9 replies to this topic

[djack77494]

Am I right in thinking that this increased flow of HP steam would result in an increase in temperature on the tube side of the exchangers?

You are wise to look at one exchanger at a time, and I urge you to stay with this approach (though I detect some wavering on your part). If you look at a single S&T exchanger with HP steam on the shell, and you now increase the steam flow (by failing open the control valve), then yes, the tube temperature will increase.

I have temperatures for all four streams under normal conditions and flowrate through the tube side and exisiting exchanger duty. From this I can calculate the log mean temp difference and thus UA, assuming that UA remains constant with increased flowrate on the shell side I am then faced with the problem on finding the two output temperatures of the heat exchanger.

I'm not sure I'm following you here. I'd suggest a simple schematic with 3 exchangers labelled A, B, and C and with 3*4=12 streams shown would go far towards presenting a clear picture of your system. Don't forget to show the controls as well.

[Zauberberg]

Am I right in thinking that this increased flow of HP steam would result in an increase in temperature on the tube side of the exchangers?

This is a tricky question. Assuming flow through the tubes remains unchanged, further opening of the steam control valve (whether it is FCV or TCV) should not result in increased steam flow. Why? In my opinion, there is no increase of flow on the cold fluid side to promote increased HP steam flow (condensation).

Regarding the other part of your query, you should post a system sketch as proposed by Doug. Then it will be more easy for all of us to visualize the problem.

[Zauberberg]

If these HP steam valves are manual valves (globe/isolation), how in the world you can get into "fail-open" or "fail-closed" position? These are not control valves actuated by instrument air.

As I said in my previous post, the flow of steam is - in my opinion - determined by the flow of cold fluid (alkylate). If there is no increase of alkylate flow, there is nothing which can promote further steam condensation (= flow) at the extent higher than the one actually happening.

[Zauberberg]

This is my understanding, and I found a reference in one of N. Lieberman's books, also in the article published by PCS as well: opening the steam control valve can cause steam flow to increase only if there is a heat sink available to condense that incremental amount of steam - meaning that as long as there is a driving force (flow and temperature) of cold fluid, additional condensation will take place.

I recall similar experiences with operating steam-heated reboilers in distillation towers.

[Zauberberg]

Lieberman book is "Working guide to process equipment" while the article was "Troubleshooting steam heated exchangers"; I don't have the copy with me.

I will quote Norm's sentence from that chapter: "Because once the steam pressure in the channel head (or shell) rises to the steam header pressure, no further increase in steam flow is possible, regardless of the position of the inlet control valve." Steam pressure in exchanger is the function of condensing temperature, which is again the function of cold fluid temperature. So the flow will increase only up to that moment when the cold fluid cannot consume any more heat from the steam.

[Art Montemayor]

Rywo:

Your thinking (or description) of what is occurring in a steam-heated exchanger is skewed or flawed. And so is your schematic flow diagram.

Your schematic flow diagram shows a condensate recycle from the second heater going back to the inlet of the same heater. This is not possible without a mechanical device (a pump) to furnish the pressure driving force. (Water cannot not flow uphill by itself!)

I believe Zauberberg has hit the nail on the head when he has brought out the simple engineering concept of a required Driving Force in order to establish heat flow. The same basic logic applies to fluid flow as well. This is all in accordance with the Fourier equation for heat flow and analogous applications to fluid flow.

I believe you are missing out on analyzing the heat transfer and related fluid flow occurring in a heat exchanger where condensing steam is used as the heat medium. First, and foremost, you must establish or have a driving force (a difference in pressures in the case of fluid flow, and a difference in temperatures in the case of heat flow). Another factor that occurs in an application where the heating medium is changing phases (your specific case) is that the rate of resulting liquid phase liquid must be equal to the entry of the vapor. If that does not happen and the condensate accumulates, you essentially cut off further heat transfer (and further condensation) due to a reduced heat transfer area. The situation that exists when your supposed steam control valve fails open is that while your cool stream flow rate is constant, the amount of steam flow that enters the heat exchanger is dependent on the amount of steam condensed within the same condenser; additionally, the steam condensed rate is dependent on the fixed heat transfer area and the temperature of the cool liquid (actually, the LMTD of the exchanger). At some point, the rate of steam condensation will be maximized and no further heat increase will take place due to no further heating surface being made available. This all assumes that the required steam trap (or condensate removal) is sufficient to handle all excess steam condensate – which may not be the case. The theoretical maximum temperature that can be achieved by your cool fluid is that of the steam – but we all know that is not realistic. Increased flow of HP steam will result in an increase in temperature on the tube side of the exchanger – but it will have a theoretical maximum (which it will never achieve) of 250 oC.

Try to visualize the following:

You have a huge vessel filled with 250 oC steam. You installed a coil within this vessel in which your cool fluid flows at a constant flow rate. The temperature of the inlet cool fluid and the area of the coil are fixed – as is the flow rate of the fluid. How much steam will condense inside the vessel, assuming that the supply of steam is plentiful? It’s the same problem as your heat exchanger – except it is expressed in a different manner. Another similar application takes place when you cool a fluid in a coil submerged in a Lake of cool water kept at a constant temperature. The heat transferred due to a constant cool fluid flow is fixed by the product of U x A. this all comes from the basic equation of

Q = U A (LMTD)