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Duty Calculation


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#1 mosa

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Posted 07 September 2008 - 10:35 AM

Dear guys,

Please help me in understanding the basics of heat duty calculations.

I have a volumetric flowrate for a liquid at normal flow conditions and an operating temperature of inlet (say 150 deg C) and outlet (at 250 oC) for the exchanger. Please clarify for me how to calculate the exchanger duty. At what density is the liquid mass flow rate calculated at? Is it at normal conditions, standard conditions, or at the operating temperature? Similarly, how is the Cp value employed?

Please clarify.

regards
momsal

#2 Art Montemayor

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Posted 07 September 2008 - 12:46 PM

momsal:

You will note that I edited your original post because it was very hard to read. I believe I know the concern you have in approaching this basic, first-year chemical engineering problem. I hope this is your first year in Ch. E., because the nature of this problem is very basic and must be fully and completely understood to proceed further into engineering with success. That is why I am spending time on this post. I regard this as very important for all Ch. E. students and recognize your concern.

The basis for your calculation is the transfer of sensible heat (no latent heat) and, as such, requires the use of the simple sensible heat transfer equation:

Q = W Cp (T2 –T1)

Where,
Q = the total sensible heat transferred, btu/hr;
W = the mass flow rate of the liquid undergoing the temperature change, lb/hr;
Cp = the Heat Capacity of the fluid undergoing the temperature change, btu/lb-oF;
(
T2 – T1) = the temperature change of the fluid (250 – 150 = 100 oC, in your case), oF

Note that the fluid quantity must be in MASS units in order to apply to the Heat Capacity value. That means that you must convert your volumetric flow rate to mass flow rate. Mass flow rate doesn't change; volumetric flow rate does. Therefore, you must apply the liquid density that applies to the liquid temperature corresponding to the volumetric flow rate given. That is merely practical engineering sense and called "engineering judgment".

The value of the Heat Capacity to employ in the equation is another thing. You have two ways you can go about this:

1) In the real world, we always calculate the maximum duty expected and allow for that. If this were a practical, real problem, you could apply the highest value of the Cp over the range 250 – 150 oC and be assured that the highest heat duty is being taken care of. Or,

2) You can take a numerical average temperature over the range (200 oC) and apply the Cp at that value as a representative of the average value. If your Cp does not vary too much, your answer is sensible. If your Cp varies quite a lot over the range, then you might be short of calculated area.

For your peace of mind, I've done it both ways in the real world and have found that the calculated required area for the necessary heat exchanger has almost always come out OK when using the average value. This is probably because of various practical reasons:

1) The values of the Cp are not that accurate;
2) The contingencies usually applied to engineering calculations take care of any differences in the Cp value used;
3) The other contingencies and allowances for inaccuracies also contribute to a conservative answer – such as the allowance for fouling factor, LMTD multiple pass factors, contingency for internal baffle leakages and by-passes, etc., etc.

The other method employed to obtain the exchanger heat load is:

Q = W (h2 – h1)

Where,
(h2 – h1) = the enthalpy change of the fluid undergoing temperature change, btu/lb.

This latter value is what is usually obtained by using a simulation program in today's engineering design ambient.

I don't know what you mean by "normal" or "standard" conditions. The important thing to always bear in mind is that you are applying common, engineering judgment in applying the equation(s) and that the values employed correspond to what is actually expected to take place in the real world.

I hope this helps you out in further understanding how to do a heat balance.

#3 mosa

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Posted 08 September 2008 - 08:25 AM

Dear Art,

Thank you very much for a good explanation.As you know in text books and in the in the any heat exchabger related website, this volumetric flow relation is not expressed.I am clear with the concept of Cp.However, to complete my understanding, you mean to say that, if any volumetric flow rate say 100m3/hr of liquid is flowing through exchanger, any data sheet or design detial , will indicate the at that temeprature ? i came to know that in the chemical industry, there will be a flow indication say hydrocarbon flowrate of density 800kg/m3( at 25 deg C) indicating 100m3/hr in DCS and enters through an exchnager at temperature 250 deg C cooled to 150 deg C .

Now,to calculate duty.Should i multilpy 100m3/hr with 800kg/m3 to get mass OR should i take density of HC at 250deg C and then multpily with 100m3/hr to convert to mass.No problem of understanding Cp.

Please clarify this doubt because point of doubt at this stage will always persist till my carrer if not cleared although it is silly.I hope you will understand
and also please tell me any references relevant to the above query.

Regards
momsal

#4 djack77494

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Posted 09 September 2008 - 12:59 PM

QUOTE (mosa @ Sep 8 2008, 04:25 AM) <{POST_SNAPBACK}>
this volumetric flow relation is not expressed.


mosa,
The better way of speaking about heat transfer and associated equipment is to think in terms of mass flow. As Art has indicated, mass flows are clear and unchanging. You should always be in a position to convert your volumetric flowrate into a mass flowrate. If you cannot do that, then there is a serious problem. Having the mass flowrate is convenient since your mass and energy balances are referenced to mass flows. Thus, your heat capacity, Cp, is expressed in units of energy/mass/degree; it is never expressed in volume units. As I have seen again and again, the use of volumetric units comes at the cost of great confusion. Volumes change as temperatures, pressures, etc. change. The use of "normal" and "standard" is far less normal and standard that we might be led to believe. Your normal probably differs from my normal. Ditto for standard. Despite this, it is common for writers to assume that their readers will understand what is meant by their normal or standard. So my recommendation is that IF you are starting with volumetric flowrates, step one should be to convert those flowrates into mass flowrates. From there, everything is much clearer.

#5 mosa

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Posted 09 September 2008 - 01:25 PM

Dear Jack,

There is no contest regarding mass flow as formulae itself carried mass as variable.But my question is, how to get the mass flow ?Please look into my question again.

Regards
momsal

#6 Zauberberg

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Posted 10 September 2008 - 07:42 AM

Momsal,

You should multiply standard flowrate with a standard density of process fluid, regardless of the sampling point (150 or 250C). Both these parameters - flow and density - should refer to the same standard conditions.

Flow measurement device which is used in your plant must have its own dedicated datasheet. There you can read which conversion factors are applied for conversion of actual volumetric flow to standard volumetric flow. But as long as you are sure that you have standard volumetric flow as DCS output of your measurement device, no need to be concerned if the measurement is done at 150C, 250C, or any other temperature.

#7 mosa

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Posted 11 September 2008 - 03:58 AM

Dear Zauberg

Thank you very much and now i am clear

Regards
momsal




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