Presentation-api-521.ppt

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)

Conventional
Relief Valve
CONVENTIONAL PRV

Bellows
Relief Valve
BALANCED 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 flow Subcritical flow Steam relief
STEP-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 pressure
Pcf=P1*pressure ratio(Table-7)
STEP-2 Find correction factor
from Napier equation(KN)
if P1 <= 1500 KN = 1
if 1500< P1 < 3200
KN = ((0.1906*P1)-1000) /
((0.2292*P1)-1061)
STEP-3 Find flow type critical or
subcritical
if, Pb < Pcf , flow is critical
Pb > Pcf, flow is sub-critical
STEP-3 Find the flow type critical
or subcritical
if, Pb < Pcf , flow is critical
Pb > Pcf, flow is sub-critical
STEP-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/P1
from this find F2, Co-efficient of
sub-critical flow. F2 using Fig-35
or Eqn 18
STEP-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)^
0.5
STEP-5 Find the required orifice
area(A),
A=(W/735*F2*Kd*Kc)*(T*Z/(M*P1
(P1-P2)))^0.5
STEP-5 Find the rated capacity of
selected orifice area
type(using API 526)
type(using API 526)

PRV’s sizing procedure for
liquid service
STEP-1
Relieving pressure(P1)
STEP-2
% Back pressure = (back pressure/set
pressure)*100
STEP-3
Backpressure correction factor Kw(from fig-31)
STEP-4 (preliminary sizing)
Assuming no viscosity correction factor, (Kv=1)
STEP-5
Find A, A=(Q/(38*Kd*Kw*Kc*Kv))*(Gl/(P1-P2))
Gl=Specific gravity of liquid at flowing conditions
STEP-6
Find AR (effective orifice type),
STEP-7 Find Re using AR calculated,
Re = Q*12700/(U*(A)^0.5)
STEP-8
Find Kv, using Re calculated,(Using graph 37)
STEP-9 Find actual orifice area,
using equation, A=AR/Kv

Flashing & non-flashing Subcooled region
Step1: Determine omega parameter
W = 9*((v9/v0)-1)
Step1: calculate the saturation omega parameter
ws = 9*((v9/v0)-1)
Step2: Determine the type of flow
Pc >= Pa critical flow
Pc < Pa sub-critical flow
Pc = nc*P0 critical pressure ratio, using graph
C.1(API-520)
Step2: Determine the sub-cooling region
Ps >= nst*P0 low subcooling region
Ps < nst*P0 high subcooling region
nst = (2*ws/(1+2*ws))
Step3: Calculate the mass flux
For 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 flow
High subcooling region, Ps >= Pa critical flow
Ps < Pa sub-critical flow
Step4: Calculate the required orifice area
A=(0.04*W)/(kd*kb*kc*kv*G)
Step4: calculate nc using graph C.2(API-520)
For ns<=nst, nc=ns
For ns>nst, using graph C.2(API-520)
Step5: Select the orifice area Step5: 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))^0.5
Step6: Find the rated capacity Step6: Calculate the required orifice area
A=0.3208*((Q*plo)/(kd*kb*kv*G))
Step7: Select the Required orifice size
Step8: Find the rated capacity

Conventional
The ratio between the backpressure & set pressure or
% gauge pressure is below 10% - conventional type.
Bellows
gauge pressure is 10% - 50% - bellows type.
Pilot operated
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)
²

GAS BLOW BY
STEAM SUPPLY
CONDENSATE TO CONDENSATE
FLASH SEPARATOR
STEAM CONDENSATE RETURN
PW TO PW HYDROCYCLONE-A/B
WELL FLUID FROM PRODUCTION
HEADER
HC GAS TO LP COMPRESSION
SYSTEM
INLET GAS
SEPARATOR-A/B
CONDENSATE
HEATER
PSV
SDV
SDV LV
LT
Shell Tube
Design, pressure = 13.5 barg 13.5
Temperature = 200 °C 200°C
Operating, pressure = 9/8.1 1.1/0.461
Temperature = 181.3/179.8 21/80
Design, pressure = 68.9 barg
Temperature = 66°C
Operating, pressure =10-48barg
Temperature = 22°C

INPUT CONDITIONS:
Fluid = gas
Hydrocarbon vapour flow = 8 MMSCFD
Mol wt.. = 17.85
Relieving temperature = 21.70°C
Set pressure = 13.5 barg or 195.8 psig
Back pressure = 3.5 barg
Accumulation = 10%
Compressibility factor Z = 0.969

Relieving pressure = set pressure + overpressure + atm
pressure
= 229.2 psia
Critical pressure = relieving pressure * pressure ratio
Calculated 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 SCFM
T = relieving temperature in °R
Z = compressibility factor
M = molecular weight
C = 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
installed
P1 = relieving pressure in psia
Kc = combination correction factor
= 1, when rupture disk is not installed
Kb = capacity correction factor due to back pressure
= 1, for conventional & pilot operated valves

Designation Effective
orifice area
(square
incenses)
D 0.110
E 0.196
F 0.307
G 0.503
H 0.785
J 1.287
K 1.838
L 2.853
M 3.600
N 4.340
P 6.380
Q 11.05
R 16.00
T 26.00

Required orifice area,
A = 1.11 in²
A = 716.12 mm²
Orifice selection,(using API 526)
For the required orifice of 1.11 in², J orifice can
be used
A = 1.287 in² & 830.32 mm²
Rated capacity, Vr = ((AA*6.32*C*kd*kc*kb*P1)
/(T*Z*M)^0.5)
Vr = 6448 SCFM, 9.285
MMSCFD
Rating from API 526, Orifice = 2” J 3” (150#inlet

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

1. 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.
1. 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 40m’s 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 <= 7.6 m above the source of
flame .
 The heat flux of pool fire can be upto 100 kW/m².

 Inlet line momentum(ρv²) < 200000
kg/ms²
 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?
600 mm
AxB
RO
BDV

 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

Flares are categorised in two ways,
1.By the height of the flare tip,
Elevated flares
Ground flares
1.By enhancing mixing at the flare tip
Steam assisted
Air assisted
Pressure assisted
Non-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:
1.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 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.

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 DIAMETER
Flare 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

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 HEIGHT
Flare 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 = 17.85
Absolute pressure at flare tip = 101.3 kpa.
Flowing temperature = 114°C
Air temperature = 20°C
Wind velocity = 9 m/s
Maximum 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 = 0.091
Heat liberated Q = mass flowrate * Heat of
combustion
Q = 3827002.2 kW
Maximum allowable radiation K = 3kW/m2.
D = ((τ*F*Q)/(4*π*K))^0.5

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 m
The calculated flare stack height is h = 67 m.

Bellows plugged
in spite of sign
Anything wrong
here?
Failed
Inspection
Program
Signs of
Maintenance
Issues

Anything wrong
here?
Anything wrong
here?
Discharges
Pointing Down

Anything wrong
here?
Will these
bolts hold
in a
relief event?

1. What is the maximum fire zone area to be
considered for fire case?
a. 150 m² b. 232 m² c. 300 m²
1. 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

1. 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
1. 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

What is the emission standards
following in India?

Presentation-api-521.ppt

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