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Technical part: Heating systems


All the pump units are reversible, to get the supply way on the left side. This operation is simple and quick: in the enclosed directions all the steps are described, even in the presence of mixing valves and bypasses. The pump units can be prepared with the left supply from the purchase order receiving: it is enough to use an item code where the letter “L” (Left) replaces the letter “R” (Right). See the following example:


  Code 20355R
right supply
  Code 20355L
left supply




  Check valve: The check valve has to be placed inside the neck of the ball valve flange of the return side. Attention: to exclude the check valve, turn the handle by 45° clockwise.  
TAKE CARE: The pump units with circulating pump Wilo RSG 25/8 are not reversible. They are available only with right supply way.



The models M3 of the pump units have a by-pass valve mounted into the upper part of the unit, suitable for installations that are working with considerable flow changes, as it happens in the systems that have thermostatic radiator valves or motorized valves. The by-pass enables a flow recycling proportional to the number of valves that close and it reduces the maximum value of the differential pressure made by the circulating pump. The here below diagram shows a situation in which all the adjustment valves of the circuit are closed. The by-pass (in the example of the position 2) reduces the maximum pressure at 47,5 kPa. The flow showed is the one that is flowing through the by-pass.


Right supply

Left supply

Click on the drawing to download pdf version in high-res

Pict. 1 Pict. 2
Adjusting of the by-pass

The data of the diagram can be used to set the valve.

Picture 1. The notch reference of the setting is the top of the nut (A).
Picture 2. Example of the setting at the value of 0,2 bar.



   TAKE CARE:  The M3 pump units must not be used with electronic self-adjustment circulating pumps.



The selection of the right circulating pump is determined by the need to provide the installation with a flow suitable to develop the power fixed in the planning stage. Knowing this datum and taking into consideration the temperature differencet between the supply and the return, we can calculate the flow in kg/h. It is also important to take into consideration the kind of pump unit that is used, that is known in advance because it has been selected on the basis of the kind of installation to be realized.
EXAMPLE: For an installation with a M2 pump unit that requires a power P = 50 kW with a temperature difference t = 20 Kthe flow is calculated as follows:

Now we have to claculate the total head loss of the installation, to be able to select a circulating pump that is not under-dimensioned. As concerns the pump unit we know the head losses looking to the diagram the curve of the used model. In this case we found that, for the model M2 with a flow of 2150 kg/h the head loss is 0,75 m of water column.

Click on the drawing to download pdf version in high-res; click here to download DXF/DWG version of the drawing


To this head loss we have to add the total head loss of the installation (pipes, connections, radiant elements, etc): this is a datum given by the planner.
As we can see from the diagram the circulating pump Yonos Para RS 25/6 at a flow of 2150 kg/h has a lifting power of 3,44 m: taking into consideration that the pump unit absorbs 0,75 m il will left 2,69 m (as 3,44 - 0,75 = 2,69 m) of water column available to compensate the head losses of the installation. Therefore we have to see if this datum is sufficient, in that case we can use the Yonos Para RS 25/6, otherwise we have to use another circulating pump provided with a bigger lifting power.

ATTENTION: if necessary it is possible to calculate by a mathematical calculus the pressure drop (at the required flow) produced by the presence of an hydraulic device, if we know its Kvs; therefore, with a good approximation, assuming a standard temperature of 20°C and overlooking the effects of viscosity of the fluid, it follows that:

Where the flow Q is expressed in m3/h e h, the pressure difference at the outlets of the device (pressure drop) is expressed in bar. Then, reversing the previous formula, we obtain:


in the example above:


As 1 bar is about 10,198 mH2O, then the pressure drop is 0,73 mH2O; this value, unless the approximations, is the same as the value of the curve.

Typical curves of Energy pump unit and circulating pumps

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