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PSV Calculation and Philosophy
1
Role of Pressure Safety Valve (PSV)
Role of Pressure Safety Valve (PSV)
PSVs are installed to make sure that
Accumulated Pressure ≤ Maximum Allowable Accumulated Pressure
as dictated by applicable code & standard
PressureVessels
(ASME SectVIII,API 520 & 521)
Unfired Boilers
(ASME Sect I)
Piping
(ASME B16.5 and 31.3)
2
Design Code & Standard (Pressure Vessel)
Design Code & Standard (Pressure Vessel)
- ASME Section VIII
- API 520 Sizing, Selection and Installation of Pressure-Relieving
Devices in Refineries
Devices in Refineries
Source : API 520
Set pressure = Pressure at which PSV is set to open
3
Design Code & Standard (Unfired Boiler)
Design Code & Standard (Unfired Boiler)
ASME Section I
106 106
106
103
106
106
Kerosene Pumparound
LP Steam Generator
LP Steam Generator
4
PROCEDURES FOR PSV CALCULATION
PROCEDURES FOR PSV CALCULATION
LOCATE PSV and SPECIFY
RELIEF PRESSURE
RELIEF PRESSURE
DEVELOP SCENARIOS
(WHAT CAN GO WRONG?)
CALCULATE PSV SIZE
Required Information
CHOOSE WORST CASE
DESIGN OF RELIEF SYSTEM
(Flare Header, etc)
5
PSV SCENARIOS
(Refer API 521)
(Refer API 521)
FOCUS ON COMMON CASES :
- Closed Outlets onVessels
- External Fire
- Failure of Automatic Controls
- Hydraulic Expansion
Hydraulic Expansion
- Heat ExchangerTube Rupture
- Total Power Failure
P i l P F il
- Partial Power Failure
- CoolingWater Failure
- Reflux Loss
- Failure of Air-Cooled Heat X
DOUBLE JEOPARDY NOT
CONSIDERED
6
(Simultaneous occurrence of two or
more unrelated causes of
overpressure) Source : API 521
Closed outlets on vessels
Closed outlets on vessels
Cause
Outlet valve is blocked while there is
i i l f hi h
continuous inlet from high pressure
source
Effects
Outlet valve closed
Effects
Pressure built-up in vessel
Calculation
Can PSV be opened in Closed Outlet Case? R li f
Pressure source (pump, compressor,
high pressure header)
No
Calculation
Can PSV be opened in Closed Outlet Case?
(Is maximum inlet pressure > PSV set pressure?)
For pump :
Is maximum pump shut off pressure ≥ PSV set
Relief case not
considered
Relief rate =
Yes
No
Is maximum pump shut-off pressure ≥ PSV set
pressure?
Relief rate
maximum inlet flow
7
External Fire (1/4)
External Fire (1/4)
Cause
External pool fire caused by accumulated hydrocarbon on the
ground or other surfaces
Effects
Effects
- Vaporization of liquid inside the vessel,
leading to pressure building up within the vessel
g p g p
Calculation
Refer next slides
Refer next slides
8
External Fire - Liq. Vessel (2/4)
External Fire Liq. Vessel (2/4)
Relief rate (W) = Heat absorbed by liquid from external fire (Q)
Latent Heat ofVaporization of liquid (λ)
Case 1 :
If adequate drainage necessary to control the spread of major
ill f t th d t t l f d i
spills from one area to another and to control surface drainage
and refinery waste water.
Q = 43,200 x F x A0.82
C 2
7 6 m Case 2 :
If adequate drainage and firefighting equipment do not exist.
Q = 70,900 x F x A0.82
7.6 m
Q = Heat absorbed by liquid from external fire(W)
F = Environment Factor
A = Wetted Surface Area (m2)
9
External Fire - Liq. Vessel (3/4)
External Fire Liq. Vessel (3/4)
Wetted Surface Area (A)
Source : API 521
10
External Fire
Liq. Vessel (4/4)
Environment Factor (F)
Liq. Vessel (4/4)
Latent Heat ofVaporization of liquid (λ)
for multi-component mixture
5 wt% flashed
λ = Dew PointVapor Enthalpy
Relief Pressure
Bubble point T
– Bubble Point Liquid Enthalpy
For column, use composition of
For column, use composition of
1. Second tray from top (or reflux
composition if unavailable)
2. Bottoms
Ch th t i l PSV i
Source : API 521
Choose one that require larger PSV size
11
Failure of Automatic Control (1/3)
Failure of Automatic Control (1/3)
Cause
- Failure of a single automatic control valve Consider this control valve fail
- Control valves are assumed to fail to non-favorable
position (not necessarily to their specified fail
position).
Consider this control valve fail
in full-open although it is
specified as “fail-close”
Effects
- Control valve fail open : maximum fluid flow
through valve
N2
Header
Flare
FC FO
SPLIT-RANGE
g
- Control valve fail close : no fluid flow
- Effect of control valve fail open or close to be
considered on case-by-case basis
FC FO
PIC
Calculation
Calculation of maximum fluid flow in control valve
fail open case, refer next slide
p ,
12
Failure of Automatic Control (2/3)
Failure of Automatic Control (2/3)
CALCULATION OF MAX. FLOW THROUGH CONTROL VALVES
1. Find Valve CV value (from manufacturer).
2. If by-pass valve is installed, consider possibility that by-pass
valve may be partially open Add 50% margin to CV value in
valve may be partially open. Add 50% margin to CV value in
1.
3. For Calculate maximum flow through control valve (refer
calculation sheet)
4. Find relief rate (to consider on case-by-case basis)
13
Failure of Automatic Control (1/3)
Failure of Automatic Control (1/3)
EXAMPLE :
1 FEED SURGE DRUM
N2
Header
Flare
PV01 PV02
SPLIT-RANGE
1. FEED SURGE DRUM
Relief rate = maximum flow through PV01 –
flow through PV02
Header PV01 PV02
PIC
flow through PV02
2. LPGVAPORIZER
Relief rate =
LPG Generated by max. steam flow
– normal LPG outlet flow
14
Hydraulic expansion (1/2)
Hydraulic expansion (1/2)
Cause
Liquid is blocked in and later heated up (by hot fluid steam
Liquid is blocked-in and later heated up (by hot fluid, steam
tracing / jacket or by solar radiation).
Effects
Liquid expands upon heating, leading to pressure build-up in
vessel or blocked in section of piping/pipeline.
15
Hydraulic expansion (1/2)
Hydraulic expansion (1/2)
Calculation
Refer calculation sheet for relief rate calculation
If applicable (e.g. in
cooling circuit) consider
cooling circuit), consider
administrative control in
place of relief valve.
16
Heat Exchanger Tube Rupture (1/3)
Heat Exchanger Tube Rupture (1/3)
Cause
Tube rupture in shell & tube heat exchanger exposing lower
Tube rupture in shell & tube heat exchanger, exposing lower
pressure side to high pressure fluid.
Effects
Effects
Lower pressure side is exposed to high pressure fluid
Note : No need to consider if design pressure of lower pressure
g p p
side is 10/13 or more of design pressure of high pressure side.
Calculation
Use orifice equation with
double cross-sectional area.
17
Heat Exchanger Tube Rupture (2/3)
Heat Exchanger Tube Rupture (2/3)
ρ
18
Heat Exchanger Tube Rupture (3/3)
Heat Exchanger Tube Rupture (3/3)
ρ
19
Total Power Failure (1/5)
Total Power Failure (1/5)
Cause
Di i i l l di l i l f il f h
Disruption in power supply, leading to electrical power failure of the
whole site.
Effects
- Loss of operation for pumps, air-cooled heat exchangers, all electrically-
driven equipments
- For Fractionating Column worst case design, assume steam system
continues to operate
Calculation
For Fractionation Column : Enthalpy Balance Method
Note
Usually controlling case for flare capacity
20
Total Power Failure (2/5)
Total Power Failure (2/5)
Enthalpy balance around Fractionator Column to find
excess heat (Q), which would cause vapor generation.
QC
FEED DISTILLATE
HF HD
F D
QR
BOTTOMS
HB
B
Excess Heat (Q) = HFF – HDD – HBB – QC + QR
Note : All values are taken from relieving condition
21
Total Power Failure (3/5)
Total Power Failure (3/5)
Excess Heat (Q) = HFF – HDD – HBB – QC + QR
All t  l f f d di till t d b tt
All pumps stop  loss of feed, distillate and bottoms
Condenser Duty (QC)
1. Water-cooled (QC = 0)
2 Air-cooled
2. Air-cooled
May consider credit
for natural draft effects
(20 30% of normal duty)
(20-30% of normal duty)
22
Total Power Failure (4/5)
Total Power Failure (4/5)
Reboiler Duty (QR)
1. Thermosyphon using steam
(Q = Normal Duty)
2. Fired Heater
No flow to fired heater, but consider the
possibility that remaining fluid inside tube is
(QR = Normal Duty) p y g
heated up by heat from refractory surfaces
(QR = 30% of normal duty)
Steam
High Integrity Pressure Protection System (HIPPS)
2. Fired Heater
Heat from refractory surfaces
1. Thermosyphon using steam
(QR = 0)
(QR = 30% of normal duty)
( R )
FUEL
23
Total Power Failure (5/5)
Total Power Failure (5/5)
Relief load =V
Vapor cannot be condensed
(loss of condenser duty)
Relief Load = Vapor generated by excess heat
= Excess Heat (Q)
Latent Heat ofVaporization of 2nd tray liquid
Vapor generated (V)
Latent Heat ofVaporization of 2 tray liquid
Excess Heat (Q)
24
Partial Power Failure (1/3)
Partial Power Failure (1/3)
Cause
Disruption in a single feeder, bus, circuit or line, leading to partial power
failure
Effects
- Varies, pending on power distribution system
- For Fractionating Column, worst case considered for Partial Power
Failure is simultaneous loss of reflux pump and air-cooled condenser,
while there is continuous heat input into column
while there is continuous heat input into column.
Calculation
For Fractionating Column : Enthalpy Balance Method
g py
Internal Reflux Method (alternative)
Note
Usually controlling case for column PSV sizing
Usually controlling case for column PSV sizing
25
Partial Power Failure (2/3)
Partial Power Failure (2/3)
Worst case : simultaneous loss of reflux pump and air-
cooled condenser
QC
FEED DISTILLATE
HF HD
F D
QR
BOTTOMS
HB
B
Excess Heat (Q) = HFF – HDD – HBB – QC + QR
26
Partial Power Failure (3/3)
Partial Power Failure (3/3)
Temp (TH)
Latent Heat of Vaporization (λH)
Mass Flow (mH)
( H)
Reflux
Specific heat (Cp,R)
Mass flow (mR) , Temp (TR)
Internal Reflux
Mass flow (mIR) Alternative : Internal reflux method
Relief load = mH + mIR
mRCp,R(TR-TH) + mIRλH = 0
m m C (T -T )
Relief load mH mIR
mIR = mRCp,R(TH-TR)
λH
27
Cooling Water Failure (1/3)
Cooling Water Failure (1/3)
Cause
Cooling Water Pump failure loss of make up water etc
Cooling Water Pump failure, loss of make-up water, etc.
Effects
Loss of duty for water cooled heat exchangers
- Loss of duty for water-cooled heat exchangers
- Operation of pumps that require cooling water for lube oil cooling may
also be effected
Calculation
Calculation
For Fractionating Column : Enthalpy Balance Method
Internal Reflux Method (Alternative)
28
Cooling Water Failure (2/3)
Cooling Water Failure (2/3)
QC
HF
HD
FEED DISTILLATE
F
D
Q
BOTTOMS
HB
B
QR
Excess Heat (Q) = HFF – HDD – HBB – QC + QR
Note : Need to recalculate D and HD
Alternative : Internal Reflux Method (refer Partial Power Failure case
(
with re-calculated reflux temp, flowrate and specific heat)
29
Cooling Water Failure (3/3)
Cooling Water Failure (3/3)
Temp (TH)
Latent Heat of Vaporization (λH)
Mass Flow (mH)
( H)
mRCp,R(TR-TH) + mIRλH = 0
mIR = mRCp R(TH-TR)
Alternative : Internal reflux method
Reflux
Specific heat (Cp,R)
Mass flow (mR) , Temp (TR)
IR R p,R( H R)
λH
Internal Reflux
Mass flow (mIR)
1. Find internal reflux without considering cooling
water failure (mIR,normal)
2. Recalculate reflux flowrate, temperature and
specific heat for cooling water failure case
3. Find internal reflux considering cooling water
failure (mIR,CWFail)
4 Relief load = overhead vapor normal +
30
4. Relief load overhead vapor_normal +
(mIR,normal - mIR,CWFail )
Reflux Loss(1/2)
Reflux Loss(1/2)
Cause
Failure of reflux pumps
Failure of reflux pumps
Effects
Loss of reflux to column
- Loss of reflux to column
- Liquid level in overhead receiver rises, ultimately flooding the condenser,
causing loss of condensing duty
Calculation
Calculation
For Fractionating Column : Enthalpy Balance Method
Alternative : Internal Reflux Method
31
Reflux Loss (2/2)
Reflux Loss (2/2)
Q
H
QC
FEED
HF
F
HD
D
DISTILLATE
HB
B
BOTTOMS
QR
Excess Heat (Q) = HFF – HDD – HBB – QC + QR
B
BOTTOMS
Alternative : Internal Reflux Method (refer Partial Power Failure case)
32
Failure of air-cooled heat exchanger (1/2)
Failure of air cooled heat exchanger (1/2)
Cause
l f d d l l d h h
Failure of individual air-cooled heat exchanger
Effects
- Loss of condensing duty in fractionating column
Calculation
For Fractionating Column : Enthalpy Balance Method
Internal Reflux Method (Alternative)
33
Failure of air-cooled heat exchanger (2/2)
Failure of air cooled heat exchanger (2/2)
QC
H H
QC
FEED DISTILLATE
HF
F
HD
D
BOTTOMS
HB
B
QR
Excess Heat (Q) = HFF – HDD – HBB – QC + QR
B
Note : Need to recalculate D and HD
Alternative : Internal Reflux Method (refer cooling water failure case)
34
THANK YOU
February 3, 2014
35

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Psv scenario-and-calculation

  • 1. PSV Calculation and Philosophy 1
  • 2. Role of Pressure Safety Valve (PSV) Role of Pressure Safety Valve (PSV) PSVs are installed to make sure that Accumulated Pressure ≤ Maximum Allowable Accumulated Pressure as dictated by applicable code & standard PressureVessels (ASME SectVIII,API 520 & 521) Unfired Boilers (ASME Sect I) Piping (ASME B16.5 and 31.3) 2
  • 3. Design Code & Standard (Pressure Vessel) Design Code & Standard (Pressure Vessel) - ASME Section VIII - API 520 Sizing, Selection and Installation of Pressure-Relieving Devices in Refineries Devices in Refineries Source : API 520 Set pressure = Pressure at which PSV is set to open 3
  • 4. Design Code & Standard (Unfired Boiler) Design Code & Standard (Unfired Boiler) ASME Section I 106 106 106 103 106 106 Kerosene Pumparound LP Steam Generator LP Steam Generator 4
  • 5. PROCEDURES FOR PSV CALCULATION PROCEDURES FOR PSV CALCULATION LOCATE PSV and SPECIFY RELIEF PRESSURE RELIEF PRESSURE DEVELOP SCENARIOS (WHAT CAN GO WRONG?) CALCULATE PSV SIZE Required Information CHOOSE WORST CASE DESIGN OF RELIEF SYSTEM (Flare Header, etc) 5
  • 6. PSV SCENARIOS (Refer API 521) (Refer API 521) FOCUS ON COMMON CASES : - Closed Outlets onVessels - External Fire - Failure of Automatic Controls - Hydraulic Expansion Hydraulic Expansion - Heat ExchangerTube Rupture - Total Power Failure P i l P F il - Partial Power Failure - CoolingWater Failure - Reflux Loss - Failure of Air-Cooled Heat X DOUBLE JEOPARDY NOT CONSIDERED 6 (Simultaneous occurrence of two or more unrelated causes of overpressure) Source : API 521
  • 7. Closed outlets on vessels Closed outlets on vessels Cause Outlet valve is blocked while there is i i l f hi h continuous inlet from high pressure source Effects Outlet valve closed Effects Pressure built-up in vessel Calculation Can PSV be opened in Closed Outlet Case? R li f Pressure source (pump, compressor, high pressure header) No Calculation Can PSV be opened in Closed Outlet Case? (Is maximum inlet pressure > PSV set pressure?) For pump : Is maximum pump shut off pressure ≥ PSV set Relief case not considered Relief rate = Yes No Is maximum pump shut-off pressure ≥ PSV set pressure? Relief rate maximum inlet flow 7
  • 8. External Fire (1/4) External Fire (1/4) Cause External pool fire caused by accumulated hydrocarbon on the ground or other surfaces Effects Effects - Vaporization of liquid inside the vessel, leading to pressure building up within the vessel g p g p Calculation Refer next slides Refer next slides 8
  • 9. External Fire - Liq. Vessel (2/4) External Fire Liq. Vessel (2/4) Relief rate (W) = Heat absorbed by liquid from external fire (Q) Latent Heat ofVaporization of liquid (λ) Case 1 : If adequate drainage necessary to control the spread of major ill f t th d t t l f d i spills from one area to another and to control surface drainage and refinery waste water. Q = 43,200 x F x A0.82 C 2 7 6 m Case 2 : If adequate drainage and firefighting equipment do not exist. Q = 70,900 x F x A0.82 7.6 m Q = Heat absorbed by liquid from external fire(W) F = Environment Factor A = Wetted Surface Area (m2) 9
  • 10. External Fire - Liq. Vessel (3/4) External Fire Liq. Vessel (3/4) Wetted Surface Area (A) Source : API 521 10
  • 11. External Fire Liq. Vessel (4/4) Environment Factor (F) Liq. Vessel (4/4) Latent Heat ofVaporization of liquid (λ) for multi-component mixture 5 wt% flashed λ = Dew PointVapor Enthalpy Relief Pressure Bubble point T – Bubble Point Liquid Enthalpy For column, use composition of For column, use composition of 1. Second tray from top (or reflux composition if unavailable) 2. Bottoms Ch th t i l PSV i Source : API 521 Choose one that require larger PSV size 11
  • 12. Failure of Automatic Control (1/3) Failure of Automatic Control (1/3) Cause - Failure of a single automatic control valve Consider this control valve fail - Control valves are assumed to fail to non-favorable position (not necessarily to their specified fail position). Consider this control valve fail in full-open although it is specified as “fail-close” Effects - Control valve fail open : maximum fluid flow through valve N2 Header Flare FC FO SPLIT-RANGE g - Control valve fail close : no fluid flow - Effect of control valve fail open or close to be considered on case-by-case basis FC FO PIC Calculation Calculation of maximum fluid flow in control valve fail open case, refer next slide p , 12
  • 13. Failure of Automatic Control (2/3) Failure of Automatic Control (2/3) CALCULATION OF MAX. FLOW THROUGH CONTROL VALVES 1. Find Valve CV value (from manufacturer). 2. If by-pass valve is installed, consider possibility that by-pass valve may be partially open Add 50% margin to CV value in valve may be partially open. Add 50% margin to CV value in 1. 3. For Calculate maximum flow through control valve (refer calculation sheet) 4. Find relief rate (to consider on case-by-case basis) 13
  • 14. Failure of Automatic Control (1/3) Failure of Automatic Control (1/3) EXAMPLE : 1 FEED SURGE DRUM N2 Header Flare PV01 PV02 SPLIT-RANGE 1. FEED SURGE DRUM Relief rate = maximum flow through PV01 – flow through PV02 Header PV01 PV02 PIC flow through PV02 2. LPGVAPORIZER Relief rate = LPG Generated by max. steam flow – normal LPG outlet flow 14
  • 15. Hydraulic expansion (1/2) Hydraulic expansion (1/2) Cause Liquid is blocked in and later heated up (by hot fluid steam Liquid is blocked-in and later heated up (by hot fluid, steam tracing / jacket or by solar radiation). Effects Liquid expands upon heating, leading to pressure build-up in vessel or blocked in section of piping/pipeline. 15
  • 16. Hydraulic expansion (1/2) Hydraulic expansion (1/2) Calculation Refer calculation sheet for relief rate calculation If applicable (e.g. in cooling circuit) consider cooling circuit), consider administrative control in place of relief valve. 16
  • 17. Heat Exchanger Tube Rupture (1/3) Heat Exchanger Tube Rupture (1/3) Cause Tube rupture in shell & tube heat exchanger exposing lower Tube rupture in shell & tube heat exchanger, exposing lower pressure side to high pressure fluid. Effects Effects Lower pressure side is exposed to high pressure fluid Note : No need to consider if design pressure of lower pressure g p p side is 10/13 or more of design pressure of high pressure side. Calculation Use orifice equation with double cross-sectional area. 17
  • 18. Heat Exchanger Tube Rupture (2/3) Heat Exchanger Tube Rupture (2/3) ρ 18
  • 19. Heat Exchanger Tube Rupture (3/3) Heat Exchanger Tube Rupture (3/3) ρ 19
  • 20. Total Power Failure (1/5) Total Power Failure (1/5) Cause Di i i l l di l i l f il f h Disruption in power supply, leading to electrical power failure of the whole site. Effects - Loss of operation for pumps, air-cooled heat exchangers, all electrically- driven equipments - For Fractionating Column worst case design, assume steam system continues to operate Calculation For Fractionation Column : Enthalpy Balance Method Note Usually controlling case for flare capacity 20
  • 21. Total Power Failure (2/5) Total Power Failure (2/5) Enthalpy balance around Fractionator Column to find excess heat (Q), which would cause vapor generation. QC FEED DISTILLATE HF HD F D QR BOTTOMS HB B Excess Heat (Q) = HFF – HDD – HBB – QC + QR Note : All values are taken from relieving condition 21
  • 22. Total Power Failure (3/5) Total Power Failure (3/5) Excess Heat (Q) = HFF – HDD – HBB – QC + QR All t  l f f d di till t d b tt All pumps stop  loss of feed, distillate and bottoms Condenser Duty (QC) 1. Water-cooled (QC = 0) 2 Air-cooled 2. Air-cooled May consider credit for natural draft effects (20 30% of normal duty) (20-30% of normal duty) 22
  • 23. Total Power Failure (4/5) Total Power Failure (4/5) Reboiler Duty (QR) 1. Thermosyphon using steam (Q = Normal Duty) 2. Fired Heater No flow to fired heater, but consider the possibility that remaining fluid inside tube is (QR = Normal Duty) p y g heated up by heat from refractory surfaces (QR = 30% of normal duty) Steam High Integrity Pressure Protection System (HIPPS) 2. Fired Heater Heat from refractory surfaces 1. Thermosyphon using steam (QR = 0) (QR = 30% of normal duty) ( R ) FUEL 23
  • 24. Total Power Failure (5/5) Total Power Failure (5/5) Relief load =V Vapor cannot be condensed (loss of condenser duty) Relief Load = Vapor generated by excess heat = Excess Heat (Q) Latent Heat ofVaporization of 2nd tray liquid Vapor generated (V) Latent Heat ofVaporization of 2 tray liquid Excess Heat (Q) 24
  • 25. Partial Power Failure (1/3) Partial Power Failure (1/3) Cause Disruption in a single feeder, bus, circuit or line, leading to partial power failure Effects - Varies, pending on power distribution system - For Fractionating Column, worst case considered for Partial Power Failure is simultaneous loss of reflux pump and air-cooled condenser, while there is continuous heat input into column while there is continuous heat input into column. Calculation For Fractionating Column : Enthalpy Balance Method g py Internal Reflux Method (alternative) Note Usually controlling case for column PSV sizing Usually controlling case for column PSV sizing 25
  • 26. Partial Power Failure (2/3) Partial Power Failure (2/3) Worst case : simultaneous loss of reflux pump and air- cooled condenser QC FEED DISTILLATE HF HD F D QR BOTTOMS HB B Excess Heat (Q) = HFF – HDD – HBB – QC + QR 26
  • 27. Partial Power Failure (3/3) Partial Power Failure (3/3) Temp (TH) Latent Heat of Vaporization (λH) Mass Flow (mH) ( H) Reflux Specific heat (Cp,R) Mass flow (mR) , Temp (TR) Internal Reflux Mass flow (mIR) Alternative : Internal reflux method Relief load = mH + mIR mRCp,R(TR-TH) + mIRλH = 0 m m C (T -T ) Relief load mH mIR mIR = mRCp,R(TH-TR) λH 27
  • 28. Cooling Water Failure (1/3) Cooling Water Failure (1/3) Cause Cooling Water Pump failure loss of make up water etc Cooling Water Pump failure, loss of make-up water, etc. Effects Loss of duty for water cooled heat exchangers - Loss of duty for water-cooled heat exchangers - Operation of pumps that require cooling water for lube oil cooling may also be effected Calculation Calculation For Fractionating Column : Enthalpy Balance Method Internal Reflux Method (Alternative) 28
  • 29. Cooling Water Failure (2/3) Cooling Water Failure (2/3) QC HF HD FEED DISTILLATE F D Q BOTTOMS HB B QR Excess Heat (Q) = HFF – HDD – HBB – QC + QR Note : Need to recalculate D and HD Alternative : Internal Reflux Method (refer Partial Power Failure case ( with re-calculated reflux temp, flowrate and specific heat) 29
  • 30. Cooling Water Failure (3/3) Cooling Water Failure (3/3) Temp (TH) Latent Heat of Vaporization (λH) Mass Flow (mH) ( H) mRCp,R(TR-TH) + mIRλH = 0 mIR = mRCp R(TH-TR) Alternative : Internal reflux method Reflux Specific heat (Cp,R) Mass flow (mR) , Temp (TR) IR R p,R( H R) λH Internal Reflux Mass flow (mIR) 1. Find internal reflux without considering cooling water failure (mIR,normal) 2. Recalculate reflux flowrate, temperature and specific heat for cooling water failure case 3. Find internal reflux considering cooling water failure (mIR,CWFail) 4 Relief load = overhead vapor normal + 30 4. Relief load overhead vapor_normal + (mIR,normal - mIR,CWFail )
  • 31. Reflux Loss(1/2) Reflux Loss(1/2) Cause Failure of reflux pumps Failure of reflux pumps Effects Loss of reflux to column - Loss of reflux to column - Liquid level in overhead receiver rises, ultimately flooding the condenser, causing loss of condensing duty Calculation Calculation For Fractionating Column : Enthalpy Balance Method Alternative : Internal Reflux Method 31
  • 32. Reflux Loss (2/2) Reflux Loss (2/2) Q H QC FEED HF F HD D DISTILLATE HB B BOTTOMS QR Excess Heat (Q) = HFF – HDD – HBB – QC + QR B BOTTOMS Alternative : Internal Reflux Method (refer Partial Power Failure case) 32
  • 33. Failure of air-cooled heat exchanger (1/2) Failure of air cooled heat exchanger (1/2) Cause l f d d l l d h h Failure of individual air-cooled heat exchanger Effects - Loss of condensing duty in fractionating column Calculation For Fractionating Column : Enthalpy Balance Method Internal Reflux Method (Alternative) 33
  • 34. Failure of air-cooled heat exchanger (2/2) Failure of air cooled heat exchanger (2/2) QC H H QC FEED DISTILLATE HF F HD D BOTTOMS HB B QR Excess Heat (Q) = HFF – HDD – HBB – QC + QR B Note : Need to recalculate D and HD Alternative : Internal Reflux Method (refer cooling water failure case) 34