5 replies to this topic

[Daniel James]

Hi all,

 

     Currently, we have an improvement initiative to substitute the heating medium for an existing heat exchanger from Medium Pressure steam (15 barg) to Low pressure steam (7.5 barg). Obviously with lower pressure steam, the saturation temperature is lower too. Is there a way to calculate the time required for the different steam to heat the process fluid to the target temperature? I would like to know how much longer it takes for the low pressure steam to heat the process fluid.

 

 

Note: This is a continuous flowing process, not batch

Edited by Daniel James, 13 June 2016 - 10:51 PM.

[elstasso]

everything you know about thermodynamics en heat/energy calculations but do it with mass flow instead of massand the flow will give you the time abscis

[latexman]

A P&ID and more details of the process description would help us help you.

[Daniel James]

A P&ID and more details of the process description would help us help you.

 

Please see this simplified drawing

 

 

 

Currently, the heat exchanger is controlled by varying the level of condensate in the drum thereby changing the number of tubes exposed to the steam for condensing which changes the heating duty.

 

We want to substitute the 16 barg steam with 8barg lower pressure steam. My plant operator is concerned that it would take much longer to heat up the process fluid to operating temperature during startup.

 

From my understanding, as long as the current heat exchanger has sufficient surface area, it will consume the required amount of steam to bring the process fluid temperature to operating temperature. Am I wrong to assume that?

[latexman]

If the HX is too big (excess tubes flooded) with 16 barg steam and not too small with 8 barg steam, you are correct.  That's about a 30^o C drop in approach.  Why would operations heat up the batch slower than the equipment is capable?  Is a different, parallel step the rate determining step?

[S.AHMAD]

Hi Dan

The LMTD will change. Let assume that the existing duty is Q₁ that is:

Q₁ = UA₁LMTD₁

The new duty  Q₂ is given by:

Q₂ = UA₂LMTD₂

 

U is about the same for the two cases.. Supposing that yo want to maintain the same time required then Q₁ = Q₂ that means you must adjust the condensate level such that it corresponds to surface area of:

A₂ = A₁.LMTD₁/LMTD₂ You can also determine the time by using the LMTD ratio.if you are maintaining the existing condensate level.

 

Hope the above short comments triggered your lateral  thinking. Good luck. 

 

S.Ahmad