A Presentation on integration of Rankine and Brayton Cycle

Presented by:Manish Kumar JaiswalUpendra YadavVikas UpadhyayPushpendra MishraVamshi KanugantiAmit srivastave

Instructor :Dr Laltu Chandra

Outline• Block & T-s diagram of combined cycle• Calculations for inputs of heat exchanger• Working Principle for heat exchanger• Design Principles for heat exchanger• Calculations for dimensions

Exhaust gases from Brayton at 650ċ is used to superheat the saturated steam coming out of boiler at 311.1ċ because irreversiblity in superheater is lesser when compared to other parts.

From energy balanceHeat lost by exhaust gasses=heat gain by steamExhaust temprature of gases comes to be 613k.So this exhaust can be used in regeneration of

Brayton Cycle

T1C

SG &E

T2

CDCP

HI =HEAT INPUTT1= BRAYTON TURBINESH=SUPER HEATERRG=REGENERATORC= COMPERSORIC= INTER COOLER

T2=RANKINE TURBINECD=CONDENSORCP=CONDESATE PUMPSG&E=STEAM GENERATOR AND ECONOMISER

HI

RG

SH

IC

T

3 4

T2=47.89ᴼC

1 6

5

f

egh

a

b c i

d

a-b=isentropic compressiona-c=non isentropic compression c-i=heat recovered in regenerationi-d=heat added in heat exchangerd-e= isentropic expansiond-f= non isentropic expansion f-g= heat transfer to saturated steam in super heaterg-h= heat transfer from hot fluid in regeneratorh-a=heat rejected to inter cooler

1-2= isentropic pump 2-3= heat added in economizer 3-4= heat added in steam generator4-5= heat added in super heat exchanger 5-6= turbine expansion 6-1= isobaric heat rejection

Temperature ‘C Enthalpy(KJ/Kg) 1 45.8 191.8 2 45.8 201.89 3 311.11 1407.6 4 311.11 2724.7 5 400 3096.5

6 45.8 1966.38

Rankine cycle Design Parameters

Parameter ValueCompressor Type Radial Centrifugal Compressor

Pressure Ratio 4.8:1 (Optimum)

Compressor Inlet Temp. 339K

Compressor Outlet Temp. 578.6K

Isentropic efficiency of Compressor 80% (assumed)

Fuel type Natural Gas

Calorific Value 12,500Kcal/kg

Turbine Type Radial Turbine (ABB MT100)

Turbine Inlet Temp. 1223K

Turbine Outlet Temp. 923K

Isentropic efficiency Of Turbine 85% (Assumed)

Brayton Cycle Design Parameter

CALCULATIONSHeat required to produce 1000kw by

Rankine cycle =2573kwHeat supplied in superheater

section=330.64Heat supplied in steam generator

section=1170.9019Heat supplied in economiser

section=1205.8

Working principle of superheated steam heat exchanger

Design procedure for heat exchangerSteps to be followed

STEP 5 Calculate heat transfer area (A) required

STEP4 Decide tentative number of shell and tube passes . Determine the LMTD

STEP1Obtain the required thermophysical properties of hot and cold fluids at the arithmetic mean temperature

STEP 2find out the heat duty of the exchanger. Q

STEP3 Assume a reasonable value of overall heat transfer coefficient . The value of Uo,assm with respect to the process hot and cold fluids can be taken from the standards

STEP 7 Decide type of shell and tube exchanger (fixed tubesheet, U-tube etc.). Select the tube pitch (PT), determine inside shell diameter ( s D ) that can accommodate thecalculated number of tubes .

STEP 8 Assign fluid to shell side or tube side

STEP 9 Determine the tube side film heat transfer coefficient using the suitable form of Sieder-Tate equation in laminar and turbulent flow regimes

STEP 10 Calculate overall heat transfer coefficient U based on the outside tube area including dirt factors

STEP 6 Select tube material, decide the tube diameter (ID , OD ), its wall thickness (in terms of BWG or SWG) and tube length . Calculate the number of tubes required to provide the heat transfer area.

IF calculated error is less than

30 %

YES

NO

Yes then go to next step 11

Then go back to step 5 and re calculate the area

using calculated U

STEP 11 Calculate % overdesign. Overdesign represents extra surface area provided beyond that required to compensate for

fouling. Typical value of 10% or less is acceptable.

Design Calculations:Mean Temprature of hot fluid=556.80ċMean Temprature of cold fluid=355.55ċThermophysical Property at mean

tempratureProperty Hot (Air T=550Ċ ) Cold fluid (Steam

p=100b;t=355Ċ)Viscosity 2.849*e-5 [pa s] 2.23 887791*e-5[Pa s]Thermal conductivity 4.357 *e-5[KW/m K] 0.067790711[W/m K]Constant Pressure Specific heat

1.0398[kJ/kg K] 3.862395 [kJ/kg K]

Density 0.6418[kg / m3] 43.6832023 [kg/m3]

Step 2: Heat duty of heat exchangerm˚(h5-h4)=330.708kwStep 3:we assume overall heat transfer cofficient to be 65

w/m²c step4:LMTD=155.88K

∆T2=89K

∆T1=250K

Step 5:A=Q/(U*LMTD*CF)CF=Correction factor=0.95(taken from hmt data book)A=34.268m²Step 6:Brass is selected essentially as tube material(K=109 w/mk)1 shell and two tube pass is essentially assumed.Considering 14 BWG OD=30 cmLength=37.5 cmId=21.83 cmNo of tubes=total area /surface area of pipe =34.268/(π*d*l)=49Step7: Calculated U comes to be 5w/m²k% error =100*(65-5)/65=92.35%Now we will go to step 3 and will proceed further