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ESRL Module 4.

Real Gas Performance (RGP) of a Cogeneration System

Prepared by F. Carl Knopf, Chemical Engineering Department, Louisiana State University

Documentation Module Use Expected Learning Outcomes/Objectives Upon completion of the module, students will be

able to: IGP Video (for review) Cogeneration Real Gas Performance Module.pdf 4 Examples files 2 Combustion Library dlls 3 Student Assignments

Thermodynamics Course Laboratory Course

(1) All the objectives of ESRL 1. (2) Explain the limitations of ideal fluid vs. real fluid assumptions in analysis of thermodynamic cycles.

You should complete, or least understand the concepts, in Module 1 - Ideal Gas Performance of a Cogeneration System before trying Module 4. This module is an extension of ESRL Module 1 – Ideal Gas Performance (IGP) of a Cogeneration System. In Module 1 we used the ideal gas assumption to determine the performance of a cogeneration system as shown in Figure 1. We also introduced costing of the cogeneration system using power law costing equations which were embedded in the provided Excel sheets. Real time data from a system with this configuration are available at www.esrl.lsu.edu or www.cogened.lsu.edu. The system consists of an air cooler, air compressor, combustion chamber, gas turbine, power generating turbine and heat recovery steam generator (HRSG). In this module we again determine the performance of the cogeneration system, but here we develop solutions with improved gas side properties. We also cost the cogeneration system – but the same as ESRL Module 1, you are not required to perform any costing calculations as they are simply embedded in the provided Excel sheets. For the gas side we will develop two alternatives. First we assume that the gas side has the properties of air throughout the entire system (the air standard assumption) – this is clearly correct prior to the combustion chamber. For the second alternative we will use the actual gas properties throughout the entire system – this will be air properties before the combustion chamber and properties for a mixture of combustion products with any unreacted air after combustion. Results from these alternatives will be compared with our ideal gas solution of Module 1. In order to generate the needed gas side properties Excel callable functions will be provided. Function calls will also be provided for the water / steam side allowing all the developed models (including the ideal gas model) to be utilized for other design conditions and operating conditions including supplemental firing.

2

barP

RT

013.1

67.519

1

1

barP

RT

20

67.536

8

8

barP

RT

Fsteam

20

874

9

9

barP

RT

013.1

17.547

0

0

Figure 1 Cogeneration Power Cycle with Air Chiller and HRSG

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Cogeneration System Performance Calculations Table 1 provides design information for the system shown in Figure 1 and we are asked to determine the amount of power and steam this system will generate. At design conditions, the incoming air is cooled to 60 F to help maintain efficiency in the aeroderivative turbine. The compressor efficiency ( ) is 84%, the gas turbine efficiency ( ) is 86% and the power turbine efficiency ( ) is 81%. The gas turbine supplies power to the compressor and the power turbine generates electricity. The compressor discharge pressure is 16.8 bar. The combustion chamber efficiency ( ) is 98%, the combustion chamber temperature is 2500 R and pressure drop in the combustion chamber is negligible. In the HRSG there is a 5% pressure drop, the pinch temperature difference ( ) is 30 R, the approach temperature difference ( ) is 60 R. Table 2 lists system variables and parameters. We will need to solve for the work done in the air compressor, / , and the natural gas fuel flow rate, / which will give a combustion chamber temperature of 2500 R. We will also need to determine the pressure leaving the gas turbine which allows sufficient power generation for the air compression ( = ). Finally, we will determine how much electricity can be produced from the power turbine, / , and the flow rate of steam, / , which can be raised in the HRSG. Table 1 Cogeneration System Design Information

/ 154.6

1 519.67 0.84

2 16.8 0.98

2500 0.86 0.81

30 60

8 536.67 8 20.0 9 874

9 20.0 Table 2 Cogeneration System Variables and Parameters

Variables: / ; / ; / ; ; / ; /

Parameters: 5% ; 21501 /

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Solution Assumptions: There are several assumptions we can make when assembling the needed material and energy balances to predict the cogeneration system performance. We have shown the use of the ideal gas assumption for the prediction of air and combustion product properties in Module 1. Here we want to explore two alternative methods which should result in improved accuracy when compared to the ideal gas-based solution. The first assumes the gas in the system (both air and product gases) can be taken as having the physical properties of air – this is sometimes referred to as the air standard system. A second alternative accounts for the actual physical properties of both the air and the combustion product gases. In both these cases we will account for the mass of fuel added to the system and real water / steam properties will be used in the HRSG. The results from the performance calculations will be used to cost the cogeneration system and explore possible options in steam generation. We will also compare system performance results to those found from the commercial utility system design program GateCycle (from General Electric).

I Air Standard System Solution

The air standard system assumes that the gas side of the cogeneration system can be entirely represented as air. The mass of fuel used is accounted for, and water and steam properties are used in the HRSG. The solution is provided in the Excel file Cogeneration Air Standard.xls. We assemble the needed equations below.

Physical Properties

The physical properties of air and steam, over the temperature and pressure ranges for cogeneration projects, were assembled as dlls (Dynamically Linked Libraries) for use within Excel. These properties as determined using the equations of state provided in Reynolds (1979) are listed in Table 3; properties for available cogeneration species are given in the Table of Function Names.pdf provided in this module. Physical properties for the combustion product gases have also been assembled and these are given in Tables 7 and 8 (below). Table 3a Pure species properties, EOS from Reynolds (1979). Units T (R ); P (psia);

h (Btu/lb-mol); s (Btu/lb-mol-R); (ft3/lb-mol); (lb-mol/ft3) Species

, , , , , , , ,

Air H_Air P_Air P_HT_Air P_ST_Air S_Air T_HP_Air T_SP_Air V_Air Steam H_Steam P_Water P_HT_Steam P_ST_Stea

m S_Steam T_HP_Steam T_SP_Steam V_Steam

Water H_Water P_Water P_HT_Water P_ST_Water S_Water T_HP_Water T_SP_Water V_Water Table 3b Pure species properties in SI units, EOS from Reynolds (1979). Units T

(K); P (MPa); h (kJ/kg-mol); s (kJ/kg-mol-K); (m3/kg-mol); (kg-mol/m3) Species , , , , , Air_SI H_Air_SI P_Air_SI P_HT_Air_SI P_ST_Air_SI S_Air_SI Steam_SI H_Steam_SI P_ Water_SI P_HT_Steam_SI P_ST_Steam_SI S_Steam_SI

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Water_SI H_Water_SI P_Water_SI P_HT_Water_SI P_ST_Water_SI S_Water_SI

The use of these functions follows the same procedure. For example, the enthalpy of air with known pressure (psia) and temperature ( R) is found as = H_Air(pressure, temperature). The solution procedure for the cogeneration system using these real physical properties is outlined below with the Excel function calls to the dlls indicated in blue. For the air standard system we will need air, water and steam properties – note that these functions are per mole.

Incoming Air Stream:

The air steam to the compressor is 1 = 14.696 psia, and 1 = 519.67 R and here the specific entropy and enthalpy of air can be found as: ̂ = S_Air(P1, T1)/MW_Air = 1.0735 Btu/lb-R (1)

, = H_Air(P1, T1)/MW_Air = 192.7660 Btu/lb (2)

The subscript, a, is used with the enthalpy to indicate this is the actual value determined from the Excel function. Recall that enthalpy values are relative (see Module 1) and it will prove convenient to modify the air enthalpy by inclusion of the reference state for the fuel lower heating value – which is = 14.696 psia and = 536.67 R. For air, = 196.8487 Btu/lb and we can subtract from 1, , giving: 1, = – 4.0837 Btu/lb

We will need to keep track of in order to calculate actual enthalpy and entropy values when using the functions in Tables 2 and 3. We address the accounting for when we determine the physical properties for the combustion products.

1. Air Compressor (AC):

The minimum or ideal work required for gas compression is found by using an adiabatic and reversible process – this would be an isentropic process. Here following Figure 1: ̂ , ̂ , , , (3)

Species , , , Air_SI T_HP_Air_SI T_SP_Air_SI V_Air_SI Steam_SI T_HP_Steam_SI T_SP_Steam_SI V_Steam_SI Water_SI T_HP_Water_SI T_SP_Water_SI V_Water_SI

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For the cogeneration process, 1 and 2, , are the entropies of air (Btu/lb-R) at the inlet and exit of the compressor and 2, , is the isentropic outlet temperature of the air. 2 is a known design variable. 2, can be found from 2, and 2 as: , = T_SP_Air(s2,isen * MW_Air , P2) (4)

2, is the enthalpy at 2, and 2, found using, , = H_Air(P2, T2,isen)/MW_Air. (5)

With known (design variable), 2, can be found as:

, ,

, , (6)

where 2, is the actual air enthalpy at 2 and 2. 2 is the actual air temperature leaving the compressor, found from, = T_HP_Air(h2,a * MW_Air, P2). (7)

The work (Btu/s) in the air compressor is:

, , (8)

where the air flow rate (lb/s) is a known design variable. To simplify our notation we allow the symbols 1, and 1to be used interchangeably. But we will use 1, when it is important to emphasize that the actual enthalpy as determined from the Excel function call is being used or calculated.

2. Combustion Chamber (CC): The combustion chamber mass and energy balances can be written as, (9)

, , (10)

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where is the combustion products gas mass flow rate (lb/s); 2is the enthalpy of the inlet air to the combustion chamber; 3is the enthalpy of the outlet stream from the combustion chamber; LHV, is the lower heating value of fuel. For methane the LHV = 21501 Btu/lb. is the rate of heat loss from the combustion chamber, 1 . Recall from our earlier discussion that we must account for the reference state for LHV of the fuel. In order to do this we have subtracted the reference state enthalpy for air, , , and the reference state enthalpy of the combustion gas products, , , at the known LHV reference conditions. We can directly solve the energy balance (equation 10) for 3,

, , (10)

In equation (11) an immediate problem is that in order to determine , we must know all the species and their concentrations in the combustion gas. We address this problem in the Real Gas System solution, but for the Air Standard System solution we simply assume the product gases and 3 have the properties of air and , = , which for air = 196.8487 Btu/lb. Now for any chosen we can solve equation (11) for 3, and using the Excel air functions we can determine 3 for that 3. As 3 is known (design variable) we can use Goal Seek in Excel (or any iterative process) to vary until 3 = 2500R. Fuel Lower and Higher Heating Values: The fuel lower heating value is defined as the heat liberated when one lb of fuel is mixed with stoichiometric oxygen at 77 F and completely burned at 1 atm. The products are then cooled to 77 F with the water remaining in the vapor state. The higher heating value of the fuel may also be provided. The higher heating value is defined as the heat liberated when one lb of fuel is mixed with stoichiometric oxygen at 77 F and completely burned at 1 atm. The products are then cooled to 77 F with the water formed from combustion condensed to the liquid state. Fuel lower and higher heating values are provided in Table 4. Table 4: Fuel lower and higher heating values at 77 F (25C) and 1 atm, from Bathie (1996)

LHV LHV HHV HHV Compound Formula State Btu/lb kJ/kg Btu/lb kJ/kg Methane CH4 Gas 21,501 50,012 23,860 55,499 Ethane C2H6 Gas 19,141 44,521 20,911 48,638 Propane C3H8 Liquid 19,927 46,351 21,644 50,343 n-Butane C4H10 Liquid 19,493 45,342 21,121 49,128 n-Heptane C7H12 Liquid 19,155 44,555 20,666 48,069 n-Octane C8H18 Liquid 19,098 44,422 20,589 47,890

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It is important to remember that fuel is generally priced based on the higher heating value, but manufacturers generally report turbine performance data based on the fuel lower heating value. This can cause problems in an economic analysis where costs are based on fuel flow rates.

3. Gas Turbine (GT):

An isentropic gas turbine would obey: ̂ , ̂ , , , (11)

For the gas turbine 3, and 4, , are the entropies of the combustion exhaust gas (Btu/lb-R) at the inlet and exit of the turbine and 4, , is the isentropic outlet temperature of the exhaust gas. For any chosen 4, we can find 4, from: , = T_SP_Air( ̂ , ∗ , ) (12)

Using 4, we can solve for 4, and 4, , using: , = H_Air( , , )/ (13)

The work done in the air compressor is supplied by direct shaft coupling with the gas turbine. We can determine 4, by using the known work done in the air compressor as:

, , ,∗ , ,

, , ,∗ , ,

(14)

Finally we can determine the gas turbine efficiency for the chosen 4 as: , ,

, , (15)

We know from our design conditions that = 0.86. We can simply vary 4 using Excel Goal Seek (or any iterative process) until the design efficiency is obtained.

4. Power Turbine (PT):

An isentropic power turbine would obey: ̂ , ̂ , , , (16)

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We know there is a 5% pressure drop in the HRSG, so 5 = 1.066 bar. We can then solve for , and , using the Excel functions. The actual exhaust enthalpy 5, at 5 will be, , , , , , , (17)

and 5can be found from the Excel functions. The work (Btu/s) produced by the power turbine is:

, , (18)

5. Heat Recovery Steam Generator (HRSG): The HRSG consists of an economizer or preheat section and an evaporator section as shown in Figure 2. A recommended approach to HRSG design (Ganapathy, 1991) is to fix both the pinch and approach temperature differences and allow the amount of steam generated to be determined by material and energy balance around the evaporator section of the HRSG. Design values for pinch temperature difference and the approach temperature difference are provided in Table 1. These values are chosen to produce a HRSG of reasonable size (area) and cost and allow acceptable HRSG performance in off-design cases. The most important HRSG off-design case is supplemental firing, where additional fuel is added to the exhaust gas and combusted just prior to the HRSG evaporator. This allows for the rapid generation of additional steam. We have allowed for this possibility with the 5 / 6 notation in Figure 1, but we do not consider supplemental firing in this module; in other words, conditions at 5 = conditions at 6.

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Figure 2 Heat Recovery Steam Generator with expanded pinch point From Figure 2, the temperature equations that govern the HRSG design are, (19)

(20)

Recalling Figure 1 and using Figure 2, the overall material and energy balance on the evaporator section allows the steam flow rate to be determined,

(21)

Here the exhaust gas enthalpy is available from the power turbine calculations. The exhaust gas enthalpy at ( , ) can be evaluated using H_Air(P7P, T7P). The HRSG gas-side pressure drop is known, and can be taken as the average of and . The steam-side enthalpies,

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and can be evaluated using H_Steam(P9, T9)/MW_Water and H_Water(P8, T8P)/MW_Water. To properly use the steam functions, the state of the steam (vapor or liquid) must be known. Here we are using º º , the saturation pressure of the steam. The rate that heat is transferred from the gas side to the steam side in the evaporator section is: , , (22)

For this problem we assume the flow rate of water equals the flow rate of steam; i.e. no blowdown.

(23)

The rate that heat is transferred from the gas side to the water side in the economizer section is,

, , (24)

where can be evaluated using H_Water(P8, T8)/MW_Water

Finally, we can calculate the exhaust gas temperature leaving the HRSG from an overall energy balance on gas side of the HRSG:

, ,

, ,

,, ,

(25)

can be found from T_HP_Air (h7, P7). Equations (11) – (17) provide the “HRSG design case”.

An important aspect of our Excel-based solution is our ability to vary the values for the design variables and obtain solutions to the material and energy balances. Table 9 (below) provides performance results for the design variables provided in Table 1. From Table 9 and the Excel file Cogeneration Air Standard.xls we find: / = 2.550; = 19.169;

/ = 10297.3; and / = 23.878.

Costing the Cogeneration System (NOT REQUIRED):

Here we again (see also ESRL Module 1.) introduce system costing. You will not need to cost any system for this module but it is instructive to see how a simple cost estimate can be obtained; the costing equations are embedded in the provided Excel sheets. We can utilize information from the CGAM problem (Valero et al., 1994) for capital and operating costs for the cogeneration system. The total cost rate of fuel and equipment, CTotal ($/s), can be found from

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3600 (26)

where: is the fuel cost per energy unit (based on the fuel LHV); LHV is the lower heating value of the methane fuel (Btu/lb); is the fuel flow rate (lb/s); i indexes the five equipment items: air compressor, combustion chamber, turbine (gas and power turbine, here use = ), air cooler, and HRSG; Zi, is the purchase cost of the ith component ($) (see Table 4); CRF is the annual capital recovery factor (CRF =18.2%); N is the number of hours of plant operation per year (N = 8000 hr and 3600 s/hr); and is the maintenance factor ( = 1.06). For the cogeneration system, the fuel and equipment costs are being brought to an equivalent $/s basis through the capital recovery factor (see ESRL Module 3.). For cogeneration systems fuel costs often dominate. Here = $4.2204/MMBtu based on the LHV of the fuel. Natural gas is actually priced based on the higher heating value of the fuel which is ~110% the LHV. Therefore fuel costs in this problem would be ~ $4.60 natural gas which would be ~ $4.60/ MM Btu or ~ $4.60/1000 SCF (standard cubic feet).

Table 5: Cost Equations (Valero et al., 1994) Component Capital Investment Costs ($)

Compressor

Combustion

Chamber

1

Turbine 1

Air Cooler

.

HRSG . .

.

In Table 4, Cij are cost coefficients (see Table 5); , is the overall heat transfer coefficient in the air cooler,

= 9.478

1000

2 , where the 1000 in the denominator is a correction needed in the original formulation;

and , is the log mean temperature difference in the specified heat recovery unit.

Table 6: Cost coefficients for the cogeneration components (Valero et al., 1994). Compressor C11 = 17.95 $/(lbm/s), C12 = 0.9

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Combustion Chamber C21 = 11.64 $/(lbm/s), C22 = 0.045, C23 = 0.01 (1/R), C24 = 26.4 Gas Turbine C31 = 212.0 $/(lbm/s), C32 = 0.92, C33 = 0.02 (1/R), C34 = 54.4 Air Cooler C41 = 2290 $/(m1.2) HRSG C51 = 6097.2 $/(Btu/(s R))0.8, C52 = 5361.5$/(lbm/s), C53 = 254.8$/(lbm/s)1.2

In the costing equation, the use of the log-mean temperature difference in the air cooler and HRSG represents the appropriate average temperature driving force of the heat exchanger, and is calculated as:

, (27)

, (28)

, (29)

The results in the Excel file Cogeneration Air Standard.xls show that the cogeneration system will have a total installed capital cost of $3,961,003 and a yearly fuel cost of $6,664,373. The capital cost equations (from 1994) should be updated as the installed cost of this cogeneration system in current dollars would be ~ $15 - 20MM.

II Real Gas System Solution All calculations and results for the Real Gas System calculations can be found in the Excel file Cogeneration Real Gas.xls.

The problem we are now facing is that the “pure” components, air and fuel, will be combusted to give CO2, H2O, N2 and Ar and unreacted O2. We need to determine the amount of fuel needed to obtain the combustion chamber temperature and the physical properties of the exhaust gas. It quickly becomes cumbersome to calculate the amounts of each combustion species and determine the physical properties for the mixture in each stream after the combustion chamber. The process for generating physical properties of the combustion products properties is detailed in Knopf (2012) and in Table 7 we list the function calls that were developed as dlls (Dynamically Linked Libraries) for use within Excel.

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Table 7a Combustion products properties, with units: T (R ); P (psia); (Btu/lb); (Btu/lb-R);

;

Species , , , ̂ , , , , , , , ̂ , ,

Products H_Products S_Products TfromH_Products TfromS_Products

Table 7b Combustion products properties with units: T (K ); P (MPa); (kJ/kg); (kJ/kg-K);

;

Species , , , ̂ , , , , , , , ̂ , ,

Products_SI H_Products_SI S_Products_SI TfromH_Products_SI TfromS_Products_SI For cogeneration problems it is generally convenient to use lbm as the flow basis. These functions all contain the key word Products and for field units use T (R), P (psia), h (Btu/lb), s (Btu/lbR) and for SI units use T (K), P (Mpa), h (kJ/kg), s (kJ/kg-K). Use of these functions is not restricted to just the combustion chamber, they can be used anywhere in the cogeneration process (where exhaust gas is present). Table 8 supplies details for each of the functions in Table 7. Table 8 Exhaust gas “Products” properties available in either Field or SI units H_Products (Pressure, Temperature, H/C Ratio, DAR) this function calculates the enthalpy of the exhaust gas in Btu/lb at Pressure in psia, Temperature in R and subtracts the enthalpy of the exhaust gas at the reference state (14.696 psia, 536.67 R). TfromH_Products (Pressure, Enthalpy, H/C Ratio, DAR) this function calculates the temperature of the exhaust gas in oR using the enthalpy of the exhaust gas Enthalpy in Btu/lb at Pressure in psia. The Enthalpy parameter accounts for the reference state (14.696 psia, 536.67 R). S_Products (Pressure, Temperature, H/C Ratio, DAR) this function calculates the entropy of the exhaust gas in Btu/lb-R at Pressure in psia, Temperature in oR, and subtracts the entropy of the exhaust gas at the reference state (14.696 psia, 536.67 R). TfromS_Products (Pressure, Entropy, H/C Ratio, DAR) this function calculates the temperature in of the exhaust gas in oR using the entropy of the products Entropy in Btu/lb-R at Pressure in psia. The Entropy parameter accounts for the reference state (14.696 psia, 536.67 R). H_Products_SI (Pressure, Temperature, H/C Ratio, DAR) this function calculates the enthalpy of the exhaust gas in kJ/kg at Pressure in MPa, Temperature in K and subtracts the enthalpy of the exhaust gas at the reference state (0.101326 MPa, 298.15 K). TfromH_Products_SI (Pressure, Enthalpy, H/C Ratio, DAR) this function calculates the temperature of the exhaust gas in oK using the enthalpy of the exhaust gas Enthalpy in kJ/kg at Pressure in MPa. The Enthalpy parameter accounts for the reference state (0.101326 MPa, 298.15 K). S_Products_SI (Pressure, Temperature, H/C Ratio, DAR) this function calculates the entropy of the exhaust gas in kJ/kg-K at Pressure in MPa, Temperature in oK, and subtracts the entropy of the exhaust gas at the reference state (0.101326 MPa, 298.15 K). TfromS_Products_SI (Pressure, Entropy, H/C Ratio, DAR) this function calculates the temperature in of the exhaust gas in oK using the entropy of the products Entropy in kJ/kg-K at

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Pressure in MPa. The Entropy parameter accounts for the reference state (0.101326 MPa, 298.15 K). MW_Products this function calculates the molecular weight of the exhaust gas. This function is used internally by the C program combustion library

Solving the Real Gas System - Here the same calculations as developed for the Air Standard System can be used for the air compressor. 2 Combustion Chamber (CC): For the combustion chamber we again utilize equation (11), which is repeated here,

, ,

(30)

For a chosen we can determine 3 , . As 3 is known we can determine 3 using: = TfromH_Products (P3, , , H/C Ratio, DAR) (32)

For methane fuel, the H/C ratio = 4 and DAR, the dry air ratio is . . The dry

air ratio is the ratio of the air actually used, , to the theoretical minimum air requirement. Do recall when using the Products functions that the reference state is included. As 3 is known (design variable) we can use Goal Seek in Excel (or any iterative process) to vary until 3 = 2500R.

3. Gas Turbine (GT):

An isentropic gas turbine would obey: ̂ , ̂ , , , (31)

For the gas turbine 3, and 4, , are the entropies of the combustion exhaust gas (Btu/lb-R) at the inlet and exit of the turbine and 4, is the isentropic outlet temperature of the exhaust gas. Since 3 and 3 are known from our combustion chamber calculations, 3 can be found using the entropy function for the exhaust gas, S_Products(P3, T3, H/C ratio, DAR). 4, can be found from 4, and a chosen 4, TfromS_Products(P4, s4,isen, H/C ratio, DAR). Using 4, we can solve for 4, as 4, , using, , , = H_Products(P4, T4,isen, H/C ratio, DAR) (32)

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The work done in the air compressor is supplied by direct shaft coupling to the gas turbine. We can determine 4, by using the known work done in the air compressor:

, , ,∗ , ,

(33)

Finally we can determine the gas turbine efficiency as, , ,

, , (34)

We know from our design conditions that = 0.86. We can simply vary 4 using Excel Goal Seek (or any iterative process) until the design efficiency is obtained. 4can now be found from the Excel function TfromH_Products.

4. Power Turbine (PT):

An isentropic power turbine would obey: ̂ , ̂ , , , (35)

We know there is a 5% pressure drop in the HRSG, so 5 = 1.066 bar. We can then solve for , and , using the Excel functions. The actual exhaust enthalpy 5, at 5 will be: , , , , , , & (36)

and 5can be found from the Excel functions. The work (Btu/s) produced by the power turbine is:

, , (37)

5. Heat Recovery Steam Generator (HRSG): For the HRSG we are not considering supplemental firing so the conditions at 5 and 6 are the same. The exhaust gas enthalpy is available from the power turbine calculations. The exhaust gas enthalpy at ( , ) can be evaluated using H_Products(P7P, T7P, H/C ratio, DAR); the HRSG gas-side pressure drop is known and can be taken as the average of and . The HRSG calculations then follow those developed for the Air Standard System solution.

From Table 9 and the Excel file Cogeneration Real Gas.xls we find / = 2.727; = 20.339; / = 10377.2; and, / = 26.196.

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Ideal Gas System Solution (from ESRL Module 1)

This cogeneration problem, with the ideal gas assumption, was solved in Module 1. In that module we used = 1.4, , = 0.24 Btu/lb-R, = 1.33, and , = 0.28 Btu/lb-R. This solution,

with the inclusion of water/ steam properties by function calls, can be found in the Excel file, Cogeneration Ideal Gas.xls. Key results have been added to Table 9.

Comparison of Solution Methods

In Table 9 we compare system performance results to those found using the utility system design program GateCycle (from General Electric). The Ideal Gas System assumption provides reasonable estimates, when compared to the GateCycle or Real Gas System solution, for key values of fuel flow rate, net power generation, and steam flow rate. It is also interesting to compare the solution accuracies based on costs - cogeneration system costs are often dominated by fuel costs and from Table 9 the ideal gas assumption again does a reasonable job at predicting fuel costs when compared to the more accurate solutions. Table 9 Cogeneration System - Comparison of Solution Methods

Variable / Parameter

Ideal Gas System Ideal Air and

Ideal Product Gas Real Steam Properties

Air Standard System

Real Air Properties and

Real Steam Properties

Real Gas System Real Air

Real Product and Real Steam Properties

GateCycle

/ 154.6 154.6 154.6 154.6

519.67 519.67 519.67 519.67 0.84 0.84 0.84 0.84

1281.19 1259.64 1259.64 1248.5 16.8 16.8 16.8 16.8

0.98 0.98 0.98 0.98 / 2.795 2.550 2.727 2.75

2500 2500 2500 2500 0.86 0.86 0.86 0.86

1858.86 1860.88 1892.28 1902.41 4.032 3.978 4.121 4.22

0.81 0.81 0.81 0.81 1435.65 1429.45 1456.03 1460.06

R 30 30 30 30 ΔT R 60 60 60 60 F lb/s 26.72 23.878 26.196 26.376

T R 733.81 729.10 719.49 719.38 19.67 19.169 20.339 20.41 /

10998.4 10297.3 10377.2 10294.5

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$/ 7,305,335 6,664,373 7,126,088 7,189,009 $ 4,007,215 3,961,003 3,999,300

Acknowledgements This work was completed as part of the National Science Foundation Phase II grants: NSF Award 0716303 “Integrating a Cogeneration Facility into Engineering Education” (September 2007 – 2012); and NSF Award 1323202 “Collaborative Proposal: Energy Sustainability Remote Laboratory” (September 2014 – 2016).

References

Bathie, W.W., Fundamentals of Gas Turbines (2nd Edition), Wiley, New York, NY (1996). Bolland, O. “Thermal Power Generation (compendium).” Norwegian University of Science and Technology, September (2008). Ganapathy, V., Waste Heat Boiler Deskbook, The Fairmont Press, Liburn GA (1991). Knopf, F. Carl., Energy Modeling, Analysis and Optimization of Process and Energy Systems, Wiley & Sons Inc., New Jersey (2012). Reale, M.J., and J. K. Prochaska, New High Efficiency Simple Cycle Gas Turbine – GE’s LMS 100TM. Paper No. 05-IAGT-1.2, Presented at the 16th Symposium on Industrial Applications of Gas Turbines (IAGT), Banff, Alberta, Canada, October 12-14 (2005). Reynolds, W.C., Thermodynamic Properties in SI, Stanford University, Stanford, CA (1979). Valero, A.; Lozano, M.A.; Serra, L.; Tsatsaronis, G.; Pisa, J.; Frangopoulos, C.; von Spakovsky, M., CGAM problem: definition and conventional solution, Energy – The International Journal Vol. 20(3), 279 (1994).

Student Assignments You are responsible for Problem 1 (3 parts) and Problem 2. Problem 1a. Cogeneration System with Air Preheat There are several common cogeneration configurations one of which is shown in Figure P1. This is a regenerative system as the exhaust gas from the turbine is used to help heat the incoming air stream prior to combustion.

19

barP

RT

013.1

17.547

1

1

barP

RT

20

67.536

8

8

barP

RT

Fsteam

20

874

9

9

Figure P1 Cogeneration Power Cycle with Air Preheater and HRSG The system in Figure P1 consists of an air compressor, air preheater, combustion chamber, a single stage gas and power generating turbine and heat recovery steam generator (HRSG). The turbine can be viewed as 4 parts. First ambient air is compressed. The air then further heated in a counter current heat exchanger by exchanging energy with the exhaust gas from the gas and power turbine. Then fuel is added to the compressed air and combusted. The hot exhaust gas passes through a single stage gas and power generating turbine. The power from the turbine is used to both compress the incoming air and produce electricity. The exhaust gas from the air preheater passes through a countercurrent heat exchanger – the heat recovery steam generator – where condensate returned from the process is converted to steam.

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Solve for the performance and cost of this cogeneration system using Table P1a and where appropriate, our previous developments. At design conditions, the incoming air is 87.5 F. The compressor isentropic efficiency ( ) is 84%, and the gas and power turbine efficiency ( & ) is 86%. The gas and power turbine generates electricity with is used to both power the air compressor and supply electricity to the process. The combustion chamber efficiency ( ) is 98% and pressure drop in the combustion chamber is negligible. The exhaust from the gas and power turbine first passes through a countercurrent heat exchanger where energy is exchanged with air from the air compressor. There is a 5% pressure drop on both the air and exhaust gas sides of this air preheater heat exchanger. In the air preheater the provided design specification is

= 10 R. In the HRSG there is also a 5% pressure drop on the exhaust gas side, the pinch temperature difference ( ) is 30 R, and the approach temperature difference ( ) is 60 R. Pinch and approach temperature differences have been previously explained. Water enters the HRSG at 536.67 R and 20 bar and steam is produced at 874 R and 20 bar. We neglect pressure drop on the water / steam side of the HRSG. For the cogeneration system performance we are assuming ideal gas properties and here we use for air, = 1.4, , =

0.24 Btu/lb-R, and for combustion products, = 1.33, , = 0.28 Btu/lb-R. Do note that some of the parameters in the costing equations of Tables 5 and 6 will need to be adjusted for this system (Figure P1). For example, for the air preheater use:

.and assume = .

Solve for the performance and cost of this cogeneration system under: 1.) the ideal gas assumption; 2.) the air standard assumption; and, 3.) the real gas assumption. Design information is provided in Table P1a. Supply your results to Table P1b. Table P1a Air Preheat Cogeneration System Design Information

/ 154.6

1 547.17 0.84

2 16.8 0.98

2500

& 0.86 10

30 60

8 536.67 8 20.0 9 874

9 20.0

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Table P1b Cogeneration System - Comparison of Solution Methods

Variable / Parameter

Ideal Gas System

Ideal Air and Ideal Product

Gas Real Steam Properties

Air Standard System Real Air

Properties and Real Steam Properties

Real Gas System Real Air

Real Product and Real Steam Properties

GateCycle

/ 154.6 154.6 154.6 154.6

547.17 547.17 547.17 547.17 0.84 0.84 0.84 0.84

1348.99 1310.07 16.8 16.8 16.8 16.8

1446.99 1466.33 0.98 0.98 0.98 0.98

/ 2.496 2.296

2500 2500 2500 2500

& 0.86 0.86 0.86 0.86 1456.99 1476.33

1.10 1.10 1374.32 1327.99 1.07 1.05

30 30 30 30 60 60 60 60

/ 23.59 19.74 753.44 765.23

17.01 17.93

/ 11354.9 9913.21 $/ . 6,521,904 5,999,777 $ 5,963,276

Solution Hint: When solving the energy balance for the air preheater you will need to utilize the approach temperature difference provided in Table P1a, = 10 R. But you will not know until downstream units (combustion chamber and gas and power turbine) are solved. This is a common problem. Remember you can always guess a needed value to allow the calculations to continue and then make corrections later. Ideal Gas Assumption: We solved this problem in Module 1 as Problem #1. The solution can be found in the provided Excel file, Problem Cogeneration Ideal Preheat.xls and here we have allowed for water/ steam properties to be determined by Excel function calls. Air Standard Assumption: Start with the provided Excel file, Problem Cogeneration Ideal Preheat.xls; here we have allowed for air, and water/ steam properties to be determined by Excel

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function calls as documented in the module text and the Table of Function Names.pdf provided in this module. Real Gas Assumption: Again start with the provided Excel file, Problem Cogeneration Ideal Preheat.xls; here we have allowed for air, combustion products, and water/ steam properties to be determined by Excel function calls as documented in the module text and the Table of Function Names.pdf provided in this module. Problem 1b. Heat Rate Compare the cogeneration configurations in Figure 1 and Figure P1. With air preheating (as in Figure P1) we would anticipate a lower heat rate compared to the configuration in Figure 1. Discuss this. Problem 1c. Excess Oxygen Using your final results, determine the amount of excess oxygen in the streams leaving the combustion chamber in Figures 1 and P1. Problem 2. The GE LMS100 shown in Figure P2 uses an air cooler (intercooler) between low- and high-pressure air compressors; these are compressors 1 and 2 in Figure P2. General Electric reports the following benchmarks for the LMS100 when using an air flow rate of 209 kg/s at design conditions: net power 98.7 MW; a heat rate of 7509 Btu/kW-hr; thermal efficiency of 46%; and, an exhaust temperature 410 C. The pressure ratio in the LMS is reported as 42:1.

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Fuel

1Compressor Turbine

Generator

gasExhaust

AirInlet

00 , PT

11 , PT er_intercool

33 , PT

55 , PT

Combustor

44 , PT

2Compressor

11 , PT

22 , PT

rIntercoole

Figure P2 Overview Schematic of the GE LMS 100 (Reale and Prochaska, 2005; used with permission). Schematic for GE LMS 100 Performance Calculations (nomenclature used in provided Excel solutions) Using the component efficiencies and properties in Table P2a determine the performance of the LMS 100 and compare these values to those reported by GE. Use: = 209 kg/s; a compression ratio, 2/ 0 = 42; 1_ = 15 C; a 4% pressure drop in the combustion chamber; 3= 1250 C; and 5 = 1.023 bar. Use the optimum pressure ratio for 1/ 0 and 2/ 1; show yourself this

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will occur when 1/ 0 = 2/ 1. Remember you will have to use Goal Seek in Excel to vary the fuel flow until 3= 1250 C. In Table P2a complete the both real gas solution and the ideal gas solutions. Table P2a Design Turbine Performance Calculations using Ideal Gas and Real Gas Properties (see also Bolland, 2008)

Variable / Parameter GE Reported Design

Design Ideal Gas

Design Real Gas

Air gamma, 1.38 Air specific heat capacity, , (kJ/kg-K) 1.046 Compressor inlet flow, 0 (kg/s) 209 209 209 Compressor inlet temperature, 0 (K) 288.15 288.15 288.15 Compressor inlet pressure, 0 (bar) 1.013 1.013 1.013 Compressor isentropic efficiency, 0.89 0.89 Compressor outlet temperature, 2 (K) Compressor outlet pressure, 2 (bar) 42.546 42.546 42.546 Compressor work (kJ/s) Fuel lower heating value (kJ/kg) 50010 50010 Fuel Flow, (kg/s) Exhaust gas gamma, 1.31

Exhaust heat capacity, , (kJ/kg-K) 1.237 Turbine inlet flow, (kg/s) Turbine inlet temperature, 3 (K) 1523.15 1523.15 Turbine inlet pressure, 3 (bar) 40.844 40.844 Turbine isentropic efficiency, 0.90 0.90 Turbine outlet temperature, 5 (K) 683.15 Turbine outlet pressure, 5 (bar) 1.023 1.023 Turbine work (kJ/s) Turbine net work (kJ/s) 98,700 Heat rate (kJ/kW-hr) 7922.4

Student Assignments - Laboratory Course

You should complete the Student Assignments (above) before attempting this laboratory course student assignment.

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Laboratory Course – Student Assignment - Go to the web site www.esrl.lsu.edu or www.cogened.lsu.edu and access operational data from the LSU cogeneration facility found in the folder Cogeneration Operational Data. The performance calculations for this facility have been performed in the first part of the module. Using data from the operational facility and real gas properties, determine the efficiencies: , , and .