PSV Flare system PSV Flare system PSV Flare system

◼ 1 General
◼ 2 Typical Flare System
◼ 3 Potentials of Overpressure
◼ 4 Determination of Individual rates
◼ 5 Sample Calculations-Flare System
◼ 6 Pressure Safety Valves

◼ 1.0 General
◼ • Main focus of presentation is on estimate of flare load for
various contingencies & sizing of flare headers (with in various
units as well as outside header ) connecting load from various
units to main flare stack through main KOD and water seal
drum.
◼ • Causes of overpressure & detailed discussion for various
contingencies for flare load calculations.
◼ • Detailed discussion and sample calculations for fire and
tube rupture cases.
◼ Types of P S Vs used in a typical flare system & relieving
pressure estimate for single/multiple valve for operating and fire
case.

◼ Temperature profile of flare header (ISBL & OSBL) is essential in
having optimum provision of expansion loops in flare headers. This
calculation is usually done with software program. For manual
estimate, formula used with sample calculations is presented.
◼ • Flare header sizing is done by considering very small sections of
the flare header at a time to achieve accurate results by minimizing
variation in gas density for each section during pressure drop
estimate), this is done conveniently by using software program.
Manual procedure with sample calculations will be discussed.
◼ • Flare stack, main flare KOD and water seal drum are a part of
flare system, typical manual calculations of these has been included.

◼ 2.0 Typical Flare System
◼ 2.1 Controlling Loads
◼ • Controlling case for refinery or petrochemical complexes are
usually cooling water or power failure.
◼ During the preparation of process package for various units, process
licensors provide flare load details for various possible failure cases.
Typical data consist of flare load (kg/hr or t/hr), relieving temperature
(0C), molecular weight of relieved gas, and back pressure( kg/cm2g).
◼ By using the flare load for various contingencies, process engineer
has to calculate the maximum relieving load for a particular failure
case for design of flare system.
◼ Some discussion on cooling water and power failure is as below:

◼ 2.1.1 Cooling Water failure
◼ • Whenever there is one cooling water system, cooling water
failure loads for various units are added to calculate total flare load of
the complex.
◼ • For a large complex with very high flare load for cooling water
failure case, complex may be segregated in to two or more cooling
water systems and then flare load of one (controlling ) cooling water
system may be considered for further design.
◼
During revamp, further reduction of load may be possible when
cooling water header is in two distinct separate directions. Here by
considering pumps and CT fans for both directions from two different
sources of power may serve to reduce flare load. IOCL Mathura &
Baroda have used this concept during revamp to use their existing
flare system.

◼ 2.1.2 Power Failure
◼ • Power distribution may be from 6-8 substations which are fed
from the main source of power supply.
◼ • Main source of power could be captive, grid power or captive in
parallel with grid power.
◼ • There are three possible situations of power failure.
◼ • Local power failure of a reflux pump for a column in a
particular unit causing reflux failure of that column.
◼ Substation power failure-here power failure load of various units
connected with that particular substation are considered. Here utility
systems like CWS or Steam/Power may or may not be connected to
that particular substation. Flare load has to be accordingly estimated
after detailed analysis.

◼ • Total power failure
◼ -It gives rise to loss of reflux to towers which results in
◼ relief of gross overheads from each tower.
◼ - It mayl give rise to cooling water as well as steam failure.
◼ During tripping of steam generation /BFW system,
◼ supply of steam to unit reboilers is limited to system hold
◼ up/surge available.
◼ Other considerations are described below:
◼ 2.2 Miscellaneous
◼ In certain cases, by suitably changing the design pressure of a few
equipment, flare load can be brought down.

◼ • In case of high flare loads, a review is to be done to
check, if by adopting suitable control philosophy, flare
load can be brought down. For example, tripping source of
heat to reboiler by sensing pressurization of column
overhead with 2 out of 3 voting of pressure switches, CRL
reduced their CDU flare load.
◼ • For high temp relief from Hydrogen unit etc ISBL
header is sized as per actual temperature during relief but
OSBL header is maintained of CS because CS can
withstand expected high temperature for the duration of
relief.

• Minimum heating value of gas being flared is between 200-300
BTU/SCF, if value drops below this value, special flare system is
provided, heat value should be maintained.
◼ • In certain cases, by segregating high and low pressure loads by
separate headers , flare system can be made economically attractive.
◼ Possibility of reducing trussel (flare support) height by providing
intermediate KOD may be considered when flare system is very
far(about 2 KM) from process units. Typical system at stack side will
have elevation equivalent to diameter of main KOD+ its elevation
from grade + piping network connecting flare header to KOD. OSBL
header from various units( after their unit KOD s )is with 1:500 slope.
With intermediate KOD having dual slope (part towards intermediate
KOD and rest towards main KOD ), structure cost on support may be
reduced. It is confirmed by comparing cost of additional KOD Vs
reduction in support cost.

◼ • Significant reduction in flare loads is possible with application of
high integrity protection system.
◼ • Reflux pump and cooling water system pumps may be
considered steam turbine driven to avoid flare load for reflux failure
and reduce flare load for cooling water failure case. In such cases,
availability of steam during power failure needs to be assured.
◼ While discussing flare job with client, there may be a need to estimate
flare header size, it may be done for low pressure flare system with
vapor sizing formulae, friction factor chart and procedure given
below:

◼ 2.3 Order of magnitude Flare Header Calculations
◼ Pressure drop calculation of flare header depends on total
equivalent length of flare header after considering
expansion loops and naturalbends as per plot plan.
Estimate of expansion loops is based on design
temperature, this requires temperature profile for various
failure cases. Based on temperature,expansion loops are
decided by engineering. Each loop consists of 4x 900 bends
( long radius(20D) ) and 20-30 meter straight length.
◼ For quick estimate, expansion loop after every 70 –80
meters of straight length may be considered.

◼ • When no data is available, design temperature of flared gas
going out of stack and main KOD may be taken 100 and 150 0C for
estimated design.
◼ • Pressure drop of ISBL header may be limited to 10% of back
pressure.
◼ • For initial calculation, pressure upstream of main flare KOD
may be considered 0.5-0.7 kg/cm2g.
◼ • For initial guess of main flare header size, total flare load at
average pressure and density (average of pressure @ upstream of
main flare KOD & OSBL) & viscosity 0.01 cp.
◼ This is for low pressure flare systems (up to1.7 kg/cm2g).
◼ For high pressure (3.5 kg/cm2g and above) flare systems, each
section of header will have significant change in density and
pressure drop & hence estimate can not be done with average
values.

◼ • Estimated properties of gases in the headers can be calculated with following
mixture relationships:
◼ MW= SWi / S(W/MW)I
◼ T=S Wi TI / S WI
◼ 3.0 POTENTIALS FOR OVERPRESSURE
◼ Overpressure is the result of an unbalance of the normal
flows of material and energy that causes buildup in some
part of the system.
UOP has simplified it with variation in heat or mass i.e.
increase in mass input( failure to open position of a control
valve which is upstream of a vessel or accidental opening of
a valve upstream of a vessel , tube rupture in a re-boiler),
decrease in mass removal (failure to a close position of a
control valve which is downstream of a vessel, accidental
closing of a valve downstream from a vessel, failure of a
compressor downstream from a vessel,

◼ failure of pump downstream of a vessel) , increase in
heat input (failure of a valve to shut off fuel to a fired
heater,increase in heat transfer caused by increase in
LMTD in a re-boiler and fire), and decrease in heat
removal(loss of cooling water, air condenser failure, loss
of heat removing circulating side stream).
◼ Accumulation ={IN – OUT } (heat or mass )
◼ Pressure relieving devices are installed to ensure that they
are not subjected to pressures
◼

◼ Detailed causes of overpressure are as follows:
◼ 3.1 Closed outlets on vessels
◼ -The inadvertent closure of a block valve/ control valve on the outlet
of a pressure vessel (while the plant is on stream ) may expose the
vessel to a pressure that exceeds MAWP. Pressure relief is required
unless lock open facility are in place.
◼ 3.2 Inadvertent valve opening
◼ -The inadvertent opening of any valve from a source of higher
pressure , such as HP steam or process fluids. This action requires
pressure relieving unless provision are made for locking or sealing the
valve .

◼ 3.3 Utility failure
◼ -The consequences that may develop from the loss of any
utility must be carefully evaluated. Typical utility and
affected equipment are given in table-1.

◼ 3.4 Partial failure
◼ -- Effect of overpressure due to loss of a utility should be
studied with chain of developments that could occur and
reaction time involved. The situation in which the
equipment fails but operates in parallel with equipment
that has different energy source, operating credit may be
taken for unaffected and functioning equipment.
◼ -An example of CWS with two pumps in parallel with
unrelated energy sources, if one of two sources of energy
fails, partial credit may be taken for other source that
continues to function. Quantity of excess vapor generated
depends on quantity of CW lost.

◼ -Another example with two pumps in parallel with one
pump providing full flow and second as standby,Second
pump has separate source of energy and is equipped with
controls for automatic start up if first pump fails. No
protective credit is taken for stand by pump because stand
by device is not considered totally reliable.
◼ -Manual cut in of auxiliaries is operator and time
dependent and must be carefully analyzed before it is used
as insurance for overpressure.

◼ 3.5 Electrical or Mechanical failure
◼ - Failure of electrical or mechanical equipment that
provides cooling or condensation in process streams can
cause overpressure in process vessels.
◼ 3.6 Loss of fans
◼ –Fans on air coolers or cooling towers become inoperative
because of loss of power (or a mechanical breakdown).On
cooling towers and air coolers, credit for cooling effect
may be obtained by convection and radiation in still air at
ambient conditions.

◼ 3.7 Loss of heat in series fractionation system
◼ – In series fractionation (bottoms from first column feed in to second
column, and the bottoms from the second feed in to the third), loss of
heat input to first column can overpressure the following column.
◼
-Loss of heat results in some of the light ends mixing with bottoms
and being transferred to the next column as feed. Under this
circumstance, overhead load of second column may consist of its
normal vapor load plus light ends from first column. If second
column does not have condensing capacity for additional load,
excessive pressure could occur.

◼ 3.8 Loss of instrument air or Electric Power
◼ - The complexity of instrument automation require
continuous sources of air or electric power. To minimize
likelihood of overpressure, fail safe position of each
control valve should be established as an integral part of
design.
◼ Failure position of each control valve is not considered
adequate relief protection as other failures in an instrument
system can cause a control valve to move in direction
opposite its failure position.

◼ 3.9 Reflux failure
◼ –The loss of reflux as a result of pump or instrument
failure can cause overpressure in a column because of
condenser flooding or loss of coolant in fractionating
process.
◼ 3.10 Abnormal heat input from reboilers
◼ –Reboilers are designed with specified heat input. When
they are new or recently cleaned, additional heat input
above normal design can occur. In the event of failure of
temperature controller, vapor generation build up pressure.

◼ 3.11 Heat Exchanger tube failure
◼ –In shell & tube exchangers, tubes are subject to failure
from thermal shock, vibration and corrosion. Whatever the
cause, HP steam will overpressure equipment on the low
pressure side of the exchanger. The possible pressure rise
need to be ascertained to determine whether additional
relief would be required if flow from tube rupture were to
discharge in to the low pressure system.

◼ 3.12 Transient Pressure Surges
◼ 3.12.1 Water Hammer
◼ -The water hammer is occurring in any liquid filled
system. It is a type of overpressure that can not be
controlled with pressure relief valve because response of
PSV is normally slow. The oscillating peak pressures,
measured in milliseconds, can rise to many times the
normal operating pressure.These pressure waves damage
the pressure vessels and piping. Water hammer is
frequently caused by quick closing valves. Where water
hammer can occur, use of pulsation dampeners should be
considered.

◼ 3.12.2 Steam Hammer
◼ -An oscillating pressure surge, called steam hammer, occur
in piping that contains compressible fluids. Most common
occurrence is initiated by rapid valve closure. The
oscillating pressure surge occurs in milliseconds, with a
possible pressure rise to many times the normal operating
pressure, resulting in vibration & violent movement of
piping and possible rupture of equipment. Avoiding use of
quick closing valves may prevent steam hammer.

◼ 3.13 Plant Fires
◼ –Fire is a cause of overpressure in plant equipment.
◼ -A provision for controlled shut down or de-pressure
system for the units can minimize overpressure caused
from exposure to external fire.
◼ -To limit vapor generation and spread of fire, facilities
should also include removal of liquids from the systems.
◼ - Normally operating product withdrawl systems are
considered superior and more effective for removing
liquids from a process unit, compared with separate liquid
pull down systems.

◼ - Liquid holdup required for normal plant operations, including
refrigerants or solvents, can be effective in keeping the vessel wall
cool & does not require removal system.
◼ -Provision may be made either to insulate the vessel, s vapor space
and apply external water for cooling or to de pressure the vessel
using a vapor de pressure system.
◼ - Area design should include adequate surface drainage
◼ facilities and a means for preventing spread of flammable
◼ liquids from one operating area to another.
◼ - Easy access to each area and to process equipment for
◼ fire fighting and their equipment.
◼ -Fire hydrants, fire fighting equipment, and fire monitors
◼ should be placed in readily accessible locations.

◼ 3.14 Process changes/Chemical reactions
◼ -In some reactions and processes, loss of process
◼ control may result in significant change in temperature
◼ and/or pressure .The result could exceed intended limit
◼ of the material selected.
◼ -For cryogenic fluids, pressure reduction can lower temperature
below minimum allowable design temp of equipment, causing low
temp brittle failure.
◼ -For exothermic reaction cases, excessive temp and/or pressure
associated with run away reactions may reduce allowable stress level
below design point, or increase pressure above MAWP.
◼ -Where normal PSV s can not protect, controls are needed to warn of
changes outside intended temp/press limits to provide corrective
action.

◼ 4.0 Determination of Individual relieving rates
◼ 4.1 Principal sources of overpressure
◼ Table-2 lists some common occurrences that may require
overpressure protection. Further descriptive analysis is also provided.
◼
◼ 4.2 Sources of Overpressure
◼ -Liquid or vapor rates to establish relief requirements are developed
by heat input(indirect pressure input through vaporization or thermal
expansion) and direct pressure input from higher pressure sources.
Overpressure may result from one or both of these sources.
◼ The peak relieving rate is the maximum rate that must be relieved to
protect equipment against overpressure due to any single cause.
◼ Probability of two unrelated failures occurring together is very remote
& normally not considered.

◼ 4.3 Effects of Pressure, Temperature , and
◼ Composition
◼ – Pressure and temperature affect the volume and
◼ composition of liquid and vapor. Vapor is generated
◼ when heat is added to liquid. The rate at which vapor is
◼ generated changes with equilibrium conditions.
◼ -When liquid is a mixture of components with different
◼ boiling points, heat introduced produces vapor that
◼ contains more low boiling components. With more heat
◼ input, successively heavier components are generated in
◼ the vapor.

◼ -During pressure relieving, the changes in vapor rates and
◼ MW at various time intervals should be investigated to
◼ determine peak relieving rate and composition of vapor.
◼ We have to do calculations for each time interval from
◼ normal operating condition to relieving condition & check
◼ peak rate.
◼ -Relieving pressure may exceed critical pressure of
◼ components then refer to compressibility co-relations
◼ to compute density-temperature- enthalpy relationship for
◼ the system fluid.
◼ - Typical sketch 24-23 in GPSA for CO2 storage in bullet,
◼ it is used to calculate relieving load during fire case.

◼ 4.4 Effect of Operator Response
◼ -The response to take credit for operator response in determining max
relieving conditions requires consideration of those who are
responsible for operation and an understanding of the consequences
of an incorrect action. Commonly accepted time range for response is
between 10-30 minutes, depending on complexity of plant.
Effectiveness of this response depends on process dynamics.
◼ 4.5 Closed Outlets
◼ -To protect a vessel or system from overpressure when all
◼ outlets are blocked, capacity of relief device must be
◼ capacity of source of pressure.

◼ -If all outlets are not blocked, capacity of unblocked may
◼ properly be considered.
◼ -Sources of overpressure include pump, compressor, HP
◼ headers, stripped gases from rich absorbent, and process
◼ heat. In case of heat exchanger, closed outlet can cause
◼ thermal expansion or vapor generation.

◼ -Quantity to be relieved should correspond to set pressure
◼ plus overpressure instead at normal conditions.
◼ -If source of liquid is pump then flow-head pump
◼ characteristics should be used to establish relief
◼ flow. Frictional losses at relieving flow and static head
◼ must also be considered to match pump curve.
◼ -The compressor flow- pressure characteristics should
◼ be considered in vapor service.

◼ 4.6 Cooling or Reflux failure
◼ 4.6.1 General
◼ - Required relieving rate is determined by heat and material balance
of the system at relieving pressure.
◼ In a distillation system, rate may require calculation with or without
reflux.
◼ Instead of detailed calculations, simple bases is as follows:
◼ 4.6.2 Total Condensing
◼ -Relief load is total incoming vapor to condenser recalculated at temp
that corresponds to new vapor composition at set pressure plus
overpressure, and heat input prevailing at the time of relief.

◼ - Reflux drum capacity is about 5 minutes, if cooling water
failure exceeds this time, reflux is lost then calculations
have to be performed without reflux i.e. tray-2 vapor.
◼ -Second tray vapor may be considered when overhead
receiver holding time is less than five minutes.
◼ 4.6 .3 Partial Condensing
◼ - Relief is difference between incoming and out going
vapor rate @ relieving conditions (or quantity in total
condensing is adjusted downwards by quantity of vapor
normally leaving reflux drum).

◼ 4.6.4 Fan Failure
◼ - Because of natural convection effects, credit for a partial condensing
capacity of 20 to 30 percent of normal duty is often used. Capacity of
relief valve is based on 70-80 % dutty.
◼
◼ 4.6.5 Louver Closure
◼ -Louver closure is considered total failure of coolant with resultant
capacity similar to total condensing or partial condensing cases.
◼ Louver failure may result from automatic control failure, mechanical
linkage failure or destructive vibration on a manually positioned
louver.

◼ 4.6.6 Overhead Circuit
◼ - In many cases reflux failure due to pump shut down or valve closure
can cause flooding of condenser, which is equivalent to total loss of
coolant with capacity established with above mentioned method.
Compositional changes caused by loss of reflux may produce
different vapor properties that affect capacity.
◼
4.6.7 Pump Around
◼ -Relief requirement is vaporization rate caused by an amount of heat
equal to heat removed by pump around circuit. Latent heat of
vaporization would correspond to latent heat at relieving conditions
of temp and pressure.

◼ 4.6.8 Overhead Circuit plus pump around
◼ -An overhead circuit plus pump around is arranged so that
simultaneous failure of pump around and overhead condenser will not
occur, part failure of one with complete failure of other is quite
possible. Required relieving capacity is similar to above mentioned.
◼
◼ 4.6.9 Side stream reflux failure
◼ –Principles similar to overhead circuit and pump around apply for
condenser flooding or changes in vapor properties resulting from
change in composition. Relieving capacity should be enough to
relieve vaporization rate caused by amount of heat removed from
system.

◼ 4.7 Absorbent Flow Failure
◼ -In a typical acid gas removal unit case, 25 % or more inlet vapor
may be removed, loss of absorbent could cause overpressure to relief
as downstream may not be adequate to handle increased flow. Each
individual case must be studied for its process and instrumentation
characteristics.
◼
◼ 4.8 Accumulation of non-condensables
◼ - Normally noncondensables do not accumulate, but with certain
piping configuration, noncondensables accumulate to the point that
condenser is blocked. Effect is equal to total loss of coolant.

◼ 4.9 Entrance of volatile material in to system
◼ –It covers water (or hydrocarbon) in to hot oil, expansion from liquid
to vapor is so fast and large that PSV can not be provided. So proper
design and operation are essential. Avoid water pockets, provide
steam condensate traps, and double block with bleed on water
connections to hot process lines.
◼
4.10 Failure of process stream automatic controls
◼ 4.10.1General
◼ Automatic control devices, directly actuated from process or
indirectly from a process variable (P, T, Flow, Level) are used at inlets
or outlets of vessels or systems. When transmission signal (or
operating medium to final control element (valve operator) ) fails,
control should assume either fully open or fully closed position.
Design of control valve has to be kept in mind for relief evaluation.

◼ 4.10.2 Capacity Credit
◼ -For relief evaluation due to any cause, control valve is
assumed in normal condition at relieving conditions.
◼
4.10.3 Inlet Control Devices
◼ -There may be single or multiple inlet lines fitted with
control valves, scenario is to consider one inlet valve in
fully open position regardless of the control valve failure
position. Opening of this control valve may be caused by
instrument failure or misoperation.

◼ If the system has multiple inlets, the position of any control device in
those remaining lines may be assumed in normal operating position.
Therefore relief capacity is difference between max expected inlet
flow and normal outlet flow adjusted for relieving conditions,
assuming that other valves in the system are still in operating position
at normal flow.
◼ -If one or more of the outlet valves are closed or more inlet valves are
opened by the same failure that caused the first valve to open,
required relief capacity is difference between max expected inlet flow
and normal flow from outlet valves that remain open. All flows
should be calculated at relieving conditions.

◼ -Typical example of a pressure vessel where liquid bottom
on level control discharge in to low pressure system.
◼ - Usually when liquid is let down from high to low
pressure, only flashing effect is of concern in the event,
low pressure has a closed outlet.
◼ -However designer should consider vapor will flow in to
low pressure system if loss of liquid level occurs in HP
vessel. In this case, if volume of incoming vapor is large
compared with volume of low-pressure system,
overpressure may occur, relief device on low pressure side
should be able to handle full vapor through liquid control
valve.

◼ -When process systems involve significant differences in pressure
level and volume of vapor contained by high pressure equipment side
is less than volume of low pressure side, additional pressure may get
absorbed in some cases without overpressure.
◼
◼ -In the event of loss of liquid level, vapor flow in to low pressure
system depends on Interconnecting system, which usually consists of
wide open valves and piping, passes with a differential pressure based
on normal operating pressure on upstream and relieving pressure
downstream equipment, this DP at initial conditions result in critical
flow & may cause rate to be several times higher than normal flow to
HP side. Unless make up equals outflow, as this will be for short
duration, upstream reservoir is depleted, relief should be sized to
handle peak flow.

◼ 4.10.4 Outlet Control Devices
◼ - Each outlet control valve should be considered both in
fully open or fully closed position for purposes of flare
load determination. This is regardless of control valve
failure position and may be caused by instrument failure or
misoperation.. If one or more of inlet valves are opened
by same failure that caused outlet valve to close, pressure-
relieving device may be required to prevent overpressure.

◼ Relief capacity is difference between max inlet & max
outlet flows @ relieving conditions. Also effect of
inadvertent closure by operator has to be considered.
◼ -For single outlet with control fail close, relief is max inlet
flow at relieving conditions.
◼ -For application with more than one outlets & one outlet
with fail close control device, relief is difference between
max expected inlet flow and design flow @ relieving
conditions.
◼ -For applications with more than one outlet each with
control device, relief is max expected flow @ relieving
conditions.

◼ If normal instrument operation results in increased
relieving rate then consider instrumentation to operate, if
normal instrument operation results in decrease of
relieving rate then consider instrumentation fail to operate.
◼
4.10.5 Fail Stationary Valves
◼ -Even though some control devices are designed to remain
stationary in the last controlled position, one can not
predict the position of the valve during failure. Designer
should consider such devices fully open or closed : no
reduction in relief capacity should be considered when
such devices are used.

◼ 4.10.6 Special Capacity Considerations
◼ -Although control valves are specified and sized for
normal operating conditions, they are expected to operate
during upset conditions, including periods when PSV s are
relieving.
◼ -Valve design and valve operator capability should be
selected to position valve plug properly in accordance with
control signals during abnormalconditions.
◼ -Since the capacity at pressure relieving are not same as
normal conditions, capacity of control valves should be
calculated for relieving conditions of temp and pressure in
determining required relief capacity.

◼ -In extreme cases , state of fluid controlled may change (from liquid
to vapor or from vapor to liquid).
◼
-Wide open capacity of control valve selected to handle liquid may
differ greatly when it handles gas. This becomes a matter of
particular concern where loss of liquid can occur, causing the valve to
pass high pressure gas to a system sized to handle only the vapor
flashed from normal liquid entry. .
◼ 4.11 Abnormal Process Heat Input
◼ - Required capacity is max rate of vapor generation at relieving
conditions (including non-condensables produced from overheating)
less the rate of normal condensation or vapor outflow.

◼ -Designer should consider potential behavior of system and
each of its components .For example built in over capacity
of burners capable of 125 % of heater, design input must
be considered.
◼ -Where limit stops are installed, wide open capacity should
be used. However if mechanical stop is installed and is
adequately documented, use of limited capacity may be
appropriate.
◼ -In shell and tube heat exchangers, heat input should be
calculated on the basis of clean conditions.

◼ 4.12 Internal Explosion
◼ -Where overpressure against internal explosions caused by ignition of
vapor air mixture is to be provided, rupture discs or explosion vent
panels should be used. Relief valve react too slowly to protect the
vessel against the extremely rapid pressure build up caused by
internal flame propagation.
◼ -Vent area required is a function of initial T/P/composition, Flame
propagation properties, volume of vessel, pressure at which vent
activates, and maximum pressure that can be tolerated during vented
explosion.
◼ -Peak pressure reached during vented explosion is usually higher than
pressure at which vent device activates.
◼ -Design should follow NFPA 68 and 69.
◼ -Inert gas purging may be used when explosion is due to air
◼ contamination during shut down or start up.

◼ 4.13 Chemical Reaction
◼ -Typical methodology for chemical reactions need pressure relief
where possible or other design strategies like automatic shut down
systems, inhibitor injection, quench, de-inventorying, alternative
power supply, and de pressure.
◼
◼ 4.14 Hydraulic Expansion
◼ -It is due to increase in liquid volume due to increase in temperature.
It may be due to blockage of cold liquid in exchanger with hot liquid
on other side, heating of long pipelines with solar radiation.
◼ Typical relieving device is ¾ inch x 1 inch relief valve.
◼ If this size appears to be inadequate then procedure given in API-521
(Section3.14.3 ) may be used.

◼ 4.15 External Fire
◼ 4.15.1 General
◼ 4.15.1.1 Effect of fire on wetted surface of a vessel
◼ - The surface area wetted by internal liquid contents of a pressure
vessel is effective in generating vapor when exposed to fire. To
determine vapor generation , only that portion of the liquid which is
wetted by its internal liquid and is equal to or less than 25 ft above the
source of flame.
◼ - Credit can be taken for insulation, if it is fireproof and project
specification allows for this credit. It should be certain that basic
insulating material to function effectively up to 904 0C during a fire.
This period of fire may be up to two hours, depending on fire fighting
provisions. Although jacketing and coating may burn off or
disintegrate, insulation must retain its shape. Hence no credit is taken
for any insulation.

◼ Wetted surface of typical vessels-
◼ Horizontal Vessel- Aw=[p L D +2 p D2 x 1.66/4] Fwp
◼ L-vessel T-T length, ft
◼ D-vessel diameter, ft
◼
% Tank Volume % Tank Diameter Fwp
◼ 0 0 0
◼ 20 25.4 0.34
◼ 40 42.8 0.45
◼ 60 57.8 0.55
◼ 80 74.6 0.66
◼ 100 100 1.0

◼ • Vertical Vessel
◼ Aw = p L D + N p D2 x1.66/4
◼ L- Length of vessel exposed to fire,ft
◼ D-Diameter of vessel,ft
◼ N-Number of heads exposed to fire (assumes
◼ elliptical head)
◼ • Liquid- vessels-(such as treating vessels )
◼ Wetted surface of vessels that operate liquid full should
◼ be total surface with in 25 ft height.
◼ Fractionating Column-Wetted surface will be based on high liquid
level (API-normal liquid level plus tray draw off capacity) in the
bottom plus wetted surface corresponding to four(API-weir height+2
inch) inches of liquid for any tray with in a height of 25 ft. from
grade.

◼
◼ • Surge Drums-Wetted surface will be up to high level but at least
50% of total vessel surface.
◼ • Knock out drum-Usually operates with only a small amount of
liquid, use surface up to high liquid alarm.
◼ • Working Storage tanks Usually 50% of surface. If material of
construction is such that tank collapses during a fire, PSV not
provided. Large capacity cone roof or floating roof tanks are not
provided PSV, however dome roof tank is provided PSV & VRV.
◼ • Vessel in pit-use entire area
◼ • Shell & Tube Exchanger

◼ Shell side –wetted surface area as liquid full vessel excluding channel
area
◼ Tube side-wetted surface of channel
◼ • Air Cooled Exchanger-Being designed for ambient inlet air
conditions, during fire, lose all cooling and condensing ability.
Assume exchanger as vessel, relieving load is calculated by using
bare tube area because most fins are constructed of aluminum and are
destroyed with in first few minutes of fire.
◼ • Condensing without sub cooling-wetted surface area equal to
0.3 times bare tube area (based on bottom 30% of circumference
being wetted by condensate layer.
◼ Condensing with sub cooling-Condensing section should be treated so
that wetted surface area is equal to bare tube area.

◼ • Gas Cooling-Surface area is equal to bare tube area.
◼ • Liquid Cooling-Wetted surface area is equal to bare
tube area. Fire loading in case of liquid coolers may
become extremely large and in some cases ,dominant
loading in sizing major portion of relief system.
◼ Exposed Height -Since air coolers tend to produce a
chimney effect by drawing hot combustion products in to
their plenums, height limit on fire exposure may not apply.

◼ 4.15.2 Sizing
◼ 4.15.2.1 Heat Absorption Equations
◼ -Amount of heat absorbed by a vessel exposed to open fire is affected
by type of fuel feeding the fire, degree to which vessel is enveloped
by flames(function of vessel size and shape) and fire proofing
measures. Following formula is used to evaluate these conditions
where there are prompt fire fighting efforts and drainage of
flammable materials away from vessel.
◼ Q = 21000 F A 0.82
◼ Q-total heat absorption to wetted surface in BTU/hr ft2
◼
◼ F=1(No credit for insulation)
◼ A-Total wetted surface in ft2

◼ 4.15.2.2 Molecular Weight, Latent heat of Vaporization and Relieving
Temperature
◼ -For pure components where P.H. diagram is available, same shall be
used for calculating relieving temp and latent heat of vaporization.
Refer GPSA, for other pure components, refer fig A-1 of API-521.
◼ Example- C2 Splitter Column
◼ Set pressure-20.5 kg/cm2g
◼ Overpressure-20%
◼ Relieving pressure=20.5x1.2 x 14.22 + 14.7 =364.5 say
365psia
◼ From P.H. diagram,
◼ Relieving temp 40 0F- (approximated on higher side)
◼ Latent heat of vaporization –125 BTU/lb (approximated to lower
side)

◼ Example-Hexane Column
◼ Set pressure-3.5 kg/cm2g
◼ Relieving Pressure-3.5x1.2 x14.22 +14.7 =75 psia
◼ Relieving Temp 270 0F
◼ (From API521 fig A-1)
◼ Latent heat of vaporization 121 BTU/lb
◼ -Mixture of pure components & mixtures containing pseudo
components (light and heavy petroleum fractions)
◼ • MW of relieving vapor =0.5 x MW of liquid + MW of vapor in
equilibrium with liquid
◼ or
◼ • MW of lighter 10% of liquid for blanketed surge drums
◼ -Relieving Temp

◼
◼ • For mixtures of pure components, use API-520 D-3
figure and average MW of liquid mixture to calculate
relieving temp.
◼ • For mixtures containing pseudo components, use
Technical Data Book.

◼ Example-Naphtha Splitter reflux drum
◼
◼ Normal operating pressure 0.5 kg/cm2 g
◼ Normal operating temp 77 0C
◼ set pressure 3.5 kg/cm2 g
◼ Relieving pressure 3.5x 1.2 + 1 = 5.2 atm
◼
◼ From Technical Data Book, 5-21- With VP 1.5 atm. and operating
temp of 170 0F, Normal boiling point for mixture
◼ is 195 0F
◼ Relieving temp @ 5.2 atm is 285 0F.

◼ -Latent Heat of Vaporization
◼ For mixtures of pure components, refer API-520 fig D-3
◼ and average MW of liquid mixture to determine latent heat of
vaporization
◼
-For mixtures containing pseudo components, Use Technical Data
Book, Figures from 7-122 to 7-129.
◼
◼ Example-Naphtha Splitter reflux drum
◼ 0API of overhead liquid 63
◼ K watson = KUOP =11.833 say 11.8
◼ From fig 7-126,for K=11.8, 0API liquid=60,
◼ Relieving temp = 260 0F

◼ Latent heat of vaporization =350- 220=130 BTU/lb.
◼ For lighter fractions than given in these charts,
◼ consider 50 Kcal/kg
◼ -Water Systems-use steam table to determine saturation
temp (relieving temp) at relieving pressure (set pressure +
accumulation) and obtain latent heat from steam tables.

◼ Example- set pressure 7 kg/cm2 g
◼ Accumulation -20%
◼ Relieving pressure = 7 x 1.2 + 1 = 9.4 atm
◼ From steam table, Relieving temp = 177 0C
◼ Latent heat of vaporization 485.8 kcal/kg
◼
◼ -System containing H/C and water

◼ For vessels with hydrocarbon and water as immiscible
phases, both phases exert their individual pressure and
whenever sum of vapor pressures equals relieving
pressure, relief valve opens and temp of liquid at this point
is relieving temp. Vapor composition relieved would be
such that molar ratio of hydrocarbon vapors and water
vapor would be in the ratio of their vapor pressure.

◼ -For Gas Filled Vessels, relieving capacity need not be worked out by
Process engineer ,as instrument engineer calculates size of relief
valve based on vessel wall temp, and vessel surface area. Vessel wall
temp shall be specified as:
◼ 5930C for CS vessels
◼ 649 0C for alloy steel and 400 series SS
◼ 705 0C for 300 series SS
◼ Vessel surface area is total surface area of vessel up to 25 ft from
source of flame.
◼ Overpressure of 20% as per API shall be specified for such valves.

◼ Process engineer to calculate max relieving temp , if it
exceeds 400 0C,sprinkler system to be provided. Following
equation may be used for relieving temp calculations,
◼
T2=P2 x T1/P1
◼ P2- 1.2 x Set pressure, kg/cm2 a
◼ P1-Operating pressure, kg/cm2a
◼ T1- Gas Operating temp, 0K
◼ T2-Max gas temp at relieving pressure, 0K

◼ Example-For a gas storage vessel(CS) operating at 20-40 0C,
Specification for relief is as below ;
◼ Vessel wall temp 593 0C
◼ Gas Operating temp 20-40 0C
◼ Operating pressure 5.35 kg/cm2 g
◼ set pressure 38.5 kg/cm2g
◼ Vessel surface area Total area including shell and
◼ dished ends
◼ Maximum gas temp @ relieving pressure
◼ P1/T1=P2/T2
◼ T2=1.2x39.5 x(273+40)/6.35
◼ =2336 0K or 2063 0C > 400 0C
◼ Provide automatic sprinkler system besides relief valve

◼ 4.16 Opening Manual Valves
◼ - When a manual valve is inadvertently opened, causing pressure
build up in a vessel, vessel should have pressure relief large enough to
pass a rate equal to flow through open valve with pressure in the
vessel at relieving conditions.
◼
◼ 4.17 Electric Power Failure
◼ Determination of relieving requirements from power failures
require a careful plant or system analysis to evaluate what equipment
is affected by power failure and how failure of equipment affects
plant operation.

◼ • Careful study and consideration should be given to material
presented in table on possible utility failures and equipment affected
& also on partial failure.
◼ • Automatic stand by is an excellent device for maximizing units
on stream time, minimizing unit upsets, and ensuring unit production
rates but circuitry, sequence and components involved are not yet
considered sufficiently reliable to permit credit in establishing
individual relieving requirements.
◼
Power failure may be analyzed in three different ways i.e. local
failure for one equipment, intermediate failure of one substation
affecting one distribution center or one MCC or one bus, and total
power failure in which all electrically operated equipment is
simultaneously affected.

◼ • Effect of local power failure such as pump failure can cause loss
of cooling water(due to flooding of condenser) or loss of reflux.
◼ • Intermediate power failure may cause more serious effects than
other two failures. Depending on the method of dividing various
pumps and drivers among electrical feeders, it is possible to lose all
the fans of an air cooler at the same time that the reflux pumps are
lost. This can flood condenser and may void any credit for natural
convection of air condenser.
◼ Total power failure requires additional study to analyze and evaluate
combined effects of multiple equipment failures. Special
consideration should be given to effect of simultaneous opening of
relief valves in several services to a closed system.

4.18 Heat Transfer Equipment Failure-Tube Rupture
◼ 4.18.1 General
◼ An internal failure can vary from a leaking tube or tube sheet to a
complete tube rupture where a sharp break occurs in one tube. For
relatively low pressure equipment, tube failure is not a contingency
when design pressure of low pressure side is equal to or greater than
two thirds the design pressure of high pressure side.
◼ Tube rupture is a contingency when design pressure of low pressure
side is less than ten by thirteen of design pressure of high pressure
side. If high pressure side of exchanger operates at 1000 psig or more
and contains a vapor or liquid that can flash or result in vaporization,
complete tube failure should be considered.

◼ 4.18.2 Flow through ruptured tube
◼ • The maximum flow through a ruptured tube occurs
when the break is at the tube sheet. If a break occurs at that
location, high pressure fluid will enter low pressure side by
two routes: 1) through tube sheet (assumed to be square
edged orifice) and 2) through ruptured tube.
◼ • By using equation for flow through square edged
orifice, an orifice coefficient of 0.7, and a pseudo orifice
area equal to 1.5 times cross sectional area of one tube,
flow rate of HP fluid is approximated.
◼ From Crane (Technical paper No 410),flow through an
orifice is:

◼ W=1891 do
2 C(DP.r)1/2
◼ Where:
◼ W= mass flow rate, lb/hr
◼ do= internal diameter of the orifice, inches ( This is taken
to be tube ID )
◼ C= flow coefficient of an orifice ( Use C=0.7)
◼ DP= pressure differential , p s i
◼ r=upstream density,lb/ft3
◼ By combining 1.5 factor, orifice coefficient, and Crane
formula constant(1891)

◼ Flow of non-flashing liquid,
◼ W=1985 d2 (DP r)1/2
◼ d= tube ID, inches
◼ DP=pressure differential, p s i
◼ r= upstream density, lb/ft3
◼ Flow of vapor or two phase mixture
◼ W=1985 Y d2 (DP r )½
◼ Y=net expansion factor for compressible flow, Refer figure-A21 from
Crane , or as conservative approach, use 0.8
◼ d= tube ID, inches
◼ r=vapor density, lb/ft3 (for two phase flow, use no slip density)

◼ 4.18.3 Effect of LP flow on downstream fluid
◼ 4.18.3.1 No heat transfer occurs
◼ If no heat transfer occurs, high pressure fluid will initially displace an
equal volume of low pressure fluid. As the casualty progresses, high
pressure fluid will be relieved directly. The relieving requirements for
both time intervals should be calculated and the largest requirement
chosen.
◼ 4.18.3.2 Heat transfer occurs
◼ If heat transfer occurs, the effect of both heat transfer and volume
displacement must be considered. (If the effect of heat transfer is
large when compared to displacement, effect of volumetric
displacement may be ignored).

◼ -Heat transfer with no phase change
◼ Initially, the relieving requirement will be the same as for
4.18.3.1 case. The high pressure fluid will displace an
equal volume of low pressure fluid. However, the high
pressure vapors may thoroughly mix with low pressure
vapors. This may result in a larger relieving requirement
and its effect may be calculated as follows:

◼ Wr = We[ Mo/Me +Ce/Co +((Te-To)/To]
◼ where:
◼ M= molecular weight
◼ C= heat capacity, BTU/lb/0F
◼ T= temperature, 0R
◼ Subscripts:
◼ r= relieving requirement
◼ e= entering
◼ o= originally present
◼ Note: This mixing effect does not have to be considered unless both
the high pressure and low pressure fluids are vapor phase and a
significant temperature difference exists between the two fluids.

◼ -Heat transfer with phase change
◼ Usually the volumetric displacement is insignificant in this situation.
◼ -Calculate bubble temp of liquid phase at relieving conditions.
◼ -Calculate change in enthalpy in the hot phase as it is
◼ cooled down to bubble point of liquid phase.
◼ -Use enthalpy change of hot phase to vaporize bubble point
◼ liquid.
◼ 4.18.4 Example
◼ Hot oil is HP fluid, From exchanger specification sheet, tube
diameter=3/4 inch,14 BWG
◼ d= 0.584 inch

◼ Design pressure of low pressure side=55psig
◼ DP across all trays= 10 psi
◼ Pacc @ column bottom=1.1(55)+14.7 +10 =85.2 psia
◼ P1=112 psig or 126.7 psia
◼ Sp Gr=0.815
◼ r=0.815 x 62.36=50.82 lb/ ft3
◼
◼ Since hot oil is non flashing HP fluid, equation for non
flashing liquid should be used.

◼ W=1985 d2 ( DP r)1/2
◼ =1985(0.584)2 [(126.7-85.2) 50.82]1/2
◼ =31090 lb / hr
◼ Calculate effect of this mass flow on downstream fluid,
◼ Heat transfer will occur with phase change of LP liquid,
◼ Bubble point of column bottom at Pacc= 85.2psia,
◼ MW-108.2
◼ From Cox chart, T=360 0F
◼ Heat change in hot oil as it is cooled to 360 0F,
◼ Cp=0.56 BTU/lb/0F
◼ Q=31090 x 0.56 x (565-360) =3.569 x106 BTU/hr

◼ -Extra vaporization of liquid inventory
◼ From Cox chart, LHV= 108 BTU/lb @ 360 0F
◼ and 85.2 psia
◼ Wvap =Q/LHV=3.569 x106/108 = 33046 lb/hr
◼ Since reboiler LMTD would be reduced at relieving
pressure, an adjusted heat and material balance shows that
this rate is absorbed and would not result in relieving load.

◼ 5.0 Sample Calculations –Flare System
◼ 5.1 Flare Header
◼ The flare header collects the material relieved by safety
valves and other sources for safe discharge to the flare.
◼ Basis of sizing the header is that back pressure at the outlet
of PSV should not exceed maximum allowable back
pressure for that valve. Also the velocity in the header are
limited to a fraction of mach number.
◼ Before header sizing, it is necessary to calculate
temperature drop to optimize flare header expansion loops
to be provided on flare header.

◼ 5.2 Temperature Drop Calculations
◼ 5.2.1 Procedure
◼ This calculates heat loss from the main header to the
environment and gives temperatures at various points of
the header These calculations are done for segments of the
header so that the assumption of mean gas temperature as
average of inlet and outlet of a segment is valid.
◼ Equation used
◼ Q=0.86[0.548E{(Tms /55.55)4-(Tma/55.55)4 }
◼ +1.957(Tms-Tma)1.25 {(196.85Vm+68.9)/68.9 }0.5 ]A

◼ Tms –mean surface temperature between two points (0K)
◼
◼ =(T1+T2)/2
◼ Tma- mean ambient temperature (0K )
◼
◼ Vm - wind velocity (m/sec)
◼
◼ E - Emissivity (0.9)
◼
◼ A - Surface area of pipe ( p D L), M2
◼ D - Diameter of flare header pipe, M

◼ 5.2.2 Typical Sample calculations
◼ Basis: Flow rate - 78311 kg/hr
◼ Cp - 0.5 Kcal/kg 0K
◼ Mean ambient temperature 25 0C or 298 0K
◼ Gas inlet temperature T1 143 0C or 416 0K
◼ Wind Velocity 30 f/sec or 9.143 m/sec
◼ Flare header diameter - 15.376 inch or 0.39 M
◼ Emissivity - 0.9
◼ LHS Equation
◼ m Cp (T1-T2) =78311 x 0.5 x(416- 403.21)
◼ =500798.85 Kcal/hr

◼ RHS of Equation
◼ 0.86[0.548E{(Tms/55.55)4 –(Tma/55.55)4 }
◼ +1.957(Tms-Tma) 1.25 x {(196.85Vm+68.9)/68.9} 0.5 ] A
◼ T1 - inlet temperature of the section 0 K 416 0K
◼ T2 -outlet temperature of the section (0K) 403.21 0K (From
computer output)
◼ Tma – mean ambient temperature 0K
◼ A- Surface area of pipe-p D L
◼ Vm – wind velocity, m/sec
◼ Tms -Mean surface temperature (0K )
◼ =(T1+T2)/2 ( Assumed )

◼ For this assumption to be valid, it is advised to take
suitable
◼ length for a particular section, with 100 m section length,
◼ exit temperature-403.21 0K
◼ RHS=0.86[0.548x0.9{((416+403.21)/55.55x2)4 –
◼ (298/55.55)4 } + 1.957{(416+403.21/2)-298}1.25
◼ {(196.85x9.14+68.9)/68.9}0.5 x 39 p
◼ =500697.58
◼
◼ So temperature after 100 meter section is 403.21 0K

◼ 5.3 Pressure Drop Calculations
◼ 5.3.1 Procedure
◼ Flare load for header calculation consists of very negligible
condensate so calculations are close to vapor sizing method.
◼ Basis of vapor sizing
◼ Gas density =11.795xMWx Pressure(Kg/cm2a)/Temp(0K)
◼ Re=35.4xM/m D DP=6.37 M2 f / D5 rg
◼ Re = Reynold Number M=Mass flow rate,kg/hr
◼ rg= Density in kg/M3
◼ f= Friction factor(From chart of friction factor Vs Reynolds
◼ Number for clean commercial steel and wrought iron pipe)
◼ D= inside diameter of pipe in cm
◼ m = viscosity in centipoise

◼ For computer program,
◼ (1-f)1/2=-4log[e/(3.7 D) +1.256/(Re .(f)1/2 ]
◼ e= Absolute roughness,0.0018 inch
◼ Re =Reynold No.
◼
◼ DP= Pressure drop in bar for one kilometer of pipe
◼
◼ Pipe Schedule=40
◼ 5.3.2 Sample Calculations
◼ Flow =78311 kg/hr
◼ Mol. Wt.=70.79
◼ T = 143 0C

◼ Flare header / Stack diameter = 17.376 inch
◼
◼ Viscosity = 0.01 c p
◼
◼ Stack height = 100M (With margins for pipe sections between KOD
& WSD & WSD and stack)
◼ Back pressure @ plant B/L= 2.8 kg/cm2 a
◼ Header Length =830M
◼ Stack Losses
◼
◼ r (Gas Density) = 11.795 x 70.79/416=2.007 kg/m3
◼ G(Mass Flow Rate) =78311/{3600 x (p x D2)/4}
◼ = 142.269 kg/s/m2

◼ Exit loss =G2/(2 . r . 9.81 x 104 ) =0.0498 kg/cm2 g
◼ Flare tip loss By Flare tip vendor 0.05
◼ Molecular Seal loss = Normally four 900 bends 0.05 kg/cm2
◼ Pressure at mol seal inlet =1.033+ 0.05+0.05 +0.0498 = 1.1828
◼ Density of gas, =11.795 x 70.79 x 1.1828/ 416
◼ = 2.3747 kg/m3
◼ Friction loss for 100 M pipe
◼ Re =35.4 x M/m D =35.4 x 78311/0.01 x 2.54 x 17.376
◼ =628,1198
◼ From graph, f = 0.012
◼ DP =6.37 x (78311)2 x 0.012/( 2.54x17.376)5 x2.3747
◼ = 1.178 kg /cm2/KM

◼ DP=0.1178 kg/cm2 (for 100 meter )
◼ Static Head = hr / 10 4 = 100 x 2.3747/10 4 =0.0237 kg/cm2
◼ Entry Loss = 0.5 G2 /(2.r.9.81.104 )= 0.0249 kg/cm2
◼
◼ Pressure at base of stack =1.1828 + 0.1178 + 0.0237 +.0249
◼ =1.3492 kg/cm2
◼
◼ Water Seal Drum Losses
◼ Entry and exit losses = 1.5x velocity head=1.5x 0.0498
◼ =0.0747 kg/cm2
◼ Seal loss =0.03 kg/cm2(based on 300 mm water seal height )
◼ Frictional loss- based on 10 m pipe length + one 900 bend
◼ Equivalent Length=10+30x17.376/39.37 =23.2 M

◼ Gas Density @ 1.3492 kg/cm2 pressure
◼ =11.795x70.79x1.3492/416 = 2.708 kg/cm2
◼
◼ DP water seal = 0.1178x2.3747 x 0.232/2.708 =0.024 kg/cm2
◼
◼ Pressure at WSD inlet = 1.3492+ 0.0747+0.03+0.024
◼ =1.4779 kg/cm2
◼
◼ KOD Losses
◼ Density of gas =11.95 x 70.79 x 1.4779/416
◼ = 2.966 kg/m3

◼ Number of bends = 3 x 900 bends
◼
◼ Equivalent length=3x 30d = 90 d
◼
◼ DPKOD= 90x 0.44x2.3747x0.1178/100x2.966
◼ =0.0374 kg/cm2
◼
◼ Inlet/outlet losses = 1.5xG2/2 x r x9.81 x104
◼ =0.05178 kg/cm2
◼
◼ Pressure @ KOD inlet = 1.4779+0.0374+0.05178
◼ = 1.5671 kg/cm2 a

◼ Flare Header Design
◼
◼ Density of gas = 2.007 x 1.5671=3.1451 kg/m3
◼ For DP calculations, header may be divided in to 50 meter
sections and density will be modified after the calculation of each
section.Pressure has to be selected in such a manner that pressure
@ B/L comes close to 2.8 kg/cm2a. Typical calculations done are
as follows:
◼ For each section, DP x Density=5.5 has been taken. These
calculations are done without any change in temperature.
◼ Section –1, DP=50x5.5/1000 x 3.1451=0.0874
◼ P=1.567+0.0874=1.6545
◼ r= 3.3206 kg/m3

◼ Section –2, DP=5.5/3.3206 x50/1000 =0.0828 kg/cm2
◼
◼ P=1.6545+0.0828= 1.7373 kg/cm2
◼ r= 3.4868 kg/m3
◼ Similar calculations for all sections are done as follows:
◼ Section –3,DP=0.07887,P=1.8162, r=3.6451
◼ Section –4,DP=0.07554, P=1.8917,r=3.7967
◼ Section-5,DP=0.07243,P=1.9641,r=3.942
◼ Section –6,DP=0.06976,P=2.034, r=4.082
◼ Section-7,DP=0.06737,P=2.1014,r=4.217
◼ Section-8,DP=0.0652,P=2.1666,r=4.3484
◼ Section-9 ,DP=0.0632,P=2.229,r=4.474
◼ Section 10,DP=0.0615,P=2.2905,r=4.597

◼ Section –11,DP=0.05982,P=2.3503,r=4.7171
◼ Section-12, DP=0.0583,P=2.4086,r=4.834
◼ Section-13,DP=0.0569,P=2.4655,r=4.9482
◼ Section-14, DP=0.556,P=2.5211,r=5.0598
◼ Section-15,DP=0.0543,P=2.5754,r=5.1689
◼ Section-16,DP=0.0532,P=2.6297,r=5.277
◼
◼ Last section of 30 meters,
◼ Section-17,DP=0.0313,P=2.66,r=5.34
◼ Final figure @ B/L is close to back pressure available.

◼ 5.4 Main Flare KOD Design
◼ 5.4.1 General
◼ • It traps any liquid droplets in the flare load or any liquid
condensed with in the flare header. It prevents hazards associated
with burning liquid droplets escaping from flare stack.
◼ • Condensation is expected in the flare header when temperature
of gases flowing inside the header is high and there is considerable
amount of heat loss to environment.
◼ • If design load is about 300 t/hr, 5M3 liquid hold up (or) 600 mm
between HLL and LLL shall be considered for design, even if no
condensation is expected as per calculations.

◼ Condensate in the KOD should be considered as subcooled for NPSH
(available) calculations. Typical NPSH for condensate pump is 2
meters and KOD shall be elevated 1.5 meter (min) from grade.
Higher elevation of KOD results in increased elevation of flare
header, thus increasing the system cost.
◼
◼ • KOD design shall be as per API 521 ( Drag coefficient method
),API code specifies 300-600 microns (irrespective of flare load) , it
has been decided to design main flare KOD with 600 micron size
particles & ISBL KOD with 400 micron particle size, adequacy of
main KOD shall be checked considering 400 micron size particles.
◼ • ISBL KOD may be a horizontal vessel to minimize unit flare
header elevation.

◼ Basic Data-
◼ Relief Flow=1,000,000 lb/hr
◼ Liquid content = 25 percent by weight
◼ Operating conditions in drum = 2 psig , 300 0F
◼ rL =40 lb/ft3
◼ rV=0.18 lb/ft3
◼ m=0.01 cp
◼ D=400 micron or 0.00133 ft

◼ KOD pump starts automatically
◼ Assume horizontal drum,
◼ Vapor flow=0.75 x 1,000,000 =750,000 lb/hr
◼ =750,000 /(3600 x 0.18) =1157 acfs
◼ Liquid flow=0.25 x 1,000,000 =250,000 lb/hr
◼
◼ C(RE)
2 =0.95 x 108 x rv x(D)3 x(rL-rV )/m2
◼ =0.95.108 .0.18 .0.001333 .(40-0.18)/ 0.012
◼ = 16020
◼ From figure-20 from API-521, page 64,C=0.9
◼ Drop out velocity Uc = 1.15 [g D(rL-rV)/C rV]0.5
=1.15[32.2.0.00133.(40-0.18)/0.9 . 0.18]0.5
◼ = 3.7 ft/sec

◼ Liquid surge time=15 minute ( automatic start of pump)
◼ Liquid Storage capacity=250,000 x 15/40 x 60 =1563 ft3 Assume
drum diameter= 12 ft ,TTL=24 ftAL = 1563/24 =65 ft2
◼ For determining area of a segment of a circle,
◼ AL =R2 cos-1{(R-h)/R}- (R-h)(2Rh- h2)0.5
◼ By trial and error, with AL=65 ft2, R=6 ft,h=6.7 ft
◼ Vertical height for liquid drop out =12-6.7= 5.3 ft
◼ Liquid drop out=5.3/3.7=1.4 sec
◼ AV=AT-AL
◼ = 122 /4 –65=48 ft2
◼ Vapor velocity=1157/48=24.1 ft/sec
◼ Required drum length=24.1 x 1.4 = 34.2 ft

◼ Greater than assumed 24 ft, assumed drum size low
◼ For next trial, with vapor velocity 3.7 ft/sec and vapor flow
◼ 1157 a c f s ,
◼ AT=1157/3.7=312 ft2
◼ Drum diameter =20 ft
◼ Liquid volume=1563 ft3
◼ Liquid height=1563/312=5 ft
◼ Height of vapor space above liquid level
◼ = drum diameter=20 ft
◼ Drum ID 20 ft
◼ TTL 25 ft

◼ 5.5 Water Seal Drum
◼ 5.5.1 Procedure
◼ • It is a safeguard against pulling of vacuum and ingress of air.
Vacuum gets created due to draft effect when flow is either small or
stack is filled with low molecular weight or high temperature gases,
or condensation of residual gases in header when no flaring is taking
place.
◼ • Vacuum pull= Atm press(1-amb.temp(K)/gas temp.(K0))
◼ • Dip pipe length = Vacuum pull x 10 (m)
◼ (excluding seal)
◼ If dip pipe length is less than 3M,take 3M

◼
◼ • Dip pipe length(including seal) = Dip pipe length
(excluding seal) + seal height
◼ • Vessel free area for gas flow above liquid level
should be at least three times inlet pipe cross sectional area
to prevent surges of gas flow to flare .
◼ Consider vertical vessel with area pD2/4 and an inlet pipe
with area pd2/4,annular area is (p/4) (D2-d2),since ratio is
1:3,D2-d2=3d2 ,D=2d
◼ • Dwsd = dia of header [1+dip pipe length/seal height]0.5
◼ (excluding seal)

◼ If dia of W.S.D.<2 x header diameter
◼ then take 2x diameter of header
◼ • Height of side seal=(back pressure-1.033)x1.75 x 10 m
◼ • Disengagement space =0.5 x dia. of WSD (M)
◼ If disengagement space < 1M,take it 1M
◼ • Total height of WSD= dip pipe length +header dia +liquid seal
height from bottom + disengagement space
◼ Back pressure at WSD exit =1.3492
◼ Mean ambient temperature=20 0C or 293 0K
◼ Max gas temp.=143 0C or 416 0K
◼ Vacuum pull=1.033(1-293/416) =0.305 kg/cm2

◼ Dip pipe length=0.305 x 10=3.05 m
◼ Dip pipe including seal = 3.05 + 0.3 = 3.35 m
◼ Diameter calculation
◼ p/4(D2-d2) x 300/1000 =p/4(d2) x 0.305
◼ D=d(3.35/0.3)0.5 =15.376(11.166)0.5 =51.38 inch
◼ =1.305m
◼ Dia=1.305m
◼ Side seal height =(1.3492-1.033) x10 x 1.75 =5.53m
◼ Liquid seal height from bottom = 1m(assumed)
◼ Header dia=15.376 inch = 0.39m
◼ Disengagement space=0.5 D=0.652 m
◼ If disengagement space < 1m take 1m
◼ Total height = 1+3.35+0.39+1=5.44m

◼ 5.6 Stack Design
◼ 5.6.1 API-521 procedure
◼ • Based on flare load and related information, stack diameter with
0.4 mach is calculated( API-521 suggests 0.2 for normal & 0.5 for
maximum load).
◼ • Flame length may be calculated after estimating heat liberated
with fig 8 and 9 of API-521.
◼ • Calculation of flame distortion caused by wind velocity, by
considering ratio of wind velocity and flare tip velocity and fig 10 .
◼ Exposure times necessary to reach pain threshold are given in table-
7.Minimum distance from a flare to an object whose exposure to
thermal radiation must be limited is calculated by considering 1500
BTU/hr ft2 at a distance of 90m.Calculations need fig 10 and C-1(to
understand various dimensions) .

◼ •
◼ • In case of problem to provide safe distance, standard
personnel protective measures or unmanned equipment
may be considered.
◼ • Another rigorous calculations with Brzustowski,s and
Sommer,s approach are given in API.

◼ • Table-8 provides recommended design for radiation situations.
For 1500 BTU/hr.ft2 – heat intensity in areas where emergency
actions lasting several minutes may be required by personnel without
shielding but with appropriate clothing.
◼ 5.6.2.1 Basis –same as for header sizing
◼ Vsonic= 39.3(g k T)0.5 / mol wt where T- 0R
◼ Dstack= [max flow (ft3/s) x 4/(p x Vsonic x Mach no) ] 0.5
◼ k ratio of Cp and Cv
◼ mol wt- molecular weight
◼ max flow- Maximum load,kg/hr k = 1.08
◼ Gas temp.=143 0C or 749.4 0R
◼ Gas Density = 11.795 x 70.79 x 1.033/416 = 2.073 kg/m3
Volumetric flow rate=78311/(2.073 x 3600) = 10.49 m3/sec

◼ Sonic Velocity=39.3[32.2x1.08 x 749.4 /70.79]0.5
◼ =754.05ft/sec or 229.81 m/sec
◼
◼ Tip area at 0.4 mach= 10.49/(229.81x 0.4) = 0.114 m2
◼
◼ Diameter= (0.114x4/p ) =0.381m=15.01 inch
◼ Next dia available is 15.376 in ,To reduce pressure drop,next higher
17.376 inch finalized.
◼
◼ 5.6.2.2 Stack height Calculation
◼ Stack heat release hc = 50 x mol wt +100
◼ = 3639.5 BTU/scf

◼ Q= W(pounds/hr)x hc x 359/mol wt
◼ = 78311x2.2046 x3639.5 x 359/70.79
◼ =3.18789x 109 BTU/hr
◼ Refer fig 8 of API-521, flame length=200ft or 61m
Hydrocarbon exit velocity =
◼ 10.49 (m3/s) / [(p/4 ){17.376 x 0.0254}2] =68.5675 m/sec
◼ m2
◼ or 224.97 ft/s
◼ mwind/mHC =30/224.97 = 0.13335
◼ From API-521 fig 10, SDx/l=0.84
◼ S D y/l=0.38

◼ Dy/Dx=0.452
◼
◼ Dy/Dx = [5.02 x d/(mw/mhc)]{1/l-1/L}
◼
◼ 0.452=[5.02x 17.376 x 0.0254/0.1333]{1/l-1/61}
◼ (m)
◼ l=22.942 m,
◼ Dx=19.271 m
◼ Dy=8.718 m

◼ Height of stack for specified radiation level
◼ H= [e.Q/4.p.R]0.5-Dy
◼ Assume 3000 BTU/hr/ft 2 below eye of flame
◼ R=H+8.718(m)
◼ =(H+28.603) ft
◼ Radiation level= e Q/4p R2
◼ g=80%-relative humidity
◼ D=170 ft-distance from flame to illuminated area,ft
◼ F- Fraction of heat radiated
◼ Emissivity estimate
◼ e=e xF
◼ e =0.79 (100/g)1/16 (100/D)1/16
◼ e=0.79 (100/80)1/16 100/170)1/16 =0.774

◼ From table-4 API-521, F=0.299
◼ e=0.774 x 0.299 =0.231
◼ For flare stack calculations, consider e=0.1
◼ 3000=0.1 x 3.18789x109 /4.p .(H+28.603)2
◼ 91.957=H+28.603
◼ H=63.354 ft
Taking 1500 BTU/hr ft2 at 90m distance,
◼ d=90x 3.281=295.29
◼ H=[e Q/(4 p Rad. val ) –(d-Dx)2 ]0.5 - Dy
◼ = [{0.1 x 3.364x109 /(4.p.1500)}- (295.29-19.271x 3.281)2]0.5 –
28.603
◼ =(-36006)0.5 –28.603

◼ Negative sign with in square root indicates distance for
desired radiation level is less than specified distance.
◼
◼ Trying 1500 BTU/hr ft2 radiation level at 45 meter,
◼ ht stack= [e . Q /(4 . p . Rad val) -(d-DX)2]1/2 -DY
=[0.1 x 3.364x109 /(4.p.1500) –(147.645-63.228)2 ]1/2 -
28.603
◼
◼ =103.5-28.603=74.9 ft

◼ 5.7 Molecular Seal/ Fluidic Seal
◼ If air enters flare stack, it will mix with combustible gases resulting in
an explosive mixture Entry of air is prevented by using continuous
purge with oxygen free gas and use of gas seal. Seal gas , inert gas or
nitrogen are commonly used as purge gases.
◼ Various types of gas seal like fluidic or molecular seal are available.
Molecular seal is commonly used. In this seal, gas is forced to make
two 1800 bends as it flows through the seal( air thus encounters a trap
caused by difference in mol. wt. between air and purge gas. If purge
gas is lighter than air, trap is formed in the upper portion of outer
cylinder.

◼
◼ Ignition System
Commonly used Flame Front Generator system requires
compressed air (15 psig) , fuel gas (15 psig ),and electricity
(110V,220V or 440 V) in an ignition panel. Air and gas are
mixed and mixture is ignited to generate a spark that
travels through ignition piping (stainless steel) to the pilot.

◼ 5.8 Design criteria for utility
◼ • No of burners required-If dia < 24 inch; burners=2
◼ 24inch < dia <36 inch,burners=3
◼ dia>36 inch, burners=4
◼ • Fuel gas requirement (Nm3/hr) 4x Number of burners
◼ • Purge gas requirement(Nm3/hr) –based on 0.1 ft/sec
for molecular seal and 0.35 ft/sec for fluidic seal
◼ • Steam requirement, kg/hr
◼ = Hydrocarbon load x(0.68-10.8/ Mol Wt)

◼ 6.0 Pressure Safety Valve
◼ 6.1 Types of PSV
◼ Generally three types of safety valves are used:
◼ • Conventional: back pressure is 10% of set pressure
◼ • Balance bellow type: Back pressure varies from 40-
50% of set pressure
◼ • Pilot Operated valves: back pressure is 70% of set
pressure

◼ •
◼ • Relief valves are usually sizes for failure condition which results
in maximum load, however during other mild upsets, fraction of that
amount is discharged through the valve. The fluid volume under such
condition may be insufficient to sustain large load and valve
operation would be cyclic. This type of service results in chattering &
relief valve, s ability to reseat may be affected. When such variation
is frequent, use of
◼ multilple valves of smaller capacity with staggered
◼ setting. With this arrangement valve with lowest setting
◼ will be able to handle minor upsets.
◼

◼ 6.2 Relieving pressure for single/multiple valve:
◼ 6.2.1Determination of Relieving Pressure for a Single Valve
◼ • Valve set pressure less than MAWP
◼ Operating case Fire case
◼ Protected vessel, MAWP, psig 100 100
◼ Max accumulated pressure, psig 110 121
◼ Valve set pressure, psig 90 90
◼ Allowable overpressure, psi 20 31
◼ Relieving pressure, psia 124.7 135.7
◼

◼
◼ • Valve set pressure equal to MAWP
◼ Max accumulated pressure,psig 110 121
◼ Valve set pressure,psig 100 100

◼ 6.2.2 Determination of Relieving pressure for a Multi-valve
◼ Installation:
◼ First Valve First Valve
◼
◼ ( set pressure equal to MAWP)
◼ Protected vessel,MAWP, psig 100 100
◼

◼
◼ Additional valve
◼ (Set pressure equal to 105 percent of MAWP)
◼

◼ 6.2.3 Determination of Relieving pressure for a supplemental valve
◼ Supplemental valves are used only in addition to valves sized for
operating(non fire) contingencies.
◼
◼ Supplemental valve
◼ (Set pressure equal to 110 percent of MAWP)
◼ Fire case
◼ Protected vessel, MAWP, psig 100
◼ Max accumulated pressure, psig 121
◼ Valve set pressure, psig 110
◼ Allowable overpressure, psi 11
◼ Relieving pressure, psia 135.7

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