HRSG. Combined Cycle HEAT RECOVERY STEAM GENERATOR

Transcription

1 Combined Cycle Fuel Q in Topping Cycle Gas Turbine Power Plant WT HRSG QChimney Bottoming Cycle Steam Power Plant WB Qout HEAT RECOVERY STEAM GENERATOR Heat Recovery Steam Generators (HRSG's) absorb heat from the exhaust of gas turbines to produce steam. Functionally they produce steam like any other boiler but with some differences. Combined heat and power, or co-generation as it is sometimes known, is simply the simultaneous generation of heat and electricity. The generation of electricity using gas turbine or diesel engine produces a large amount of heat which, in conventional methods, is often wasted. Co-generation uses Heat Recovery Boiler to recover this waste heat and convert it into a useful form such as steam or hot water which can be used for a wide variety of process, heating and sometimes cooling applications. HRSG Dr. Kalim- pg. 1

2 What is an HRSG? Heat Recovery Steam Generator's (HRSG's) are waste-heat boilers. The steam turbine or a downstream process uses the steam. The term HRSG refers to the HRSG Dr. Kalim- pg. 2

3 waste-heat boiler in a Combined Cycle Power Plant. In its basic form these are bundles of water or steam carrying tubes paced in the hot gas flow path. These recover the heat from the gas to generate superheated steam, hence the name Heat Recovery Steam Generator. The water steam circuit of an HRSG consists of an economizer, an evaporator, and a Super-heater placed in the flue gas duct. The evaporator section consists of a drum to which the coils are connected to create the circulation. The Differences HRSG is only a heat transfer area. There is no furnace even though the sections like economizer, evaporator, and super heaters are present. An exception is the supplementary-fired HRSG or an additional boiler.. The exhaust gas temperature from a Gas Turbine is about 600 C. Higher exhaust temperatures will reduce the efficiency of the gas turbine. This temperature head available for heat transfer is very low compared to a conventional boiler. In a conventional boiler where fuel burning takes place, the temperature head available for heat transfer is in the range of 1300 C. Because of this low temperature head, proper positioning and apportioning of the heat transfer surfaces is very important. In conventional boilers the evaporation and super-heating takes place in a single pressure level. To get the optimum heat transfer from HRSG in large Combined Cycle plants operate on a triple pressure format. Water-steam conversion takes place in three different pressure levels in three independent circuits. Unlike the power plant boilers, there are no air or gas handling fans. The flow of gases is due to the exhaust gas pressure from the Gas turbine. Pressure drops in the gas path of a HRSG is at a minimum to avoid higher exhaust gas pressure, which will affect the performance of the GT. This also eliminates the need of air preheaters. The output of the HRSG is solely dependent on the performance and load of the gas turbine. Steam temperature control is limited to the use of super-heaters. Performance at part loads depends on the design of heat transfer surfaces. HRSG Dr. Kalim- pg. 3

4 Most of the HRSG's are internally insulated so that expansion does not affect the external surfaces and the structures. Conventional boilers on the other hand are top supported and free to expand so the insulation is on the outside. An HRSG recovers the heat energy from the gas turbine exhaust. It consists of a number of heat exchangers that heat water, evaporate it, and superheat steam. HRSG s for combined-cycle applications are natural, unfired and forcedcirculation, often include supplementary fired steam boilers that cover the turbine power range up to 280 MW and have main steam pressures up to 140 bars (2030 psia) and temperatures up to 545 C (1010 F) (Croonenbrock et al., 1996). The exhaust gas temperature of advanced gas turbines is 600 C. With higher exhaust gas temperatures, the HRSG becomes even more complex, and thus dual-pressure and triple-pressure HRSGs are required. (Bannister et al., 1995). Cold feedwater is fed from the feedwater tank to the economizer. The feedwater temperature in the HRSG of a combined-cycle power plant must be above the dew point temperature to prevent low-temperature corrosion. HRSGs consist of three major components: the Evaporator, Superheater, and Economizer. The different components are put together to meet the operating requirements of the unit. In evaporator section water is converted to steam. This steam then passes through superheaters to raise the temperature and pressure past the saturation point. Economizers are mechanical devices intended to reduce energy consumption. Triple pressure HRSGs consist of three sections: an LP (low pressure) section, a reheat/ip (intermediate pressure) section, and an HP (high pressure) section. Two parameters are very critical in the design of the HRSG. They are the Approach point and Pinch Point. The approach point is the difference between saturation and the water temperature leaving the economizer HRSG Dr. Kalim- pg. 4

5 The Pinch Point PP is the minimum temp difference between exhaust gas and water/steam in the HRSG. It should exceed some minimum design value (about 10 K) for all operating conditions of the system to make effective use of entire HRSG heat transfer surface. Note that the smaller PP values would substantially increase the heat transfer surface area needed from HRSG. It is T 2 -T 5 and T 4 -T 7 in the heat flux and entropy diagrams below. These are not applicable to conventional boilers. HRSG Dr. Kalim- pg. 5

6 HRSG Dr. Kalim- pg. 6

7 Key Components: Gas Turbine Gas Turbine Generator Superheater Steam Drums Blow-down Vessel Evaporator Gas Treatment Unit Water Treatment Unit Economizer Problem 1 Calculate the rate of steam production, the stack gas temperature, and the stack gas heat rejection rate for a single-pressure HRSG of a combined-cycle power plant (see Figures above). The conditions are as listed below: Gas turbine exhaust gas mass flow rate m g = 540 kg/s Exhaust gas temperature T 1 = 610 C Superheated steam pressure P 6 = 60 bars and temperature T 6 = 500 C Feedwater temperature T 4 = 110 C Assume that the HRSG s PP is 10 K and that the exhaust gas specific heat C pg is 1.05 kj/(kg K). Stack Solution 1. Superheated steam enthalpy h 6 = 3425 kj/kg, Feedwater enthalpy h 4 = C pw T fw = x 110 = kj/kg Saturation temperature at P = 60 bars is T 5 = C, and Saturated liquid enthalpy h f5 = kj/kg HRSG FeedWater Exhaust Gas from Topping Cycle Steam Exhaust Gas Temperature 1 Temperature 1 7 Superheater Exhaust Gas Temperature - HRSG 6 Evaporator 2 5 Economizer 3 4 Temperature 3 4 Economizer Pinch Point 2 5 Evaporater 6 Superheater Heat Flux Entropy Temperature-Heat flux and Temperature-Entropy diagram of single- HRSG Dr. Kalim- pg. 7

8 pressure HRSG 2. Designating the states as shown in Figure, we first calculate the rate of steam production per unit mass of the exhaust gas. The flue gas temperature at state 2 is = T 5 + PP = = C Apply energy balance for the superheater and evaporator of the HRSG, m s (h 6 - h 5 )= m g C pg (T 1 - T 2 ) steam production rate is m s = m g C pg (T 1 - T 2 )/ (h 6 - h 5 ) = 540 x 1.05 x ( )/( ) = kg/s 3. So that the rate of heat recovery in the HRSG is Q HRSG = m s (h 6 - h 4 ) = x ( ) = 246,574 kw 4. The stack gas temperature is given by applying the energy balance to HRSG so that Q HRSG = m g C pg (T 1 - T 3 ) which T 3 = T 1 - Q HRSG / m g C pg = ,574/(540 x 1.05) = C 5. The stack gas heat rejection rate is Q stack = m g C pg T 3 = 540 x 1.05 x = 99,296 kw Problem 2 For the single-pressure combined-cycle power plant of problem 1, the following conditions are given (states are designated as shown in Figure): Live steam parameters: T 1 = 500 C, P 1 = 60 bars, enthalpy h 1 = 3425 kj/kg, mass flow rate m s = kg/s Steam turbine isentropic efficiency η t = 0.9 Condenser pressure P 2 = 0.04 bar Gas turbine exhaust temperature t exh = 610 C and Gas mass flow rate m g = 540 kg/s Calculate the actual steam turbine specific work, steam turbine power output, and thermal efficiency of the steam cycle. HRSG Dr. Kalim- pg. 8

9 Solution 1. For the isentropic expansion, the exhaust steam enthalpy is taken from table: h 2 = 2075 kj/kg 2. The actual specific work of the steam turbine is W t,a = (h 1 - h 2 ) η t = ( ) x 0.9 = 1215 kj/kg 3. The steam turbine power output is P t, a = m s W t,a = x 1215 = 101,064 kw 4. The steam cycle heat input rate is equal to the heat of the gas turbine exhaust gas (HRSG) i.e., Q in = m g C pg T exhaust = 540 x 1.05 x 610 = 345,870 kw 5. The thermal efficiency of the steam cycle is η th = P t, a /Q in = 101,064/345,870 = = 29.2 % Dual- and Triple-Pressure Combined-Cycle Power Plants To enhance the utilization of the gas turbine exhaust heat, dual- and triple-pressure combined- cycle power plants are employed. Consider the dual-pressure combined-cycle power plant shown in Figure below. Corresponding figure shows a temperature HRSG Dr. Kalim- pg. 9 LP Economizer LP Evaporator HP Economizer HP Evaporator HP Superheater Stack HRSG FeedWater from condenser and condensate LP Drum HP Drum To Condenser HP Steam Exhaust Gas from Topping Cycle

10 heat flux diagram for the dualpressure combined- cycle power plant. 1 Exhaust Gas Temperature Temperature 10 Superheater 9 Pinch Point HP Evaporater 140 bar High Pressure Approach HP Economizer Superheater 4 7 LP Evaporater 8 bars LP Economizer Low Pressure 5 Stack 6 Heat Flux The energy balance for the high-pressure (HP) evaporator and superheater in the dual-pressure HRSG is given by m g C pg (T 1 T 2 ) = m HP (h 10 h 9 ) kj/s Where m HP is the steam mass flow in the HP section of the HRSG in kg/s. h 10 is the enthalpy of superheated steam leaving the HRSG in kj/kg, and h 9 is the enthalpy of saturated liquid at high pressure in kj/kg. Then the rate of HP steam production in the HRSG is m HP = m g C pg (T 1 T 2 )/(h 10 h 9 ) kg/s From the energy balance for the HP economizer, the gas temperature at the HRSG high-pressure section exit is T 3 = T 2 m HP (h 9 h 8 )/(m g C pg ) C Where h 8 is the enthalpy of HP feedwater in kj/kg. Similarly, the rate of low-pressure (LP) steam production in the HRSG is m LP = m g C pg (T 3 T 4 )/h fg, LP kg/s Item Evaporator type Inlet gas temp, F bare Pinch Point, F finned Approac h, F HRSG Dr. Kalim- pg. 10

11 The heat output rate of the HRSG is Q HRSG = m g C pg (T 1 T 5 ) = m HP (h 10 h 8 ) + m LP (h 8 h 6 ) kw From above, the stack gas temperature is T 5 = T 1 Q HRSG /(m g C pg ) C Alternatively, it is also T 5 = T 3 m LP (h 8 h 6 ) /(m g C pg ) C With single pressure or even multiple pressure HRSGs, an option to improve energy recovery is to use lower pinch and approach points. Problem 3: For a triple pressure combined cycle with the temperature-heat flux diagram is shown in the following figure with the following conditions: Gas turbine exhaust gas flow rate= 540 kg/sec Temperature T 1 = 610 o C, Specific heat Cpg = 1.05 kj/(kg K) Steam cycle parameters: Steam turbine isentropic efficiency ηt = 0.9 De-aerator pressure Pd = 1.5 bars HP live steam condition: P14 =120 bars, T14 =545 C Low-press steam production rate m Lp =10.48 kg/s Steam consum. for the de-aerator m d =12.59 kg/s Pinch point PP = 10 K IP steam pressure: P 2 = 20 bars LP steam pressure: P 3 = 4 bars Cond. pressure Pc = 0.04 bar Feed-water temperature = 110 C Calculate (1) steam production rate, in kg/s (2) steam turbine power output, in kw, and (3) steam plant efficiency. Temperature 1 14 Pinch Point HP Evaporater 120 bars High Pressure Exhaust Gas Temperature - HRSG 4 IP Economizer 20 bars 11 Intermediate Pressure Low Pressure 4 bars 7 8 Stack Heat Flux HRSG Dr. Kalim- pg. 11

12 Solution Steam and water conditions (from water/steam table and h-s diagram): State State Condition Pressure bars HRSG Temperature C Specific Enthalpy kj/kg 8 Feed water Sat, liquid Sat. steam Sat. liquid Sat. steam Sat, liquid Live steam Sat., saturated Condition Pressure bars STEAM POWER PLANT Temperature C Specific Enthalpy kj/kg 14 Live steam s s Specific Entropy kj/kgk Flue gas temperatures are T 1 = 610 C, T 2 = T 13 + PP = = C, T 4 = T 11 + PP = = C, and T 6 = T 9 + PP = = C HP steam production rate is from topping cycle m HP = m g C pg (T 1 T 2 )/(h 14 h 13 ) = 540 x 1.05 x 13 Pump Topping cycle and HRSG T14 = 545 C P14 = 12 MPa ηt= 90% Steam Turbine Condenser 17 P 16= P 17 = 4 kpa Bottoming Cycle T15 = 400 K P15 = 0.4 MPa s15 = ηt= 90% 16 HRSG Dr. Kalim- pg. 12

13 ( )/( ) = kg/s Actual steam exhaust enthalpy is calculated as below: s14 = s15s = bars, s = x15= ( )/(5.1191) = h15 = *2133 = 2731 kj/kg (taken as 2840 from h-s diagram) h 16 = h 15 (h 15 h 16s ) η t = 2731 ( )*0.9 = 2192 kj/kg s s 16 8 Actual specific work for the HP and the LP steam turbine are w HP = h 14 h 15 = = 585 kj/kg w LP = h 15 h 16 = = 648 kj/kg Steam mass flow rate for the HP and the LP steam turbine are m HP = 79.12kg/s m LP = m HP + m LP m d = = kg/s Steam turbine power output is P ST = m HP w HP + m LP w LP = x x 648 = 96, kw Steam plant efficiency is η ST = P ST /Q in = P ST / m g C pg T 1 = 96, 188.8/ 540 x 1.05 x 610 = The plant net heat rate = 3600/ηth HRSG Dr. Kalim- pg. 13

14 Example Fuel Qin Topping Cycle WT Bottoming Cycle WB Qout Net work of the topping cycle - We can write WT = Qin ηt, Q η T = in Qout Qin So that Heat rejected from topping cycle is WT = QT,out = Qin (1-ηT) Similarly Net work of the bottoming cycle WB = QT,out ηb = Qin (1-ηT) ηb Thus Total Net work of the combined cycle WCC = WT +WB = Qin ηt + Qin (1-ηT) ηb Therefore thermal Efficiency of the combined cycle is, ηcc = W CC Q in Or ηcc = {Qin ηt + Qin (1-ηT) ηb}/ Qin = ηt + (1-ηT) ηb = ηt + ηb -ηtηb HRSG Dr. Kalim- pg. 14

15 Topping COMBINED CYCLE POWER PLANT Combustor 2 3 Gas Turbin T3 = 1400 K P3 = P2 = 1200 kpa ηt= 88% 1 4 T5 = 400 K P5 = P4 = 100 kpa 5 HRS Heat-recovery steam generator Steam Turbin 8 Condenser P 9= P 8 = 100 kpa T7 = 400 K P7 = 8 MPa ηt= 90% Bottoming A combined gas turbine-vapor power plant has a net power output of 45 MW. Air enters the compressor of the gas turbine at 100 kpa, 300 K, and is compressed to 1200 kpa. The isentropic efficiency of the compressor is 84%. The condition at the inlet to the turbine is 1200 kpa, 1400 K. Air expands through the turbine, which has an isentropic efficiency of 88%, to a pressure of 100 kpa. The air then passes through the interconnecting heat-recovery steam generator and is finally discharged at 400 K. Steam enters the turbine of the vapor power cycle at 8 MPa, 400 C, and expands to the condenser pressure of 8 kpa. Water enters the pump as saturated liquid at 8 kpa. The turbine and pump of the vapor cycle have isentropic efficiencies of 90 and 80%, respectively. (a) Determine the mass flow rates of the air and the steam, each in kg/s; the net power developed by the gas turbine and vapor power cycle, each in MW; and the thermal efficiency. (b) Develop a full accounting of the net rate of exergy increase as the air passes through the gas turbine combustor. Assume T o = 300 K, P o = 100 kpa. Each component on the accompanying sketch is analyzed as a control volume at steady state. HRSG Dr. Kalim- pg. 15

16 Asumptions: The turbines, compressor, pump, and interconnecting heat-recovery steam generator operate adiabatically. Kinetic and potential energy effects are negligible. There are no pressure drops for flow through the combustor, heatrecovery steam generator, and condenser. An air-standard analysis is used for the gas turbine. Gas Turbine Vapor Cycle State h (kj/kg) S (kj/kgk) State h (kj/kg) s (kj/kg Determine the mass flow rates of the vapor, my, and the air, m g begin by applying mass and energy rate balances to the interconnecting heat-recovery steam generator to obtain HRSG Dr. Kalim- pg. 16

20 Problem 4. Consider a gas turbine based combined cycle. Calculate the power output of the simple-cycle gas turbine and that of the combined cycle if the rate of heat supply to the gas turbine combustor is Q GT = 600 MW. Compare the increases in the power output of the simple cycle and combined cycle power plants if the initial values of the thermal efficiencies of the gas turbine cycle η GT = 0.31 and that of the bottoming steam process η ST = 0.29 are changed to η GT = 0.38 and η ST = Solution 1. The power output of the simple-cycle gas turbine power plant with the initial and changed thermal efficiency of the gas turbine, respectively, is (i) (ii) P GT = η GT Q GT = 0.31 x 600 = 186 MW P GT = η GT Q GT = 0.38 x 600 = 228 MW 2. The overall thermal efficiency of the combined cycle with the initial and changed thermal efficiencies of the gas turbine and of steam turbine, respectively, is (i) η CC = η GT + η ST (1- η GT ) = (1 0.31) = 0.51 (ii) η CC = η GT + η ST (1- η GT ) = (1 0.38) = Correspondingly, the overall power output of the combined-cycle power plant is (i) (ii) P CC = η CC Q GT = 0.51 x 600 MW = 306 MW P CC = η CC Q GT = 0.54 x 600 MW = 324 MW 4. An increase in the overall power output of the combined-cycle power plant is P CC = P CC - P CC = = 18MW or 100% x P CC /P CC = 5.55%. 5. The power output of the bottoming steam process is (i) P ST = P CC P GT =306 l86 = l20 MW (ji) P ST = P CC P GT = = 96 MW Hence there is an increase in the proportion of the overall power output being provided by the gas turbine, thus increasing the efficiency of the bottoming steam process. 6. The plant net heat rate = 3600/ηth HRSG Dr. Kalim- pg. 20