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How To

How To Size A Gas Control Valve

Sizing a control valve means selecting a valve with the correct size orifice to allow good control of flow

rate within a required range.

There are other important factors to consider when selecting a control valve, such as valve type and

valve characteristic but this article will concentrate on valve sizing.

The procedure explained in this article applies to gas and vapour control valves including steam control

valves. The method for sizing gas control valves is based on the method for sizing liquid control valves.

More information on sizing a liquid control valve can be found here: “How To Size A Liquid Control Valve”.

Sizing a control valve for a particular duty is governed by the required flow rate the valve must pass and

the pressure drop that can be allowed across the valve.

Steps To Accurately Size A Gas Control Valve

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Steps To Accurately Size A Gas Control Valve

Specify the required design flow rate1.

Specify the allowable pressure drop across the valve2.

Choose a valve type and body size from the manufacturers’ tables3.

Calculate the first estimate of the piping geometry factor and pressure drop ratio factor4.

Determine if the flow through the valve will be sub-critical or critical5.

Calculate the effective pressure drop ratio across the valve6.

Calculate the expansion factor7.

Calculate the first estimate of the required valve Cv8.

Check that the calculated Cv is less than the actual Cv of the selected valve (re-select suitable

valve from manufacturers’ tables if required)

9.

Check that valve control range is OK10.

If the Cv and control range are suitable the valve is correctly sized. If not re-select another

valve and repeat the sizing procedure from Step 3

11.

Sizing a control valve accurately is an iterative process requiring manufacturer’s information and

knowledge of the piping system in which the valve is to be installed.

The procedure is a little bit more complicated than sizing a liquid control valve but is certainly not

difficult. For preliminary estimates of control valve size it is usually OK to assume that the piping

geometry factor is 1.

Calculate Control Valve Cv

Gas control valves are sized using a modified version of the liquid control valve equation. As for liquid

control valves, the valve orifice size as given as a “valve flow coefficient” or Cv.

The Cv is defined as the flow rate of water in US gallons per minute that can pass through a valve with a

pressure drop of 1 psi at a temperature of 60F.

The equation for calculating the Cv for a gas control valve using metric units is:

Effective Pressure Drop


The effective pressure drop across a gas control valve depends on the properties of the gas flowing

through the valve and the valve design.

If the pressure downstream of the valve is lower than a critical value, the flow through the valve will be

choked. Choked flow is also known as critical flow.

The flow is sub-critical if:

For sub-critical flow:

For critical flow:

Where:

Pressure Drop Ratio Factor, XT

The pressure drop ratio factor is the pressure drop ratio required to produce critical flow through the

valve when Fk is equal to 1.

The valve pressure drop ratio is measured experimentally and is tabulated in valve manufacturers

catalogues.

If the valve has fittings connected directly upstream and/or downstream, the pressure drop ratio factor

must be modified to account for expansion and contraction of the fluid through the fittings.

The modified pressure drop ratio factor, X is calculated using:

If the valve has fittings connected directly upstream and/or downstream, the pressure drop ratio factor

must be modified to account for expansion and contraction of the fluid through the fittings.

The modified pressure drop ratio factor, XTP is calculated using:

Where:

For valves installed with a reducer installed upstream, the inlet fittings head loss coefficient becomes:

Expansion Factor, Y

The expansion factor accounts for the expansion of gas flowing through the valve as the pressure

reduces from inlet to outlet. The expansion factor is the ratio of flow coefficients for a gas to that for a

liquid at the same Reynolds number.

The expansion factor must be less than or equal to a value of 0.667. The following equation defines the

expansion factor:

Piping Geometry Factor, Fp

The piping geometry factor is an allowance for the pressure drop associated with fittings that may be

connected directly upstream and/or downstream of the valve.

If no fittings are connected to the valve, the piping geometry factor is 1.



The piping geometry factor is often listed in valve manufacturers catalogues. Alternatively, it can be

calculated using:

Most commonly, the fittings connected to a control valve are upstream and downstream reducers. In

this case the sum of the fittings factors for the reducers is:

Note:

Determining the control valve Cv becomes an iterative process when the piping geometry factor doesn’t

equal 1.

Estimate the required Cv1.

Select an appropriate valve from the manufacturers’ tables2.

Calculate Fp and XTP using the actual Cv of the selected valve3.

Re-calculate the required Cv using the values of Fp and XTP determined in Step 34.

Check that the re-calculated Cv is less than the actual Cv of the selected valve5.

If the re-calculated Cv is less than the actual Cv, the selected valve is adequately sized6.

If the re-calculated Cv is greater than the actual Cv of the selected valve, select another valve

with a larger Cv and return to Step 3

7.

Control Valve Sizing Rules Of Thumb

There are many rules of thumb designed to help with control valve sizing. The following guidance is taken

from “Rules of Thumb For Chemical Engineers” by Carl Brannan and the author’s personal notes.

Set the design flow as the greater of:1.

1.3 x normal flow rate

1.1 x maximum required flow rate

Set the control pressure drop to equal 50% – 60% of the frictional pressure loss of the piping

system

2.



system

2.

Limit the maximum flow rate : minimum flow rate turndown to 5:1 for linear trim valves and 10:1

for equal percentage trim valves

3.

The valve should be able to control the required range of flow rates between 10% and 80% of valve

opening

4.

Ideally select a valve that has a body size 1 pipe size smaller than the pipe in which it is to be

installed (e.g. select a 3” valve for a 4” pipe)

5.

Never select a valve larger than the pipe in which it is to be installed6.

How To Size A Gas Control Valve Example

The following example has been adapted from the Emerson Control Valve Handbook.

We need to size a valve in steam duty. The required information is given below:

Design flow rate = 125000 lb/hr = 56689 kg/hr

Upstream pressure = 500 psig = 35.50 bara

Downstream pressure = 250 psig = 18.26 bara

Pressure drop across valve = 250 psi = 17.24 bar

Upstream steam temperature = 500F

Density of steam at upstream conditions = 1.0434 lb/ft3

= 16.71 kg/m3

Steam ratio of specific heat capacities = 1.28

Pipe size = 6 inch

Calculation

Following the steps given at the start of this article:

Design flow rate = 56689 kg/hr1.

Allowable pressure drop across the valve = 17.24 bar2.

We will choose an Emerson 4” ED globe valve with linear cage as the preliminary selection. From

the valve table we can see that the actual Cv is 236 and XTP is 0.688

3.

We will assume that 6” x 4” reducers will be used to install the selected 4” valve in the 6” pipe.

In this case the piping geometry factor is:

ΣK = 1.5 (1 – (42

/ 62))

2= 0.463FP = [1 + (0.463 / 890)(236 / 4

2)2]-0.5

= 0.95

The pressure drop ratio factor, XT = 0.688

The inlet fittings head loss coefficient is:

Ki = 0.5 (1 – (42

/ 62))

2+ 1 – (4 / 6)

4= 0.957

So the modified pressure drop ratio factor is:

XTP = 0.688 / 0.952[1 + 0.688 x 0.957 / 1000 (236 / 4

2)2]-1

= 0.667

4.

Check if flow through the valve is sub-critical or critical:

Fk = 1.28 / 1.4 = 0.91Fk XTP = 0.91 x 0.667 = 0.607(P1 – P2 ) / P1 = (35.50 – 18.26) / 35.50 = 0.486

Therefore: (P1 – P2 ) / P1 < Fk XTP so the flow is sub-critical

5.

The effective pressure drop ratio across the valve is (P1 – P2 ) / P1 because the flow is sub-

critical:

Xeff = 0.486

6.

Calculate the expansion factor:

Y = 1 – 0.486 / (3 x 0.91 x 0.667) = 0.733

7.

Calculate the first estimate of Cv:

Cv = 56689 / (27.3 x 0.95 x 0.733 (0.486 x 35.50 x 16.71)0.5

) = 175.6

8.

The calculated required Cv of 175.6 is less than the actual Cv of the selected valve of 236 so the

valve is large enough

9.

Check if the control valve range is OK:

From the valve table, the selected valve will be a just less 70% open to give the required Cv of

175.6. This is within the acceptable control range of 10% to 80% of valve opening.

10.

The selected 4” linear cage valve is correctly sized for the specified duty11.

Result

The selected valve is an Emerson 4” ED globe valve with linear trim and a maximum Cv of 236.

Blackmonk Engineering Calculator Result

The selected valve is an Emerson 4” ED globe valve with linear trim and a maximum Cv of 236.


The output from the Blackmonk Engineering Liquid Control Valve Calculator is attached below for

comparison. As you can see the calculated Cv is virtually identical to the hand calculation (175.4

compared to 175.6).

How To

How To Size A Liquid Control Valve

Sizing a control valve means selecting a valve with the correct size orifice to allow good control of flow

rate within a required range.






Sizing a control valve for a particular duty is governed by the required flow rate the valve must pass and

the pressure drop that can be allowed across the valve.

Steps To Accurately Size A Liquid Control Valve

Specify the required design flow rate1.

Specify the allowable pressure drop across the valve2.

Choose a valve type and body size from the manufacturers’ tables3.

Calculate the first estimate of the piping geometry factor4.

Determine if the flow through the valve will be sub-critical or critical. That is, will some of the

liquid vaporise causing flashing or cavitation?

5.

Calculate the effective pressure drop across the valve6.

Calculate the first estimate of the required valve Cv7.

Check that the calculated Cv is less than the actual Cv of the selected valve (re-select suitable

valve from manufacturers’ tables if required)

8.

Check that valve control range is OK9.

If the Cv and control range are suitable the valve is correctly sized. If not re-select another

valve and repeat the sizing procedure from Step 3

10.

Sizing a control valve accurately is an iterative process requiring manufacturer’s information and

knowledge of the piping system in which the valve is to be installed.

However, the procedure is fairly simple and straightforward. It becomes even easier if it is known that

the liquid will not flash or cavitate as it flows through the valve. For preliminary estimates of control

valve size it is usually OK to assume that the piping geometry factor is 1.

However, the procedure is fairly simple and straightforward. It becomes even easier if it is known that

the liquid will not flash or cavitate as it flows through the valve. For preliminary estimates of control

valve size it is usually OK to assume that the piping geometry factor is 1.

Calculate Control Valve Cv

Traditionally, control valves are sized using a special form of the orifice equation which gives the valve

orifice size as a “valve flow coefficient” or Cv.

The Cv is defined as the flow rate of water in US gallons per minute that can pass through a valve with a

pressure drop of 1 psi at a temperature of 60F.

The equation for calculating the Cv in US units is:


The effective pressure drop across a liquid control valve depends on the nature of the liquid flowing

through the valve and the valve design.

If the pressures upstream, inside and downstream of the control valve are greater than the vapour

pressure of the liquid at the flowing temperature, the effective pressure drop is equal to the actual

pressure difference between the upstream and downstream sides of the valve. In this case, the flow is

said to be “sub-critical” and the fluid remains in the liquid phase throughout the system.

In the vast majority of cases it is preferable to maintain sub-critical flow as it reduces valve damage,

improves controllability and requires simpler, less expensive valve designs.

However, if the liquid vapour pressure exceeds the system pressure inside or downstream of the valve,

vaporisation will occur and the flow will become “critical”. In this case, the effective pressure drop

across the valve will be limited by the valve design and the physical properties of the liquid. When the

flow is “critical”, the pressure downstream of the valve does not affect the flow rate.

The flow is sub-critical if:



Where:

For critical flow:

Valve Liquid Pressure Recovery Factor, FL

The valve liquid pressure recovery factor is the ratio of effective pressure drop to the pressure

difference between the upstream pressure and the vena contracta pressure.

The valve liquid pressure recovery factor is usually measured experimentally and is tabulated in valve

manufacturers’ catalogues.

Liquid Critical Pressure Ratio Factor, FF

The liquid critical pressure ratio factor is a means of estimating the pressure at the vena contracta of

the valve under critical flow conditions.

Piping Geometry Factor, Fp

The piping geometry factor is an allowance for the pressure drop associated with fittings that may be




The piping geometry factor is often listed in valve manufacturers catalogues. Alternatively, it can be

calculated using:

Most commonly, the fittings connected to a control valve are upstream and downstream reducers. In

this case the sum of the fittings factors for the reducers is:

Note:

Determining the control valve Cv becomes an iterative process when the piping geometry factor doesn’t

equal 1.

Estimate the required Cv1.

Select an appropriate valve from the manufacturers’ tables2.

Calculate Fp using the actual Cv of the selected valve3.

Re-calculate the required Cv using the value of Fp determined in Step 34.

Check that the re-calculated Cv is less than the actual Cv of the selected valve5.

If the re-calculated Cv is less than the actual Cv, the selected valve is adequately sized6.

If the re-calculated Cv is greater than the actual Cv of the selected valve, select another valve

with a larger Cv and return to Step 3

7.

Control Valve Sizing Rules Of Thumb

There are many rules of thumb designed to help with control valve sizing. The following guidance is taken

from “Rules of Thumb For Chemical Engineers” by Carl Brannan and the author’s personal notes.

Set the design flow as the greater of:1.

1.3 x normal flow rate



system

2.



system

2.

Limit the maximum flow rate : minimum flow rate turndown to 5:1 for linear trim valves and 10:1

for equal percentage trim valves

3.

The valve should be able to control the required range of flow rates between 10% and 80% of valve

opening

4.

Ideally select a valve that has a body size 1 pipe size smaller than the pipe in which it is to be

installed (e.g. select a 3” valve for a 4” pipe)

5.

Never select a valve larger than the pipe in which it is to be installed6.

How To Size A Liquid Control Valve Example

The following example has been adapted from the Emerson Control Valve Handbook.

We need to size a valve in liquid propane duty. The required information is given below:

Design flow rate = 800 US gpm

Upstream pressure = 314.7 psia

Downstream pressure = 289.7 psia

Pressure drop across valve = 25 psi

Liquid temperature = 70F

Propane specific gravity = 0.5

Propane vapour pressure = 124.3 psia

Propane critical pressure = 616.3 psia

Pipe size = 4 inch

Calculation

Following the steps given at the start of this article:

Design flow rate = 800 US gpm1.

Allowable pressure drop across the valve = 25 psi2.

We will choose an Emerson 3” ES globe valve with linear trim as the preliminary selection. From

the valve table we can see that the actual Cv is 135 and FL is 0.89

3.

We will assume that 4” x 3” reducers will be used to install the selected 3” valve in the 4” pipe.

In this case the piping geometry factor is:

ΣK = 1.5 (1 – (32

/ 42))

2= 0.287

FP = [1 + (0.287 / 890)(135 / 32)2]-0.5

= 0.96

4.

Check if flow through the valve is sub-critical or critical:

FF = 0.96 – 0.28 (124.3 / 616.3) = 0.83

DPmax = (0.89)2

(314.7 – 0.83 x 124.3) = 167.6 psi

P1 – P2 = 314.7 – 289.7 = 25 psi

Therefore: P1 – P2 < DPmax so the flow is sub-critical

5.

The effective pressure drop across the valve is P1 – P2 because the flow is sub-critical

DPeff =25 psi

6.

Calculate the first estimate of Cv:

Cv = (800 / 0.96)(0.5 / 25)0.5

= 117.9

7.

The calculated required Cv of 117.9 is less than the actual Cv of the selected valve of 135 so the

valve is large enough

8.

Check if the control valve range is OK:

From the valve table, the selected valve will be about 75% open to give the required Cv of 117.9.

This is within the acceptable control range of 10% to 80% of valve opening.

9.

The selected 3” linear trim valve is correctly sized for the specified duty10.

Result

The selected valve is an Emerson 3” ES globe valve with linear trim and a maximum Cv of 135.


The output from the Blackmonk Engineering Liquid Control Valve Calculator is attached below for

comparison. The values used in the example have been converted to metric units. As you can see the

calculated Cv is virtually identical to the hand calculation (118.2 compared to 117.9).

How To

How To Size A Pump

To size a pump, you must define:

The flow rate of liquid the pump is required to deliver

The total differential head the pump must generate to deliver the required flow rate

This is the case for all types of pumps: centrifugal or positive displacement.

Other key considerations for pump sizing are the net positive suction head available (NPSHa) and the

power required to drive the pump.

Pump System Diagram


Flow Rate

Usually, the flow rate of liquid a pump needs to deliver is determined by the process in which the pump is

installed. This ultimately is defined by the mass and energy balance of the process.

For instance the required flow rate of a pump feeding oil into a refinery distillation column will be

determined by how much product the column is required to produce. Another example is the flow rate of

a cooling water pump circulating water through a heat exchanger is defined by the amount of heat

transfer required.

Total Differential Head

The total differential head a pump must generate is determined by the flow rate of liquid being pumped

and the system through which the liquid flows.

Essentially, the total differential head is made up of 2 components. The first is the static head across the

pump and the second is the frictional head loss through the suction and discharge piping systems.

Total differential head = static head difference + frictional head losses

Static Head Difference

The static head difference across the pump is the difference in head between the discharge static head

and the suction static head.

Static head difference = discharge static head – suction static head

Discharge Static Head

The discharge static head is sum of the gas pressure at the surface of the liquid in the discharge vessel

(expressed as head rather than pressure) and the difference in elevation between the outlet of the

discharge pipe, and the centre line of the pump.

Discharge static head = Discharge vessel gas pressure head + elevation of discharge pipe outlet –

elevation of pump centre line

The discharge pipe outlet may be above the surface of the liquid in the discharge vessel or it may be

submerged as shown in these 3 diagrams.

The discharge pipe outlet may be above the surface of the liquid in the discharge vessel or it may be

submerged as shown in these 3 diagrams.

Pump Discharge Above Liquid Surface

Submerged Pump Discharge Pipe

Discharge Pipe Enters The Bottom Of The Vessel

Discharge Pipe Enters The Bottom Of The Vessel

Suction Static Head

The suction static head is sum of the gas pressure at the surface of the liquid in the suction vessel

(expressed as head rather than pressure) and the difference in elevation between the surface of the

liquid in the suction vessel and the centre line of the pump.

Suction static head = Suction vessel gas pressure head + elevation of suction vessel liquid surface

– elevation of pump centre line

Note: gas pressure can be converted to head using:

Gas head = gas pressure ÷ (liquid density x acceleration due to gravity)

Pump Suction

Frictional Head Losses

The total frictional head losses in a system are comprised of the frictional losses in the suction piping

system and the frictional losses in the discharge piping system.

Frictional head losses = fric tional losses in suction piping system + frictional losses in discharge

piping system

The frictional losses in the suction and discharge piping systems are the sum of the frictional losses due

to the liquid flowing through the pipes, fittings and equipment. The frictional head losses are usually

calculated from the Darcy-Weisbach equation using friction factors and fittings factors to calculate the

pressure loss in pipes and fittings.

Darcy-Weisbach equation:

In order to calculate the frictional head losses you therefore need to know the lengths and diameters of

the piping in the system and the number and type of fittings such as bends, valves and other equipment.

Net Positive Suction Head Available

The net positive suction head available (NPSHa) is the difference between the absolute pressure at the

pump suction and the vapour pressure of the pumped liquid at the pumping temperature.

It is important because for the pump to operate properly, the pressure at the pump suction must exceed

the vapour pressure for the pumped fluid to remain liquid in the pump. If the vapour pressure exceeds

the pressure at the pump suction, vapour bubbles will form in the liquid. This is known as cavitation and

leads to a loss of pump efficiency and can result in significant pump damage.

To ensure that the pump operates correctly the net positive suction head available (NPSHa) must exceed

the net positive suction head required (NPSHr) for that particular pump. The NPSHr is given by the pump

manufacturer and is often shown on the pump curve.

Net positive suction head available = absolute pressure head at the pump suction – liquid vapour

pressure head

Pump Power

Pumps are usually driven by electric motors, diesel engines or steam turbines. Determining the power

required is essential to sizing the pump driver.

Pump power = flow rate x total differential head x liquid density x acceleration due to gravity ÷

pump effic iency

How To Size A Pump Example

Let’s look at an example to demonstrate how to size a pump.

30000 kg/hr of water needs to be pumped from one vessel to another through the system shown in the

diagram below. The water is at 20C, has a density of 998 kg/m3

, a vapour pressure of 0.023 bara and a

viscosity of 1cP. We’ll assume that the pump efficiency is 70%.

Calculation

The calculation is presented below:

Results

Pump flow rate = 30 m3/hr

Pump total differential head = 134.8 m

Net positive suction head available = 22.13 m

Pump power = 15.7 kW

Technical Information

Pressure Equipment Directive FAQs

6 months ago we published our free “Process Engineers Guide to the Pressure Equipment Directive“.

Since then the guide has been downloaded by readers from all around the world. Many people have

emailed us with further questions related to the PED or have requested help with classifying fluids and

categorising equipment, so, based on this feedback we have compiled a list of frequently asked questions

and answers to share this knowledge.

If you have more questions, please add them to the list.

Does the PED apply to vessels under vacuum?1.

How is a valve classified under the PED?2.

How is a heat exchanger classified under the PED?3.

How should a vessel that contains both a liquid and a gas be classified?4.

Can the ASME Boiler and Pressure Vessel Code Section VIII be used to design pressure vessel to

comply with the PED?

5.

Where can I find a list of notified bodies?6.

How do I classify a pressure accessory?7.

Is PED classification required if we already have ISO, API or ASME certification?8.

When is equipment required to carry CE marking?9.

Are all dangerous fluids classified as Group 1 fluids?10.

How should a mixture of fluids be classified?11.

Are replacements, repairs or modifications to pressure equipment covered by the PED?12.

Are pipes considered to be “piping” under the PED when they are placed on the market as

individual components?

13.

Can you give some examples of pressure assemblies?14.


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Are pipes considered to be “piping” under the PED when they are placed on the market as

individual components?

13.

Can you give some examples of pressure assemblies?14.

Is on site assembly of pressure equipment by the user covered by the PED?15.

How is the PED enforced in the UK, compared with national legislations in other EU countries?16.

Does the PED apply to vessels under vacuum?

Equipment with maximum allowable working pressures of less than 0.5 barg

are exempt from the Pressure Equipment Directive. Such equipment should be designed, built

and tested to appropriate standards but this equipment is not covered by the PED.

How is a valve c lassified under the PED?

Valves are usually classified as pressure accessories. However, the PED

category of a valve is usually determined based on the valve nominal diameter in which case the

classification charts for piping can be used. If the valve has a significant internal volume, the

classification should be carried out using the classification charts for both piping and vessels

and the higher category selected.

How is a heat exchanger c lassified under the PED?

Heat exchangers are generally considered to be pressure vessels.

However, the following type of heat exchanger is treated as piping:

Heat exchangers consisting of straight or bent pipes which may be connected to common

circular headers also made of pipe providing that air is the secondary fluid, they are used in

refrigeration systems, in air conditioning systems or in heat pumps and that the piping aspects

are predominant.

For more details see Guideline 2/4

How should a vessel that contains both a liquid and a gas be

c lassified?

The vessel should be classified on the basis of the fluid which requires the higher category. The

total volume of the vessel should be used to determine the category – not the actual volumes

occupied by the individual fluids.

Can the ASME Boiler and Pressure Vessel Code Section VIII be used

to design pressure vessel to comply with the PED?

National standards and professional codes (including ASME VIII) can be used for the design and

manufacture of pressure equipment. However, a notified body may be required to validate the

selected approach if the equipment is categorised as Category II, III or IV.

See Guideline 9/5

Where can I find a list of notified bodies?

A list of all EU approved notified bodies is given here: Notified Bodies

How do I c lassify a pressure accessory?

A pressure accessory should be classified based on its characteristic

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How do I c lassify a pressure accessory?

A pressure accessory should be classified based on its characteristic

dimension – diameter or volume. If both diameter and volume are relevant, the equipment should

be classified according to whichever gives the higher category.

For example, a valve is usually classified using diameter as the characteristic dimension

whereas a filter is usually classified using volume as the characteristic dimension.

Is PED classification required if we already have ISO, API or ASME

certification?

PED classification is required in addition to other certification.

When is equipment required to carry CE marking?

The PED requires equipment that is classified as Category I, Category II,

Category III and Category IV to carry CE marking. Equipment classified as SEP must not carry

CE marking.

Are all dangerous fluids classified as Group 1 fluids?

No. Only fluids classified as:

• Explosive

• Extremely flammable

• Highly flammable

• Very toxic

• Toxic

• Oxidising

How should a mixture of fluids be c lassified?

If a mixture of fluids contains at least one Group 1 fluid, the mixture

should be classified as a Group 1 fluid. The exception to this is if the safety datasheet for the

mixture allows it to be classified as a Group 2 fluid.

Are replacements, repairs or modifications to pressure equipment

covered by the PED?

Complete replacement of an item of pressure equipment by a new one is covered by the PED.

Repairs are not covered by the PED but may be covered by national regulations.

Pressure equipment that has been modified to change its original characteristics, purpose

and/or type after it has been put into service is covered by the PED.

Are pipes considered to be “piping” under the PED when they are

placed on the market as individual components?

Individual piping components such as pipes, tubing, fittings, expansion bellows or other

pressure bearing components are not considered to be “piping” under the PED until they are

assembled into a system. However, a single pipe or system of pipes for a specific application can

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Individual piping components such as pipes, tubing, fittings, expansion bellows or other

pressure bearing components are not considered to be “piping” under the PED until they are

assembled into a system. However, a single pipe or system of pipes for a specific application can

be classed as “piping” under the PED if all appropriate manufacturing operations such as

bending, forming, flanging and heat treatment have been completed.

On this basis, PED classification of general piping stock would not be carried out by the piping

supplier. The pipes and components would be classified by the organisation responsible for the

“manufacture” of the piping system.

Can you give some examples of pressure assemblies?

Examples of pressure assemblies given in the PED guidelines include

pressure cookers, portable extinguishers, breathing apparatus, skid mounted systems,

autoclaves; air conditioner, compressed air supply in a factory, refrigerating system, shell

boilers, water tube boilers, distillation, evaporation or filtering units in process plants, oil

heating furnaces.

Is on site assembly of pressure equipment by the user covered by

the PED?

Pressure equipment assembled on site under the responsibility of the user is not covered by the

PED. Usually the separate components of the system being assembled by the user – such as

pressure vessels, valves, piping systems – are covered by the PED. The completion of pressure

assemblies on site by the manufacturer is covered by the PED.

How is the PED enforced in the UK, compared with national

legislations in other EU countries?

The PED is enforced in the UK by the Pressure Equipment Regulations 1999. These regulations

make compliance with the Pressure Equipment Directive a legal requirement in the UK. Failure

to comply with these regulations can result in prosecution and penalties on conviction of a fine,

imprisonment or both. Similar legislation has been enacted in all member states of the European

Economic Area.

The central purpose of the PED is to harmonise the national laws of the member states

regarding the design, manufacture, testing and conformity assessment of pressure equipment

and to remove technical barriers to trade. Therefore, compliance with the PED under any

member state's legislation entitles a manufacturer to sell pressure equipment throughout the

European Economic Area.

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Overall Heat Transfer Coefficients

To carry out quick heat exchanger calculations, an estimate of the overall heat transfer coefficient

usually needs to be made. Some typical values for the overall heat transfer coefficient for a variety of

different types of heat exchanger are listed below.

The values can be used in the Blackmonk Heat Exchanger Calculator.

The reference values have been taken from Coulson & Richardson’s Chemical Engineering Vol. 6, 3rd

Edition, R K Sinnot.

Shell & Tube Heat Exchangers

Hot fluid Cold fluid Overall HTC (W/(m2.K)) Overall HTC (Btu/(hr.ft2.F))

Water Water 800 - 1500 141 - 264

Organic solvents Organic solvents 100 - 300 18 - 53

Light oils Light oils 100 - 400 18 - 70

Heavy oils Heavy oils 50 - 300 9 - 53

Gases Gases 10 - 50 2 - 9

Shell & Tube Coolers


Organic solvents Water 250 - 750 44 - 132

Light oils Water 350 - 900 62 - 158

Heavy oils Water 60 - 300 11 - 53

Gases Water 20 - 300 4 - 53

Organic solvents Br ine 150 - 500 26 - 88

Water Br ine 600 - 1200 106 - 211

Gases Br ine 15 - 250 3 - 44

Shell & Tube Heaters


October 22, 2009 No comments yet

Shell & Tube Heaters


Steam Water 1500 - 4000 264 - 704

Steam Organic solvents 500 - 1000 88 - 176

Steam Light oils 300 - 900 53 - 158

Steam Heavy oils 60 - 450 11 - 79

Steam Gases 30 - 300 5 - 53

Dowtherm Heavy oils 50 - 300 9 - 53

Dowtherm Gases 20 - 200 4 - 35

Flue gas Steam 30 - 100 5 - 18

Flue gas Hydrocarbon vapours 30 -100 5 - 18

Shell & Tube Condensers


Aqueous vapours Water 1000 - 1500 176 - 264

Organic vapours Water 700 - 1000 123 - 176

Organics (some non-condensibles) Water 500 - 700 88 - 123

Vacuum condensers Water 200 - 500 35 - 88

Shell & Tube Vaporisers


Steam Aqueous solutions 1000 - 1500 176 - 264

Steam Light organics 900 - 1200 158 - 211

Steam Heavy organics 600 - 900 106 - 158

Air-Cooled Exchangers


Water Air 300 - 450 53 - 79

Light organics Air 300 - 700 53 - 123

Heavy organics Air 50 - 150 9 - 26

Gases (5 - 10 bar ) Air 50 - 100 9 - 18

Gases (10 - 30 bar) Air 100 - 300 18 - 53

Condensing hydrocarbons Air 300 - 600 53 - 106

Gases (10 - 30 bar) Air 100 - 300 18 - 53

Condensing hydrocarbons Air 300 - 600 53 - 106

Immersed Coils – Natural Circulation

Coil Pool Overall HTC (W/(m2.K)) Overall HTC (Btu/(hr.ft2.F))

Steam Dilute aqueous solutions 500 - 1000 88 - 176



Aqueous solutions Water 200 - 500 35 - 88


Immersed Coils – Agitated

Coil Pool Overall HTC (W/(m2.K)) Overall HTC (Btu/(hr.ft2.F))




Aqueous solutions Water 400 - 700 70 - 123


Jacketed Vessels

Jacket Vessel Overall HTC (W/(m2.K)) Overall HTC (Btu/(hr.ft2.F))


Steam Light organics 250 - 500 44 - 88

Water Dilute aqueous solutions 200 - 500 35 - 88

Water Light organics 200 - 300 35 - 53

Plate Heat Exchangers

Hot Fluid Cold Fluid Overall HTC (W/(m2.K)) Overall HTC (Btu/(hr.ft2.F))

Light organic Light organic 2500 - 5000 440 - 880

Light organic Viscous organic 250 - 500 44 - 88

Viscous organic Viscous organic 100 - 200 18 - 35

Light organic Process water 2500 - 3500 440 - 616

Viscous organic Process water 250 - 500 44 - 88

Name

Light organic Process water 2500 - 3500 440 - 616

Viscous organic Process water 250 - 500 44 - 88

Light organic Cooling water 2000 - 4500 352 - 792

Viscous organic Cooling water 250 - 450 44 - 79

Condensing steam Light organic 2500 - 3500 440 - 616

Condensing steam Viscous organic 250 - 500 44 - 88

Process water Process water 5000 - 7500 880 - 1321

Process water Cooling water 5000 - 7000 880 - 1233

Dilute aqueous solutions Cooling water 5000 - 7000 880 - 1233

Condensing steam Process water 3500 - 4500 616 - 792

Blackmonk News

Web-Based Process Calculators Released

We are pleased to announce that our first collection of web-based process calculators has just been

released.

The calculators are available to use right now. We’re even offering a 1 month trial for only £1 to all

subscribers complete with a 30 day money back guarantee.

To find out more and get access to the full collection of calculators click here.


Free Pressure Equipment Directive Guide

Sign up for the free Pressure EquipmentDirective guide

October 18, 2009 No comments yet

August 7, 2009 10 comments

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Over the last couple of years I’ve worked on a number of projects that have involved the European Union

Pressure Equipment Directive (or PED as it’s sometime known). The Directive is legislation which aims to

ensure that pressure equipment used within the EU is safe. For the process industries, this most often

means vessels and piping.

A key part of complying with the PED is to ensure that equipment has been classified correctly. Basically

this classification categorises equipment according to the degree of hazard should the equipment fail.

Equipment in the most hazardous applications, for instance large vessels containing toxic or flammable

gases at high pressure, is required to have extensive quality assurance procedures throughout the

design, manufacture and testing stages. Equipment in low hazard applications, such as a small storage

vessel for water at low pressure, has less onerous quality assurance requirements.

In the projects I’ve been involved with, equipment classification has been the responsibilty of the

process engineers although the information is required by the other disciplines, especially piping

and control/instrumentation engineers. Higher classification requirements also tend to affect equipment

cost and delivery, so project managers also have a keen interest!

Given that the PED is a legal requirement, along with potential cost and delivery implications, it is

essential that equipment classification is carried out thoroughly and accurately. To help you classify

equipment correctly, I have written a free guide to the Pressure Equipment Directive which is available

to download to all Blackmonk email subscribers.

To get your free copy just enter your name and email details into the boxes above or on the right hand

side of this page.

I hope you find the guide useful and would appreciate any comments you might have.

Regards,

Simon.

I hope you find the guide useful and would appreciate any comments you might have.

Regards,

Simon.

Blackmonk News

Save Up To 80% of Project Costs

Blackmonk Engineering has been approved as a registered supplier on the Business Link North East

England Service Providers Register. This means that our customers could be eligible for up to 80% funding

for qualifying projects. To find out more please contact us.

Blackmonk News

Updated Website

We’ve been updating our website over the past couple of weeks to make it more user friendly. We hope

you like it. Comments can now be left directly via the site on various pages and the navigation has been

improved. The free calculators are still there and now they should be easier to find!

In addition we’ve added a blog where we’ll be writing articles relevant to those of us in the process

industries.

As always, we would appreciate your feedback on the new website.

Best regards,

Simon.

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July 31, 2009 No comments yet

July 31, 2009 No comments yet

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