14 replies to this topic

[ADR]

Good day to All,

 

I am a budding Chemical Engineer recently been given an assignment to design a PFD for one our new project.

 

Generating a PFD for our packages is usually quite simple; however for this particular job I am finding it difficult because the dosing is by gravity. I have read and understood Art Montemayor’s workbook about gravity flow but still I need some guidance about sizing the control valve from you all.

 

We have  a storage tank, the inlet flowrate is 10 m³/hr and the outlet flow rate is 10 m³/hr. There will always be an ample amount of product left in the tank because we are maintaining a 5 minutes residence time to disentrain Hydrogen present in the product. Therefore the discharge line is always be flooded (A level transmitter monitors the level inside the tank, when a low level is detected, dosing will be stopped). Dosing flow control is by means  of a proportional flow control valve- mag. flowmeter loop.

 

Our scope ends after the control valve; we don’t have any details about the piping layout after our limits. However, we have been informed that the dosing point is 15 m below the tank outlet.

 

My question is about sizing the control valve. Could anyone help me to determine the ∆P for the following two cases? (please see attached schematics)

1.       Discharge pipe submerged in water

2.       Dosing at sea level 

 

Product density:1028 kg/m3

dynamic viscosity:1.37 cP

Product temp: 12 deg C

 

 

Many thanks

AR

 

Attached Files

•  Book1.xlsx   15.06KB   84 downloads

[Art Montemayor]

Allen:

 

There are some discrepancies / errors / omissions with your data information.  Please refer to your workbook, Rev1, as attached.  Note my sketch.

 

You say your responsibility (battery limits) ends at the control valve’s outlet.  So why are you concerned with anything else downstream?  You must obtain the pressure drop downstream of the CV by those parties responsible for that design.

 

You must identify the static level of liquid above the CV.  That, and the pressure drop through the tank’s outlet and associated piping and meter will give you the pressure at the CV’s inlet.  You now know the pressures on both sides of the valve.  If your drop at the downstream side results in your not having sufficient static head in the tank, you are in trouble.  You have to extend the height of the tank so you can have a reasonable valve size to handle the 10 m3/hr.  The static head is your only driving force involved in developing the required flow rate.

 

Another thing: you show a back-pressure CV.  The process can’t work that way.  You must use a flow controller acting directly with the CV.  You are trying to control the flow at a set point, not the pressure upstream of the CV.

 

Although you have gravity drop due to the static head, your discharge is totally flooded, so you have a conventional liquid flow pressure drop problem.  This should be a rather simple - although detailed - flow calculation resolved by using Darcy’s formula + entrance, exit, and fittings’ losses.

 

 Book1 Rev1.xlsx   51.65KB   96 downloads

[ADR]

Hi Art,

 

Many thanks for your prompt response and for the detailed explanation.

 

First of all, apologies for the inaccuracies in my drawing, it is a result of getting things done in a hurry.

 

Regarding the battery limits, yes of course it is our customer’s responsiblity to give the pressure drop on the downstream side of the CV.  However, I thought its better to go well prepared for the actual techical clarification meeting.

 

Looking through the project documentation this morning, I relaised that dosing is at about 25 metres below mean sea level and the tank outlet nozzle is 27 m above the sea level (Please see attached Book1 Rev2). With these information I calcualted the static pressure and the line loss using Darcy’s friction loss method and determined that the allowable ∆P across the control valve have to be 0.1173 bar.

 

The customer requests for a globe type FCV which I believe requires fairly high ∆P to operate. I sized a globe type FCV using Fisher’s sizing program (∆P =0.1173 bar) and got the following results

 

C_V = 34.821.

cavitation ∆P =1.05 bar

 

Please have a look through my calcs and advise what I am doing is correct?

 

Many thanks

Allen

Attached Files

•  Book1 Rev2.xlsx   68.62KB   68 downloads
•  Control Valve Specification Sheet- Chem resources.docx   36.39KB   50 downloads

[samayaraj]

Hi allen,

 

You can refer the below post

 

http://www.cheresour...ng-calculation/

 

The excel calculation in above post is nearly equal to what you have posted. Look at it.

[ADR]

Hi Samayaraj,

 

Thanks for sharing, it looks really impressive.

 

Could you advise the values I have to enter in H1 (Elevation of LL in storage tank) & H2 (Elevation of Pipe connecting Receiver)?

 

I entered H1= 1.1 m (minimum product level in the tank), H2= 50 m(Height from the tank's outlet nozzle to the dosing point-below sea level) and getting runtime error '1004'. when I click 'debug' the following is displayed.

 

Sub Macro1()

'

' Macro1 Macro

'

 

'

    Range("E40").GoalSeek Goal:=1, ChangingCell:=Range("E39")

    Range("E43").GoalSeek Goal:=1, ChangingCell:=Range("E42")

    Range("H40").GoalSeek Goal:=1, ChangingCell:=Range("H39")

End Sub

 

Allen

[ADR]

Hi Sam,

 

I have redone the calculation, could you please check.

Attached Files

•  New Gravity flow calculation by Sam.xlsm   79.2KB   70 downloads

[samayaraj]

Hi allen,

 

The error is due to negative elevation difference.

 

I've seen your calculation. I have following doubt.

 

1. The elevation difference shown in Book1Rev2 is different (Refer attached PDF). What is the exact elevation of dosing liquid level and dosing point?

 

 

Our scope ends after the control valve; we don’t have any details about the piping layout after our limits. However, we have been informed that the dosing point is 15 m below the tank outlet.

 

My question is about sizing the control valve. Could anyone help me to determine the ∆P for the following two cases? (please see attached schematics)

1.       Discharge pipe submerged in water

2.       Dosing at sea level 

 

2. You said dosing point is 15m below the tank. I could not understand what is the actual elevation difference.

3. Pressure you have shown is 3 kg/cm^2 + for both storage tank and dosing point. Both having nearly same pressure? Are they not operating in atmospheric pressure? Anyways, pressure is not a question here and it will be taken care by the head of liquid connecting the dosing point. 

Attached Files

•  Book1 Rev2.pdf   6.74KB   44 downloads

[ADR]

Hi Samayaraj,

 

Please disregard the data I posted on the 26th of July.

 

Please see attached my calculation notes in which I have specified the correct elevation and pressure.

 

Attached Files

•  20150728104254.pdf   233.57KB   72 downloads

[samayaraj]

Hi Allen,

 

Steps you have followed is correct.

 

If the storage tank is operating at atmospheric pressure and density of dosing liquid and sea water is nearly same, the total available head is (1.1 m dosing liq + 27 m down - 25 m below sea) nearly 3 m to drive the dosing liquid.

 

Total available head - losses = ∆P required across CV.

Edited by samayaraj, 28 July 2015 - 05:48 AM.

[ADR]

Hi Samayaraj,

 

Yes it is a atmospheric tank and the density of the product is almost the same as seawater.

 

Thank you so much for helping me out.

[ADR]

Hi Art / Samayaraj,

 

Sorry for digging up this topic again.

 

Please could you tell me what the pressure would be at the inlet of the control valve ? (Please see attached Book1 Rev 3)

 

Thanks

AR

Attached Files

•  Book1 Rev 3.xlsx   70.46KB   46 downloads

[samayaraj]

Hi allen,

 

The pressure at the inlet of control valve is nearly 0.11 bar g.

[ADR]

Thanks Samayaraj,

 

Could you also explain why the pressure wont be 2.8 barg (elevation from sea level is 27m+1.1 m of water in the tank)?

 

Thanks in advance.

[samayaraj]

Dear Allen,

 

As per the sketch shown, the center line elevation of tank outlet nozzle and control valve is at +27m, where as the dosing liquid is 1.1m above the outlet nozzle. This is what you have shown in the sketch. So you will get nearly 0.11 bar g pressure at the inlet of control valve.

 

If the control valve is placed at the sea level, what you have said above will apply.

[ADR]

Thank you Samayaraj.