THERMAL RADIATION

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                            IN THE FLARE DESIGN, THE EFFECT OF HEAT RADIATING FROM THE COMBUSTION OF LARGE QUANTITIES OF FLAMMABLES ON PERSONNEL AND EQUIPMENT MUST BE EVALUATED.

                            FOR THE SAFE OPERATION, THE FLARE HEIGHT SHOULD BE DESIGNED WITH CONSIDERATION GIVEN TO POTENTIAL RADIATION LEVELS AROUND THE FLARE.  THE POTENTIAL RADIATION LEVELS SHOULD NOT EXCEED ALLOWABLE RADIATION LEVELS ON THE GROUND.

                            THERMAL RADIATION IMPACTS CAN BE ESTIMATED USING THE API APPROACH. SEE ATTACHED FIGURE FOR DIMENSIONS OF FLARE GO TO FIGURE

IN THE FIGURE :

H = FLARE STACK HEIGHT

Lf = FLAME LENGTH

dX = X ( HORIZONTAL) DIMENSION OF FLAME

dY =  Y (VERTICAL) DIMENSION OF FLAME

Xc = HALF OF THE X (HORIZONTAL) DIMENSION OF FLAME

      = dX / 2

Yc = HALF OF THE Y (VERTICAL) DIMENSION OF FLAME

      = dY / 2

D = DISTANCE FROM THE CENTER POINT OF THE FLAME TO THE OBJECT FOR WHICH THE RADIATION LEVEL IS TO BE CALCULATED

R = DISTANCE FROM THE CENTER POINT OF THE BASE OF THE FLARE TO THE OBJECT FOR WHICH RADIATION FROM THE FLARE IS TO BE CALCULATED.

R' = R - Xc

Heff = EFFECTIVE FLAME HEIGHT FROM THE GROUND

         = H + Yc = H+ ( dY / 2 )

ALSO FROM FIGURE WE CAN SEE THAT

          D2 = (R')2 + Heff2

NOW WE WILL SEE HOW TO CALCULATE FLAME LENGHT Lf :

TO ESTIMATE THE FLAME LENGTHS, THE FOLLOWING EQUATION WAS DERIVED USING THE FLAME LENGTH AND HEAT RELEASE DATA PROVIDED BY API :

Lf = 0.01288 * HBTU0.44

WHERE   Lf = FLAME LENGTH, ft

                  HBTU = HEAT RELEASE PER HOUR ( BTU/Hr.)

FLAME DISTORTION DUE TO WIND :

FLAME WILL OFTEN BE TILTED DUE TO WIND AS SHOWN IN THE FIGURE.  IF WIND SPEED, u AND FLARE EXIT VELOCITY Vex ARE KNOWN, THEN dX / Lf AND dY / Lf CAN BE DETERMINED FROM THE BELOW TABLE  ONCE dX / Lf AND dY / Lf ARE ESTIMATED, Yc AND Xc CAN BE DETERMINED USING Xc = dX/2 AND Yc=dY/2

TILTED FLAME PARAMETERS WITH RESPECT TO WIND SPEED
V/Vex dX / Lf dY / Lf
0.00 0.00 1.00
0.05 0.6 0.6
0.10 0.78 0.44
0.20 0.88 0.32
0.30 0.93 0.24
0.40 0.95 0.20
0.90 0.97 0.10

RADIATION LEVEL :  RADIATION LEVEL IN THE VICINITY OF THE FLARE CAN BE ESTIMATED USING THE EQUATION GIVEN BELOW:

Kr = ( If * F * Qh ) / ( 4 * ¶ * D2 )

WHERE

          Kr = RADIATION LEVEL , BTU / ( Hr. ft2 )

          If = FRACTION OF HEAT INTENSITY TRANSMITTED, ~1.0

          F = FRACTION OF HEAT RADIATED

              = 0.15 - 0.30 FOR MOST CASES

          D = MINIMUM DISTANCE FROM CENTER POINT OF THE  FLAME TO THE OBJECT BEING CONSIDERED.

           Qh = HEAT RELEASE RATE, BTU / (Hr)(ft2)

NOTE THAT ABOVE EQUATION, IN CONJUNCTION WITH EQUATIONS Xc = dX / 2 & Yc = dY / 2  AND Heff = H + Yc , CAN BE USED TO CALCULATE EITHER THE HEAT RADIATION LEVEL AT A GIVEN DISTANCE FROM THE FLAME, OR THE DISTANCE, D FOR A GIVEN HEAT RADIATION LEVEL.

Example :  IF THE HEAT RELEASE RATE FOR A FLAME IS 4.5 * 109 BTU/Hr., AND ALLOWABLE RADIATION LEVEL IS 1500 BTU/(Hr.)(ft2) ESTIMATE THE DISTANCE AROUND THE FLARE STACK WITHIN WHICH THE RADIATION LEVEL WOULD EXCEED 1500BTU/(Hr.)(ft2). THE FLARE IS 150 ft HIGH.  ASSUME A WIND SPEED OF 15 ft/Sec. AND FLARE GAS EXIT VELOCITY OF 300 ft/Sec.

Solution : Assuming F ( fraction of heat radiated ) = 0.3

WE KNOW Kr = ( If * F * Qh ) / ( 4 * ¶ * D2 )

REARRANGING  D = V=====================                                        AS                                                  (If * F*Qh) / (4*¶*Kr

GIVEN :   Qh = 4.5 *109 BTU/(Hr)(ft2)

                  I = 1.0

                  Kr = 1500 BTU/(Hr)(ft2)

                  U = 15 ft/Sec

                  Vex = 300 ft/Sec.

SUBSTITUTING THESE VALUES IN THE EQUATION FOR D WE GET

D = 268 ft

FLAME LENGTH, Lf = 0.01288 * HBTU0.44

                                          Lf = 0.01288 * ( 4.5 * 109 )0.44 = 228 ft

NOW  ,   U/Vex  = 15/300 =0.05

FOR U/Vex  = 0.05  , FROM ABOVE GIVEN TABLE

                          dX / Lf = dY / Lf = 0.60

                         SO, dX = dY = 0.6 * 228 = 136.8 ft ( Lf =  228 )

                         Xc = Yc = dX / 2 = dY / 2 = 136.8/2 = 68.4 ft ~ 69ft

NOW, Heff  = H + Yc = 150 + 69 = 219 ft

WE KNOW THAT  D2 = (R')2 + Heff2

                      SO , 2682 = (R')2 + 2192

                                  R' = 154 ft.

NOW, WE KNOW R' = R - Xc

SO R = R' + Xc = 154 + 69 = 223 ft.

IT MEANS THAT WITHIN 223 ft OF FLARE, THE RADIATION LEVEL WILL BE ABOVE 1500 BTU/(Hr)(ft2) LEVEL.