Pressure Drop across the Mist Eliminator Vessel with Internal Cyclone Bundles.

Have you asked the multicyclone bundle vendor ? What max permissible diff press did you allow for this bundle in your specification to this vendor ? Do you have a DP gauge across this bundle on this vessel - does it meet your spec ?

Hi Georgeverghese, Thanks for the response. We have consulted with our multicyclone bundle vendor, but they confirmed the maximum pressure drop as 1.5 psi for the operating parameters mentioned. ( That was the maximum permissible diff. pressure we requested them to follow as well.)
On the top of this pressure drop, based on our calculation of pressure drop at entrance and exit nozzles using the formula given below we are getting a total pressure drop of 1.7 psi. But client is getting pressure drop around 2.8 ( Approx. 3 psi) during the actual operation. Their actual current operating flow rate is 1,750 m3/hr and operating pressure is 189 psig. (Note: These are not the values for which we designed the cyclone bundle and pressure drop). However based on these new values our pressure drop should ideally reduce further, but to everyone's surprise it increases. What could be the reasons for this higher pressure drop?

There is an enhanced cooling air cooler ( mostly an evaporative cooler) situated upstream of this Mist eliminator vessel and a compressor (downstream of Mist eliminator) is sucking the air through this Enhanced air cooler through ,mist eliminator vessel and finally supply to the system. If the air is cooled with introducing moisture into the air and then if this moisture is being eliminated from the air while it passes through the cyclone bundle, then is there any impact in the densities of the inlet and outlets stream ? Will this scenario causes this extra pressure drop? Client is unable to provide the composition of the Air + Water condensate upstream and downstream of the Mist eliminator vessel.

Note:
1. It seems, our cyclone bundle supplier also consider this air as normal air with molar mass as 28.96 g/mol. Will the composition of the water condensate in the upstream of the vessel have any significant impact on this pressure drop calcs?

2. Nozzle Pressure drop calculation based on reference: Carl Branan, "Rules of Thumb for Chemical Engineers", 2nd edition, Gulf Publishing,1998.page 131
The Pressure drop over inlet nozzle is appr. 0.5 x density x velocity2
Pressure drop over vapor outlet nozzle is appr. 0.25 x density x velocity2

Where: pressure drop is in Pascal (=0.00001 bar)
density is in kg/m3
velocity is in m/s

Errr, your client is operating at a higher flowrate ( is that sm3/ hr??) and a lower pressure, so this would be a much higher actual flowrate, assuming your flow figures are standard volumes.

Yes density will increase if it is essentially at 100%RH.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

Is 1750 m3/hr referencing actual press/temp conditions or some standard condition?

Hi,
If you really want to know the delta P, let you measure it and install calibrated gauges at the inlet and outlet of your equipment.
Everything else is useless to validate the performance of an equipment, especially your calculation.

About operating conditions and references you are mixing apples and carrots. Make sure you, your client and the bundle vendor are using the same references. From your post it's unclear, actual or standard values.
Of course, if the flow rate is bigger than design, le head loss will increase. Delta P is proportional to v^2 , the reason why we need the reference point (delta P @ a given flowrate).

Note: The mist eliminator could also be partially plugged and or damaged, creating head loss.

Pierre

Hi @LittleInch, Thanks for the response. The flow rates mentioned are all Actual m3/hr not at standard conditions. In fact client is providing these values which we believe they are measuring at a downstream side of this Mist Eliminator vessel, not at the upstream side. Site has minimum and maximum humidity values as 40% and 100% respectively. It seems Client is providing the flow rates based on ideal gas equation PV=nRT. Please also do comment, for the temperature and pressure which I have mentioned, will the ideal gas law holds good ? or should we take any compressibility factor(z) for the calculations?

HI @Georgeverghese, 1,750m3/hr is based on actual condition. Client has given the values as follows.

Operating Temperature: 142C
Operating Pressure: 189psig
Mass Flow Rate: 20486 kg/hr
Volumetric Flow Rate: 1753 Am3/hr

It seems Client is providing the flow rates based on ideal gas equation PV=nRT. Please also do comment, for the temperature and pressure which I have mentioned, will the ideal gas law holds good ? or should we take any compressibility factor(z) for the calculations?
Kindly let me know your further comments or questions.

Hi @pierreick, Thanks for the response. We have pressure transmitter across the inlet and outlet of of this Vessel. Client is measuring the values in the same pressure transmitter which shows higher values than what we have designed. (Eg: Maximum of 1.5 psi as per confirmation from internal cyclone bundle supplier and our nozzle entrance and exit will introduce further 0.2 psi maximum ( based on our calculations = Total 1.7 psi)

Currently client is operating the equipment at a different operating parameters which is given as below.
Operating Temperature: 142C
Operating Pressure: 189psig
Mass Flow Rate: 20486 kg/hr
Volumetric Flow Rate: 1753 Am3/hr

For this new operating parameters our internal cyclone vendor confirmed the max pressure drop across their bundle is 1.2 psi and with our inlet and exit nozzle pressure drop calcs we are getting a total of 1.33 psi. However client is getting 2.28 psi ( 15.748 kPa).
We are curious and investigating why this much change (71%) than the designed value) occurs.

I agree on your point that with the increasing flow rate the pressure drop will increase. But density is reducing for the new operating parameters, which also will have an effect on the pressure drop.( Right ?). All the values mentioned here are Actual flow rates, both the client and cyclone bundle supplier has the same understanding.

We have checked for clogs or damages in the bundle, but found nothing. (Bundles are made of 2" pipes x 42 numbers.)

The big reduction in operating pressure has a large impact on pressure drop. Given the "composite resistance coeff K" of the vessel remains constant, press drop ratio dP2/dP1 = (K.rho_2 . v2^2)/(K.rho_1 . v1^2)
pressure drop ratio varies linearly in proportion to density, which is related to pressure (p2/p1 = 205/330 = 0.62)
pressure drop ratio varies as the square of velocity ( v2/v1)^2

In turn, velocity ratio v2/v1 varies linearly in proportion to actual flow (q2/q1 = 1750/1471 = 1.19)
Velocity ratio also varies inversely with pressure (p1/p2 = 330/205 =1.61)

So, dp2/dp1 = 0.62 x (1.19x1.61)^2 = 2.27
So dp2 = 1.7x2.27 = 3.8 psi

If dp2 is actually 2.8psi, then this tells me the vessel is operating much better than design expectation. So, extrapolating to min operating pressure of 313psig and 1471am3/hr, pressure drop will be (2.8/3.8) x 1.7 = 1.3psi total dp for this vessel

Humidity has a small impact on dp, and is neglected here when compared to the much bigger effect of change in pressure and flow.

Hi Georgeverghese,

Great.. That's an amazing explanation. In fact we missed this velocity's inverse proportionality with the pressure. We have considered only the flow rate dependency. We will further study and work based on this. Thanks for the supports.

Hi @Georgeverghese, I believe temperature also needs to be considered into the calculations. Original temp was 120 C and current operating is 140C. Should we consider density as inversely proportional to temp and velocity as directly proportional to square root of temperature as well ? This will vary the results slightly. What is your thoughts?

Also just to re-confirm the concept of dependence of velocity on both flow rate and pressure, we are not duplicating the factor right? Both the effects of flow rate and pressure should be independently taken into account (As you have multiplied 1.19 and 1.61 to get the velocity change. ) Could you please explain a bit on this. Thanks in advance.

Please review your ideal gas laws.
Ideal Gas
Volume varies directly with temperature referenced to absolute temperature.
Density must therefore vary inversely with absolute temperature.
Density varies directly with absolute pressure.

You may account for both Pressure and Temperature change simultaneously with
V2 = V1 x P1 /T1 x T2/P2

Hence flow rate and velocity both change with temperature as follows,
1421 m3/h at 120°C becomes
1421 x (273+140)/(273+120) = 1421 × 1.05 = 1493m3/h
Velocity is also V2 = 1.05 x V1

Frictional pressure drop varies linearly with density, decreasing at 140C and increases with the square of velocity.
Pd2 = Pd1 x (1.05 x V1)^2 / (1/1.05 x V1)
Pd2 = Pd1 x (1.05 x V1)^2 / (0.952 x V1)

Now include the increase of density with absolute pressure.



--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

There is an error in my calcs, and it is to do with the fact that the flows 1471m3/hr and 1750m3/hr do not reference the same P and T. So correcting the 1750m3/hr back to the design case P of 330psia, and this time correcting for temp also, I get a total expected dp = 1.9psi for operation at 1750m3/hr and 205psia, assuming that the vessel produces a dp of 2psi at design case operation.

If this vessel actually sees a dp of 2.8psi at 205psia / 1750am3/hr, extrapolating to design case operation at 1470am3/hr / 330psia, this vessel will see a dp = 3.3 psi, clearly well beyond design limits.

So further questions:
a) Are any swirltubes in the swirldeck blanked off deliberately for turndown operations ?
b) How do you know flow is 1750am3/hr ? Details of flow meter. Pressure compensation included for density correction ? Is the FE installed correctly; do you have a P&I D ?

Hi 1503-44, Thanks for the response. While you calculated the flow rate difference and velocity difference, you have accounted for temperature and pressure already. So are you suggesting that we don't need to consider the reduction in operating pressure separately to arrive at the velocity change. Could you please reconfirm if we don't need to consider the velocity increase due to reduced pressure separately.

Please clarify the statement "Now include the increase of density with absolute pressure"
My understanding is that along with pressure we should take absolute temperature also for finding the change in density. Please let me know your response.

Hi Georgeverghese, Thanks for the reply and pointing out the corrections. We will further study on this. Answer to your questions as below.

a) Are any swirltubes in the swirldeck blanked off deliberately for turndown operations ? : None of the swirltubes are blanked. All are open to the flow.
b) How do you know flow is 1750am3/hr ? Details of flow meter. Pressure compensation included for density correction ? Is the FE installed correctly; do you have a P&I D ?
Client advised the flow rate as 1,750 am3/hr based on the site measured values as below.
Operating Temperature: 142C = 415K (This is the inlet temp of the incoming air)
Operating Pressure: 189psig. (It is bit unclear for us where this is measured from.)
Mass Flow Rate: 20,486 kg/hr (This flow rate is measured from a compressor outlet which is downstream of this mist eliminator vessel)
Volumetric Flow Rate: 1753 Am3/hr (based on PV=nRT, this value is calculated by client with referring to the above values, We rechecked this is correct if the molar mass is 28.71 g/mol. Please do comment on this as well )

If you don't mind, Could you please provide me your personnel contact details.

A lot of water vapor would have dropped out at the compressor due to compression, so wet air flow through this upstream vessel should include water vapor also. So wet air flow at this separator is > 20500 kg/hr. If you do a heat balance over the evaporative cooler (is this a quench cooler?), you may be able to get some idea of how much water vapor has gone into the air stream ?

Hi GeorgeVerghese, Thank you for further valuable remarks. We also had the same concern and informed to client to measure the flow rate before the inlet to the Mist eliminator vessel. Still awaiting that information. Similarly to the water vapor loss in the Compressor, we can expect some moisture loss in the cyclone bundle itself as well right?( that is what the purpose of this "mist eliminator", right? ) So this flow rate also should be taken into account or not?

Client datasheet always mentions the fluid as "Air+ Water condensate". But as of now, we haven't got any firm information about the composition of this "Air + Water condensate" mixture.
There is a cooler upstream of this mist eliminator vessel, which they shows as S&T HEx in some P&IDs and some other projects shows as "Radiator type". But they haven't mentioned it as Quench coolers.

Sruthish,

I was just telling you that you can correct Volume-flow for pressure and temperature simultaneously, but since that is not what you did, I preceded to follow your calculation sequence, first correcting only for temperature. I left the pressure conversion up to you. But I include that now.

Simultaneous Temperature and Pressure Corrections
1421 m3/h x (273+140)/(273+120) x (14.7+323)/(14.7+189)
1421 x 1.051 x 1.658
2476 m3/h

After correcting volume for both temperature and pressure,
Volume increased by a factor of 2476/1421 = 1.742
Density decreased by 1/1.742 = 0.574

Including the pressure conversion changed the previous 1.05 factor to 1,742
V2 = 1.742 V1
Den2 = 0.574 x Den1

Frictional pressure drop varies linearly with density.
Density decreases with both the 120 to 140C temperature change and
The 323 to 189 psig pressure drop, so
for pressure drop effect of density, use a factor of 1/1.742 = 0.574

Frictional pressure drop varies with the square of velocity, so use a factor of 1.742^2
for pressure drop effects of velocity
Pd2 = Pd1 x (V2)^2/(V1)^2 x Den2/Den1

Pd2 = Pd1 x ((1.742 x V1)^2)/(V1)^2 x 0.574 x Den1/Den1
Pd2 = Pd1 x 1.742^2 x 0.574
Pd2 = Pd1 x 1.742

Note that, since volume and density are inversely related, the net effect on pressure drop makes pressure drop directly proportional to the inverse of density ratio, or the velocity ratio, or the flow rate ratio. All are numerically equal.
George uses the flow rate ratio 1750/1470 = 1.19 in his estimate of
dP =2.8 psi x 1750m3/h / 1470m3/h = 3.33 psi







--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."

Hi 1503-44,
Thanks for the clarification. Just one more doubt. If we are taking 1421 m3/hr @ Operating pressure of 313 psi.g (327.7 psi.a) and temp of 120C(393K) as the original volumetric flow rate for designing the cyclone bundle, now based on the new Operating pressure (189 psi.g = 203.7 psi.a) and operating temperature (140C = 413K) the volumetric flow rate should be 2498.9 m3/hr, given that mass flow rate is being constant. I just wonder then how could Client is getting flow rate as 1753 m3/hr. ?