me ii hmt seminar swapnil vanjara

7/29/2019 Me II Hmt Seminar swapnil vanjara

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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


28/42

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


31/42

HRSG simulation-fired case

Eff f bi


32/42

Effect of ambient temperature on

HRSG performance

Multiplication factor on steam flow is 0.1


33/42

Evaluating HRSG performance

HRSG performance is evaluated at different gas flow,exhaust

temperature conditions to see if the performance is reasonable.


34/42

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

me ii hmt seminar swapnil vanjara

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