me ii hmt seminar swapnil vanjara
TRANSCRIPT
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Heat Transfer Analysis
ofForced Circulated HRSG
Prepared by Guided by
Swapnil Vanjara Dr. P. Prabhakaran
Professor, msu
Mechanical Depart ment
Thermal Design Ideas for Systematic Development of
Heat Recovery Steam Generators
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H R S G
What?:The heat recovery steam generator (HRSG) is a heat exchanger designedto recover the exhaust waste heat from power generation plant prime
movers, such as gas turbines or large reciprocating engines, thus improving
overall energy efficiencies.
Why?: HRSGs help bring overall plant efficiency to 85%90%, andthe economic and environmental benefits are well recognized. In a recent
Chinese triple-pressure HRSG application, an exhaust gas flow of 702 kg/s
was cooled from an inlet temperature of 596C to 119C at the HRSG outlet
before exhausting to stack. Total heat removed was 371 MWth.
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Role of SG in Rankine Cycle
Perform Using Natural resources of energy .
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Steam Generation Theory
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HRSG Plant Layout
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A Physical View
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Classification
Use basis
cogenration/combined cycle
Circulation basis
naturally/forced
orvertical tube/horizontal tube
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Circulation Basis
Naturally circulated
forced circulated
Vertical tubes can handle much higher heat
fluxes than horizontal tubes, up to40% to 50% more1. Known
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Use Basis
cogeneration Combined cycleefficiency Upto 90 % 6065%
M h i f H T f
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Mechanism of Heat Transfer :
Generalized Newtons Law of
Cooling Rate of heat transfer from hot gas to cold steam is proportional to:
Surface area of heat transfer
Mean Temperature difference between Hot Gas and Cold Steam.
meansur TAQ NEWTON
Thot gas,in
Tcold steam,in
Thot gas,out
Tcold steam,out
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Pinch and Approach PointsNote: This is the Design
mode ..We cannot pre-select pinch and
approach points in off-
design mode!
Pinch Point:- Difference between the gas temperature
leaving the evaporator and temperature of saturatedsteam
Approach Point:- Difference between the temperature
of saturated steam and the temperature of the water
the evaporator
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Pinch and Approach points are selected in unfired mode at Design gas
flow, exhaust gas temperature. These are called design pinch and
approach points
Once selected, they fall in place in other cases of gas flow/inlet gas
temperature/steam conditions, whether unfired or fired.
Pinch/approach points increase with inlet gas temperature
They cannot be arbitrarily selected
Facts about Pinch and Approach
Points
F t b t Pi h d A h
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Facts about Pinch and Approach
Points
They cannot be arbitrarily selected
-temperature cross can occur
-low pinch point may not be physically feasible unless extended surfaces are used
-affected by inlet gas temperature
-economizer steaming is a concern ;suggest minimum approach at coldest ambient
HRSG conditions
-steam temperature can be achieved in fired conditions if it is achieved in unfired
conditions
HRSG surfaces are determined once design pinch/approach points are selected
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Why HRSG exit gas temperatures cannot be
assumedExit gas temperature cannot be assumed as in conventional fired steam generators as
temperature cross can occur.Looking at the superheater and evaporator,we have:
WgxCpgx(tg1-tg3)=Ws(h5-h7) (1) Looking at the entire HRSG,
WgxCpgx(tg1-tg4)=Ws(h5-h8) (2) [blow down and heat loss neglected]
Dividing (1) by (3) and neglecting effect of variations in
Cpg with temperature,we have:
(tg1-tg3)/ (tg1-tg4)= (h5-h7)/ (h5-h8)=K (3)
For steam generation to occur and
for a thermodynamically feasible
temperature profile,two conditions
must be met: If pinch and
approach points are arbitrarilyselected,one of these may not be
met.
tg3>ts and tg4>tw1.
Psig stm temp,F sat temp,F K exit gas,F
100 sat 338 .904 300
150 sat 366 .8704 313
250 sat 406 .8337 332
400 sat 448 .7895 353
400 600 450 .8063 367
600 sat 490 .7400 373
600 750 492 .7728 398
Pinch=20F,approach=15 F,gas
inlet=900 F,feed water=230 F
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RFERANCES
U.P.B. Sci. Bull. Series D, Vol. 71, Iss. 4, 2009
ISSN 1454-2358
ADVANCED HEAT TRANSFER BY V. GANPATHY
POWER PLANT ENGINEERING BY ARORA &
DOMKUNDWR
ASME PTC 4.4-2008 [Revision of ANSI/ASMEPTC 4.4-1981 (R2003)]
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Simplified HRSG PerformanceUsing the concept that firing in a HRSG is
100 % efficient,we can evaluate the
performance in fired case for estimationpurposes.
Example:160,000 lb/h of exhaust at 950 F
enters a HRSG to generate 600 psig
steam at 750 F from 230 F
water.Determine unfired steam production
and also burner duty,firing temperatureand exit gas temperature when generating
35,000 lb/h of steam at 600 psig,750 F.
Solution:Using 25 F pinch and 20 F approach,compute energy absorbed by
SH+evap=160,000x0.27x(950-517)x0.98=18.33 MM Btu/h=Ws(1378.9-455.4) or
Ws=19,850 lb/h. Energy absorbed by HRSG=19,850x(1378.9-199.7)=23.4 MM
Btu/h=160,000x0.98x0.268x(950-tg4) or tg4=393 F.
Fired case: Energy absorbed by steam=35000x(1378.9-199.7)=41.27 MM Btu/h.
Additional fuel energy required=(41.27-23.4)=17.87 MM Btu/h.
Oxygen consumed=17.87x106/(160000x58.4)=1.91 % So there is plenty of oxygen left.
Firing temperature=17.87x106=160000x0.3x(T-950) or T=1322 F
Exit gas temperature=1322-41.27x106 /(160000x.275x.98)=364 F
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Design & Off-design calculationsDESIGN
unfired
establishes configuration
establishes surface areas indirectly
only one case
zero desuperheater spray
pinch and approach points selected
zero economizer steamingOFF-DESIGN
unfired/fired/fan mode/combination
several cases possible
computes desuperheater spraypinch and approach points computed
economizer steaming possible
WHATIF STUDIES
steam pressure variations
firing temperature restrictions
effect of fuels
performance testing
effect of gas turbine load
variations in ambient temperature
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A simple example of simulationThe energy transferred to the evaporator is given by:
Q=WgCp(T1-T2)=UST=US (T1-T2)/ln[(T1-ts)/(T2-ts)] ; simplifying,
ln[(T1-ts)/(T2-ts)]=US/WgCp . In a fire tube boiler,U Wg0.8. For a water tube
boiler,U Wg0.6 ,neglecting the effects of temperature.
Then, Wg0.2ln[(T1-ts)/(T2-ts)]=K1 for a fire tube boiler
and Wg0.4ln[(T1
-ts)/(T2
-ts)]=K2
for a water tube boiler
Example:A water tube boiler is designed to generatesteam at 250 psig with 100,000 lb/h of flue gas at 1000
F.Exit gas temperature is 500 F.What is the exit gas
temperature when 90,000 lb/h of flue gas enters the boiler
at 970 F and steam pressure is 200 psig?
Solution: First compute K2 using design conditions...1000000.4ln[(1000-406)/(500-406)]=184.4=K2
In the off-design case,900000.4ln[(970-388)/(T2-
388)]=184.4 or T2=473 F.Duty and steam generation may
be computed from this.
[406 and 388 F are saturation temperatures
corresponding to 250 and 200 psig respectively.]
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Example of a HRSG simulation
Example:140,000 lb/h of turbine exhaust gases at 980 F enter a HRSG generating sat
steam at 200 psig.Determine the steam generation and temperature profiles if feedwater temperature is 230 Fand blow down=5%.
Solution: Let us choose a pinch point of 20F and approach of 15 F.Sat
temperature=388F. Gas temperature leaving evaporator=408 F and water temperature
entering it is 373 F.Evaporator duty=140000x.99x.27x(980-408)=21.4 Mm Btu/h. [ 1%
heat loss and average specific heat of 0.27 Btu/lbF is assumed]
Enthalpy absorbed in evaporator=1199.3-345+.05x(362.2-345)=855.2 Btu/lb
[1199.3,345 and 362.2 are enthalpies of sat steam,water entering evaporator and
saturated water respectively]. Hence steam generation=21.4x106/855.2=25,000 lb/h
Economizer duty=25000x1.05x(345-198.5)=3.84 Mm Btu/h .gas temperature
drop=3840000/(140000x.253x.99)=109 F.Hence exit gas temperature=408-109=299 F
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Off-design PerformanceSimulate the HRSG performance with a 165,000 lb/h of gas flow at 880 F.Steam
pressure =150 psig.
Using the model for evaporators discussed elsewhere,ln[(980-388)/(408-388)]=Kx140000-0.4 or K=387.6 Under new conditions: ln[(880-366)/(Tg-
366)]=387x165000-0.4 =3.1724 or Tg=388 F.Evaporator duty=165000x.99x.27x(880-
388)=21.7 MM Btu/h
In order to determine the steam flow,the feed water temperature to evaporator must
be known.Try 360 F.Then steam flow=21.7x106/[1195.7-332)+.05x(338.5-332)]=25,110 lb/h. Economizer duty(assumed) Qa=25110x1.05x(332-198.5)=3.52MM
Btu/h.Compute (US)d=Q/T for economizer based on design conditions. Q=3.84x106
T =[(408-373)-(299-230)]/ln[(69/35)]=50 F.(US)d=3840000/50=76800. Correct this
for off-design case. (US)p=(US)dx(165000/140000).65=85200.The effect of variations
in gas temperature is minor and not considered. The energy transferred =(US)p xT.
Based on 360F water exit temperature,the economizer duty=3.52MM Btu/h and gastemperature drop=3520000/(165000x.99x.253)=85 F or exit gas =388-85=303
F.T=[(303-230)-(388-350)]/ln[(73/28)]=47 F or transferred duty=85200x47=4.00 Mm
Btu/h.As this does not match the assumed value of 360F and duty ,another iteration is
required. It can be shown at 366 F,the balance is obtained.
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HRSG Performance Calculations
Performance may
be obtained even if
HRSG geometry is
unknown using
simulation concept.
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Why are HRSGS inefficient?
Low steam/gas ratios
Low inlet gas temperatures(900 F vs 3300 F)
Temperature profiles depend on steam
pressure and temperature
Higher the pressure,lower the steam generation
Higher the steam temperature,lower the steam
generation (and higher the exit gas temperature)
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ImprovingHRSG EfficiencyDesign with lower pinch and approach points
Use of secondary surfaces such as condensate
heater,heat exchanger,deaerator
Consider multiple pressure HRSG
Use supplementary firing
Optimize temperature profiles by rearranging
surfaces
I i HRSG f
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Improving HRSG performance
Bottom line is tolower the exit gas
temperature!
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RESULTS OF A SIMPLE STUDY
Data base Cond htr Heatexch LPevapGas inlet temp,F 975 975 975 975
Stack gas temp,F 374 310 323 297
Steam to turbine,Klb/h 80 80 80 80
Steam to deaerator 10250 1730 3400 0
Feed water temp,F 240 240 151 240
Electric power,kw 6528 6830 6770 6890
Gas flow=550,000 lb/h pinch=20 F approach=20F,make up=60 F,condpr=2.5 in hg,steam at 620 psig,650F
HRSG simulation
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HRSG simulationKnowing gas flow,temperature,analysis and steam parameters,establish HRSG
temperature profiles,duty and steam flows.In the design case,solve for:UA=Q/T.Inthe off-design case knowing the new gas parameters,use the NTU method to establish
performance using Q=(UA)T.Correct for UA using new gas parameters. We do nothave to compute U. Hence there is no need to know the tube size,fin details,HRSG
mechanical data;anyone can perform such calculations and evaluate HRSG
performance in unfired,fired modes,evaluate burner duty,optimize temperature
profiles,predict part load performance,review performance different gas turbines...
HRSG T fil
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HRSG Temperature profile
HP stage is followed by LP section. Not a very efficient design
HRSG T fil
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HRSG Temperature profile
Using common Economizer concept,we improve energy recovery
HRSG f t L L d
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HRSG performance at Low Load
HRSG performance at 40 % load. Note steaming in
economizer and also the high exit gas temperature.
HRSG Simulation unfired case
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HRSG Simulation-unfired case
HRSG simulation fired case
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HRSG simulation-fired case
Eff f bi
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Effect of ambient temperature on
HRSG performance
Multiplication factor on steam flow is 0.1
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Evaluating HRSG performance
HRSG performance is evaluated at different gas flow,exhaust
temperature conditions to see if the performance is reasonable.
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Evaluating HRSG performance
Designbasis
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We are trying to see if a 2 pressure HRSG isrequired. Customer wants about 40,000 kg/h,30
kg/cm2 steam and 3000 kg/h steam at 6 kg/cm2
in fired mode and about 3500 kg/h LP steam in
unfired mode,which is taken off the drum and
pressure reduced..
TWO OR SINGLE PRESSURE HRSG-case 1
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Multiple Pressure Level HRSG
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Design Problem
Design a forced- circulation waste heat recovery boiler to
recover energy from the gas turbine exhaust with the following
parameter:-
Gas turbine capacity = 15 MW
Fuel = natural gas Gas quantity =900,000 lb/hr
Gas temp. entering the boiler =900dg.F
Steam pressure @ evaporator =400psia
Super heater steam temp. =600dg.F
feed water temp. entering economiser =260dg.F
Radiation losses 2%
fouling effect s may be neglected
and suggested total gas pressure is 8 in wc.
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Solution
Assume pinch point 40dg F and
Aapproch point
from steam table @ 400 psia Tsat =445dg.F
Hence, Gas temp leaving the evaporator will be
Tg3 = Tsat + Tpinch= 445 + 40
=485dg.F
Temperature leaving economiser(tw2)
T7 = Tsat - Tapproch
= 445 - 40
=400dg.F
T7(tw`2
tg1
T6
tg3
T5
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Heat Balance
Heat loss in evaporator and super heater
Q(sh+evap)=Wg*Cp*(tg1-tg3)* radiation losses
=900000*1.088*(500-245)*0.98
=95x10E6 BTU/hr
(Cp=0.26 BTU/lb dg.f)
Steam Generated
Ws=Q(sh+evap)/(h5-h7)
h5=enthalpy of super heated steam (assuming 10 psia pressuredropin SH=1308BTU/lb
h7=water enthalpy entering evaporator=375BTU/lb
Ws= 95E6/(1308-375)
=101820lb/hr
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Heat Balance (con)
Economizer Absorption
Q(economizer)=Ws*(h7-h8)
h8=Enthalpy of feed water at 260 F=229BTU/lb
Q(economizer) =101820(375-229)=14.9x10E6 BTU/hr
Super Heater Absorption
Qsh=Ws*(h5-hv)
hv=enthalpy of saturated steam at 400psia=1205BTU/lb
Qsh=101820(1308-1205)
=10.5x 10E6 BTU/hr
Gas temp. leaving Economizer
Tg4= tg3-(Q(eco)/(wg*Cp*%radiation losses))
=485-(14.9x10E6/(900000*0.26*0.98)
=420dg.F
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Heat Balance (con)
Gas temp. drop
= Q(SH)/ (wg*Cp*%radiation loss)
=10.5x10E6/(900000*0.26*0.98)
=45dg./F Gas temp. entering Evaporator
=Tg1-Gas temp drop
= 855 gd.F
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Results
Section Q (BTUx10E6/hr)
Superheater 10.5
Evaporator 84.5
Economiser 14.9