# Power needed for saturated steam compression...

1. Aug 21, 2016

### pranj5

Want to discuss a matter here. If steam remains saturated during compression, how to calculate the power consumption for that by using steam table.
Suppose, saturated steam has been compressed from pressure level A to pressure level B and it remained saturated during and after compression. We have steam table before us and we know the gross enthalpy of steam a level A and level B.
Now, to calculate the power consumption with 100% efficiency, do we deduct gross enthalpy of steam at level A from gross enthalpy of steam at level B or there is some other method to be applied here?

2. Aug 21, 2016

### Staff: Mentor

Some clarification questions:

Are we talking about a continuous flow compressor, with steam entering one end and leaving through the other, or are we talking about compression in a piston-cylinder (closed system) arrangement?

Are we talking about a mixture of saturated steam and liquid water, or are we talking about pure saturated steam at pressure levels A and B?

Chet

3. Aug 21, 2016

### pranj5

This is pure saturated steam and it's a continuous process. Saturated steam enters the system and being compressed and leaves at higher pressure. But, by some special means, the steam remains saturated during the process and also leaves as saturated steam.

4. Aug 22, 2016

### Staff: Mentor

The power required is going to depend on the specific details of your proprietary process. It is not just the change in enthalpy. For example, to maintain the steam 100% quality throughout the compressor, the only way I can think of for accomplishing this is to remove heat from the system in a very precise way along the length of the compressor (so that the steam does not become superheated). For such a compressor, the open system version of the first law of thermodynamics would reduce to: $$\dot{Q}-\dot{W_S}+\dot{m}(h_A-h_B)=0$$where $\dot{Q}$ is the rate of adding heat, $\dot{W_S}$ is the rate at which the system is doing shaft work on the surroundings, $\dot{m}$ is the mass rate of flow of steam through the system, and h is the enthalpy per unit mass of the steam. The rate of doing shaft work on the surroundings is related to the power P by:$$P=-\dot{W_S}$$. Therefore,$$P=-\dot{Q}+\dot{m}(h_B-h_A)$$ So the power will be equal to the mass flow rate times the change in enthalpy per unit mass minus the rate at which heat is being added to the system. But, since we will be removing heat from the system, $\dot{Q}$ will be negative, and the power requirement will be higher than just that associated with the enthalpy change.

If you would like me to continue this analysis to show how the steam tables can be used to precisely quantify the power requirements for this particular scheme, please let me know.

5. Aug 22, 2016

### pranj5

Well, I like to explain your maths above in this way.
If the systems will be a 100% efficient, then Q will be zero here. But, like all man made machinery, this system too is inefficient upto a degree and Q is the extra amount of work that will be done by compressor other than increasing the enthalpy hB to hA. This additional amount of heat has to be removed from the system to keep the steam saturated. Am I right?
At present, many steam compressor manufacturers try to control temperature by spraying water and I have already mentioned that the steam will be in contact with water during compression. Therefore, can we conclude that this extra amount of heat will be converted into some extra steam?

6. Aug 22, 2016

### Staff: Mentor

The part about removing heat to keep the steam saturated is correct. The rest is mostly not correct. It is still possible to operate the compressor reversibly, even if there is heat removed or added. The reversible operation assumes that there is no dissipation of mechanical energy or thermal energy (i.e., no inefficiency of this kind)

I don't quite understand this question. Are you saying that you are considering spraying liquid water into the compressor along the length of the compressor in order to make sure that the flowing steam remains saturated at 100% quality? Are you saying that you would like me to analyze such a process quantitatively?

7. Aug 22, 2016

### pranj5

If there is no inefficiency of any kind, then where this extra heat will do? As far as I can understand the process, for a 100% efficient system all the work will be added to the steam and increase its enthalpy. But, unlike other compressors where steam is superheated from the very beginning and behave as an ideal gas, in this process the steam will remain saturated and at the end, the work done can be calculated by subtracting the gross enthalpy of State B from State A. Here in the process, the steam compression always follow the saturation curve.
Kindly just tell me what the extra heat will do in such a scenario. What I want to know is nothing but the power consumption for steam compression if the steam remains saturated during the compression. I am not saying that I am spraying water to keep the steam saturated, but just want to say that the steam will be in contact with water during compression and therefore remain saturated. How? That's a different question. Just keep it in mind that the steam is saturated during the process.

8. Aug 22, 2016

### Staff: Mentor

So you're saying that you do have a saturated mixture of steam and liquid water that you are processing in the compressor? Is that correct?

9. Aug 22, 2016

### pranj5

Yes. You can say that. But, it's not just wet steam. The amount of water here is much more than the amount of steam. I hope you have remembered that in such a scenario, a part of steam will dissolve in water and raise its enthalpy and the steam will always remain saturated.

10. Aug 22, 2016

### Staff: Mentor

So it's mostly liquid water with a smaller fraction of steam (by mass, not volume)?

11. Aug 22, 2016

### pranj5

Yes, you can say that.

12. Aug 22, 2016

### Staff: Mentor

So, what are the starting and ending conditions of the process?

13. Aug 22, 2016

### pranj5

That has already been stated at the starting of this thread. There is a specific volume of saturated steam here that has been compressed from level B to level A and before, during and after the compression; the steam remained saturated and not superheated.

14. Aug 22, 2016

### Staff: Mentor

Problem Statement:

There is a saturated mixture of steam and liquid water being processed through a continuous adiabatic reversible compressor at $\dot{m}$ kg/sec. The inlet mixture (State A) is at pressure $p_A$ and water vapor mass fraction $x_A$. The mixture at the outlet (State B) is saturated at pressure $p_B$. The outlet mass fraction of water vapor $x_B$ is, as yet, unknown. The objective is to determine the mechanical power required to run the compressor. Since the compressor is operating adiabatically and reversibly, the entropy per unit mass of the outlet stream $s_B$ is equal to the entropy per unit mass of the inlet stream $s_A$.

Step 1: Determine the entropy per unit mass and the enthalpy per unit mass of the inlet stream in State A
a. From the saturated steam tables, at pressure $p_A$, determine the following:
• $h_{LA}$ = enthalpy per unit mass of liquid water
• $h_{VA}$ = enthalpy per unit mass of water vapor
• $s_{LA}$ = entropy per unit mass of liquid water
• $s_{VA}$ = entropy per unit mass of water vapor
b. Using these data, calculate the enthalpy per unit mass $h_A$ and the entropy per unit mass $s_A$ of the entering mixture in State A
$$h_A=h_{LA}+(h_{VA}-h_{LA})x_A$$
$$s_A=s_{LA}+(s_{VA}-s_{LA})x_A$$

Step 2: Determine the mass fraction of water vapor in the outlet stream (State B) from the compressor $x_B$
a. From the saturated steam tables, at pressure $p_B$, determine the following:
• $h_{LB}$ = enthalpy per unit mass of liquid water
• $h_{VB}$ = enthalpy per unit mass of water vapor
• $s_{LB}$ = entropy per unit mass of liquid water
• $s_{VB}$ = entropy per unit mass of water vapor
b. Since $s_B=s_A$, using these data, calculate the exit water vapor mass fraction $x_B$ from:
$$x_B=\frac{s_A-s_{LB}}{s_{VB}-s_{LB}}$$

Step 3 Calculate the enthalpy per unit mass of the mixture in outlet State B $h_B$
$$h_B=h_{LB}+(h_{VB}-h_{LB})x_B$$

Step 4 Calculate the mechanical power that must be supplied to the compressor
$$P=\dot{m}(h_B-h_A)$$

15. Aug 22, 2016

### Staff: Mentor

No you didn't: I'm asking for the actual numbers. What pressure, in pascals or bar, are states A and B?

Last edited: Aug 23, 2016
16. Aug 23, 2016

### pranj5

@Chestermiller, many many thanks for your valuable contribution. I will answer your questions in a few days. But, I want to be assured about a few points. In Step 4, you have considered the power consumption to be equal to the difference in gross enthalpy of steam a Pressure level B and Pressure level A multiplied by steam flow. Kindly just tell me that do you agree to that if the steam remain saturated before, during and after compression; then the power consumption would be just the same amount as per the formula in step 4. It's a very very important assumption, at least for me.

17. Aug 23, 2016

### Staff: Mentor

Yes, provided you calculate the enthalpy per unit mass in States A and B according to the equations I presented in Steps 1 and 3, and the compression operates ideally (i.e., adiabatic and reversibly).

18. Aug 23, 2016

### pranj5

In case, the compression isn't idea, what will happen to the excess amount of energy. Do this add to the system as a whole and evaporate a little more amount of water?

19. Aug 23, 2016

### Staff: Mentor

Yes the final temperature will be higher, and you will have to do more work to compress it.

20. Aug 24, 2016

### pranj5

The answer isn't clear. If the inefficient compression will evaporate more water, that means adding some more steam to the system. But how that means higher temperature and more power to compress? Though it can be easily understood that this extra amount of steam will increase the density and pressure. But, if the pressure level is the goal, then it means reaching to the desired pressure level before time. In short, the extra steam will help in increasing the pressure and if there is special valve available and will open when the pressure inside will reach a level, then it would be opened earlier.