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pressure and temp compensation formula for the mass flow calculation of superheated steam

wrong reading of flow of superheated steam

By tajinder on 19 July, 2007 - 7:12 pm

we have added the pressure and temperature compensation for calculating the mass flow of the superheated steam. The design pressure and temperature of the orifice plate are 43kg/cm2 and 420 C and the differential pressure created by the orifice plate is 10000mmWC.

kindly suggest the correct formula for the above

By Janapati Aswani Dutt on 28 March, 2008 - 12:13 am

Temperature & pressure compensation is not at all required for mass flow rate only is accountable in volumetric flow rate. As you are using differential pressure type flow element, it ensures that the transmitter is having linear and in DCS/PLC square root. And also check the specifications whether mass flow or volumetric flow is considered during the process of manfacturing.

Anyway, I have the formula, it is:

SQUARE ROOT (P+1.033 * Td+273/ Pd+ 1.033 * T+ 293)

P--- Kg/Cm2

T--- ABSOLUTE TEMPERATURE

Make sure that you are using gauge pressure transmitter.

Effect of temperature & pressure is negligible in

mass flow as it is movement of mass/time, but affects volume as density changes with pressure & temperature.

If you are unable to solve this, can you give me a clear picture or contact me at aswasidutt2 @ hotmail. com

By GP on 28 March, 2008 - 11:48 pm

I've been interested in this topic for a while as well.

Why is it you dont take into account the size of the orifice?

Also, is it possible to use this (or maybe some other that people know about) to calculate the mass flow through a pipe as it enters a main header (i.e. you have 3 x 8" pipes connected to a common 14" header and you want to know the mass flow through each of the 8" pipes)?

Glenn

By Kristian on 25 May, 2008 - 10:10 pm

Glenn, you use the orifice plate size when you calculate the flow constant. The flow (without compensation for temperature or pressure variations) is then normally calculated as

Q = C * sqrt(dp)

Where C is the flow constant that you normally find bu using an orifice plate sizing program. We use the program FlowCalc http://www.controlengineering.se/flowcalceng.htm

There are also other more accurate formulas you can use. This is described in the FlowCalc manual.

By Vinodh on 3 April, 2008 - 1:04 am

This formula may be used:

Q=Qin X Sq. root (Tr+273.14)/(Tin+273.14)X Sq root (Pin+Pabs)/(Pr+Pabs)

Q - Compensated flow

Qin - measured flow

Tr & Pr are temp & press considered during flow element design

Pabs - absolute press

Tin & Pin - measured temp & press using TT & PT in the line.

Hope that helps.

Regards,

Vinodh

By hossein on 7 July, 2010 - 2:24 pm

Dear sir

as you know we will measure flow as normal cubic meter per hour. is it possible to compensate this value by using mentioned formula?

thank you

What, exactly, is a normal cubic meter of saturated steam? Superheated steam does not exist at normal conditions!! (Normal conditions are 0 deg C temperature and 1 atm pressure.)

By hossein on 9 July, 2010 - 3:46 pm

dear sir

its true, in normal condition steam is not really exist but we consider this value only for comparison in PFD in related to other gases. any how for other gases?

is there any difference between square root in transmitter or in DCS?

thanks

Expressing flow of a gas in terms of normal cubic meters per time unit is effectively a mass flow rather than a volumetric flow. It only has meaning if the substance is a gas at normal conditions. If 1 cubic meter of the gas weighed 0.5 kg at normal conditions, then 1000 Nm3/minute is the same as 500 kg/minute for that gas.

On your 2nd question, mathematically there is no difference in having the square root operation in the transmitter instead of the DCS. However, unless the pressure and temperature compensation is also done in the transmitter you are not saving anything in terms of DCS loading - the temperature and pressure compensation should be done before the square root is extracted.

Greetings!

I have a query on pressure temp. compensation if you can help. My problem is that the vendor gave the orifice flow calculated in terms of kg/hr and not volumeric flow rate, the orifice has a DP transmitter.

I would like to know is there a way out to calculate this mass flow after pressure temperature compensation?

Or the only way is to get it calculated into volmetric flow and then calculate the compensation. It will be a big task and will take time

By bob peterson on 11 August, 2010 - 6:19 pm

I think if you look at this closer you will find that the d/p across the orifice is related to the mass of the material flowing through it and not the volume.

If you know the density (specific gravity) or specific volume of the fluid, you can convert between mass flow and volumetric flow.

Is it really required to convert it back to volumetric flow for pressure temp. conversions I think as this Mass is calculated at a fixed Temp and Press, the same can be used for pressure temp. as well without any conversion. Pl confirm.

If the mass flow has already been compensated for pressure and temperature (which is really density compensation), and you know the density or specific volume, you divide the compensated mass flow by density (or multiply by specific volume) to get actual volumetric flow.

Divide the mass flow by the density of the fluid (or multiply by the specific volume) to convert it to volumetric flow.

By jharn on 18 August, 2010 - 3:16 pm

In visual basic format:

Function StmFlow(Density, WC, PipeID, OrifID, ExpF, DisC)

StmFlow = (358.92684) * (DisC * (OrifID ^ 2) / (Sqr(1 - ((OrifID / PipeID) ^ 4)))) * (Sqr(Density)) * (Sqr(WC)) * ExpF

End Function

By Raiza Jane Isidro Casim on 7 July, 2012 - 8:54 pm

Formula for Temperature and Pressure

FIRST FORMULA:

1. The problem statement, all variables and given/known data

A quantity of gas occupies a volume of 0.5m. The pressure of the gas is 300kPa, when its temperature is 30°C. Calculate the pressure of the gas if it is compressed to half of its volume and heated to a temperature of 140°C.

2. Relevant equations

(P1 x V1)/T1 = (P2 x V2)/T2

3. The attempt at a solution

P1 = 300 kPa

V1 = 0.5m (i'm not sure what unit of volume i'm meant to be using or converting to so i'm using as it is)

T1 = 30°C (303 Kelvin)

P2 = THIS IS WHAT I NEED TO FIND OUT

V2 = 0.25m

T2 = 140°C (413 Kelvin)

I moved stuff around to make V2 after the "=" Is this right?

= (300x0.5)/303 = (P2 x 0.25m)/413

= (300x0.5)/303/0.25x413 = 817.82kPa?

SECOND FORMULA:

You need to be careful with calcs like "(300x0.5)/303/0.25" your calculator might not do them in the order you think.

It might be safer to do something like:

P1V1/T1 = P2V2/T2

so P2 = P1(V1/V2) * (T2/T1)

this also makes it obvious that units of the answer are correct.