What Is Pressure Loss?
Airflow resistance in ventilation system is mainly determined by air speed in this system. Resistance grows with speed enhancing. This phenomenon is called pressure loss. Static pressure, produced by a fan, causes air motion in the ventilation system, which has a certain resistance. The higher is the resistance of such a system, the less is the air consumption, moved by the fan. Calculation of friction losses for air in air ducts, as well as resistance of the networking equipment (a filter, muffler, heater, valve, etc.) can be done with help of tables and diagrams, mentioned in the catalogue. One can calculate general pressure drop, having summed the resistance indices of all the elements of a ventilation system.
Recommended air speed in air ducts:
|Type||Air speed, m/s|
|In main air ducts||6,0 - 8,0|
|In side branches||4,0 - 5,0|
|In air distribution ducts||1,5 - 2,0|
|Supply grills at the ceiling||1,0 – 3,0|
|Exhaust grills||1,5 – 3,0|
Determining air speed in air ducts:
V= L / 3600*F (m/s)
where L is air consumption, m3/hour;
F – is area of section, m2
Loss of pressure in the duct system can be reduced due to enlargement of duct section, which provides comparatively equal speed of air in the whole system. At the figure below it is shown how to provide comparatively equal speed of air in the duct system with the minimal loss of pressure.
In systems with big length of ducts and large number of dampers it is reasonable to locate the fan in the middle of the ventilation system. Such a solution has several advantages. On the one hand, pressure losses are reduced, on the other hand, you can use ducts with smaller section.
Example of calculation of ventilation system:
Start the calculation with designing a draft of the system, showing the location of air ducts, dampers, fans and also the length of air duct segments between T-joint, then calculate the air consumption at every segment of the network. Let’s determine pressure losses for segments 1-6, using the graph of pressure losses in round air ducts, let’s determine the necessary diameters of air ducts and pressure losses in them under condition that it is necessary to provide the permissible air speed.
Segment 1: air consumption through this segment will make 220 m3/hour. The diameter of the air duct is 200 mm, speed is 1.95 m/sec., pressure loss is 0.2 Pa/mx15 m=3 Pa (see the diagram of pressure losses for 1 m of extended air duct.
Segment 2: let’s repeat the same calculations and remember that air consumption through this segment will make 220+350=570 m3/hour. The diameter of the air duct is 250 mm, speed is 3.23 m/sec., pressure loss is 0.9 Pa/mx20 m=18 Pa.
Segment 3: air consumption through this segment will make 1070 m3/hour. The diameter of the air duct is 315 mm, speed is 3.82 m/sec., pressure loss is 1.1 Pa/mx20 m=22 Pa.
Segment 4: air discharge through this area will amount 1570 m3/h. Take up air duct diameter 315 mm, speed 5.6 m/s. Pressure loss 2.3 Pa x 20=46 Pa.
Segment 5: air discharge through this area will amount 1570 m3/h, pressure loss 2.3 Pa/m x 1=2.3 Pa.
Segment 6: air discharge through this area will amount 1570 m3/h. Take up air duct diameter 315 mm, speed 5.6 m/s. Pressure loss 2.3 Pa x 10=23 Pa. Total pressure loss in air ducts will amount 114.3 Pa.
When the last area is calculated determine pressure losses in network elements: in the noise damper CP 315/900 (16 Pa) and in the check valve KOM 315 (22 Pa). Further determine pressure loss in pipe bends to the grids (resistance of 4 pipe bends will amount 8 Pa).
Take up ceiling diffusers PF to the unit which resistance will make up 26 Pa according to the diagram.
Now sum all the values of pressure loss for straight areas of air ducts, network elements, pipe bends and grids. Target value is 186.3 Pa.
We calculated the system and determined that we need the fan which excludes 1570 m3/h of the air with network resistance 186.3 Pa. Taking into account the requested for the system operation characteristics VENTS VKMS 315 will suit us.
Diagram of pressure loss for 1 m of extended air ducts.
Diagram enables to determine the value of pressure loss in the pipe bend using the value of bending angle, diameter and air discharge.
Example: Determine pressure loss for pipe bend of 90°C with diameter 250 mm with air discharge of 500 m3/h. For this purpose find crossing of the vertical line that corresponds to our air discharge with slash which characterizes the diameter 250 mm, and on the vertical line in the left for pipe bend in 90°C and find the value of pressure loss which makes up 2 Pa.