# Compression of water

1. Jul 7, 2011

### kbka

Hi PF!

I'm modeling a system using EES in which compression of water occurs from approximately 2 bar to 200 bar.

During the compression, the temperature rises from 300K to 305K. There's no heat transfer and the pump works as an ideal machine.

I remember my thermo professor saying that water is only virtually incompressible and as a gas it will during compression experience a slight change in temperature.

I would say that 5 degrees K is a tad more than a slight increase. But not an increase that seems way out of left field.

Any knowledge on how big an increase can be expected at this pressure increase? Litterature or articles on the matter would also be deeply appreciated.

Thank you

2. Jul 8, 2011

### Mapes

Hi kbka,

If there's no heat transfer (adiabatic) and the process is reversible, then the process is also isentropic. So what we're looking for is the change in temperature with pressure at constant entropy, or $(\partial T/\partial P)_S$. If you know Maxwell relations, you know that

$$\left(\frac{\partial T}{\partial P}\right)_S=\left(\frac{\partial V}{\partial S}\right)_P=\left(\frac{\partial V}{\partial T}\right)_P\left(\frac{\partial T}{\partial S}\right)_P=\frac{\alpha VT}{C_P}=\frac{\alpha T}{c_P\rho}$$

where $\alpha$ is the thermal expansion coefficient, $c_P$ is the constant-pressure specific heat capacity, and $\rho$ is the density. These are all pretty insensitive to pressure and are approximately 2x10-4 K-1, 4000 J kg-1 K-1, and 1000 kg m-3 respectively. With the values you've given, however, I'd expect a temperature increase of only 0.2 K, and I'm not sure why there's a discrepancy.

3. Jul 8, 2011

### Q_Goest

I agree with Mapes. The database I have indicates just a few tenths of a degree K, not 5 K. You should be able to locate water/steam tables and simply pull that info directly out of those tables. Just look under the 2 bar table at 300 K, get the entropy valye, then go to the 200 bar table and see what temperature that entropy corresponds to. You'll need to interpolate the temperature value of course, but having the table is all the data you need.

4. Jul 8, 2011

### SteamKing

Staff Emeritus
Remember, you are using a pump to increase the pressure of the water from 2 bar to 200 bar. This requires work, the heat equivalent of which can be transmitted to the water.

5. Jul 8, 2011

### kbka

Hi guys!

Thanks for your replies. Still haven't solved the problem but I did come to think of something that may have to do with the large temp. increase.

The energy balance I wrote where I believe the problem lies doesn't account for water actually being compressible. I haven't accounted for the work that is being put into compressing the water. as in P*dV if this was a gas.
The only unknown of the equation is the internal energy which due to a to rapid increase gives me the high temperature.

If it is I'm thinking the work done is somehow related to the bulk modulus of water.
B=-V*dP/dV
Can't figure out how though.

Any thoughts?

6. Jul 8, 2011

### kbka

I should specify here. The energy balance is the first law of thermodynamics for a control volume. There's only mass flow into the system, and thus a change in mass over time...

d/dt*(M*u)=m*h

And as i wrote earlier, only internal energy is the unknown. The internal energy is obviously too high since the temperature comes out 5 degrees over what could be expected. If I add work on the rhs the value of u would drop -> lower temp.

Last edited: Jul 8, 2011
7. Jul 8, 2011

### Q_Goest

Hi kbka,
When you say "no heat transfer" that means adiabatic. When you say "works as an ideal machine" that means the compression of the water is an isentropic process. At that point, you've fully defined the process the water goes through. Work is put into the water isentropically until the pressure has reached 200 bar. That process will increase the temperature of the water as Mapes and I have described, but temperature rise will only be a few tenths of a degree K in this case. So if you have a computer you're working on and its coming out with a temperature increase of 5 K, something's wrong either with what has been described as the process you're modeling or there's something screwy going on with your software.

If you want to use the first law and perform an energy balance, I would draw a control volume around the pump and then reducing the first law you should find that the enthalpy in + work in = enthalpy out at the higher pressure. But that's exactly the same as an isentropic process. If on the other hand you are only trying to model a control mass (ie: the mass of water that is being compressed without flowing such as water in a cylinder fitted with a piston and that piston is moved such that water pressure increases from 2 bar to 200 bar) then the change in internal is equal to the work put in. Those are two slightly different processes, though the temperature still won't change much.

Perhaps you could provide a bit more detail regarding what you're trying to model and how?

Thanks.

8. Jul 8, 2011

### kbka

Of course :)

I have a vessel filled with air and water. There is a constant flow of water, m, into the vessel, compressing the gas. The amount of gas in the vessel is constant.

So, I first have a mass balance over time from 0 to t. So I know exactly how much water is in the vessel, and how much air is in the vessel (M_w and M_a). Assuming a constant density of the water in the vessel, I also know the volume of each medium.

Knowing the volume of each component and the initial state of the gas, I can find the pressure of the gas by using 1st law for a control mass.

M_a*(u_a-u_a0)=int(P*dV,time=0..tau)

I know the following:
M_a, u_a0, dV (mass flow rate over density of water)

Leaving the internal energy u and pressure P. I add the three equations
u_a=f(T_a;P) (int. energy from Temperature and pressure)
P=f(T_a;v) (Pressure from Temperature and spec. volume)
v=V_a/M_a (Specific volume)

The pump is as i mentioned ideal and I know the state of the water before it enters the pump, so the enthalpy after the pump just as it enters the vessel is also known.

Now, using the 1st law for a CV I write

M_w*u_w-M_w0*u_w0=int(m_dot*h, time=0..tau)

Known are:
M_w, M_w0, u_w0, m_dot and h

leaving u_w.

The temperature of the water is then a function of pressure p and internal energy of the water u_w.

As you can see the only assumption is constant density. In reality it increases 0.075 kg/m^3

Regards
kbka

9. Jul 9, 2011

### kbka

I have been thinking now that it's not the work done on the fluid, but by the fluid. So it is equal to that done on the gas, but opposite in sign. So that the equation for the CV is

M_w*u_w-M_w0*u_w0=int(m_dot*h-P*dV, time=0..tau)

dV still being mass flow rate into the system over the density of the water. The temperature rise is now in the order of 0.9 K.

So, am I right when I say that:
work done in the energybalance of the CV is not compression of the water (as in it's small change in density). It is equal, but opposite, to the work done on the gas? in other words work done BY the water.

Last edited: Jul 9, 2011
10. Jul 9, 2011

### Q_Goest

Hi kbka,
I think I understand your system better now. Does this have anything to do with one of your earlier threads regarding the liquid piston compressor?

I think that's right, or at the very least it’s a very close approximation. I'm still struggling with exactly what your system is. I'm guessing it's air in a cylinder being compressed as you pump water in, is that correct? If so, then the air is what is being compressed and the energy required to compress the air comes from the water pump, so the water pump is putting energy into the water which is going into the water and the air. The vast majority is going into the air I think. The water is at a higher pressure of course, so some small amount of energy added by the pump remains in the water and has caused it to get warmer. Here’s how I’m looking at it. See if you agree….

For a CV around the air alone:
dU = Win(air) = PdV

CV around pump:
dU = 0
1) Hin(pump) + Win(pump) = Hout(pump)

CV around water in cylinder:
2) dU = Hin2 –Wout(air)

Note that: Hin2 = Hout(pump)

So, putting 1) into 2):
3) dU = Hin(pump) + Win(pump) - Wout(air)

Equation 3 is the same as wrapping a CV around both the pump and the water in the cylinder. Note that Win(pump) is always going to be slightly larger than Wout(air) because some energy is being stored in the water. The increase in temperature of the water can always be determined using equation 1 for any instant in time, and the highest rise in temperature is going to be the couple of tenths of a degree K.

11. Jul 10, 2011

### kbka

I see. your way of setting it up does seem a bit more straight forward. I'll try that.. Thank you...

And yes, the former post was regarding the same issue :)

Q_goest, you're the man!

Regards,
kbka