15 replies to this topic

[JLMONTREAL]

HI

I AM LOST THESE DAYS, WE CANNOT FIGURE OUT WHO IS RIGHT.

AS THE GENERAL RULE, STEAM COMSUMPTION CAN BE LOWERED BY LOWERING THE OPERATION PRESSURE.

HOWEVER, THE LATENT HEAT OF LIQUID IS HIGHER AT LOWER TEMPERATURE/PRESSSURE THAN THAT AT HIGHER TEMPERATURE/PRESSURE, WHICH MEANS TO EVAPORATE THE SAME QUANTITY LIQUID, MORE STEAM IS NEEDED TO PROVIDE SUFFICIENT ENERGY AT LOWER TEMPERATURE/PRESSURE.

I CANNOT FIND THE CONSISTENCE IN TWO EXPLAINATION. CAN ANYONE HELP?

THANKS IN ADVANCE.

JOE

[JoeWong]

HOWEVER, THE LATENT HEAT OF LIQUID IS HIGHER AT LOWER TEMPERATURE/PRESSSURE THAN THAT AT HIGHER TEMPERATURE/PRESSURE, WHICH MEANS TO EVAPORATE THE SAME QUANTITY LIQUID, MORE STEAM IS NEEDED TO PROVIDE SUFFICIENT ENERGY AT LOWER TEMPERATURE/PRESSURE.

Joe,
Welcome...

I don't how you get above statement, but let see from physical characteristic of fluid at saturate condition.

Lower operating pressure would ease components leaving the liquid surface, less energy is required.

A lot of engineer has looked at latent heat per unit mass (kJ/kg). As pressure is lowered, more light end would have higher potential leaving the liquid. The energy in evaporating the light end is reduced. HOWEVER, more light end with low MW would leave the liquid. Above may leads to higher latent heat per unit mass (kJ/kg). Latent heat per unit mole (kJ/mole) would reduce.

I hope i am getting the correct track as now is late night...

Hope this help.

[JLMONTREAL]

THANKS, Joewong, I appreciate your quick response.

In fact, I got the actual experience from petrochemical plant that lowering pressure can lower steam consumption in distillation column. This can be explained as what you mentioned above.

But, now, in a column stripping water from another liquid, there is one voice that higher pressure can reduce steam consumption because the water latent heat decreases when temperature/saturation pressure increases. In PERRY's handbook, water laten heat (kj/kg) is lower at higher temperature/pressure, this means to evaporate same QTY water at higher temperature/pressure, less energy is needed, as per Q=MASS x LATENT HEAT, but this is contradictive to actual experience. Could you make more favor to explain?

Thanks.
Joe
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[JoeWong]

Latent heat of vaporisation is the energy required to overcome the molecular forces of attraction between the molecule liquid, and bring them to the vapour phase.

For your water and liquid, I don't know if there are any additional interaction and force (e.g. additional energy required to overcome the bond between water and the liquid like water-in-amine). Assuming your water and liquid (like water-in-oil) is Non miscible and non-reactive type and there are only weak interaction between them. For water-in-oil where both are get together but with NO or very low interaction force between water & oil. The water will be in aqueous phase and behave like pure water. In addition, generally water and oil would have large difference in latent heat of vaporisation.

Under pure condition, the higher the operating pressure, higher the saturation temperature. But at higher operating pressure and temperature (Sat), the internal energy is getting higher and molecular is getting "excited". So it is much easy to vaporise a mole of water at high pressure as compare to low pressure.

Hope this help.

[JLMONTREAL]

"Under pure condition, the higher the operating pressure, higher the saturation temperature. But at higher operating pressure and temperature (Sat), the internal energy is getting higher and molecular is getting "excited". So it is much easy to vaporise a mole of water at high pressure as compare to low pressure."

As the result of what you mentioned above, less energy is needed to vaporize same QTY of water at higher temperature/pressure. Applying this result in distillation, then less steam is needed under higher pressure/temperature.
This is not consistent with the common experience, i.e. lower pressure, lower steam consumption.

Where am I lost?

Joe

THANKS, Joewong, I appreciate your quick response.

In fact, I got the actual experience from petrochemical plant that lowering pressure can lower steam consumption in distillation column. This can be explained as what you mentioned above.

But, now, in a column stripping water from another liquid, there is one voice that higher pressure can reduce steam consumption because the water latent heat decreases when temperature/saturation pressure increases. In PERRY's handbook, water laten heat (kj/kg) is lower at higher temperature/pressure, this means to evaporate same QTY water at higher temperature/pressure, less energy is needed, as per Q=MASS x LATENT HEAT, but this is contradictive to actual experience. Could you make more favor to explain?

Thanks.
Joe
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Latent heat of vaporisation is the energy required to overcome the molecular forces of attraction between the molecule liquid, and bring them to the vapour phase.

For your water and liquid, I don't know if there are any additional interaction and force (e.g. additional energy required to overcome the bond between water and the liquid like water-in-amine). Assuming your water and liquid (like water-in-oil) is Non miscible and non-reactive type and there are only weak interaction between them. For water-in-oil where both are get together but with NO or very low interaction force between water & oil. The water will be in aqueous phase and behave like pure water. In addition, generally water and oil would have large difference in latent heat of vaporisation.

Under pure condition, the higher the operating pressure, higher the saturation temperature. But at higher operating pressure and temperature (Sat), the internal energy is getting higher and molecular is getting "excited". So it is much easy to vaporise a mole of water at high pressure as compare to low pressure.

Hope this help.

[Zauberberg]

Hello Joe/Joe,

Are we talking here about steam reboilers or live steam stripping?

As you increase column operating pressure, the temperature level at which heat has to be supplied to the system also increases. It means that if you use 4barg steam in tower which is operating at 1barg pressure, it is OK. But if you continue to increase the tower pressure further, you are simply reducing heat input to the tower because driving force is reduced (remember U*A*LMTD equation) - speaking of reboiled towers. Boiling temperature of water goes up as pressure is increased, right?

For live-steam stripping applications, the reduction in latent heat of water evaporation is accompanied by reduced latent heat of hot steam condensation. So where is the gain?
In your stripper tower, pressure increase leads to lower energy consumption - energy required to evaporate 1kg of water decreases. But, at the same time, relative volatility decreases, and solubility equilibrium point of component(s) that has to be removed from water is shifted towards higher concentrations of these components in water at elevated pressures (form of Henry's law). In other words, it is easier to strip-out H2S from waste water at lower pressure, if quality of steam which is used for that particular purpose is the same in both cases.

Operate the tower at lowest possible pressure at which you can achieve required condensing duty. That's the basic rule of distillation and it will not change as long as this world exists. Lower the column operating pressure and you will need less steam to achive the same process (quality) targets.

[JLMONTREAL]

Zauberberg, I am lost again. please see my confusion in red below. I completely agree lower pressure can save energy, but I need a strong explanation.

Thanks.
Joe

Hello Joe/Joe,

Are we talking here about steam reboilers or live steam stripping?

As you increase column operating pressure, the temperature level at which heat has to be supplied to the system also increases. It means that if you use 4barg steam in tower which is operating at 1barg pressure, it is OK. But if you continue to increase the tower pressure further, you are simply reducing heat input to the tower because driving force is reduced (remember U*A*LMTD equation) - speaking of reboiled towers. Boiling temperature of water goes up as pressure is increased, right? It is true that heat input is reduced as pressure increases, but higher pressure reduces energy need, so on other hand, this point may demonstrate that higher pressure is exactly saving steam because lower latent heat and lower energy need.

For live-steam stripping applications, the reduction in latent heat of water evaporation is accompanied by reduced latent heat of hot steam condensation. So where is the gain?
In your stripper tower, pressure increase leads to lower energy consumption - energy required to evaporate 1kg of water decreases. But, at the same time, relative volatility decreases, and solubility equilibrium point of component(s) that has to be removed from water is shifted towards higher concentrations of these components in water at elevated pressures (form of Henry's law). In other words, it is easier to strip-out H2S from waste water at lower pressure, if quality of steam which is used for that particular purpose is the same in both cases.

Operate the tower at lowest possible pressure at which you can achieve required condensing duty. That's the basic rule of distillation and it will not change as long as this world exists. Lower the column operating pressure and you will need less steam to achive the same process (quality) targets.

[Zauberberg]

Hi Joe,

I'm still holding my point that mixing of apples and oranges is the case here.

Here is your question: "It is true that heat input is reduced as pressure increases, but higher pressure reduces energy need, so on other hand, this point may demonstrate that higher pressure is exactly saving steam because lower latent heat and lower energy need."

How can higher pressure reduce energy consumption? Look at one simple example:

1. You are stripping H2S from waste water, by using saturated steam at 150C (and corresponding pressure 4.8 bar abs). Your tower operates at 1 bar abs and steam reboiler is used for this application. Since your reboiler surface area is fixed, heat input is governed by Q = UxAxLMTDxF. LMTD is determined by condensing temperature of reboiler steam and boiling point of water inside the tower = 100C.

2. Now, you increase your tower operating pressure up to 3 bar abs. What is your boiling point of water inside the tower? It is 133.5C and reboiler steam condensing temperature is still 150C. How much the heat input will be? You certainly agree it will be less than in the previous case, simply by following the same equation Q = UxAxLMTDxF.

3. With this, new heat input (LOWER heat input) you cannot maintain the quality of tower bottoms product - stripped water. Why? It is because the solubility of H2S in water increases as the pressure is increased (Henry's law), and you can open steam control valve as much as you want but there will be not more heat flowing into the tower. And your product is not anymore on spec, because you actually decreased heat input BUT not the heat demand for this service. And this lower heat input doesn't give you anything: your steam consumption is the same, but your product is not on spec.

Let us put things in this way: if you are asking if less heat is required to boil 1kg of water at higher pressure, the answer is certainly - yes. But, in the real world of distillation towers, it also means that temperature level of heat medium has to increase in order to maintain product quality and required heat input (it is more difficult to strip-out lighter components from bottoms at higher pressures). Increasing the temperature level of heating medium is energy consumption process - do you agree? You need more energy to make 8 bar steam than 5 bar steam, from the same feed water parameters.

If you ever visit some high mountain, try to make a tea. And you will see that water will boil more faster than on the Miami beach or in some high-pressure chamber. And you will need much less woods in the mountain than on the beach, in spite of higher vaporization enthalpy at lower pressures. Now, saving the woods - we can say that it is saving the energy, is it?

Best regards,

[JLMONTREAL]

Hi,Zauberberg
Sorry for my late response.
Qualitatively speaking, I agree with you. Do you mind to provide equations which may quantitatively specify the energy consumption differernce when lowering operation pressure in an exiting distillation column?

Thanks in advance.
Joe

[Zauberberg]

Hi Joe,

I already gave you essentials that can be applied in your and any other system. However, if you are interested in specific service, I need the following:

1. Please tell us finally what is your application? Wastewater stripper, stripped reboiler,...?
2. Feed composition and flowrate
3. Product specs
4. Tower configuration and operating parameters - available condenser and reboiler duties, pumparounds, etc.
5. Overview of initial and desired parameters/operating conditions.

I am convinced that misunderstanding here is a consequence of not knowing basic data about specific service which is under observation, and our discussion is constantly shifting between academical and practical areas/levels. I am sure that everything is easy to be explained, and that the answers you received above from all of us - including your own conclusions and observations - contain everything that you need from the beginning.

Greetings,

[JLMONTREAL]

Hi Zauberberg,

Thank you for your reply. Unfortunately I cannot provide detail information due to confidential info included.

This topic is regarding common experience about distillation, I wanted to clarify some discussion with my workmates. Your input is good enough. I appreciate your information and your time, hope you may provide more info for me for other issues.

Thanks once again.

Joe

[Ali66]

Joe:

Back to your first post.

Yes, lower pressure results in lower energy use in spite of the fact that you see higher latent heat of vaporization at the lower pressure.

Work out this example and it should help.

Start with liquid water, say 1 pound, at 60 F

Case 1 evaporation at 1 atm (absolute). This means 1) heating 1 lb from 60 F to 212 F then 2) provide the latent heat at 212 F. Add these two quantities and call the total Q1

Case 2 evaporation at lower pressure, say 0.2 atm, This requires 1) heating 1 lb from 60 F to 140 F plus 2) heat of evaporation at 140 F. Call the total Q2.

You'll find that Q1 is larger than Q2 even though the latent heat in the second case is larger than in the first. The confusion, I think, was all because of ignoring the contribution of sensible heat.

Ali

[JLMONTREAL]

HI,Ali

You are right. Now the answer is more and more clear. We can explain in two ways: volatility and heat balance. Both lead to consistent answers.

Thanks.
Joe

[Bennett Willis]

I'm an industry retiree (33 years as a chemist doing process research and plant support) and am training operators at a community college. We are having a discussion about whether running a standard distillation (as in separating a feed into overhead and bottom streams of different compositions) would take a lot less steam (25-50% less) if you ran it under a vacuum--and thus the energy savings would motivate you to run something under a vacuum.

This logic is not consistent with my experience. What I saw was that you ran the pressure on distillations to meet the needs of the overheads (condensing temperature/energy recovery) and the bottoms (product decomposition and steam pressure available). Vacuum separations seemed to require larger columns due to the higher vapor velocity caused by the much larger volumes of vapor due to the low pressures.

From my chemist's point of view, it should take about the same amount of energy (pounds of steam or BTU's) to do the same job (roughly) regardless of the pressure you are doing it.

My position is "strongly held but poorly documented." However, experience has taught me to hold all opinions gently. Could some good engineer offer some intuition to a chemist?

Thanks,
Bennett Willis
Brazosport College

[joerd]

My two cents:
Distillation is all about relative volatility. Most of the time, relative volatility increases as you decrease the pressure, so you would need less trays, less reflux, less energy for the same separation if you perform it at lower pressure. This would also decrease the vapor / liquid traffic (on a molar basis) in the column.
However, the price you pay is a lower overhead condensing temperature, which in some cases forces you into refrigeration, for example (= more energy, not less). Also, as you rightfully said, a lower pressure increases the vapor volume, and could lead to a bigger diameter column. Lastly, running a vacuum system has its own problems, energy, and cost associated, as well as the potential for air ingress.
So there is never an easy answer, one would weigh the different effects of pressure and try to come up with the optimum, based on available heating and cooling media, overall cost, decomposition temperature, separation efficiency at higher pressure, etc.

By the way, welcome to the forum. Next time, it probably would be better to start a new topic rather than tacking on to an existing discussion.

[Bennett Willis]

I was thinking about this last night (putting it up as a new topic) and will do so in a few minutes. I hope that this is not a violation of site policies.

I would really like some discussion as I have been telling my classes that it takes roughly the same amount of energy to do the same job "regardless of the path". This puts me in conflict with what they have been told in other classes. I don't feel too bad about the conflict because none of us know about how the energy required changes with pressure. I suppose that I feel that I am closer to right than the other instructors because vacuum columns were really rare where I worked except for issues related to bottom temperatures. This should not have been the case if the amount of energy drops significantly when you reduce the pressure.

Bennett Willis