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Heat Absorbed Of Otsg


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#1 ahyong

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Posted 13 February 2014 - 09:51 AM

Dear All,

I am in the middle of calculating the heat absorbed in order to vaporise incoming BFW into 80% steam.

Inlet condition:

-108°C

-52056kg/hr

 

Outlet condition:

-331.7°C

-80% steam

At 133kg/cm2

 

Required Heat Absorbed = mcdT + mHv

 

The total heat duty for the OTSG is 100MMBTU/hr and my findings regarding required heat absorbed is around 90MMBtu/hr (with specific heat capacity: 4.18J/g.K and Hv: around 1120KJ/Kg). However, the calculation I received from 2 vendors is around 97++MMBtu/hr. I've been trying with trial and error with vendors' data and realising that my data could be wrong. May I know:

1. What is the value of specific heat capacity at 133kg/cm2 (is it around 7.++)??

2. I think my value for latent heat of vaporization could be wrong too. Please let me know.

3. As per my research these few days, the properties of water and steam vary greatly with the changes of pressure. Where can I get the full table regarding properties of steam and water?

 

Thank you.



#2 PaoloPemi

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Posted 13 February 2014 - 10:44 AM

you can use classical tables or some model based for example on IAPWS 95

there are online resources as NIST site

 

http://webbook.nist....hemistry/fluid/

 

also there are free resources as

 

http://www.prode.com/en/properties.htm

 

at p = 133 Kgcm2 t = 331.715 C (saturation point)

 

cpl ~ 7.32 Kj/Kg

cpg ~ 10.09 Kj/Kg

 

calculated with IAPWS 95

 

by the way to calculate dH you can calculate

dH / Kg and then multiply by mass flow,

 

 in your example I do not see how pin is defined


Edited by PaoloPemi, 13 February 2014 - 10:53 AM.


#3 PingPong

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Posted 13 February 2014 - 02:03 PM


at p = 133 Kgcm2 t = 331.715 C (saturation point)

 

cpl ~ 7.32 Kj/Kg

cpg ~ 10.09 Kj/Kg

This does not seem correct.

 

To calculate this kind of problems always work with enthalpy data (Hv and Hl) from a steam table, not with specific heats (cp).

 


I am in the middle of calculating the heat absorbed in order to vaporise incoming BFW into 80% steam.

Inlet condition:

-108°C

-52056kg/hr

 

Outlet condition:

-331.7°C

-80% steam

At 133kg/cm2

With these data, the absorbed duty will be 28.65 MW or 97.8 MM BTU/hr.

#4 ahyong

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Posted 13 February 2014 - 06:09 PM

 


at p = 133 Kgcm2 t = 331.715 C (saturation point)

 

cpl ~ 7.32 Kj/Kg

cpg ~ 10.09 Kj/Kg

This does not seem correct.

 

To calculate this kind of problems always work with enthalpy data (Hv and Hl) from a steam table, not with specific heats (cp).

 

 


I am in the middle of calculating the heat absorbed in order to vaporise incoming BFW into 80% steam.

Inlet condition:

-108°C

-52056kg/hr

 

Outlet condition:

-331.7°C

-80% steam

At 133kg/cm2

With these data, the absorbed duty will be 28.65 MW or 97.8 MM BTU/hr.

 

Pingpong,

Can you please show your calculation here? I need to check what did I do wrong with my calculation by comparing with yours.
I also think that 97.++ should be the correct one. It would be better if you did it by hand calculation. Simulation program makes engineering become dul (my personal opinion)

Thanks a lot!!!



#5 PaoloPemi

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Posted 14 February 2014 - 02:27 AM

Ping Pong wrote

>at p = 133 Kgcm2 t = 331.715 C (saturation point)

>cpl ~ 7.32 Kj/Kg

>cpg ~ 10.09 Kj/Kg

>This does not seem correct

out of curiosity, which values  have you calculated ?

these are the values calculated at 133 Kg/cm2 for saturated liquid and vapor phase with IAPWS 95

please note that unit is Kj/Kg,

kindly review your data, you may go to NIST site or use a tool

as PRODE PROPERTIES to verify your values,
 

 

ahyong and ping pong,

as said in my previous post, you need to specify initial and final conditions to calculate dH,

in your example Pinitial has not been specified,

assuming

initial conditions Pin 133 Kg/cm2 and 108 C

final conditions  Pout 133 Kg/cm2 and 331.715 C  (saturation point)

dH for liquid is 1474.15 Kj/Kg

dH for vapor is 2596.15 Kj/Kg

of course the difference being the latent heat

you do not mention the units (volumetric fraction. mass fraction etc.) of vapor %

so I leave to you the final step

Paolo



#6 breizh

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Posted 14 February 2014 - 02:55 AM

http://www.spiraxsar...s/wet-steam.asp

 

This link is a good one for such calculation. 

=(2434070-452905)*52056/3600/1000000= 28.64 MW , similar to ping Pong's result

    ( H vap out  -H vap  in ) J/Kg

 

Breizh


Edited by breizh, 14 February 2014 - 03:44 AM.


#7 PingPong

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Posted 14 February 2014 - 04:33 AM


Pingpong,

Can you please show your calculation here? I need to check what did I do wrong with my calculation by comparing with yours.
I also think that 97.++ should be the correct one. It would be better if you did it by hand calculation.

I did it by hand, using enthalpy data from a steam table.

Simulation program makes engineering become dul (my personal opinion)
Exactly! That is why you should also do it by hand, using enthalpy values from a steam table. If you do not find the right duty, you can post your calculation here and we will point out where you went wrong.

#8 ahyong

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Posted 14 February 2014 - 09:51 AM

http://www.spiraxsar...s/wet-steam.asp

 

This link is a good one for such calculation. 

=(2434070-452905)*52056/3600/1000000= 28.64 MW , similar to ping Pong's result

    ( H vap out  -H vap  in ) J/Kg

 

Breizh

 

 

 


Pingpong,

Can you please show your calculation here? I need to check what did I do wrong with my calculation by comparing with yours.
I also think that 97.++ should be the correct one. It would be better if you did it by hand calculation.

I did it by hand, using enthalpy data from a steam table.

 

Simulation program makes engineering become dul (my personal opinion)
Exactly! That is why you should also do it by hand, using enthalpy values from a steam table. If you do not find the right duty, you can post your calculation here and we will point out where you went wrong.

 

Breizh,

Could you please explain more regarding your calculation method? I do not understand how did you get the "452905" and why is it "Hv out - Hv in". Dont we have to consider also when the fluid is at total liquid form? (before phase changing started)

 

PingPong,

I use the formula of Q=mcdT + mHv*0.8

m= 52056

c= 4.2 J/g.K (I think this is wrong)

dT= 223.7K

Hv= 1122 KJ/Kg

First, I calculate the total heat absorbed when it is in liquid form (vapor fraction: 0)--> mcdT.

Next at saturation temp (331.7°C), when phase changing happened where 80% steam generated--> mHv*0.8

Finally I get 90++MMBtu/hr. If I use c=7.3J/g.K (i think this is the correct cp at 133 kg/cm2), the total heat absorbed will be over 100MMBtu/hr.

Please let me know what's wrong with my understanding and calculation.



#9 PaoloPemi

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Posted 14 February 2014 - 11:07 AM

ahyong,

starting from 108 C 133 Kg/cm2

total dH is different from that calculated by ping pong and breizh,

the difference is more than 20 % which is the reason why I did

ask ping pong to explain the results,

(see my previous posts)

your procedure is correct but the values not, note that liquid cp is not constant

so you should estimate an average value, it would be better to

integrate (i.e. calculate enthalpy at initial point and final point)

to improve the accuracy,

once you know dH (see my previous post)

it is easy to calculate value for vapor + liquid

dH = 0.2 * dH(liquid) + 0.8 * dH(vapor)

from that and mass flow you get the duty

Paolo


Edited by PaoloPemi, 14 February 2014 - 11:10 AM.


#10 PingPong

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Posted 14 February 2014 - 11:33 AM


total dH is different from that calculated by ping pong and breizh,

the difference is more than 20 % which is the reason why I did

ask ping pong to explain the results,

(see my previous posts)

Paolo, I am sorry to say, but all your numbers so far are incorrect.

 

If I use c=7.3J/g.K (i think this is the correct cp at 133 kg/cm2), the total heat absorbed will be over 100MMBtu/hr.

Please let me know what's wrong with my understanding and calculation.

It has been said already several times, but I will repeat it once more: you should use ENTHALPY values.

That is what I did, and that is also what breizh did, so obviously we both got the same (correct) duty.

The duty is the enthalpy difference between outlet and inlet of the OTSG.

1) take a steam table and look for the enthalpy (kJ/kg) of the water and of the steam at the outlet temperature and pressure,

2) based on the given % steam at outlet, calculate the enthalpy (kJ/kg) of the water/steam mixture at the outlet,

3) look in the steam table for the enthalpy (kJ/kg) of the water at the inlet conditions,

4) calculate the enthalpy difference (kJ/kg) between 2) and 3),

5) multiply the enthalpy difference (kJ/kg) with the flowrate (kg/s) through the OTGS to obtain the duty (kW).



#11 PaoloPemi

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Posted 14 February 2014 - 02:32 PM

ahyong,

the procedure you follow is correct but you have to use average values for cp (liquid phase),

basically you have two steps,

heat the water from 108 C to 331.715 C at 133 Kg/cm2 (isobar)

the heat required can be calculated as cpm * (331.715-108)

note that there are large variations in value of cp going from 108 C (4.195 Kj/Kg) to 331 C (7.32 Kj/Kg)

the usual way to solve this problem is to use values of enthalpy instead,

however you can easily calculate an average value as

cpm = [H(331 C) - H(108 C)] / (331 C - 108 C) = 4.80 Kj/kg

in the second step you calculate the contribute of change of state,
we know the value of latent heat and with those values you can proceed with your method

Q=mcdT + mHv*0.8

Q = 52056*4.8*(331-108)+52056*1122*0.8

if you divide by 3600

Q= 28457

I must apologize for the error in post #5 where I utilized a wrong initial value for T
while values in post 1 are correct



#12 PingPong

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Posted 14 February 2014 - 03:18 PM

however you can easily calculate an average value as

cpm = [H(331 C) - H(108 C)] / (331 C - 108 C) = 4.80 Kj/kg

What is the point of calculating a non existing "average" cp from an enthalpy difference if that "average" cp is then used to calculate that same enthalpy difference? Note also that cp is in kJ/(kg.K).

 

It is sad that an in essence very simple calculation has become such a problem, only because it is not done systematically using enthalpies.


Edited by PingPong, 14 February 2014 - 03:18 PM.


#13 PaoloPemi

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Posted 15 February 2014 - 03:14 AM

in my opinion there is nothing wrong about a simplified model based on constant cp

and once you use the correct values results are the same

which is what ahyong did ask.

By the way, in these days I have not much time to access the forum

and in most cases no time to review the answer,

personally, I do allow everyone here to make some little mistake

(after all it's a free service) :-)

"Why do you see the speck in your brother’s eye, but fail to see the beam..."
 



#14 PingPong

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Posted 15 February 2014 - 06:27 AM

... there is nothing wrong about a simplified model based on constant cp and once you use the correct values results are the same ...

Maybe you missed my point: the only way to determine the "correct value" of that "average" cp is from the enthalpy difference between inlet and outlet. But the only reason that one needs that cp is to calculate that enthalpy difference, so if one already knows that enthalpy difference there is no need for the "average" cp anymore. There is no way that one could have determined the "correct" cp of 4.80 kJ/kg.K simply from cp value(s) at whatever temperature(s).

 

The methodology of using an "average" cp is flawed for this kind of calculations. That's why ahyong originally found a 10 % too low duty, and you originally found a 20 % difference with the duty calculated by me and breizh based on enthalpies. Based on cp's I could not have done it accurately either.

 

 

The generated steam will probably also be superheated in another OTSG coil. Let's assume that it is superheated to 540 oC at constant pressure (ignore coil pressure drop). Now I challenge everybody to calculate the duty of the superheat coil based on an "average" steam cp value that is obtained from whatever steam cp value(s) you wish to consider for averaging (but not calculated back from the enthalpy difference between 540 and 331 oC).


Edited by PingPong, 15 February 2014 - 06:41 AM.


#15 ahyong

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Posted 15 February 2014 - 08:27 AM

Guys, thanks a lot for the constructive opinion and guidance provided. I guessed I was too "obsessed" with the calculation provided by my boss and the terms of latent heat of vaporization that I learnt in school. I am not sure which one is the correct method in engineering and there is no right or wrong in engineering as long as in the end we got the CORRECT answer. Afterall there are so many practises in engineering and different people have different opinion. Its good to have argument because we all learn from argument and through experience. Cheers!!

Thank you



#16 PaoloPemi

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Posted 15 February 2014 - 08:53 AM

ahyong,

as far as you have been able to find the right solution I think the discussion has been useful.

 

ping pong,
no one discusses the accuracy of integration (H,S) over a range of values,
and it is very easy to do that with help of tables or a software
(mine has IAPWS 1995 integrated so no difficult at all,
wish I have used it to avoid any discussion)
however the request from ahyong was to explain the wrong results obtained by his model
and that is the reason why I calculated an average cp showing how it can
obtain equivalent results,
a simple linear model (as that based on constant cp) is calculated (from a rigorous model, as for example IAPWS 1995) exactly in that way, accuracy on intermediate points is limited due to high non linearity of H vs. T,P
in water, however, if you wish, you can add terms (second, third) to improve accuracy,
although I do not consider this procedure competitive in a range T,P with the approach based on Helmhotz energy (IAPWS 1995) or the simplified model for industrial application (97) it could be much more quick to solve and result interesting for applications where one expects limited variations of T,P,

Paolo






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