B Too much energy -- thought experiment

[Arjan82]

3/ At a flow rate of Qv , at a hydrostatic head of 2 * h1 , can I draw an electrical power of 2 * P1 ?

I think your reasoning is that the difference in density causes a larger difference in pressure if you increase the height. So (rho_hot*g*h1 - rho_cold*g*h1) < (rho_hot*g*2*h1 - rho_cold*g*2*h1), which on itself is true of course.

The thing is that for equal P0 the temperature and therefore the density of the fluid is higher at the hot side if you increase the height to 2*h1. This is because the moment you increase the pressure difference and therefore the flowrate, you decrease the temperature at equal P0. This means the driving force is lower and you do not get exactly 2 times the pressure difference.

But if you have a turbine this of course does not have to be this way. For the first case with h1, you can add a turbine that causes the flowrate to be Qv, but for the 2*h1 case you can increase the turbine size to still get the same Qv. In that case you do indeed get a pressure difference that is twice as large. This would indeed mean that P1 can increase by almost a factor of two, almost, because the plumbing is now longer and you therefore have more friction.

But what also plays a role is that now the pressure difference between the locations where you add P0 and retract Pcool is also twice as high. This means that P0-Pcool is higher and the *net* amount of energy you put in is higher. So then surely you can increase P1 as well.

What needs to be clear is that the amount of energy that you are adding to the system is P0-Pcool, not P0. So with 2*h1 you also input more power.

 

[Martin Jediny]

I don't know if I understand you all. I would like to be clear.
If I have a simple circuit without turbines.
Flow rate 1kg/s
a height of 1km with perfect thermal insulation.
And one column with water has a temperature of 90°C and the other has a temperature of 95°C,
How much power do I need down to heat the liquid and how much cooling power do I need up to cool the liquid.

 

[russ_watters]

I don't know if I understand you all. I would like to be clear.
If I have a simple circuit without turbines.
Flow rate 1kg/s
a height of 1km with perfect thermal insulation.
And one column with water has a temperature of 90°C and the other has a temperature of 95°C,
How much power do I need down to heat the liquid and how much cooling power do I need up to cool the liquid.

In order to find that answer, you need to design/solve the whole cycle. I dont think it has a simple answer and I'm skeptical that the way you are using simplifying assumptions will work. For starters:

Point 1: Top of cold column. 90C, 1Bar.

Process 1-2: Adiabatic compression through 1,000 m elevation change.

Point 2: Pressure? Temperature? I don't think you can assume constant temperature through the column. But calculating it and finding out if it is enough to matter would be a good start.

 

[Martin Jediny]

In order to find that answer, you need to design/solve the whole cycle....

sorry, maybe I found my problem.

Please, how much heat do I need to heat 1 kg of water from 90°C to 95°C, at constant pressure?

Please be accurate to 4 or 5 significant digits if it is possible

At a pressure of 1 bar
At a pressure of 100 bar

 

[Dale]

how much heat do I need to heat 1 kg of water from 90°C to 95°C, at constant pressure?

So now we are back to using water?

You can find this information at:
https://webbook.nist.gov/cgi/fluid.cgi?ID=C7732185&Action=Page

Here is the isobaric data at 0.1 MPa
https://webbook.nist.gov/cgi/fluid....nit=m/s&VisUnit=uPa*s&STUnit=N/m&RefState=DEF

Here is the isobaric data at 10.0 MPa
https://webbook.nist.gov/cgi/fluid....nit=m/s&VisUnit=uPa*s&STUnit=N/m&RefState=DEF

From this site you can piece together a complete thermodynamic cycle.

 

[Martin Jediny]

So now we are back to using water?

You can find this information at:
https://webbook.nist.gov/cgi/fluid.cgi?ID=C7732185&Action=Page
...

Thank you for the recommended tables and patience

If I'm quoting correctly, it's 90/95 for the fallout at pressure
0,1MPa I need 21,04 kJ to heat 1kg of water (just a note Cp=4,2102, dRo= 0,96531-0,96189=0,00342 g/ml)

and at a pressure of 10MPa I need 20.93 kJ for 1kg of water (just a note Cp=4.1884, dRo= 0.96978-0.96641=0.00337 g/ml)

Does this mean that if I have a lossless circuit, I heat the boiler downstairs with 20.93 kW, but the radiator 1km upstairs heats 21.04 kW?

Should I add the energy of the work to change the volume of the liquid to the boiler output?
3.6E-3 ml/g *10MPa = 36 J => 36W


It's still low. And we have 5 valid numbers, so the error is probably not in rounding.

20930+36 = 20966 < 21040

 

[Martin Jediny]

You're not giving us all the details of the cycle and making us guess. We don't like that. We can't tell if the numbers work unless we see them in the context of the rest of the cycle.

I apologize for reindexing the variables, but they make better logic for the p-V diagram.

1kg/s flow rate
Line 1-2 lower pipeline, pressure 10MPa. Heating P12= 20.93 kW ? +36W ?
Point 1: 90°C, Ro1=0.96978 g/ml, Enthalpy: 384.73 kJ/kg , Cp=4.1837 J/(g.K)
Point 2: 95°C, Ro2=0,96641 g/ml, Enthalpy: 405,66 kJ/kg , Cp=4,1884 J/(g.K)

Line 2-3 gravity hot water rise without pump. Height 1000m

Line 3-4 upper pipeline, pressure 0,1MPa, Cooling P34= 21,04 kW
Point 3: 95°C, Ro3=0,96189 g/ml, Enthalpy: 398,1 kJ/kg , Cp=4,2102 J/(g.K)
Point 4: 90°C, Ro4=0,96531 g/ml, Enthalpy: 377,06 kJ/kg , Cp=4,2052 J/(g.K)

Line 4-1 gravity descent of cold water. Height 1000m

Problem: P34 > P12
Did I make a mistake somewhere?
Is gravity bringing some energy in there ?

 

[Dale]

I apologize for reindexing the variables, but they make better logic for the p-V diagram.
View attachment 344574
1kg/s flow rate
Line 1-2 lower pipeline, pressure 10MPa. Heating P12= 20.93 kW ? +36W ?
Point 1: 90°C, Ro1=0.96978 g/ml, Enthalpy: 384.73 kJ/kg , Cp=4.1837 J/(g.K)
Point 2: 95°C, Ro2=0,96641 g/ml, Enthalpy: 405,66 kJ/kg , Cp=4,1884 J/(g.K)

Line 2-3 gravity hot water rise without pump. Height 1000m

Line 3-4 upper pipeline, pressure 0,1MPa, Cooling P34= 21,04 kW
Point 3: 95°C, Ro3=0,96189 g/ml, Enthalpy: 398,1 kJ/kg , Cp=4,2102 J/(g.K)
Point 4: 90°C, Ro4=0,96531 g/ml, Enthalpy: 377,06 kJ/kg , Cp=4,2052 J/(g.K)

Line 4-1 gravity descent of cold water. Height 1000m

Problem: P34 > P12
Did I make a mistake somewhere?
Is gravity bringing some energy in there ?

This is much better than before. So let’s look at what you have and what you are still missing (you keep jumping to the end without completing the middle)

Process 1-2 appears to be isobaric heating given the graph, but you never specifically state it. Please confirm?

Similarly process 3-4 appears to be isobaric cooling. Correct?

Now, the real confusion. You have previously described the process 2-3 as adiabatic. But now you are describing it as isothermal decompression. It cannot be both. Please clarify?

Similarly with process 4-1. Is it isothermal or adiabatic compression?

 

[Martin Jediny]

This is much better than before. So let’s look at what you have and what you are still missing ...

From the start it is 2 columns, one cold and one hot, ideally thermally insulated.
T2=T3 and T4=T1
this process of 2-3 and 4-1 is therefore isothermal

The columns are top and bottom connected. (Consider the free natural circulation of water, using the different densities of hot and cold water in the columns)
p1=p2 and p3=p4
then heating 1-2 and cooling 3-4 is an isobaric process

(if I strayed from this idea, it must have been a mistake)
(Now I only cancelled the turbine, which was supposed to take any of electricity from the natural circulation of water. Even without this complication we have an inequality of inputs and outputs.)

 

[Dale]

ideally thermally insulated.
T2=T3 and T4=T1

This is self contradictory. If you want T2=T3 (isothermal) then it cannot be thermally insulated (adiabatic). You have to choose.

 

[Martin Jediny]

This is self contradictory. If you want T2=T3 (isothermal) then it cannot be thermally insulated (adiabatic). You have to choose.

can you give me some advice, please? What is the process? (I've always solved the adiabatic plot with gas/steam only.)

Column 2-3 is full of water and is ideally insulated. The column does not trap heat, nor does it accept heat from the surroundings.
If the volume of water increases slightly, because as it rises 1000m upwards, the hydrostatic pressure decreases, then can't T2=T3 ?


Originally I didn't think to consider water compressibility, then T2 = T3. But by 1000m is consider compressibility water as realy number

 

[Dale]

Column 2-3 is full of water and is ideally insulated. The column does not trap heat, nor does it accept heat from the surroundings.
If the volume of water increases slightly, because as it rises 1000m upwards, the hydrostatic pressure decreases, then can't T2=T3 ?

So it sounds like you want adiabatic, not isothermal.

Consider a gas. As a gas expands adiabatically it reduces its temperature. And as it compresses adiabatically its temperature increases. So an adiabatic process is not isothermal.

The same happens in a liquid, but just to a smaller degree. But since you want very high precision then you must consider these things.

can you give me some advice, please? What is the process?

So what you need to do is find the fluid properties at 10 MPa and 95 C using the earlier website. Then you calculate $P V^{\gamma}$ where $\gamma=C_P/C_V$.

Once you have that number then you go to 0.1 MPa and you find the temperature with that same number.

Spoiler

I get 94.78 C. Not much lower than 95 C, but you were not looking for a big difference.

 

[Martin Jediny]

So it sounds like you want adiabatic, not isothermal.
...

Thank you, I think you have found my lost energy.