June 28, 2008

Mist Eliminator / Demister - Sizing & Efficiency Calculation

Mist Eliminators or Demister Pads are very common & important things for process engineers but they are often neglected. Here is some input from my side theory as well as practical knowledge & gains from them.

The importance of a demister pad comes from the efficiency of separation & its impact on either product recovery or the indirect impact on downstream processes. Usually later is more important than direct impacts & therefore they are generally neglected peace of equipments.

The separation of entrained liquid droplets from a vapor / gas stream is called mist elimination & the equipment used for the purpose is called De-mister. Generally, in a manufacturing plant, process engineers or production engineers do not focus on the benefits which can be derived from the improvement in demister pad or mist eliminators.

For example, suppose your outlet gas goes to a reactor with VLE amount of component i.e. Ammonia. Now if ammonia from the separator goes up due to mist carry over or poor efficinecy of de-mister pad then the conversion in the reactor will go down, resulting in overall impact on synthesis loop pressure & energy. Also the production will go down.

Similarly, if a separator in acidic gas service do not work properly the downstream equipment will corrode easily. That's why we need efficient mist separation.

What is mist.......? Mist is the fine droplets of liquids of various sizes. They may vary from 0.5 micron to few 100 of microns in size. For example, good spray system generally generate 20 - 1000 micron size droplets, while columns & tray do have ~8 - 100 micron size drops. Saturated Vapors generally have finest size of droplets depending on their velocity at generating point.

Generally we use Souder's equation as used for phase separator Or for knock out drums. That is

Vd = k x [ (L-G)/G ]^0.5

L & G are liquid & gas densities.

where k is the important part & is called the capacity design factor. It depends on type of de-mister pad. Selection of a too low or too high k is always have a negative impact in case of demisters as the efficiency greatly depends on velocities.

In case of lower velocities, droplets have low momentum to get path impingement & coalescene & therefore avoid capture into bigger drops & thus escape from the pad. At higher velocities the vapors have sufficient kinetic energy to re-entrain them. Therefore, correct range of k selection is necessary.

Based on past experiences & designs a value of k = 0.42 is most suitable for many applications. So after choosing k get the design velocity & then find out the diameter of separator. Now for predicting efficiency of de-mister pad, calculate K inertial parameter as below

K = [ (L - G)/ Vd^2 ] / ( 9 x mu x D)

L & G are liquid & gas density
Vd - velocity of gas calculated above
D - Diameter of pad

Now use following curve to get E factor for above K value.

Now calculate specific area of pad as below

A = Specific Area x Thickness x 0.67 / PI()

Now calculate % Efficiency as below

Eff = 100 - 100 / e ^ ( 0.213 x A x E )

In the next part I will consider pressure drop calculation.

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June 23, 2008

Impact of Specific Gravity on Pumps

The pump performance or characteristic curve is often referred to as a water performance curve (see Figure 1). Water is naturally the fluid of choice for testing pumps because of its availability. The question is: can the water performance curve be used for other fluids? The answer is YES with care.

The pump manufacturers will sometimes identify the unit of the total head axis as feet of water, or sometimes there will be a notice somewhere on the chart saying: tested with water at 70ºF. This is the manufacturer’s way of telling the user that the pump was tested with water and that the power curves are ONLY valid for water, but careful there are other pitfalls.

Related References

Here is the question that you need to think about. Say that you measure the shut-off head of a pump with water, will you get a different value if the fluid is a dense salt solution (assuming the viscosity is the same)? The answer is NO. Why....Because pump is going to develop the same head.

Head is what?.... Height of Liquid.....So Height of liquid column will remain same whether it is water or salt solution but the pressure exerted by the head will be different. Head remains same is the typical property of a centrifugal pump....Keep this always in mind whenever you are dealing with pumps. They are constant head machines.

Head = Pressure / SG


Pressure / SG = Constant


P1/SG1 = P2/SG2

Thus developed pressure will be = curve pressure x SG / SG of water (1.0)

Here the curve pressure is the developed pressure for water, SG is the specific gravity of fluid in question & SG of water is 1.0 at test conditions or curve conditions.

Impact on Power

Power = Q (M3/Hr) x H (Head in meters) x SG / 367.2 / Efficiency of pump

So the power will be different because of SG term in the equation. So in case of different fluid do not consider power from pump curve it is better to calculate from above formula...OR if you consider curve power correct it for SG change.

Next I am going to cover the impact of viscosity on pumping.

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June 18, 2008

Deaerators - Basic Understanding

Deaerators are simple mechanical devices that remove dissolved gases from boiler feed water using steam stripping. De-aeration protects the steam system from the effects of corrosive gases. It accomplishes this by reducing the concentration of dissolved oxygen and carbon dioxide to a level where corrosion is minimized.

A dissolved oxygen level of 5 parts per billion (ppb) or lower is needed to prevent corrosion in most medium or high pressure (>200 pounds per square inch) boilers. While oxygen concentrations of up to 43 ppb may be tolerated in low-pressure boilers, equipment life is extended at little or no cost by limiting the oxygen concentration to 5 ppb. Dissolved carbon dioxide is essentially completely removed by the de-aerator.

Please understand that the main function of a de-aerator is to remove “Dissolved” gases not free air or free oxygen. The most important gases are oxygen & CO2.

How They Work
The design of an effective de-aeration system depends upon the amount of gases to be removed and the final oxygen gas concentration desired. This in turn depends upon the ratio of boiler feed water makeup to returned condensate and the operating pressure of the de-aerator.

Deaerators use steam to heat the water to the full saturation temperature corresponding to the steam pressure in the de-aerator and to scrub out and carry away dissolved gases. Steam flow may be parallel, cross, or counter to the water flow. The de-aerator consists of a de-aeration section, a storage tank, and a vent.

In the de-aeration section, steam bubbles through the water, both heating and agitating it. Steam is cooled by incoming water and condensed at the vent condenser. Non-condensable gases and some steam are released through the vent. Steam provided to the de-aerator provides physical stripping action and heats the mixture of returned condensate and boiler feed water makeup to saturation temperature. Most of the steam will condense, but a small fraction (usually 5% to 14%) must be vented to accommodate the stripping requirements.

Normal design practice is to calculate the steam required for heating and then make sure that the flow is sufficient for stripping as well. If the condensates return rate is high (>80%) and the condensate pressure is high in comparison to the de-aerator pressure, then very little steam is needed for heating and provisions may be made for condensing the surplus flash steam

De-aerator Steam Consumption
The de-aerator steam consumption is equal to the steam required to heat incoming water to its saturation temperature, plus the amount vented with the non-condensable gases, less any flashed steam from hot condensate or steam losses through failed traps.

The heat balance calculation is made with the incoming water at its lowest expected temperature. The vent rate is a function of de-aerator type, size (rated feed water capacity), and the amount of makeup water. The operating vent rate is at its maximum with the introduction of cold, oxygen-rich makeup water.

The de-aerator section and storage tank and all piping conveying hot water or steam should be adequately insulated to prevent the condensation of steam and loss of heat. This will reduce the steam consumption in de-aerator which is an additional cost to increase the life of equipment.

Sudden increases in free or “flash” steam can cause a spike in de-aerator vessel pressure, resulting in re-oxygenation of the feed water. A dedicated pressure-regulating valve should be provided to maintain the de-aerator at a constant pressure. This also helps in reducing steam consumption.

Additional Benefits
Deaerators provide the water storage capacity and the net positive suction head necessary at the boiler feed pump inlet. Returned condensate is mixed with makeup water within the de-aerator. Operating temperatures range from 215° to more than 350°F, which reduces the thermal shock on downstream preheating equipment and the boiler.

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June 13, 2008

Now you can print posts from Chemical Professionals

Dear Visitors
I have installed printing option for printing of the individual posts, so that you can use them later on as a reference or for any other use. This will remove all the widgets from Blog & only post text is printed.

To print a post, do the following.

1. Click on th etitle of that individual post.
2. Once it is completed loaded go to the end of post (Not to the bottom of your screen).
3. Just at the end of my words of post you will find a printer icon with 'Print this post' link on the left side.
4. Click it & you will find your printer selection pop up in few seconds.
5. Print & keep them without having any sidebar, blog header etc.

Thanks for your co-operation & appreciation.

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June 12, 2008

Pumps - 50 Troubleshooting Tips

All of us as a process engineer face different problems related to pump operation. Be it a problem of low capacity, low head, pump loosing prime,
requires excessive power, stuffing box or seal leakages which are very common, bearings life, overheating etc.

All these problems are critical if the pump is in process fluid service which is either corrosive or hazardous. Therefore a step by step approach is necessary to attend to these problems. Recnetly I came across a compiled list of such 50 tips which I am sharing here for you. Note down these tips & follow them whenever you get any pump problem for troubleshooting.
Pumps are rotary devices which converts velocity or kinetic energy into pressure energy for compresible fluids to pump them from one location to the other. Being a rotary device they are prone to general wear & tear problems therefore, a process engineer should know how to handle these problems & do effective troubleshooting.

Here is the list of 50 such tips for pump troubleshooting categorized into different heads.
  • Suction Troubles

    1. Pump not primed.
    2. Pump or suction pipe not completely filled with liquid.
    3. Suction pipe lift too high.
    4. Limiting pipe size in suction or higher velocities resulting in flashing.
    5. Insufficient margin between suction pressure and vapor pressure.
    6. Excessive amount of air or gas in liquid.
    7. Air pocket in suction line.
    8. Air leaks in suction line.
    9. Air leaks into pump through stuffing box.
    10. Foot valve too small.
    11. Foot valve partially clogged.
    12. Inlet of suction pipe insufficiently submerged.
    13. Water seal pipe plugged.
    14. Seal cage improperly located in stuffing box, preventing sealing fluid entering space to form seal.

  • System Troubles

    1. Speed too low.
    2. Speed too high.
    3. Wrong direction of rotation.
    4. Total head of system higher than design head of pump.
    5. Total head of system lower than design head of pump.
    6. Specific gravity of liquid different from design.
    7. Viscosity of liquid different from design criteria.
    8. Operation at very low capacity.
    9. Parallel operation of pumps unsuitable for such operation
    10. Incorrect piping layout.

  • Mechanical Problems

    1. Foreign matter in impeller.
    2. Misalignment.
    3. Foundations not rigid.
    4. Shaft bent.
    5. Rotating part rubbing on stationary part.
    6. Bearings worn.
    7. Wearing rings worn.
    8. Impeller damaged.
    9. Casing gasket defective permitting internal leakage.
    10. Shaft or shat sleeves worn or scored at the packing.
    11. Packing improperly installed.
    12. Incorrect type of packing for operating conditions.
    13. Shaft running off center because of worn bearings or misalignment.
    14. Rotor out of balance resulting in vibration.
    15. Gland too tight resulting in no flow of ;liquid to lubricate packing.
    16. Failure to provide cooling liquid to water cooled stuffing box.
    17. Excessive clearance at bottom of stuffing box between the shaft and casing, causing packing to be forced into pump interior.
    18. Dirt or grit in sealing liquid, leading to scoring of shaft or shaft sleeves.
    19. Excessive thrust caused by a mechanical failure inside the pump or by the failure of the hydraulic balancing device, if any.
    20. Excessive grease or oil in bearing housing or lack of cooling, causing excessive bearing temperature.
    21. Lack of lubrication.
    22. Improper installation of antifriction bearings (damage during assembly, incorrect assembly of stacked bearings, use of unmatched bearings as a pair, etc.).
    23. Dirt getting into bearings.
    24. Rusting of bearings due to water getting into housing.
    25. Excessive cooling of water cooled bearing resulting in condensation in the bearing housing from moisture in the atmosphere.
    26. Wear Ring clearance.

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June 09, 2008

Guaranteed Energy Saving in Pumps

Today I'll share a very useful sure shot energy saving method in cooling water pumping system. So let me start as usual from some initial questions which are necessary for estimating the amount of efforts required in proving the concept.

How many cooling water pumps do you have in your unit? I have ~50 major such pumps.
What is the total power load of these pumps? It is ~ 6000 kW in my unit - YES whopping 6 MW load.
What is the potential of saving out of this load - May vary from 5 - 7% or more, based on actually achieved savings in my units.

Related References

How it Works
Whenever a cooling water pump is used it is generally getting corroded over a period of time due to several issues related to cooling water treatment & therefore, mostly it is pitting corrosion which makes the casing inner surface rough & it becomes tough for handling water efficiently.

In a centrifugal pump the action is to provide rotary forces to convert the velocity energy into pressure energy. These centrifugal forces when comes in contact with the rough wall of casing the frictional coefficient between fluid & pump casing wall increases. Thus, overall losses inside the pump increases & hence efficiency drop is observed over a period of time.

When we apply any ceramic or enamel coating on the inside wall of pump casing this frictional resistance reduces, therefore energy saving is achieved. In general I have used this method on different pumps ranging from samll ~100 kW to very large 1 MW pumps & found 5-7% savings actually achieved. The saving range can vary depending on the condition of your pump.

The benefit of this suggestion is that
  1. It increase efficiency for new pumps by reducing inherent friction of material & fluid - So you get more efficiency but it will be limited to ~1-2%
  2. Increase in efficiency of older pumps - the scope here is more 5-7% or higher
  3. Increase in life of older pumps
  4. Avoid early damage to new pumps
  5. Avoid drop in efficiency of new pumps
  6. Delay investment for replacement of older pumps saving capital & interest cost
  7. Restore pump head
  8. Restore pump capacity

Here are the pics of before & after coating in my unit.

In my case of 50 pumps the estimated investment was only Rs. 20 Lac & saving of 360 kW power @ 6% out of total 6 MW load. This equals to Rs. 114 Lac/year saving at Rs. 4.0 / unit power cost. Thus payback was only 2 months.

Do you still need more good reason to read 'Chemical Professionals'......NO......So get this idea, start working on it, send a proposal to your BOSS..............IT WILL NOT FAIL....................& get a promotion next year.

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