Calculations thermosiphon: power, slope, diameter and flow

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Calculations thermosiphon: power, slope, diameter and flow




by Christophe » 10/02/09, 13:07

Here is a method of calculating a thermosiphon.

This is a typical solar example, but the method is also applicable for other heating devices. The problem is that for "good" operation, a thermosyphon requires a fairly cold return so with a risk of being below the dew point in the boiler ...

Thermosiphon solar water heater.

The solar water heaters thermosyphon is an effective and costless operation since it has no need to pump or need for regulation. Hence its amortization will be faster. Only constraint this system, the sensors must be at a level below the storage tank. The thermosiphon is well known, the operation result of the difference in density of the coolant due to the temperature difference between the sensors and the water heater. The height difference between the top of the sensor and lower ball will be 0,5 m minimum to ensure proper implementation of the pressure (or load) hydro-motor.

This hydro-motor pressure is equal to:
P = H x (Mfr - Mfd)
P = pressure hydro-motor available in mmCE
H = height difference in meters between the axis of the sensors and the axis of the storage tank
Mfr = mass of fluid at the lowest temperature (return sensor)
Mfd = mass of fluid at the highest temperature (start sensors)


For clear water, see table on page: variation of the density of liquid water as a function of temperature

For the brine mass is a function of the percentage of antifreeze in water. See supplier.

But as the calculation is performed with a mass gap, clear water values ​​can be used without risk of error.

Values ​​for sizing.

Installation is usually calculated for a flow of 0,7 liter / minute per m² of collector approximately 42 l / hm.

Adding Christophe to a boiler so it would be logical to take 0,7 L / min for 1kW boiler power.

The temperature drop / return is on average 20 ° C.
Operating temperatures can be taken 80 ° C for the start and therefore with a fall of 20 ° C to 60 ° C for return, but to allow running at lower temperatures (between seasons for example) they snuff may be more unfavorably for the calculation of the hydro-motor pressure, 65 / 45 ° C.

The J / Z report (see below, calculating losses) will depend on the configuration of the installation and for a first approach to be 35 / 65% (65% for Z to take account of sensors and pressure losses ball if they are not known).

Recommendations

- To minimize the losses, the first enemy of the thermosiphon system, the sensors should be preferably Tickelmann mounting (see drawing below) rather than mount S.
Image
- No cons slope must be made because it has the effect of cutting the thermosiphon.
- The slope must always be ascending to the ball, avoid poses level.
- The air purge will be made by the open expansion tank located above the storage tank (see drawing for example).
- The ball ECS should preferably jacketed coil rather than this always to limit losses.
- The pipes should be insulated.
- Elbows should preferably be made using a bender for a the largest possible radius.

Calculation example

Image

- Starting line length sensors / ball 6,5 m
- Length of the return balloon / sensor 7 m
- Height difference axis sensor / balloon axis, 5,80 m
- J / Z Report, 35 / 65%
- Area sensors, 5m²
- Flow, 42 l / h / m²
- Flow temperature sensors, 65 ° C
- Return Temperature sensors with a fall of 20 ° C, 45 ° C
- Density of water 65 ° C 980,48 kg / m3
- Density of water 45 ° C 990,16 kg / m3

hydro-motor pressure mmCE available in:
5,8 P = x (990,16 - 980,48) = 56,14

J value in mmCE / m:
J = 56,14 0,35 x / (7 6,5 +) = 1,45

The diameter of the pipes must be chosen by successive approximations so as not to exceed 1,45 mmCE / m. To facilitate calculations we can use the Excel workbook "Losses".

So by entering the following parameters: Speed ​​= x 42 5 210 = l / h
copper pipes. flow temperature of the fluid, 65 ° C. DeltaT, 20 ° C and by process of approximation, it is the diameter of 26x28 which gives the next lower value of 1,45 0,84 MMWC / m.

With this value, the actual throughput will necessarily higher than calculated, therefore, increasing throughput by approximation, we find the flow 288 l / h, which gives a rate of 57,6 l / hm.

The higher the starting temperature sensors, the higher the density difference will be as well which will increase the hydro-motor pressure and thus the flow. The amount of heat exchange at the balloon being proportional to the average temperature difference between the average coolant temperature and the average temperature of the domestic hot water, the exchange volume will increase with the flow rate as the increase the latter will induce a lower temperature drop and thus increase the average deviation.


About Excel file for calculating losses: https://www.econologie.info/share/partag ... KPz3dP.xls

The Excel workbook "Pressure drops JZ.xls" allows you to calculate the pressure losses of a circuit and this by section. To perform the calculations for a complete installation, start with the most disadvantaged circuit, generally the radiator furthest from the boiler or the underfloor heating loop furthest from the boiler and the longest, in order to know the reference pressure drop and see if its value does not exceed that of the circulator if it is already defined (delivered with the boiler) in order to be able to make corrections. A section is a part of the circuit such as, for example, the part which goes from the boiler to the tees of the 1st radiator or the first collectors, from the tees of the 1st radiator or collectors to the tees of the next, etc ...

For ease of use the workbook, not start the boiler (1ère line) coming to the last loop radiators or underfloor heating (or the most disadvantaged (e)) to define the loss reference load, then simply to remove the sections of the end to branch off to the other in order not to have to re-enter the values ​​starting from the boiler and this to set the diameters of the other circuits.

The workbook has hidden columns in order to reduce the scope of work. It is possible to display these columns by clicking on the small cross located above the columns.


Source et Excel file
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dap35
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by dap35 » 04/05/11, 13:31

Hello

I 2 questions about this mode for a solar water heater:
At night, the thermosiphon Will not work the other way: warm the panels? How to avoid it? Check valve?
Risk of frost in the panels? Antifreeze?
Will it be possible in Britain, for example?

Thank you

Dap
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Gaston
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by Gaston » 04/05/11, 13:43

dap35 wrote:At night, the thermosiphon Will not work the other way: warm the panels?
No, the hottest water being in the flask at this time, it remains high.

dap35 wrote:Risk of frost in the panels? Antifreeze?
Yes, antifreeze is needed (depending on weather).

dap35 wrote:Will it be possible in Britain, for example?
Technically, no doubt.
You still have to consider the sunshine ...
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Mido66
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by Mido66 » 06/02/13, 11:21

Hello,

Is it necessary to add a vase expansion since the ball has a security group that will be activated in case of overpressure.
I live in Morocco in Agadir 360 days of sunshine and average temperature 20 32 ° in winter and in summer,
I made the horizontal bars in sensors with a ball of liters 300 300 liters of the problem is that it does not exist Horizontal, the ball is about 1.80m height, inclination to 45 ° sensor gives a height of more 80cm 1.80m difference in the ball and 0.50m to operate the thermosiphon 3.10m gives me my height and ball like a rocket on the start but I can not do otherwise.
The circuit is open circuit, it heats water directly from the tank.
I change the electric heating balloon replacing resistance by the entrance to hot and back to cold, this entry is on the side about 35cm from the bottom of the balloon, sending hot water ball I have to put a pipe leading from the entrance mentioned above and go inside the balloon to reach the top of this one.
The hot start and cold water return are 22mm

have you any comment or advice?
Thank you
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MJB61
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Re: Thermosyphon calculations: power, slope, diameter and flow




by MJB61 » 04/10/23, 11:54

Hello,
Thank you for this tutorial, but I didn't understand everything, I would need further explanation.

At the start of the calculation, after the diagram, you explain:
J value in mmCE / m:
J = 56,14 0,35 x / (7 6,5 +) = 1,45

I don't understand how the J value which corresponds to a charge loss can decrease when the length of the circuit increases. In this calculation, if instead of 7m. + 6m. we put 70m. + 60m. the pressure loss J would only be 0,45. so 10 times less loss with 10 times more transit length???

a little further down, you write:
So by entering the following parameters: Speed ​​= x 42 5 210 = l / h
copper pipes. flow temperature of the fluid, 65 ° C. DeltaT, 20 ° C and by process of approximation, it is the diameter of 26x28 which gives the next lower value of 1,45 0,84 MMWC / m.


in the spreadsheet, following your advice, I typed the following values: flow rate 210; copper pipe; starting temperature 65; delta T 20; ...nothing happens. I added length of the section 13,5; %J 35 ;... ... nothing happens.
I then typed 26 in column D interior in mm. and at that moment, different data is displayed.
But from there, I'm lost.
you write : the diameter of 26x28 which gives the value immediately below 1,45 of 0,84 mmCE/m.
I don't understand this last sentence.

I hope someone here can explain to me how this spreadsheet works.

cordially
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