Isentropic Compressor Efficiency With an Ideal Gas Entering Continuously

Energy and Exergy Analyses of a Geothermal-Based Integrated System for Trigeneration

Osamah Siddiqui , Ibrahim Dincer , in Exergetic, Energetic and Environmental Dimensions, 2018

4.3 Effect of Turbine Isentropic Efficiencies on Overall System Performance

The isentropic efficiency of the turbines affects the overall system performance. Fig. 5 shows the effect of the isentropic efficiencies of turbine 1 and turbine 2 on the overall system energy and exergy efficiency. The energy efficiency of the overall system changes from 26.6% to 34.1% as the isentropic efficiencies of the turbines varies from 50% to 90%. In addition, the exergy efficiency of the overall system increases from 24.2% to 39.9% as the isentropic efficiencies of the turbines increases from 50% to 90%. As the isentropic efficiency of the turbine increases, the irreversibilities decrease. Hence, a turbine with a higher isentropic efficiency is capable of producing more power. This leads to an increase in the overall system efficiencies.

Figure 5. Effect of turbine isentropic efficiency on the overall system energy and exergy efficiencies.

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Scroll compressors and intermediate valve ports

P. Ginies , ... D. Gross , in 7th International Conference on Compressors and their Systems 2011, 2011

1.5 General benefit of IDV's in isentropic efficiency

The relative isentropic efficiency (1) gives an image of the compressor behavior over the operating map. The relative isentropic efficiency was drawn versus the APR for a compressor having a two hole IDV and for a compressor without IDV. These comparative curves were drawn for a number of saturated condensing temperatures: 35  /   45   /   55   /65°C. The horizontal axis APR allows to find the evaporating temperature for each case of condensing temperature.

(1) the relative isentropic efficiency is the isentropic efficiency multiplied by a coefficient.

The Graphs 3, 4, 5, 6 show that the benefit in isentropic efficiency is large for an APR smaller than 2.0 and that there is little disadvantage for an APR greater than 3.0. The benefit is with a weight in terms of applications, which is greater than the weight of the disadvantage observed for a high compression ratio corresponding to exceptional operating conditions.

Graph 3.

Graph 4.

Graph 5.

Graph 6.

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Gas turbine performance modelling, analysis and optimisation

A.M.Y. Razak , in Modern Gas Turbine Systems, 2013

Isentropic efficiency

The isentropic efficiency of a compression process is defined as the ratio of the ideal work done to the actual work done. Therefore, the isentropic efficiency ( η _c) is given by:

[11.13] ${\eta }_c=\frac{cp\left({T^{\prime }}_2-T_1\right)}{cp\left(T_2-T_1\right)}$

where T₂′ is the ideal compressor exit temperature. T ₁ is the compressor inlet temperature and cp is the specific heat at constant pressure – we shall assume a constant specific heat, and the effect of the variation of specific heats will be discussed later in the chapter. Equation [11.13] can be represented in terms of the compressor pressure ratio as:

[11.14] ${\eta }_c=\frac{T_1\left(c-1\right)}{T_2-T_1}$

where $C_{\text{opt}}={\mathrm{Pr}}^{\frac{\gamma -1}{\gamma }}$ and Prc is the compressor pressure ratio.

Therefore the compressor exit temperature can be determined by:

[11.15] $T_2=T_1+\frac{T_1}{{\eta }_c}\left(c-1\right)$

Similarly, for an expansion process the isentropic efficiency (η_t ) is given by:

[11.16] ${\eta }_t=\frac{cp\left(T_3-T_4\right)}{cp\left(T_3-{T^{\prime }}_4\right)}$

where T ₄′ is the ideal expander/turbine exit temperature, T ₃ is the expander/turbine inlet temperature and cp is the specific heat at constant pressure – again we shall assume a constant specific heat and the effect of the variation of specific heats will be discussed later.

In a manner similar to the compressor isentropic analysis above, it can be shown that the expander/turbine exit temperature is given by:

[11.17] $T_4=T_3-T_3{\eta }_t\left(1-\frac{1}{c}\right)$

where $c={\text{Prt}}^{\frac{\gamma -1}{\gamma }}$ and Prt is the expander/turbine pressure ratio.

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Thermodynamics of gas turbine cycles

A.M.Y. Razak , in Industrial Gas Turbines, 2007

2.18.3 Third method

The third method determines the performance of the gas turbine using the enthalpies and entropies at the various salient points in the cycle. It is considered the most accurate method for calculating the design point performance of a gas turbine. The method is much more detailed and is usually carried out using a computer program developed for this purpose. However, the processes involved will be outlined.

Integrating Equation 2.44, which describes the variation of specific heat with temperature for air and products of combustion, equations for enthalpy and entropy can be developed. Therefore:

[2.65] $H=a(T-T_0)+b\frac{T^2-T_0^2}{2}-c\left(\frac{1}{T}-\frac{1}{T_0}\right)$

[2.66] $S=a\ln \frac{T}{T_0}+b(T-T_0)-\frac{c}{2}\left(\frac{1}{T^2}-\frac{1}{T_0^2}\right)-R\ln \frac{P}{P_0}$

where T and P are the temperature and pressure of air or gas, respectively, and T ₀ and P ₀ are the reference temperature and pressure when the enthalpy and entropy, respectively, are assumed to be zero, when the temperature and pressure are 273   K and 1.013 Bar-A, respectively.

The constants a, b and c are determined as follows:

$\begin{array}{l}a={\displaystyle \sum _{i=1}^{\text{noc}}{a}_i\times m{f}_i}\hfill \\ b={\displaystyle \sum _{i=1}^{\text{noc}}{b}_i\times m{f}_i}\hfill \\ c={\displaystyle \sum _{i=1}^{\text{noc}}{c}_i\times m{f}_i}\hfill \end{array}$

a _i, b _i and c _i are the constants defined in Table 2.1 for each component and noc are the number of components in air or products of combustion.

In the example, the compressor inlet pressure and temperature is 1.013   Bar and 288   K. From Equations 2.65 and 2.66 we calculate the enthalpy and entropy at the compressor inlet as:

$\begin{array}{l}{H}_1=14.876\text{kJ}/\text{kg}\hfill \\ S_1=0.053\text{kJ}/\text{kg}\text{K}\text{.}\hfill \end{array}$

For a compressor pressure ratio of 20, the compressor discharge pressure, P ₂ = 20.26 Bar-A. From Equation 2.66 the isentropic compressor discharge temperature can be determined. This is achieved by using P ₂ for the pressure term in Equation 2.66 and varying the temperature until the entropy equals 0.053 kJ/kg K. The isentropic compressor discharge temperature, $T_{2^{\prime }}$ , works out to:

$T_{2^{\prime }}=659.452\text{K}\text{.}$

Using this value in Equation 2.65, the enthalpy at compressor discharge, $H_{2^{\prime }}$ due to isentropic compression is obtained:

$H_{2^{\prime }}=402.286\text{kJ/kg}$

The isentropic efficiency Equation 2.28 for a compression process can be written in terms of enthalpies as:

${\eta }_c=\frac{H_{2^{\prime }}-H_1}{H_2-H_1}$ where H ₂ is the actual enthalpy at the discharge of the compressor which corresponds to:

$H_2=460.175\text{kJ/kg}$

Using the value for H ₂ in Equation 2.65, the actual compressor discharge temperature, T ₂, can be determined implicitly:

$T_2=713.102\text{K}$

The compressor-specific work: Wc = H ₂ − H ₁. Therefore:

$Wc=445.3\text{kJ/kg}$

The fuel—air ratio may now be computed similarly to that discussed in Method 2. The combustor inlet temperature and combustor temperature rise for this case are 702.86   K and 697.14   K, respectively. A theoretical fuel—air ratio, f, of 0.0195 is obtained. The actual fuel—air ratio, f _a = 0.0195/0.99 = 0.0197. The heat input Q _in is:

$Q_{\text{in}}=0.0197\times 43100=849.388\text{kJ/kg}$

The fuel used is kerosene and can be modelled as C₁₂H₂₄. Knowing the fuel—air ratio and the air composition, the composition of the products of combustion can be calculated, as described by Goodger. ¹³

[2.67] $\begin{array}{l}{\text{C}}_x{\text{H}}_y+m\left({\text{O}}_2+\frac{0.7809}{0.2095}{\text{N}}_2+\frac{0.0093}{0.2095}\text{Ar}+\frac{0.0003}{0.2095}{\text{CO}}_2\right)\hfill \\ =n_1{\text{CO}}_2+n_2{\text{H}}_2\text{O+}{n}_3{\text{N}}_2+n_4\text{Ar+}{n}_5{\text{O}}_2\hfill \end{array}$

The quantities 0.7809, 0.0093, 0.003 and 0.2095 are the volume-fractions or molar-fractions (mole-fraction) of N₂, Ar, CO₂ and O₂ in air, respectively, and n ₁, n ₂, n ₃, n ₄ and n ₅ are the mole-fraction of CO₂, H₂O, N₂, Ar and O₂ in the products of combustion, respectively. The terms x and y are the mole-fractions of carbon and hydrogen in the fuel. For kerosene, x = 12 and y = 24 and the term m is the excess air which is determined using the fuel—air ratio (f _a ) as follows:

$f_a=\frac{12.01x+1.008y}{\left(1+\frac{0.7809}{0.2095}+\frac{0.0093}{0.2095}+\frac{0.0003}{0.2095}\right)MW}$

where MW is the mole-weight of air and the factors 12.01 and 1.008 are the atomic weights of carbon and hydrogen, respectively.

By performing a molar balance using Equation 2.67, the mole-fraction of the products of combustion (n ₁, n ₂, n ₃, n ₄ and n ₅) can be determined in a manner similar to that discussed in Chapter 6 (Section 6.18.4).

Since the turbine entry temperature, T ₃, pressure, P ₃, and the combustion gas composition are now known, Equations 2.65 and 2.66 can be used to determine the enthalpy, H ₃ and entropy, S ₃ at turbine entry. The enthalpy at the exit due to isentropic expansion must be determined. This is achieved by using Equation 2.66 and varying the turbine exit temperature, T ₄, until the entropy equals the value determined at the inlet of the turbine, S ₃. From Equation 2.65 the enthalpy, $H_{4^{\prime }}$ at turbine exit due to isentropic expansion can be determined. The turbine isentropic efficiency in Equation 2.30 can be represented as:

${\eta }_t=\frac{H_3-H_4}{H_3-H_{4^{\prime }}}$

where H ₄ is the actual enthalpy at turbine exit.

The values for H ₃, S ₃ and $H_{4^{\prime }}$ are 1272.995 kJ/kg, 0.958 kJ/kgK and 428.005 kJ/kg, respectively. For a turbine isentropic efficiency of 0.9, the actual enthalpy at exit from the turbine is 512.504 kJ/kg and the entropy at turbine exit is 1.0768 kJ/kgK. Thus the turbine specific work, Wt, is:

$Wt=H_3-H_4=1272.995-512.504=760.491\text{kJ/kg}$

The net specific work (W _net) from the gas turbine is:

$W_{\text{net}}=Wc-Wt=760.491-445.3=315.191\text{kJ/kg}$

The thermal efficiency (η _th) is:

${\eta }_{\text{th}}=\frac{W_{\text{net}}}{Q_{\text{in}}}=\frac{315.191}{849.388}=\mathrm{0.3711.}$

The specific heats at the salient points 1, 2, 3 and 4, as shown in Fig. 2.29, correspond to 1.0011, 1.083, 1.2193 and 1.1198, respectively. The corresponding values for the ratios of specific heats, γ= c _p /c _v , at the salient points 1, 2, 3 and 4 are 1.402, 1.3607, 1.3082 and 1.345, respectively. The increase in c _p due to compression is due to the increase in temperature as described by Equation 2.44. Similarly, there is an increase in c _p at salient point 3 and a decrease at point 4. However, the increase in c _p at point 3 is also due to the increase in water vapour in the products of combustion, which is significant, as can be seen in Table 2.3. Also, note there is an increase in CO₂ content in the products of combustion, a greenhouse gas and thought to be responsible for global warming. Therefore, gas turbines operating with fuels such as natural gas or methane, which have a higher hydrogen content, will result in increased specific work due the high content of water vapour in the products of combustion. With methane as fuel, this increase in power output may be as high as 2% compared with that when using kerosene. Note that the increases in specific heats have resulted in a decrease in γ.

2.29. Turbine cycle on the temperature—entropy diagram.

Table 2.3. Composition of products of combustion

Component	Gravimetric or mass fraction
N2	0.744
O2	0.162
Ar	0.009
CO2	0.061
H2O	0.025

The above example considered dry air. The effects of humidity can also be included in the analysis. For example, given the relative humidity of the air, the specific humidity can be calculated, as discussed in Section 2.11.1, which is the mass of water vapour per unit of dry air. Therefore, the specific humidity can be added to the composition of air as shown in Table 2.2 and air/gas composition normalised to determine the gravimetric composition of moist/humid air and then repeat the above procedure. The additional heat input required to heat the water vapour from the compressor discharge temperature, T ₂, to the turbine entry temperature, T ₃, needs to be calculated. This can be determined using Equation 2.68:

[2.68] $H_s=2.232T_s+2352.623$

where H _s is the water/steam enthalpy (kJ/kg) and T _s is the water vapour/ steam temperature in Celsius.

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Introduction

S.L. Dixon B. Eng., Ph.D. , C.A. Hall Ph.D. , in Fluid Mechanics and Thermodynamics of Turbomachinery (Sixth Edition), 2010

Efficiency of Compressors and Pumps

The isentropic efficiency, η_c , of a compressor or the hydraulic efficiency of a pump, η_h , is broadly defined as

${\eta }_c(\text{or}{\eta }_h)=\frac{\text{useful (hydrodynamic) energy input to fluid in unit time}}{\text{power input to rotor}}$

The power input to the rotor (or impeller) is always less than the power supplied at the coupling because of external energy losses in the bearings, glands, etc. Thus, the overall efficiency of the compressor or pump is

${\eta }_o=\frac{\text{useful (hydrodynamic) energy input to fluid in unit time}}{\text{power input to coupling of shaft}}.$

Hence, the mechanical efficiency is

(1.45) ${\eta }_m={\eta }_o/{\eta }_c(\text{or}{\eta }_o/{\eta }_h).$

For a complete adiabatic compression process going from state 1 to state 2, the specific work input is

$\mathrm{\Delta }{W}_c=(h_{02}-h_{01})+g(z_2-z_1).$

Figure 1.9(b) shows a Mollier diagram on which the actual compression process is represented by the state change 1–2 and the corresponding ideal process by 1–2s. For an adiabatic compressor in which potential energy changes are negligible, the most meaningful efficiency is the total-to-total efficiency, which can be written as

(1.46a) ${\eta }_c=\frac{\text{ideal}\text{(minimum)}\text{work}\text{input}}{\text{actual}\text{work}\text{input}}=\frac{h_{02s}-h_{01}}{h_{02}-h_{01}}.$

If the difference between inlet and outlet kinetic energies is small, $\frac{1}{2}{c}_1^2\cong \frac{1}{2}{c}_2^2$ then

(1.46b) ${\eta }_c=\frac{h_{2s}-h_1}{h_2-h_1}.$

For incompressible flow, the minimum work input is given by

$\mathrm{\Delta }W\text{min}={\dot{W}}_{\text{min}}/\dot{m}=[(p_2-p_1)/\rho +\frac{1}{2}(c_2^2-c_1^2)+g(z_2-z_1)]=g[H_2-H_1].$

For a pump the hydraulic efficiency is therefore defined as

(1.47) ${\eta }_h=\frac{{\stackrel{.}{W}}_{\text{min}}}{{\stackrel{.}{W}}_c}=\frac{g[H_2-H_1]}{\mathrm{\Delta }{W}_c}.$

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Simulation model to predict temperature distribution along scroll wraps

M.C. Diniz , ... C.J. Deschamps , in 8th International Conference on Compressors and their Systems, 2013

ABSTRACT

The volumetric and isentropic efficiencies of scroll compressors are affected by the heat transfer that takes place inside the pockets during suction and compression processes. The present paper details a numerical model developed to predict the conduction heat transfer and temperature distribution of scroll wraps. The model was developed via the finite volume method and coupled to a thermodynamic model of the compression cycle. The results showed that the discharge temperature predicted with the solution of heat conduction through the scroll wraps was slightly lower than that obtained when a linear temperature profile was prescribed. It was also found that the heat transfer taking place in the metallic contact between the scrolls wraps acts to produce a linear temperature variation along their length.

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Turbo pumps and compressors

Philip Thomas , in Simulation of Industrial Processes for Control Engineers, 1999

17.7.2 Polytropic efficiency and polytropic head

The definition of isentropic efficiency, equation (17.64), is based on a ratio of isentropic specific work to actual specific work, across a complete section of the compressor. However, it is also possible to define a differential efficiency, assumed constant over the section, known as the polytropic efficiency, η_p , given by

(17.72) ${\eta }_p=\frac{d{w}_s}{dw}$

We may use equation (17.58) to transform equation (17.72) into:

(17.73) ${\eta }_p=\frac{d{h}_s}{dh}=\frac{c_{p}d{T}_s}{c_{p}dT}=\frac{d{T}_s}{dT}$

where we have also assumed that the specific heat is constant. The form of equation (17.73) enables us to use the mathematical treatment outlined in Section 14.3 for nozzle efficiency, except that the nozzle efficiency, η_N , used in that section will be replaced by the inverse of the polytropic efficiency, 1/η_p. Thus the actual compression may be characterized by

(17.74) $p{v}^m=\text{constant}$

where the exponent, m, is given by:

(17.75) $m=\frac{{\eta }_p\gamma }{1-\gamma \left(1-{\eta }_p\right)}$

As a consequence of equation (17.74) and the characteristic gas equation (3.2), the actual temperature ratio across a compressor section operating at a polytropic efficiency, η_p , will be given by

(17.76) $\frac{T_2}{T_0}={\left(\frac{p_2}{p_0}\right)}^{\left(m-1\right)/m}$

where, from the definition of the exponent, m, in equation (17.75)

(17.77) $\frac{m-1}{m}=\frac{1}{{\eta }_p}\frac{\gamma -1}{\gamma }$

The expression for the temperature ratio in an actual compression may be found by substituting into equation (17.61) to give the actual specific work:

(17.78) $-w=\frac{\gamma }{\gamma -1}Z{R}_w{T}_0\left({\left(\frac{p_2}{p_0}\right)}^{\left(m-1\right)/m}-1\right)$

We may use this new expression for the actual specific work together with equation (17.63) that gives the isentropic specific work in order to evaluate the isentropic efficiency:

(17.79) $\begin{array}{ll}{\eta }_s\hfill & =\frac{\frac{\gamma }{\gamma -1}Z{R}_w{T}_0\left({\left(\frac{p_2}{p_0}\right)}^{\left(\gamma -1\right)/\gamma }-1\right)}{\frac{\gamma }{\gamma -1}Z{R}_w{T}_0\left({\left(\frac{p_2}{p_0}\right)}^{\left(m-1\right)/m}-1\right)}\hfill \\ \hfill & =\frac{\left({\left(\frac{p_2}{p_0}\right)}^{\left(\gamma -1\right)/\gamma }-1\right)}{\left({\left(\frac{p_2}{p_0}\right)}^{1/{\eta }_p\left(\left(\gamma -1\right)/\gamma \right)}-1\right)}\hfill \end{array}$

From their definitions, we would expect the isentropic efficiency and the polytropic efficiency to be similar in value, and this is indeed the case. Nevertheless, it is possible to evaluate equation (17.79) over a range of pressure ratios for fixed values of polytropic efficiency, and thus highlight divergencies. Figure 17.6 shows the isentropic efficiency calculated for three typical values of polytropic efficiency over a range of pressure ratios.

Figure 17.6. Isentropic efficiency against pressure ratio for polytropic efficiencies of 0.7, 0.8 and 0.9, with γ = 1.4.

It will be seen that the isentropic efficiency of the section is the same as the polytropic efficiency at unity pressure ratio, but falls away as the pressure ratio rises, the fall being rather more marked at lower polytropic efficiencies. The isentropic effficiency is between about 2% and 7% less than the polytropic efficiency, depending on the latter's value, for the normal range of pressure ratios found on industrial plant, namely 2.5 to 4.5.

The fact that the isentropic efficiency varies with pressure ratio for a constant value of polytropic efficiency has led some to regard the polytropic efficiency as a preferable foundation on which to base their analysis of the compressor section. Instead of taking the isentropic specific work as our ideal against which to measure the actual specific work, it is possible to devise a new ideal measure, the polytropic specific work, –Wp, defined so that its ratio to the actual specific work is the polytropic efficiency, η_p :

(17.80) ${\eta }_p=\frac{w_p}{w}$

Combining equations (17.78) and (17.80), the polytropic specific work emerges as:

(17.81) $-w_p={\eta }_p\frac{\gamma }{\gamma -1}Z{R}_w{T}_0\left({\left(\frac{p_2}{p_0}\right)}^{\left(m-1\right)/m}-1\right)$

or, using equation (17.77)

(17.82) $-w_p=\frac{m}{m-1}Z{R}_w{T}_0\left({\left(\frac{p_2}{p_0}\right)}^{\left(m-1\right)/m}-1\right)$

An experimental determination of m may be made after measuring the pressure and temperature at the section inlet, p ₀,T ₀ and outlet, p _2, T ₂ and applying equation (17.76), which may be solved to give the formula:

(17.83) $m=\frac{\text{In}\frac{p_2}{p_0}}{\text{In}\frac{p_2}{p_0}\frac{T_2}{T_0}}$

The polytropic efficiency may be found similarly by solving equations (17.76) and (17.77):

(17.84) ${\eta }_p=\frac{\gamma }{\gamma -1}\frac{\text{In}\frac{p_2}{p_0}}{\text{In}\frac{T_2}{T_0}}$

In fact, an experimental determination of the exponent, m, making use of equation (17.83) (or an equivalent) will introduce a small error due to the implicit inclusion of an imperfect model of compressibility effects. To compensate for this, manufacturers sometimes introduce an additional factor, f, into the equation for polytropic specific work, although in most cases f is so close to unity as to make negligible difference. Polytropic specific work is normally given the name 'polytropic head', H_p , based on the same reasoning used for isentropic head, and so we have the final form:

(17.85) $H_p=f\frac{m}{m-1}Z{R}_w{T}_0\left({\left(\frac{p_2}{p_0}\right)}^{\left(m-1\right)/m}-1\right)$

The power absorbed by the compressor section is given by equation (17.68). We may use equation (17.80) to express the power in terms of polytropic specific work/polytropic head and polytropic efficiency:

(17.86) $\begin{array}{ll}P\hfill & =-\frac{W{w}_p}{{\eta }_p}\hfill \\ \hfill & =\frac{W{H}_p}{{\eta }_p}\hfill \end{array}$

where the final form uses the fact that H_p = −w_p.

It should be emphasized that the polytropic head is an idealization in the same way that the isentropic head was an idealization. Taken together with the polytropic efficiency, however, it provides a way of analysing compressor performance, as will be shown in the next section.

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13th International Symposium on Process Systems Engineering (PSE 2018)

Wen W. Zhang , ... Qing L. Chen , in Computer Aided Chemical Engineering, 2018

3.1.2 The model for compressors

For isentropic compression, isentropic efficiency is defined to calculate the practical enthalpy change. Hence, the required power is computed using Eq. (4).

(4) $E_{h,u}=\frac{1}{{\eta }_{h,u}}{\sum }_{cϵC}\left({\epsilon }_c{M}_{h,u^{\prime },u,c}{F}_{h,u}\right)\left[{\left(R_{h,u}\right)}^{\frac{{\theta }_u-1}{{\theta }_u}}-1\right]\forall u\in \mathit{COM}$

where, variable E _{h, u} is power consumed or generated, M is mole fraction of component c, isentropic exponent θ _u is the ratio of heat capacities of gas streams at constant pressure and temperature, ε _c is specific heat capacity, and c is a set of chemical components.

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Thermodynamics

P. Friedman , M. Anderson , in Fundamentals and Applications of Supercritical Carbon Dioxide (sCO₂) Based Power Cycles, 2017

3.3.1 Turbomachinery

The energy and exergy balances for a turbine (Fig. 3.6) are,

Figure 3.6. Adiabatic turbine analysis. Inlet and exit streams carry flow enthalpy and exergy. Specific work leaves via the turbine shaft. Destroyed exergy is shown as irreversibility. Process diagram shows real process (1–2) and ideal process (1–2s).

(3.36) $w_T=h_1-h_2$

(3.37) $w_T={\psi }_1-{\psi }_2-i_T$

The first law (isentropic) efficiency for the turbine compares actual work that would be realized in a process that exits at the same pressure as the real turbine if the process is isentropic (designated as point 2s). Second law efficiency is the ratio of realized work to exergy supplied.

(3.38) ${\eta }_{T,I}=\frac{h_1-h_2}{h_1-h_{2s}}$

(3.39) ${\eta }_{T,II}=\frac{h_1-h_2}{{\psi }_1-{\psi }_2}=1-\frac{i_T}{{\psi }_1-{\psi }_2}$

For compressor or pump (see Fig. 3.7) governing equations are similarly defined,

Figure 3.7. Adiabatic compressor or pump analysis. Inlet and exit streams carry flow enthalpy and exergy. Specific work is supplied via the turbine shaft. Destroyed exergy is shown as irreversibility. Process diagram shows real process (1–2) and ideal process (1–2s).

(3.40) $w_C=h_2-h_1$

(3.41) $w_C={\psi }_2-{\psi }_1+i$

and,

(3.42) ${\eta }_{C,I}=\frac{h_{2s}-h_1}{h_2-h_1}$

(3.43) ${\eta }_{C,II}=\frac{{\psi }_2-{\psi }_1}{h_2-h_1}=1-\frac{i_C}{h_2-h_1}$

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Energy Fundamentals

Ibrahim Dincer , in Comprehensive Energy Systems, 2018

1.7.12.1 Turbine

A fluid is expanded in a turbine to produce power. Steam and gas turbines are considered here. Turbines are normally well-insulated so that their operation can be assumed to be adiabatic. The performance of an adiabatic turbine is usually expressed by isentropic (adiabatic) efficiency.

Consider a turbine with inlet state 1 with temperature T ₁ and pressure P ₁ and an exit state 2 with temperature T ₂ (or steam quality) and pressure P ₂ as shown in Fig. 13 . The power output from this compressor would be maximum if the fluid is expanded reversibly and adiabatically (i.e., isentropically) between the given initial state and given exit pressure. The isentropic efficiency is the ratio of actual power to the isentropic power, which is the power produced by the same turbine if it had an isentropic efficiency of 100%.

Fig. 13. Schematic diagram of a turbine.

(83) ${\eta }_{\text{isen},\text{Tr}}={\dot{W}}_{\mathrm{actual}}/{\dot{W}}_{\mathrm{isen}}$

where $\dot{m}$ is the mass flow rate of fluid and h _s is the enthalpy of the fluid at the turbine outlet if the process was isentropic and the subscript Tr refers to turbine. This enthalpy may be obtained from exit pressure and exit entropy (equal to inlet entropy). KE and PE changes are neglected.

Exergy efficiency of an adiabatic turbine may be determined from "exergy recovered (produced or obtained)/exergy expended" approach. In this case, the exergy resource is steam, and exergy expended is the exergy supplied to steam to turbine, which is the decrease in the exergy of steam as it passes through the turbine. Note that the exergy recovered or produced or obtained is the shaft work. Taking state 1 as the inlet and state 2 as the outlet, the second-law efficiency (and hence exergy efficiency) is defined as

(84) ${\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}1}=\frac{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{recovered}}}{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{expended}}}=\frac{{\dot{W}}_{\mathrm{out}}}{\dot{\mathrm{E}}{\mathrm{x}}_1-\dot{\mathrm{E}}{\mathrm{x}}_2}=\frac{{\dot{W}}_{\mathrm{out}}}{{\dot{W}}_{\mathrm{rev}}}=\frac{\dot{m}\left(h_1-h_2\right)}{\dot{m}\left(h_1-h_2-T_0\left(s_1-s_2\right)\right)}$

or

(85) ${\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}1}=1-\frac{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{d}}}{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{expended}}}=1-\frac{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{d}}}{\dot{\mathrm{E}}{\mathrm{x}}_1-\dot{\mathrm{E}}{\mathrm{x}}_2}=1-\frac{{\dot{W}}_{\mathrm{rev}}-{\dot{W}}_{\mathrm{out}}}{\dot{\mathrm{E}}{\mathrm{x}}_1-\dot{\mathrm{E}}{\mathrm{x}}_2}$

The exergy efficiency definition based on the "exergy out/exergy in" approach is

(86) ${\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}2}=\frac{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{out}}}{\dot{\mathrm{E}}\mathrm{x}}=\frac{{\dot{W}}_{\mathrm{out}}+\dot{\mathrm{E}}{\mathrm{x}}_2}{\dot{\mathrm{E}}{\mathrm{x}}_1}\ne \frac{{\dot{W}}_{\mathrm{out}}}{{\dot{W}}_{\mathrm{rev}}}$

A third definition only assumes power output as the product and inlet exergy as the input:

(87) ${\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}3}=\frac{{\dot{W}}_{\mathrm{out}}}{\dot{\mathrm{E}}{\mathrm{x}}_1}=1-\frac{\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{d}}}{\dot{\mathrm{E}}{\mathrm{x}}_1}=1-\frac{{\dot{W}}_{\mathrm{rev}}-{\dot{W}}_{\mathrm{out}}}{\dot{\mathrm{E}}{\mathrm{x}}_1}$

Note that the first definition (Eq. 84) is consistent with the general definition for the second-law efficiency of WPDs (the ratio of actual work to reversible work), but the second and third definitions (Eqs. 86 and 87) are not. Also, the first definition satisfies both bounding conditions for the second-law law efficiency: it is 100% when actual work equals reversible work, and 0% when actual work is zero (and thus the entire expended exergy is destroyed).

It should be noted that isentropic efficiency and second-law efficiency are different definitions. In the isentropic efficiency, the actual process is compared to an ideal isentropic process between actual initial state and an assumed hypothetical exit state while in the exergy efficiency, the actual process is compared to an ideal reversible process between actual inlet state and actual exit state is used. Consequently, close but different values for isentropic and exergy efficiencies are obtained.

Example 7

Consider an adiabatic steam turbine, as shown in Fig. 14, with the following inlet and exit states: P ₁=10,000 kPa, T ₁=500°C, P ₂=10 kPa, x ₂=0.95. Taking the dead-state temperature of steam as saturated liquid at 25°C, determine isentropic efficiency and exergy efficiency based on different approaches.

Fig. 14. A schematic diagram of the steam turbine.

Solution

The various efficiencies are to be determined from Eqs. (83), (84), (86), and (87). Before proceeding further, there is a need to write the balance equations:

$\begin{array}{l}\text{MBE}:{\dot{m}}_1={\dot{m}}_2\hfill \\ \text{EBE}:{\dot{m}}_1{h}_1={\dot{m}}_2{h}_2+{\dot{W}}_{\mathrm{out}}\hfill \\ \text{EnBE}:{\dot{m}}_1{s}_1+{\dot{S}}_{\mathrm{gen}}={\dot{m}}_2{s}_2\hfill \\ \mathrm{ExBE}:{\dot{m}}_1{\mathrm{ex}}_1={\dot{m}}_2{\mathrm{ex}}_2+{\dot{W}}_{\mathrm{out}}+\dot{\mathrm{E}}{\mathrm{x}}_{\mathrm{d}}\hfill \end{array}$

Using the EBE to calculate the specific power produced by the steam turbine as follows:

$h_1=h_2+w_{\mathrm{out}}$

$3374=2464+w_{\mathrm{out}}$

$w_{\mathrm{out}}=909.8\text{kJ}/\text{kg}$

Then using the ExBE, the exergy destructed through the expansion process is calculated as follows:

${\mathrm{ex}}_1={\mathrm{ex}}_2+w_{\mathrm{out}}+{\mathrm{ex}}_{\mathrm{d}}$

${\mathrm{ex}}_{\mathrm{d}}=1412-151.1-909.8=\mathbf{350.7}\text{kJ}\mathbf{/}\text{kg}$

Then by using the exergy efficiency definitions these are the resulting efficiencies:

${\eta }_{\text{isen},\text{Tr}}=0.708,{\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}1}=0.722,{\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}2}=0.752,{\eta }_{\mathrm{ex},\text{Tr}\mathrm{‐}3}=0.6445$

That is, the second-law efficiency is 75.2% based on Eq. (86) and it is 72.2% based on Eq. (84). In Eqs. (86) and (87), the exergy of the steam at the turbine exit is part of the exergy destroyed by the turbine. However, the turbine should not be held responsible for the exergy it did not destroy as part the processes associated with power production. With the first definition, the difference between the exergies of the inlet and exit steams is used for the exergy expended in the system.

The effect of turbine inlet pressure on isentropic efficiency (Eq. 83) and three forms of exergy efficiencies (Eqs. (84)–(87)) is investigated while maintaining the exit conditions constant (Fig. 15). The efficiencies based on four definitions are considerably different. However, isentropic efficiency and the second-law efficiency by Eq. (84) are more appropriate forms. Interestingly, values of these two efficiencies are close to each other.

Fig. 15. Effect of turbine inlet pressure on the isentropic and second-law efficiencies.

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