B Too much energy -- thought experiment

[Martin Jediny]

1/ Have a closed hydraulic circuit. Two columns of fluid. One hot, one cold, which are connected at the bottom and top. Let's have small heat loss and small hydrodynamic resistance to fluid flow and incompressible fluid.
I heat the lower interconnection, I cool the upper interconnection. I have constant flow, constant temperature difference.
I need a heating power P0 to heat it.
There will be a flow of liquid. The movement is caused by the different density of the liquid in the hot and cold columns. At hydrostatic head h1, I get some pressure difference that replaces the pump in the system and causes the liquid to flow.
( The above description is a common thing, also used industrially in the past, in natural circulation central heating.)

2/ If I insert a small propeller with a generator into a flowing circuit with hydrostatic head h1 and draw a small electrical power P1. I will call this flow with the inserted device a constant flow Qv (m3/s) and thus supply a constant heating power P0, when the column temperatures are kept constant.
If "n" is a real number, then n * P1 = P0
(It's not our job to build anything right now, so please let's avoid answering that "n" would be a terribly large number. With a column h1=1km high, n << 1000 000. For now, it suffices that some n exists. Let's skip the design "dimensional" issues for now, I'm asking about the principle now.)

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

4/ XXX
I don't ask question (4) because usually people stop talking to me. And I would love to know if I have a mistake in the definition of (1) or (2), and I would like to know the answer to my question (3)
... But whoever can answer question 3 will also know the answer to question 4.

 

[russ_watters]

I don't see that you defined "n" anywhere, but that probably doesn't matter. The main issue is you don't mention the heat rejection or write a full conservation of energy statement: heat rejection is lower than heat addition if you are extracting mechanical work.

 

[Dale]

Have a closed hydraulic circuit. Two columns of fluid. One hot, one cold, which are connected at the bottom and top. Let's have small heat loss and small hydrodynamic resistance to fluid flow and incompressible fluid ...
insert a small propeller with a generator

So what you are describing is essentially an external combustion heat engine. I think your description is closest to a closed Brayton cycle.

https://en.m.wikipedia.org/wiki/Brayton_cycle

Using an incompressible fluid is not a good idea, but other than that it is pretty standard.

The propeller is typically called a turbine.

 

[Martin Jediny]

I don't see that you defined "n" anywhere, but that probably doesn't matter. The main issue is you don't mention the heat rejection or write a full conservation of energy statement: heat rejection is lower than heat addition if you are extracting mechanical work.

Thank you, very good view.

(n is not really important now. it only says that the power I get from the circuit P1 is much less than I need for heating, P0. Then n=P0/P1)

Your comment is much more important. The law of conservation of energy,

If P0 is the power needed to heat the circuit, then Pcooling is the power of the cooler.
I think (but don't know) that for water, P0 = Pcooling,
as long as I approach the ideal conditions of thermal insulation and laminar flow with low friction.

But then I have the problem that if I take P1 out of the circuit, because I can slow down the fluid, the outputs Pcooling + P1 > P0.

And that's the problem.

In order to keep the equality I would have to reduce Pcooling, then Pcooling + P1 = P0, but I don't know the formula according to which the water cools down after the work is done.

The other solution to the equality is Pcooling + P1 > P0 + Pxy.
I know that in calculating P1 I consider gravity, the difference in density of the medium. But where to find the source of the additional Pxy, I have no idea.

 

[Dale]

P0 = Pcooling

This is false. These cycles are well understood. They don’t work this way.

 

[Martin Jediny]

So what you are describing is essentially an external combustion heat engine. I think your description is closest to a closed Brayton cycle...

I have also considered the Brayton cycle. but as you have noticed, I do not use a pump
I'm using a hydrostatic height.

Then if I take P1 out of the circuit, at hydrostatic height h1, at flow Qv and temperature 90°C and 95°C, I will need P0 for heating. (In parallel, we are solving the question whether I need Pcooling = P0 for cooling.)

(The use of a non-compressible fluid is interesting because it does not undergo adiabatic processes.)

If I keep the Qv and temperatures 90/95°C but increase the height to 2x h1, will I still need only P0 for heating ?
Can I take 2x P1 ?

 

[Martin Jediny]

This is false. These cycles are well understood. They don’t work this way.

Ok thank you.
Please, can you help me, how much does the temperature of the water drop when it passes through the turbine?

 

[russ_watters]

In order to keep the equality I would have to reduce Pcooling, then Pcooling + P1 = P0, but I don't know the formula according to which the water cools down after the work is done.

That is the correct conservation of energy statement, showing that there is no inequality/"too much energy". The details....well, you'd need to design the cycle. It's not a simple/single equation, you need to analyze the state of the water at every point in the cycle.

Note though that cycles like this usually use low pressure water and steam because just using the density change in liquid water gives almost zero head to drive the flow. If you insist on that, you're sifting through the rounding-error trash that is usually thrown away.

Edit: also, of course you can't simultaneously assume the water is incompressible but changes density. That's an oxymoron. And this is what I mean by "trash": youre looking for excess energy in things that are discarded with simplifying assumptions.

 

[Dale]

I do not use a pump
I'm using a hydrostatic height

That is fine. It will just slow things down so that the power output will be reduced.

The use of a non-compressible fluid is interesting because it does not undergo adiabatic processes

Why would a non-compressible fluid flow at all? There is a really good reason why existing engines don’t usually use incompressible fluids.

 

[russ_watters]

Why would a non-compressible fluid flow at all? There is a really good reason why existing engines don’t usually use incompressible fluids.

He's assuming an oxymoron: that an incompressible fluid can have variable density, so the density differences (weight difference between the water columns) would drive the flow.

 

[Martin Jediny]

I know the steam cycle very well and can design steam turbines with superheated steam and regenerative cycle. (apologies for the propeller)

I'm really interested in incompressible fluid now and I'm not going to revolutionize the energy industry. An incompressible liquid is really unsuitable for that.

Can you please advise me by how much the temperature of the water drops when it passes through the turbine? Or where does the temperature drop occur so that we have the law of conservation of energy?

We are not talking about statistical errors here, but real measurable values. admittedly small, but measurable.

 

[russ_watters]

I know the steam cycle very well and can design steam turbines with superheated steam and regenerative cycle. (apologies for the propeller)...

Can you please advise me by how much the temperature of the water drops when it passes through the turbine?

No. If you can design a steam cycle you can design this one too. But if you have trouble, we can help you through it(that's PF's mission). Regardless, your diagram has no numbers so we couldn't answer that even if we wanted to!

You have four states associated with the four normal processes in a thermodynamic cycle. Since it's your cycle, you pick the states and calculate the energy flows. As I said, if you get stuck we can help you through it.

We are not talking about statistical errors here, but real measurable values. admittedly small, but measurable.

If you make a self-contradictory assumption, it's an error.

 

[Martin Jediny]

No. If you can design a steam cycle you can design this one too.
...If you make a self-contradictory assumption, it's an error.

Thank you for giving me your time.

I don't have any temperature in the formula at the turbine.

Turbine power:
P1 = dp * Qv * ni
dp pressure difference inlet/outlet
Qv Flow rate (m3/s)
ni efficiency

dp = h1 * dRo *g
h1 hydrostatic head
dRo density difference between hot and cold column
g 9,81 ms-2

then if

n= P0 / P1

Then there must be a height n * h1 where n * P1 = P0 ?

Where am I wrong?

 

[Arjan82]

The heating of the fluid is at a higher pressure than the cooling due to hydrostatics. This means that the work done by the heating is higher than by the cooling. This is why Pcool has to be smaller than P0.

A turbine will not cool down the fluid, it will convert some mechanical energy into heat. If you assume incompressibility (due to pressure that is, heat will of course change the density) the turbine cannot have a cooling effect on the flow.

 

[Arjan82]

You cannot use the full dp due to difference in rho for the turbune power. Because the system has to overcome a small amount of friction as well.

If you don't include that friction it becomes a problem where you do not need any energy at all to keep the flow going (Newton's first). In that case the heating causes a force imbalance leading to indefinite acceleration, which is not realistic of course.

n= P0 / P1

Then there must be a height n * h1 where n * P1 = P0 ?

I don't get this, first you define n = P0/P1, then by definition n*P1 = P0? Why would the height influence this relationship?

 

[Dale]

I'm really interested in incompressible fluid now and I'm not going to revolutionize the energy industry. An incompressible liquid is really unsuitable for that.

Can you please advise me by how much the temperature of the water drops when it passes through the turbine?

For an incompressible fluid there is no temperature drop. The efficiency of an ideal Brayton cycle is given by $$1-\frac{T_1}{T_2}=1-\left(\frac{P_1}{P_2}\right)^{(\gamma-1)/\gamma}$$ Since for an incompressible fluid $\gamma=1$ the efficiency is zero, the temperatures and pressures are equal and there is only heat transfer with no work performed.

 

[Martin Jediny]

The heating of the fluid is at a higher pressure than the cooling due to hydrostatics. This means that the work done by the heating is higher than by the cooling. This is why Pcool has to be smaller than P0.

I don't know if wiki is a good source, but it says that I need 4215 J / K / kg for cooling, while for heating I need 4165 J / K / kg , just for hydrostatic pressure for example 200bar. (https://en.wikipedia.org/wiki/Heat_capacity, April, 2024)

And flow is the same.

 

[russ_watters]

then if

n= P0 / P1

Then there must be a height n * h1 where n * P1 = P0 ?

Where am I wrong?

Well you haven't connected that in any way to the first part. You didn't say anything about P0.

I'm skeptical that you know how to design a thermodynamic cycle, because you haven't even started/aren't trying. You're just pointlessly manipulating disembodied equations.

 

[russ_watters]

I don't know if wiki is a good source, but it says that I need 4215 J / K / kg for cooling, while for heating I need 4165 J / K / kg , just for hydrostatic pressure for example 200bar. (https://en.wikipedia.org/wiki/Heat_capacity, April, 2024)

And flow is the same.

Where does it say that? That's either an incomplete statement or a contradiction.

 

[Martin Jediny]

Where does it say that? That's either an incomplete statement or a contradiction.

Heat_capacity_of_water_2.jpg

 

[russ_watters]

Heat_capacity_of_water_2.jpg

How did you get two different answers for the same point?
(Edit) you're assuming different temperatures and/or pressures, right? Did you specify them?

 

[Arjan82]

I don't know if wiki is a good source, but it says that I need 4215 J / K / kg for cooling, while for heating I need 4165 J / K / kg , just for hydrostatic pressure for example 200bar. (https://en.wikipedia.org/wiki/Heat_capacity, April, 2024)

And flow is the same.

The Cp for cooling is exactly equal to the Cp for heating at the same pressure and temperature, I don't know how you got two different answers....

 

[Martin Jediny]

How did you get two different answers for the same point?
(Edit) you're assuming different temperatures and/or pressures, right? Did you specify them?

Let's create a random example. Let's say we have a continuous continuous flow of 1kg/s (Q=0.00104 m3/s (hot side)) of water in our plant. The cold water down entering the heater will have a temperature of 90°C. The water coming out of the heater will have a temperature of 95°C. The density of water in the hot column will be 961.9kg/m3 and in the cold column 965.3kg/m3. At a hydrostatic height of h1=1000m the pressure difference will be 34kPa. 10kPa I need for flow. Electric produce with turbine: 24kPa * 0,00104m3/s * 1000m *0,6 = 15W.

I have good thermal insulation and laboratory conditions throughout the height +90°C for cold column and 95°C for hot column. Zero heat flow zero heat loss.
Funding the laboratory is not a question of this thought experiment.

Heating:
5°C * 1kg / s * 4170 J /kg /K = 20850 W (by 100 bar pressure)

Cooling:
5°C * 1kg / s * 4210 J / kg /K = 21050 W (by 1 bar pressure)

*************************************
For double high:
Electric produce with turbine: 30W
Heating:
5°C * 1kg / s * 4165 J /kg /K = 20825 W (by 200 bar pressure)
Cooling:
5°C * 1kg / s * 4210 J / kg /K = 21050 W (by 1 bar pressure)

I round off my entire professional practice and use proven rules.
Now I am solving a thought experiment and questions are coming that I am trying to understand.

 

[Dale]

The density of water in the hot column will be 961.9kg/m3 and in the cold column 965.3kg/m3.

So now you do not want to use an incompressible fluid? You want to use water?

 

[russ_watters]

Heating:
5°C * 1kg / s * 4170 J /kg /K = 20850 W (by 100 bar pressure)

Cooling:
5°C * 1kg / s * 4210 J / kg /K = 21050 W (by 1 bar pressure)

I round off my entire professional practice and use proven rules.
Now I am solving a thought experiment and questions are coming that I am trying to understand.

[Edit] deleting while I rewrite. You changed the examples and I missed that.

 

[Martin Jediny]

So now you do not want to use an incompressible fluid? You want to use water?

I apologize if I caused confusion with an imprecise definition.
In any case, I want to use a real liquid and not water in the end result. but I don't want to complicate it now. the only problem is that the required transmission height is many times higher than for other liquids. But this is not a problem for a thought experiment.

 

[Dale]

I don't want to complicate it now

Trying to make a cycle with an incompressible fluid is complicating things

Again, Why would a non-compressible fluid flow at all?

 

[Martin Jediny]

The heating and cooling values have to be the same otherwise you've made a mistake, and someone with professional experience with this subject matter should know this.

By this point the main flaw is pretty well established: you're picking and choosing when to consider it compressible or incompressible. Your approach is wrong: you can't use Cp * delta T here. You need to use states and subtract the enthalpies at the two states.

I guess I didn't understand what you wanted from me in the previous posts.

Yes, that was the first idea that heating and cooling are the same. Through enthalpy, it is 21.06 kJ / kg. Then I heat and cool 21.06 kW.

But if I have the same heating and cooling, then we will have to find another power source for my 15W electrics.

 

[russ_watters]

Heating:
5°C * 1kg / s * 4170 J /kg /K = 20850 W (by 100 bar pressure)

Cooling:
5°C * 1kg / s * 4210 J / kg /K = 21050 W (by 1 bar pressure)

What is your next step? It would be better if you gave us a diagram with all the states and processes labeled. Those two processes alone don't tell us much in isolation.

 

[russ_watters]

I guess I didn't understand what you wanted from me in the previous posts.

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.

 

[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.