26}
$$

$$\left(\nabla \dot{\mathbf{E}} + \dot{\mathbf{W}}\_{\rm C}\right) = \dot{\mathbf{m}} \cdot \left[\mathbf{e}(\mathbf{P}\_{1}, \mathbf{T}\_{1}) - \mathbf{e}(\mathbf{P}\_{1}, \mathbf{T}\_{0})\right] + \dot{\mathbf{W}}\_{\rm C} = \dot{\mathbf{m}} \cdot (\nabla \mathbf{e}\_{\Gamma})\_{\rm P\_{1}} + \dot{\mathbf{W}}\_{\rm C} \tag{27}$$

ΔE represents the increase of thermo-mechanical exergy due to the rise in pressure from P to P \_ <sup>2</sup> and the rise in temperature from T0 to T2. ∇E represents the destruction of thermal exergy due \_ to the rise in temperature from T1 to T0 under conditions of constant pressure P1, plus the consumed power <sup>W</sup>\_ C. The ratio <sup>Δ</sup>E/( \_ <sup>∇</sup>E\_ <sup>þ</sup> <sup>W</sup>\_ <sup>C</sup><sup>Þ</sup> gives the value of exergy efficiency <sup>η</sup>tr.

#### Example 5

An adiabatic compressor of a refrigeration plant is supplied with the working fluid R152a at the rate m = 1 kg/s at P \_ <sup>1</sup> = 142.9 kPa and T1 = 263 K (superheated). The fluid is compressed to a pressure of P2 = 615.1 kPa. The ambient temperature T0 = 298 K. Calculate the variation of E\_ tr, ΔE and \_ ∇E and \_ ηtr as a function of the isentropic efficiency η<sup>C</sup> in the range 0.75–0.90.


Table 4. Variation in exergy metrics with the cold mass fraction for a vortex tube expansion process.

#### Solution

Solution

exergy is:

Example 5

T2 (K)

μ (–)

fraction increase. The E\_

74 Energy Systems and Environment

The results are shown in Table 4. It is illustrated that E\_

opposite effect is observed for the hot stream, where ΔE\_

T2 = T0. As a result, for the cold stream ΔE\_

3.4. Compression across ambient temperature

The produced and consumed exergies are:

<sup>∇</sup>E\_ <sup>þ</sup> <sup>W</sup>\_ <sup>C</sup> 

consumed power W\_ C. The ratio ΔE/(

T3 (K) ∇E\_ <sup>C</sup> (kW) tr, H decreases with the cold mass

<sup>C</sup> increases strongly with μ. An

<sup>H</sup> decreases. As a

tr,

ηtr (%)

tr,<sup>C</sup> is zero, because it is defined by ambient conditions P2 = P0 and

<sup>H</sup> increases and ∇E\_

etr ¼ e Pð Þ <sup>1</sup>; T0 (25)

<sup>Δ</sup>E\_ <sup>¼</sup> <sup>m</sup>\_ � <sup>½</sup>e Pð Þ� <sup>2</sup>; T2 e Pð Þ <sup>1</sup>; T0 � ¼ <sup>m</sup>\_ � ð Þ <sup>Δ</sup>eP,<sup>T</sup> (26)

<sup>¼</sup> <sup>m</sup>\_ � <sup>½</sup>e Pð Þ� <sup>1</sup>; T1 e Pð Þ <sup>1</sup>; T0 � þ <sup>W</sup>\_ <sup>C</sup> <sup>¼</sup> <sup>m</sup>\_ � ð Þ <sup>∇</sup>eT P1 <sup>þ</sup> <sup>W</sup>\_ <sup>C</sup> (27)

\_ <sup>∇</sup>E\_ <sup>þ</sup> <sup>W</sup>\_ <sup>C</sup><sup>Þ</sup> gives the value of exergy efficiency <sup>η</sup>tr.

ΔE\_ <sup>H</sup> (kW) E\_ tr,<sup>H</sup> (kW) D\_ (kW)

<sup>C</sup> decreases but ∇E\_

result, ηtr increases, despite the rise in the exergy losses with the increasing cold mass fraction. This can be explained by the fact that the rise in exergy produced in the hot stream surpasses

In most refrigeration plants and heat pumps compression starts at T < T0 and ends at T > T0. The process is presented on an e-h diagram (see Figure 8). According to Eq. (9c), transiting

ΔE represents the increase of thermo-mechanical exergy due to the rise in pressure from P to P \_ <sup>2</sup> and the rise in temperature from T0 to T2. ∇E represents the destruction of thermal exergy due \_ to the rise in temperature from T1 to T0 under conditions of constant pressure P1, plus the

An adiabatic compressor of a refrigeration plant is supplied with the working fluid R152a at the rate m = 1 kg/s at P \_ <sup>1</sup> = 142.9 kPa and T1 = 263 K (superheated). The fluid is compressed to a pressure of P2 = 615.1 kPa. The ambient temperature T0 = 298 K. Calculate the variation of E\_

ΔE and \_ ∇E and \_ ηtr as a function of the isentropic efficiency η<sup>C</sup> in the range 0.75–0.90.

E\_ tr,<sup>C</sup> (kW)

0.35 268.0 329.5 61.9 0.6 0.0 93.1 0.9 21.9 153.5 0.97 0.60 273.0 360.4 106.2 0.7 0.0 57.3 2.3 13.5 160.5 1.83 0.75 278.0 397.6 132.7 0.5 0.0 35.8 3.4 8.4 164.6 2.32

∇E\_ H (kW)

ΔE\_ <sup>C</sup> (kW)

Table 4. Variation in exergy metrics with the cold mass fraction for a vortex tube expansion process.

the increase in exergy losses. The exergy efficiency of the vortex tube is relatively low.

The results are shown in Table 5. The transiting exergy does not change with the isentropic efficiency ηC, because is a function of constant parameters T0 and P1. The exergy consumed does not change either. The produced exergy decreases. This drop in ΔE is explained by the reduction \_ in "compression reheat". The decrease in ΔE is offset by the greater decrease in \_ W\_ C, and as a result ηtr increases.

#### 3.5. Compression and expansion in a one phase ejector

A combination of the processes of vapor expansion and compression takes place within a onephase ejector presented in Figure 9a. A primary (pr) stream at high pressure P1 and temperature

Figure 8. Compression process across ambient temperature (T0) on an exergy-enthalpy diagram.


Table 5. Variation in exergy metrics with the isentropic efficiency for a compression process.

T1 expands and entrains a secondary stream (s) at low pressure P2 and temperature T2 < T0. The ratio m\_ s/m\_ pr gives the value of the entrainment ratio (ω) of the ejector. The mixed stream with the parameters P2 < P3 < P1 and T2 < T3 < T1 leaves the ejector. The exergy balance around the ejector is:

$$\left(\frac{1}{1+\omega}\right)\cdot\mathbf{e}\_1 + \left(\frac{\omega}{1+\omega}\right)\cdot\mathbf{e}\_2 = \mathbf{e}\_3 + \mathbf{d}\tag{28}$$

The processes of expansion of the primary stream and compression of the secondary stream are presented on an e-h diagram (Figure 9b). The secondary stream is compressed across T0, meaning that Eq. (9c) is applied to calculate (etr)s. As a result, the transiting exergy for secondary and primary streams are:

$$(\mathbf{e}\_{\rm tr})\_s = \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_0) \tag{29}$$

$$\mathbf{e}(\mathbf{e}\_{\rm tr})\_{\rm pr} = \mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_3) \tag{30}$$

4. Environmental life cycle analysis and exergy efficiency of cooling

Life Cycle Analysis (LCA) is an important tool to analyze environmental problems associated with the production, use, and disposal of products or systems [14]. For every product produced within a system the total inflow and outflow of energy and materials are evaluated. The

systems

ω (–)

diagram.

∇E\_ pr (kW) ∇E\_ <sup>s</sup> (kW) ΔE\_ <sup>s</sup> (kW) D\_ (kW) E\_ tr,pr (kW)

Figure 9. One phase ejector (a) and presentation of expansion and the compression processes (b) on an exergy-enthalpy

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature

0.15 9.99 0.017 1.189 8.818 4.286 �0.546 3.740 11.9 1000 418 0.17 10.07 0.019 1.334 8.755 4.205 �0.619 3.586 13.2 1000 418 0.20 10.19 0.023 1.547 8.666 4.090 �0.729 3.361 15.2 1000 418 0.23 10.29 0.026 1.754 8.562 3.983 �0.838 3.145 17.0 1000 418 0.25 10.36 0.028 1.890 8.498 3.915 �0.911 3.004 18.2 1000 418

Table 6. Variation in exergy metrics with the entrainment factor for compression-expansion processes in an ejector.

E\_ tr, <sup>s</sup> (kW) E\_ tr (kW) ηtr (%)

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P1 (kPa) T1 (K)

This means that the exergies produced and consumed may be computed as:

$$
\Delta \dot{\mathbf{E}}\_s = \dot{\mathbf{m}}\_s \cdot \left[ \mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_3) - \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_0) \right] = \dot{\mathbf{m}}\_s \cdot (\Delta \mathbf{e}\_{\mathbf{P}, \mathbf{T}})\_s \tag{31}
$$

$$\begin{aligned} \nabla \dot{\mathbf{E}}\_{\rm s} &= \dot{\mathbf{m}}\_{\rm s} \cdot \left[ \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_2) - \mathbf{e}(\mathbf{P}\_2, \mathbf{T}\_0) \right] \\ \nabla \dot{\mathbf{E}}\_{\rm pr} &= \dot{\mathbf{m}}\_{\rm pr} \cdot \left[ \mathbf{e}(\mathbf{P}\_1, \mathbf{T}\_1) - \mathbf{e}(\mathbf{P}\_3, \mathbf{T}\_3) \right] = \dot{\mathbf{m}}\_{\rm pr} \cdot \left( \nabla \mathbf{e}, \mathbf{r} \right)\_{\rm pr} \end{aligned} \tag{32}$$

ΔE\_ <sup>s</sup> is the increase of thermo-mechanical exergy of the secondary stream due to the compression from P2 to P3 and the rise in temperature from T0 to T3. The exergy consumed ∇E\_ <sup>s</sup> within the secondary stream represents the decrease of thermal exergy component because of the temperature rise from T2 to T0 (the partial cold destruction). The exergy consumed within the primary stream ∇E\_ pr is the decrease of thermo-mechanical exergy The ratio ΔE\_ s/(∇E\_ <sup>s</sup> + ∇E\_ pr) gives the value of exergy efficiency ηtr. The detailed analysis of efficiencies for different parts of an ejector is given in [13].

#### Example 6

An ejector of a refrigeration plant is supplied with the working fluid R141b. The parameters of the secondary stream are: P2 = 22.3 kPa, T2 = 268 K. The pressure of the mixed stream is P3 = 91 kPa. The ambient temperature T0 = 289 K. Calculate the variation of (E\_ trÞpr, (E\_ trÞs, <sup>Δ</sup>E\_ and ∇E and \_ ηtr as a function of the entrainment ratio ω = m\_ s/m\_ pr in the range 0.15–0.25.

#### Solution

The calculation results are shown in Table 6. The transiting exergy in the secondary flow is negative because the parameters P2 and T0 define the state of the flow under vacuum conditions. The exergy produced and exergy consumed increase with the entrainment factor. The increase in ΔE\_ <sup>s</sup> offsets the increase in (∇E\_ <sup>s</sup> + ∇E\_ pr), and as a result ηtr increases.

Exergy Flows Inside Expansion and Compression Devices Operating below and across Ambient Temperature http://dx.doi.org/10.5772/intechopen.74041 77

T1 expands and entrains a secondary stream (s) at low pressure P2 and temperature T2 < T0. The ratio m\_ s/m\_ pr gives the value of the entrainment ratio (ω) of the ejector. The mixed stream with the parameters P2 < P3 < P1 and T2 < T3 < T1 leaves the ejector. The exergy balance around the

> ω 1 þ ω

The processes of expansion of the primary stream and compression of the secondary stream are presented on an e-h diagram (Figure 9b). The secondary stream is compressed across T0, meaning that Eq. (9c) is applied to calculate (etr)s. As a result, the transiting exergy for secondary

<sup>s</sup> ¼ m\_ <sup>s</sup> � ½e Pð Þ� <sup>2</sup>; T2 eðP2; T0Þ� ¼ m\_ <sup>s</sup> � ð Þ ∇eT P2

pr ¼ m\_ pr � ½e Pð Þ� <sup>1</sup>; T1 eðP3; T3Þ� ¼ m\_ pr � ð Þ ∇eP,<sup>T</sup> pr

<sup>s</sup> is the increase of thermo-mechanical exergy of the secondary stream due to the compres-

pr is the decrease of thermo-mechanical exergy The ratio ΔE\_

the secondary stream represents the decrease of thermal exergy component because of the temperature rise from T2 to T0 (the partial cold destruction). The exergy consumed within the

gives the value of exergy efficiency ηtr. The detailed analysis of efficiencies for different parts of

An ejector of a refrigeration plant is supplied with the working fluid R141b. The parameters of the secondary stream are: P2 = 22.3 kPa, T2 = 268 K. The pressure of the mixed stream is

The calculation results are shown in Table 6. The transiting exergy in the secondary flow is negative because the parameters P2 and T0 define the state of the flow under vacuum conditions. The exergy produced and exergy consumed increase with the entrainment factor. The

<sup>s</sup> + ∇E\_

P3 = 91 kPa. The ambient temperature T0 = 289 K. Calculate the variation of (E\_

<sup>s</sup> offsets the increase in (∇E\_

and ∇E and \_ ηtr as a function of the entrainment ratio ω = m\_ s/m\_ pr in the range 0.15–0.25.

sion from P2 to P3 and the rise in temperature from T0 to T3. The exergy consumed ∇E\_

� e2 ¼ e3 þ d (28)

ð Þ etr <sup>s</sup> ¼ e Pð Þ <sup>2</sup>; T0 (29)

ð Þ etr pr ¼ e Pð Þ <sup>3</sup>; T3 (30)

pr), and as a result ηtr increases.

s

(32)

<sup>s</sup> within

<sup>s</sup> + ∇E\_ pr)

trÞs, <sup>Δ</sup>E\_

s/(∇E\_

trÞpr, (E\_

<sup>s</sup> ¼ m\_ <sup>s</sup> � ½e Pð Þ� <sup>3</sup>; T3 e Pð Þ <sup>2</sup>; T0 � ¼ m\_ <sup>s</sup> � ð Þ ΔeP,<sup>T</sup> <sup>s</sup> (31)

1 1 þ ω 

� e1 þ

This means that the exergies produced and consumed may be computed as:

ΔE\_

∇E\_

∇E\_

ejector is:

ΔE\_

primary stream ∇E\_

Example 6

Solution

increase in ΔE\_

an ejector is given in [13].

and primary streams are:

76 Energy Systems and Environment

Figure 9. One phase ejector (a) and presentation of expansion and the compression processes (b) on an exergy-enthalpy diagram.


Table 6. Variation in exergy metrics with the entrainment factor for compression-expansion processes in an ejector.
