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pinch analysis hen heat exchanger composite curve

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### #1 chemical_teo

chemical_teo

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Posted 06 November 2018 - 05:17 PM

Hi all,

Let's consider the flowsheet below.

Stream S1 is heated to 180°C and is entering reactor 1. The product leaving the reactor at 250°C is split into two streams. Stream S3 is cooled to 40°C, whereas stream S4 is cooled down to 200°C and then enters reactor 2 together with stream S5, which in turn heated from 100°C to 200°C. Finally, the outlet stream of the isothermal reactor 2 at 200°C is cooled to 80°C. No phase changes take place in the described streams so we can assume constant FCp

I need to compute and plot the hot and cold composite curves for the problem and determine the amount of cold and hot utilities. Here are data I will need:

So, to build my composite curve, I firstly took an approach temperature of 10°C (DeltaT_min). I'll put every temperature in the cold reference (ie. subtracting the hot streams temperatures by 10°C). Afterwards, I divided the problem into intervals (7 in my case) following this YouTube video . By doing so, I end up with Q1 = 12 MW (1 stands for interval 1 which is between 240-200°C). Everything is fine for all intervals except interval 2 (ie. between 200 and 190°C) where I find Q2 = 0 MW. Does it mean it is the pinch point?

Finally, I don't see how to built my composite curve from this data... Can I say if Q > 0 -> Cold composite and if Q < 0 -> Hot composite ?

NB: Below are full results of Q's for each temperature interval:

Interval 1: 240-200°C , Q1 = 12 MW

Interval 2: 200-190°C , Q2 = 0 MW

Interval 3: 190-180°C , Q3 = 1 MW

Interval 4: 180-100°C , Q4 = -8 MW

Interval 5: 100-70°C , Q5 = 6 MW

Interval 6: 70-30°C , Q6 = -2 MW

Interval 7: 30-20°C , Q5 = -2 MW

### #2 PingPong

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Posted 13 November 2018 - 10:32 AM

Everything is fine for all intervals except interval 2 (ie. between 200 and 190°C) where I find Q2 = 0 MW. Does it mean it is the pinch point?
No. To determine the pinch temperature you need to also calculate the Residual dH which you did not yet do. See the video at 6:23 onwards.

You will find however that using dT = 10 oC in your process scheme will result in a negative Hot Utility. In other words: you would not need to add heat above the pinch but remove heat, for example by generating LP steam, or making warm water, or .......

### #3 chemical_teo

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Posted 15 November 2018 - 10:35 AM

Thank you for your reply. I found a pinch point of 110 °C in the hot scale and, as you said, no hot utility is needed.. Found I need a cold utility of 5 MW. I don't know if you've done the calculations but - if so -  do you find the same?

### #4 PingPong

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Posted 15 November 2018 - 03:10 PM

For a pinch dT of 10 oC the hot/cold pinch is indeed 110 / 100 oC

It is however not clear to me how you get that 5 MW.

I estimate that the hot utility would then be -5 MW and the cold utility 2 MW, so a total of 7 MW needs to be removed (2 MW below the pinch, 5 MW above the pinch.

For a pinch dT of 22.5 oC the hot utility would be 0 MW and the cold utility 7 MW.

### #5 chemical_teo

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Posted 16 November 2018 - 11:22 AM

Basically, I found -5 MW above the pinch as you but I thought since it was negative it cannot be a hot utility but it is a cold utility "above the pinch" of 5 MW. Can't we say we have in total 7 MW of cold utilities ?

I'm asking so because the next question of this exercise was asking  to redistribute 5 MW (out of the total you found before so 7 MW I guess) to the very top of the cascade and recalculate the composite curves as well as utility requirement and pinch temperature...

Thanks!

EDIT:

Actually, I think I found the pinch with a bit of luck... I tried to do the calculations following the tutorial again. If I take the values in my initial post, here are the residual enthalpies:

Interval 1: 240-200°C , Q1 = 12 MW -> Redisual enthalpy: 12 MW

Interval 2: 200-190°C , Q2 = 0 MW -> Redisual enthalpy: 12 MW

Interval 3: 190-180°C , Q3 = 1 MW -> Redisual enthalpy: 13 MW

Interval 4: 180-100°C , Q4 = -8 MW  -> Redisual enthalpy: 5 MW

Interval 5: 100-70°C , Q5 = 6 MW  -> Redisual enthalpy: 11 MW

Interval 6: 70-30°C , Q6 = -2 MW  -> Redisual enthalpy: 9 MW

Interval 7: 30-20°C , Q5 = -2 MW  -> Redisual enthalpy: 7 MW

After, I decided to remove 5 MW but I kinda this by chance and without reasonning so much..

Giving:

12 - 5 = 7 MW
etc.
2 MW (bottom => cold utility)

But with this data, I don't see how to draw a proper composite curve (what data will be on the hot curve and what data on the cold curve?)

Edited by chemical_teo, 17 November 2018 - 06:45 AM.

### #6 PingPong

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Posted 17 November 2018 - 11:20 AM

Can't we say we have in total 7 MW of cold utilities ?
Depends on what exactly you mean by cold utility.

If that is cooling water and aircooling (which both basically throw the heat away) then it makes more sense to increase the pinch dT to 22.5 minimum because then you need much less heat exchanger area in the HEN.

If you can use the available 5 MW above the pinch as useful heat (steam generation, hot water heating, .....) then one might give it another name than cold utility.

I'm asking so because the next question of this exercise was asking  to redistribute 5 MW (out of the total you found before so 7 MW I guess) to the very top of the cascade and recalculate the composite curves as well as utility requirement and pinch temperature...
It is not clear to me what that means.

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