The most important aspect of every petroleum industry or energy sector is safety. Various types of protective measures are followed to give a safe environment. 1. Using relief devices 2. External insulation 3. Proper maintenance.

A Pressure Relief Device designed to open and relieve over pressure and to reclose and prevent the further flow of fluid after normal conditions have been restored

When inflow > outflow, Inlet upstream control valve fails, downstream control valve blocked.

1. Reclosing type Conventional PRV Balanced bellows PRV Pilot operated PRV

2.Nonreclosing type Rupture disk device Pin actuated device

Conventional PRV: Spring loaded PRV. Its operation is directly affected by backpressure. It can be used as relief,safety & thermal safety valve.

Balanced bellows PRV: Spring loaded PRV. Incorporates a bellows for minimizing the effect of backpressure.

Pilot operated PRV: It is a pressure relief valve in which the major relieving device or main valve is combined with and controlled by a self actuated PRV(pilot)

ConventionalRelief ValveCONVENTIONAL PRV

BellowsRelief ValveBALANCED BELLOWS PRV

FIRECASE THERMAL EXPANSION BLOCKED DISCHARGE TUBE RUPTURE GAS BLOW BY CONTROL VALVE FAILURE

Fire is one of the least predictable events which may occur in a gas processing facility, but is a condition that may create the greatest relieving requirements.

Formula selection varies with the system and fluid considered.

Fire conditions may overpressure vapor-filled, liquid-filled, or mixed-phase systems.

If any equipment item or line can be isolated while full of liquid, a relief valve should be provided for thermal expansion of the contained liquid.

Low process temperatures, solar radiation, or changes in atmospheric temperature can necessitate thermal protection.

The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error.

When a large difference exists between the design pressure of the shell and tube sides of an exchanger, provisions are required for relieving the low pressure side.

In practice, the control valve may not fail in the desired position.A valve may stick in the wrong position, or a control loop may fail.Relief protection for these factors must be provided.

Critical flowSubcritical flowSteam reliefSTEP-1 Relieving pressure(P1)STEP-1 Relieving pressure(P1)STEP-1 Relieving pressure(P1)STEP-2 Find critical flow pressure & back pressure, Pcf=P1*pressure ratio(Table-7)STEP-2 Find critical flow pressure & back pressurePcf=P1*pressure ratio(Table-7)STEP-2 Find correction factor from Napier equation(KN) if P1 Pcf, flow is sub-criticalSTEP-3 Find the flow type critical or subcriticalif, Pb < Pcf , flow is critical Pb > Pcf, flow is sub-criticalSTEP-3 Find Required orifice area (A)A = W/ (51.5*P1*Kd*Kb*Kc*KN*KSH)STEP-4 Find the specific heat ratio(k) & C(co-efficient), (using Table-7 & Fig-33)STEP-4 Find the specific heat ratio(k) & pressure ratio r=P2/P1from this find F2, Co-efficient of sub-critical flow. F2 using Fig-35 or Eqn 18STEP-4 Select the orifice type(using API 526)STEP-5 Find the required orifice area(A), A=(W/C*Kd*Kc*Kb*P1)*(T*Z/M)^0.5STEP-5 Find the required orifice area(A), A=(W/735*F2*Kd*Kc)*(T*Z/(M*P1(P1-P2)))^0.5STEP-5 Find the rated capacity of selected orifice areaSTEP-6 Select the orifice type(using API 526)STEP-6 Select the orifice type(using API 526)STEP-7 Find the rated capacity of selected orifice areaSTEP-7 Find the rated capacity of selected orifice area

PRVs sizing procedure for liquid serviceSTEP-1Relieving pressure(P1)STEP-2% Back pressure = (back pressure/set pressure)*100STEP-3Backpressure correction factor Kw(from fig-31)

STEP-4 (preliminary sizing) Assuming no viscosity correction factor, (Kv=1)STEP-5Find A, A=(Q/(38*Kd*Kw*Kc*Kv))*(Gl/(P1-P2)) Gl=Specific gravity of liquid at flowing conditionsSTEP-6Find AR (effective orifice type),STEP-7 Find Re using AR calculated, Re = Q*12700/(U*(A)^0.5)STEP-8Find Kv, using Re calculated,(Using graph 37)STEP-9 Find actual orifice area,using equation, A=AR/Kv

Flashing & non-flashing Subcooled regionStep1: Determine omega parameterW = 9*((v9/v0)-1)Step1: calculate the saturation omega parameter ws = 9*((v9/v0)-1)Step2: Determine the type of flow Pc >= Pa critical flowPc < Pa sub-critical flowPc = nc*P0 critical pressure ratio, using graph C.1(API-520)Step2: Determine the sub-cooling regionPs >= nst*P0 low subcooling regionPs < nst*P0 high subcooling regionnst = (2*ws/(1+2*ws))

Step3: Calculate the mass fluxFor critical flow, G=68.09*nc*(P0/(v0*w))For subcritical flow, G=68.09*{-2[w*ln(na)+(w-1) (1-na)]}^0.5*(P0/v0)^0.5/ (w*((1/na)-1)+1)Step3: Determine the type of flow, Low subcooling region, Pc >= Pa critical flow Pc < Pa sub-critical flowHigh subcooling region, Ps >= Pa critical flow Ps < Pa sub-critical flowStep4: Calculate the required orifice areaA=(0.04*W)/(kd*kb*kc*kv*G)Step4: calculate nc using graph C.2(API-520)For nsnst, using graph C.2(API-520)

Step5: Select the orifice areaStep5: Calculate the mass flux,In low subcooling region, if the flow is critical, use nc for n, G=68.09{2(1-ns)+2[ws*ns*ln(ns/n)-(ws-1) *(ns-n)]}^0.5*(P/vl0)^0.5/(ws((ns/n)-1)+1)In high subcooling region, G=96.3*(pl0(Po-P))^0.5Step6: Find the rated capacityStep6: Calculate the required orifice areaA=0.3208*((Q*plo)/(kd*kb*kv*G))Step7: Select the Required orifice sizeStep8: Find the rated capacity

ConventionalThe ratio between the backpressure & set pressure or % gauge pressure is below 10% - conventional type.

BellowsThe ratio between the backpressure & set pressure or gauge pressure is 10% - 50% - bellows type.

Pilot operatedThe ratio between the backpressure & set pressure or gauge pressure is above 50% - pilot operated.

21%, for Fire case 10%, for other than fire case, vessels equipped with a single PRV. 16%, for other than fire case vessels equipped with multiple PRV.

PSV inlet line criteria pressure drop < 3% of set pressure PSV outlet line criteria velocity limited to 0.5 mach(API) or 0.7 mach (norsok)

Shell TubeDesign, pressure = 13.5 barg 13.5 Temperature = 200 C 200COperating, pressure = 9/8.1 1.1/0.461 Temperature = 181.3/179.8 21/80 Design, pressure = 68.9 barg Temperature = 66COperating, pressure =10-48barg Temperature = 22C

INPUT CONDITIONS:

Fluid = gasHydrocarbon vapour flow = 8 MMSCFDMol wt.. = 17.85Relieving temperature = 21.70CSet pressure = 13.5 barg or 195.8 psigBack pressure = 3.5 bargAccumulation = 10%Compressibility factor Z = 0.969

Relieving pressure = set pressure + overpressure + atm pressure = 229.2 psiaCritical pressure = relieving pressure * pressure ratioCalculated pressure ratio = 0.585(using Table 7, API 520) Pcf = 229.2 * 0.58 Pcf = 129.116 psia The calculated critical pressure > back pressure (Pcf>Pb), so the flow is critical. The back pressure < 50% & >10% of set pressure, so the balanced bellows PRV can be used. Critical flow formula, A = ((v*(T*Z*M)^0.5)/(6.32*C*kd*kc*kb*P1))

Where,V = Vapor flow rate in SCFMT = relieving temperature in R Z = compressibility factorM = molecular weightC = coefficient (calculated using fig-33)Kd = effective co-efficient of discharge = 0.975, when PRV installed with or w/o rupture disk = 0.62, when PRV is not installed & rupture disk is installedP1 = relieving pressure in psiaKc = combination correction factor = 1, when rupture disk is not installedKb = capacity correction factor due to back pressure = 1, for conventional & pilot operated valves

DesignationEffective orifice area (square incenses)D0.110E0.196F0.307G0.503H0.785J1.287K1.838L2.853M3.600N4.340P6.380Q11.05R16.00T26.00

Required orifice area, A = 1.11 in A = 716.12 mm Orifice selection,(using API 526)For the required orifice of 1.11 in, J orifice can be used A = 1.287 in & 830.32 mm

Rated capacity, Vr = ((AA*6.32*C*kd*kc*kb*P1) /(T*Z*M)^0.5) Vr = 6448 SCFM, 9.285 MMSCFDRating from API 526, Orifice = 2 J 3 (150#inlet & 150#outlet)

It is an manual operation valve, used to depressurise the plant or a equipment, before it reaches the abnormal condition(i.e., fire case..) & also for maintenance purpose.

The main difference between PSV & BDV is the mode of operation, BDV is operated by pneumatic action (instrument air).

2-types of pneumatic action: 1. Fail open 2. Fail closed

Fail open: Instrument air is sent continuously from the top of the diaphragm to close the valve at normal operating condition. At abnormal condition instrument air supply is closed to make valve open.

Fail closed: Instrument air is sent from the bottom of the stem, to make the valve open. If the instrument air supply is stoped, then the valve will be closed.

BDV should be considered where the large equipment operating at a gauge pressure of 1700 kpa (250 psi) or higher. (ref API 521 5.20.1 page 57)

Depressuring to a gauge pressure of 690 kpa is commonly considered when the depressuring system designed to reduce the consequences..

FIRECASE(hot blowdown)ADIABETIC CASE(cold blowdown)

Two types of fire: Jet fire Pool fire

Jet fire can happen when combustible fluid in pressurized system is released to atmosphere.

Jet fire can cause vessel failure in < 5min.

The heat flux of jet fire can be as high as 300 kW/m^2.

Pressure relief devices fails at jet fire.

Turned off through isolation & depressuring of jet fire source.

Hydrocarbon fire can exceed 40ms in height.

To determine the vapour generation, it is necessary to recognize only that portion of the vessel that is wetted by its internal liquid and is

Inlet line momentum(v) < 200000 kg/ms Outlet line mach no 0.7

When pool fire exposes the unwetted wall of a large vessel fabricated from ASTM A 515 Grade 70 carbon steel, it will take about 15 min to heat the vessel walls to around 649C to reach its rupture temperature.

It can be overcome, if the vessel is depressurized within the 15 min heat-up time to, 50% of the initial pressure, then the time to rupture increase to about 2-3hr. (Ref., API 521, Section 5.20)

Adiabatic case is considered at low temperature. Blowdown from the minimum ambient temperature is done only if the gas inventory may be contained for extended durations, blowdown at minimum operating temperature is done if minimum operating temperature is lower than minimum ambient. Vapour load is depends on liquid quantity & liquid properties in the system. Depressurisation from operating pressure to atmospheric pressure.

Why restriction orifice is some distance from blowdown valve?

Major pressure drop will take place at restriction orifice during blowdown. Joule-Thompson effect results fluid temperature downstream of RO drops below zero C. Decrease in temperature will leads to hydrate formation. This coldness will travel back to upstream of RO & probably reaches BDV. It potentially cause the upstream of BDV body temperature drops below sub zero. Moisture from atmosphere will freeze at the BDV body & potentially cause the stem stuck at its position. Operator may not possible to close the BDV after blowdown activities & may potentially leads to backflow.

Flares are categorised in two ways,

By the height of the flare tip,

Elevated flaresGround flares

By enhancing mixing at the flare tip

Steam assistedAir assistedPressure assistedNon-assisted

These are the most common type of flare used at present.

Elevated flare can prevent potentially dangerous conditions at ground level where the open flame is located near a process unit.

Further, the products of combustion can be dispersed above working areas to reduce the effects of noise, heat, smoke, & objectionable odors.

These are classified into 3-types:SELF-SUPPORTED.

Self supported stacks are normally the most desirable.These are the most expensive because of greater material requirements needed to ensure structural integrity over the anticipated conditions(wind & seismic).These are normally limited to a stack height of 60-90m.

These are the least expensive but require largest land area for the guy-wire radius.

Guy-wire radius is equal to one-half the overall stack height.

Guyed stacks of heights of 180-250 m have been used.

These are used only on larger stacks where self-supported design is not practical or available land area excludes a guy-wire design.

These can be designed as high as possible.

In locations where land is not available, the multi flare stack can be used.

Increasingly strict requirement regarding flame visibility, emissions & noise, enclosed flares can offer the advantages of hiding flames, monitoring emissions & lowering noise.Advantages:Reduced flame visibility.Minimal noise.Minimal heat radiation due to ceramic insulation.Smokeless combustion.

Disadvantages:Potential accumulation of a vapour cloud in the event of flare malfunction.Most expensive because of the size of the shell.Significantly less capacity than elevated flares.

Steam injected to the flame zone to create a turbulence.Improved air distribution allows the air to react more rapidly with flare gas.Another factor is the steam water-gas shift interaction where the CO & H2O vapor react to form a CO2 & hydrogen, which can easily burned.

Steam is injected through, A single pipe nozzle located in the centre of the flare, A series of steam injectors in the flare,A manifold located around the periphery of the flare tip, to get a smokeless combustion.

High pressure air can also be used to prevent smoke formation.Less common because compressed air more expense than steam.It can be preferable were low temperature applications.

Disadvantage:

The mass quantity required is approximately 200% > steam.No water-gas shift reaction that occurs with steam.

Pressure-assisted flares use the vent stream pressure to promote mixing at the burner tip.If sufficient vent stream pressure is available, these flares can be applied to streams previously requiring steam or air assist for smokeless operation.

It is used where smokeless burning assist is not required.The non-assisted flare is just a flare tip without any auxiliary provision for enhancing the mixing of air into its flame.

FLARE DIAMETERFlare diameter is sized based on the velocity basis & also pressure drop to be considered.It is desirable to permit a velocity of 0.5 mach for a peak, short term, infrequent flow, with 0.2 mach maintained for normal condition for LP flares.Sonic velocity is desirable for HP flare.

Other factors:Flare tip velocity should be maintained with mach no 0.8 or higher for assisted & non-assisted.Too low a tip velocity can cause heat & corrosion damage.Low-pressure area on the downwind side of the stack can cause the burning gases to be drawn down along the stack for 3m or more

Too low a tip velocity will leads to the propagation of flame into the flare stack, most common method to prevent propagation of flame, 1. Install a seal drum at bottom of the flare stack 2. continuous introduction of purge gas. 3. purge reduction seal

Continuous introduction of purge gas will prevent the propagation of flame. Purge gas rate can be reduced by the use of a purge-reduction seal.

Purge gas rate calculated from this equation, for lighter than air, Q = 190.8*D^0.36* (1/y)ln(20.9/O2)(Ci^0.65 * Ki) standard criteria to limit the oxygen volume fraction to 6% at a distance of 7.62m down the flare stack, Q = 31.25*D^3.46*K

FLARE STACK HEIGHTFlare stack height is based on the radiant heat intensity generated by the flame.Flame radiation is considered to a point of interest is to consider the flame to have a single radiant epi centre. This is origin of total radiant heat intensity level.Effect of thermal radiation: Investigations have been undertaken to determine the effect of radiation on human skin.

The quality of combustion affects the radiation characteristics. This radiation characteristics will affect the flare stack height.Smoke free operation can be attained by various methods, 1. steam assisted 2. high-pressure waste-gas 3. forced draft air , etc..

Smokeless flares can be obtained in the form of no operator shall allow the flare emissions to exceed 20% opacity for more than 5 min in any consecutive 2-h period (Ringleman 1 performance).

DESIGN BASIS:

Flare stack diameter is sized based on the velocity basis by allowing the mach no of 0.5 for a peak, short term, infrequent flow, with 0.2 mach for normal operating conditions for LP flares.

Sonic velocity is appropriate for HP flares.

Flare stack height is based on the radiant heat intensity generated by the flame.

AVAILABLE DATA:

Mass flow rate = 340 MMSCFD, 302783 kg/hr.Avg relative molecular mass = 17.85Absolute pressure at flare tip = 101.3 kpa.Flowing temperature = 114CAir temperature = 20CWind velocity = 9 m/sMaximum allowable radiation = 3 kW/m2.Mach no = 0.5.

Step 1: Flare diameter

Flare dia is calculated by fixing the mach no of 0.5.

Ma2 = 3.23*10^(-5)*(qm/P2*d^2)*(Z*T/M)^0.5 d = 0.94 m

Step 2: Location of flame centre,

Isothermal sonic velocity = 91.2*(Tj/Mj)^0.5 = 430.1 m/s Tip velocity, uj = 0.5*isothermal sonic velocity uj = 215 m/s

To find out the horizontal & vertical distances from flare tip to the flame centre we need 2 parameters,

LEL concentration parameter CL, Jet & Wind thrust (dj*R). LEL concentration parameter for the flare gas CL, CL = CL*(uj/ua)*(Mj/Ma) CL = 0.729

Parameter for Jet & Wind thrust, dj*R = dj*(uj/ua)*((Ta*Mj)/Tj)^0.5 dj*R = 81.5

From fig: C.2 CL bar =0.729 & dj*R = 81.5 xc = 12 m

From fig: C.4 CL bar =0.729 & dj*R = 81.5 yc = 29 m

Step:3 calculation of the distance from flame centre to the object being considered.

Design basis for this calculation,

The fraction of heat radiated F = ? From thesis, emphirical equation relating Fraction of heat radiated & jet velocity by Cook et al.. (1987) follows, F = 0.321-0.418*10^-3*uj F = 0.091Heat liberated Q = mass flowrate * Heat of combustion Q = 3827002.2 kWMaximum allowable radiation K = 3kW/m2.

D = ((*F*Q)/(4**K))^0.5 D = 95.6 m

Assume a grade level r = 22 m, with K of 3 kW/m2.

h = h + yc r = r xc D^2 = r^2 + h^2 (h+29)^2 = (95.6^2)-(10^2)

h = 66.4 mThe calculated flare stack height is h = 67 m.

Bellows pluggedin spite of signAnything wronghere?FailedInspectionProgramSigns ofMaintenanceIssues

Anything wronghere?

Anything wronghere?Will thesebolts holdin arelief event?

What is the maximum fire zone area to be considered for fire case? a. 150 m b. 232 m c. 300 m

Where depressuring for fire scenario should be considered? a. for equipments operating at a gauge pressure of 205 psi b. for equipments operating at a gauge pressure of 250 psi

What is the final pressure for depressuring the system? a. 690 kpa b. 50% of the initial pressure c. 650 kpa d. both a & b

What is the flow opening % to be considered for PSV sizing for check valve leakage? a. 1 % of nominal dia of check valve b. 20 % of nominal dia of check valve c. 10 % of nominal dia of check valve d. 15 % of nominal dia of check valve

What is the criteria should be followed for PSV sizing for tube rupture? a. (3/2) design pressure of HP side = design pressure of LP side b. (13/10) design pressure of HP side = design pressure of LP side c. (10/13) design pressure of HP side = design pressure of LP side

What is the emission standards following in India?

*********Venting arrangements must be carefully selected and designed to meet the following requirements:1.Manufacturers shipping plugs must be removed from the bonnet vent holes before a new valve is commissioned.2.Each PR valve must be installed so that the bonnet vent does not allow released vapors to impinge on lines or equipment,or towards personnel walkways. Where necessary, a short nipple and elbow should be added to direct flow away from such areas. In these cases, the vent piping should discharge horizontally to avoid entry of rainwater and debris, and should terminate in a position which is accessible for leak testing.3.In cases where bellows failure would release flammable, toxic or corrosive liquids through the vent, a short nipple and elbow should be used to direct leakage to an open funnel which is piped to grade and ties into a catch basin or manhole through a sealed inlet connection.4.Although venting to the atmosphere is preferred, an alternative is to tie into a closed low pressure system, if available. ***Fire Exposure Fire is one of the least predictable events which may occur in a gas processing facility, but is a condition that may create the greatest relieving requirements. If fire can occur on a plant-wide basis, this condition may dictate the sizing of the entire relief system; however, since equipment may be dispersed geographically, the effect of fire exposure on the relief system may be limited to a specific plot area. Vapor generation will be higher in any area which contains a large number of un insulated vessels. Various empirical equations have been developed to determine relief loads from vessels exposed to fire. Formula selection varies with the system and fluid considered. Fire conditions may overpressure vapor-filled, liquid-filled, or mixed-phase systems.

**Thermal Expansion If isolation of a process line on the cold side of an exchanger can result in excess pressure due to heat input from the warm side, then the line or cold side of the exchanger should be protected by a relief valve. If any equipment item or line can be isolated while full of liquid, a relief valve should be provided for thermal expansion of the contained liquid. Low process temperatures, solar radiation, or changes in atmospheric temperature can necessitate thermal protection. Flashing across the relief valve needs to be considered.

**Blocked Discharge The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error. In this case, the relief load is usually the maximum flow which the pump, compressor, or other flow source produces at relief conditions.

**

Tube Rupture When a large difference exists between the design pressure of the shell and tube sides of an exchanger (usually a ratio of 1.5 to 1 or greater),provisions are required for relieving the low pressure side. Normally, for design, only one tube is considered to rupture. Relief volume for one tube rupture can be calculated using appropriate sizing equations in this section. When a cool media contacts a hot stream, the effects of flashing should be considered. Also the possibility of a transient over-pressure caused by the sudden release of vapor into an all-liquid system should be considered. **Control Valve Failure The failure positions of instruments and control valves must be carefully evaluated. In practice, the control valve may not fail in the desired position.A valve may stick in the wrong position, or a control loop may fail.Relief protection for these factors must be provided. Relief valve sizing requirements for these conditions should be based on flow coefficients (manufacturer data) and pressure differentials for the specific control valves and the facility involved.

**Blocked Discharge The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error. In this case, the relief load is usually the maximum flow which the pump, compressor, or other flow source produces at relief conditions.

**Blocked Discharge The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error. In this case, the relief load is usually the maximum flow which the pump, compressor, or other flow source produces at relief conditions.

**Blocked Discharge The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error. In this case, the relief load is usually the maximum flow which the pump, compressor, or other flow source produces at relief conditions.

**Blocked Discharge The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error. In this case, the relief load is usually the maximum flow which the pump, compressor, or other flow source produces at relief conditions.

****Blocked Discharge The outlet of almost any vessel, pump, compressor, fired heater, or other equipment item can be blocked by mechanical failure or human error. In this case, the relief load is usually the maximum flow which the pump, compressor, or other flow source produces at relief conditions.

******Sizing for Gas or Vapor Relief The rate of flow through a relief valve nozzle is dependent on the absolute upstream pressure (as indicated in Eq 5-1, Eq 5-2, and Eq 5-3) and is independent of the downstream pressure as long as the downstream pressure is less than the critical-flow pressure.However, when the downstream pressure increases above the critical flow pressure, the flow through the relief valve is materially reduced (e.g., when the downstream pressure equals the upstream pressure, there is no flow). The critical-flow pressure, PCF, may be estimated by the perfect gas relationship shown in Eq 5-5. As a rule of thumb if the downstream pressure at the relief valve is greater than one-half of the valve inlet pressure (both pressures in absolute units), then the relief valve nozzle will experience subcritical flow. Critical Flow Safety valves in gas or vapor service may be sized by use of one of these equations: 131.6W (T1)(Z) A = Eq 5-1 (C1)(Kd)(P1)(Kb)(Kc) MW

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