Static Pressure and Velocity Pressure
Notes on Static pressure and Velocity Pressure
The above mentioned pressures always cross our paths while performing Fan/Pump Sizing. In this article we will look at how they interchange when a flow occurs and how to calculate them.
All examples assume air at 1.2 kg/m³.
1. Pressure in a duct
Three pressures describe air at any point in a duct.
Static pressure (Ps) is the pressure acting on the duct walls. It is the pressure required to overcome friction and fitting losses.
Velocity pressure (Pv) is the energy held in the moving air, equal to ½ρv². It exists because the air has mass and is travelling at a given speed. At standard density that is 0.6v², so 3 m/s gives 5.4 Pa and 10 m/s gives 60 Pa. It rises with the square of velocity, which is why high-velocity systems need to be designed carefully.
Total pressure (Pt) is simply the two added together: Pt = Ps + Pv.
The three pressures at one point
Static pressure pushes on the walls, velocity pressure travels with the air, and total pressure is the two added together.
At 3.5 m/s the velocity pressure is only 7.35 Pa, so it looks ignorable. Take the velocity to 7.0 m/s and it becomes 29.4 Pa.
A fan raises the total pressure of the air passing through it. Fan total pressure (FTP) is the total pressure at the outlet minus the total pressure at the inlet. Fan static pressure (FSP) is that figure minus the velocity pressure at the fan outlet:
Since there are two pressures involved, why do we specify the fans using external static pressure instead of total pressure? This is because the velocity pressure is automatically created when the fan creates the flow.
Lets say we specify a flow rate of 100 l/s with an external static pressure requirement of 50 Pa. When the fan creates a flow of 100 l/s at 50 Pa static pressure and lets say the velocity at the exit of the fan is 5 m/s - So it has created 0.5 x 1.2 x 5 x 5 = 15Pa as well. So the total pressure created by the fan is 65Pa. We don’t care about this velocity pressure as long as we are getting the right flow at the static pressure specified. Velocity pressure is a by product of the flow created by the fan.
Fan total pressure against fan static pressure
The fan develops both, but only the static half is what we specify. Use the interactive model to understand the difference.
We design for 100 L/s at 50 Pa static and we get 15.0 Pa of velocity pressure whether we wanted it or not, so the fan is developing 65.0 Pa total. That is 23 per cent of its duty created automatically. We do not need to specify it, because the flow and the outlet size decide it between them.
2. Static - Velocity pressures interchange
Total pressure falls as air moves downstream, and it falls only due to friction and turbulence.
Static and velocity pressure, on the other hand, trade freely against each other. Whenever the duct cross-section changes, the air changes speed and one converts into the other. If the duct is narrowed, static pressure converts into velocity pressure as velocity increases and vice versa.
Static and velocity pressure trade, total pressure only falls
One duct with a waist in the middle. Look at the pressure gradients to see the effect.
The waist takes the air to 8.2 m/s, so velocity pressure climbs and static pressure gives up the same amount. Nothing is lost in that exchange. What is lost is the steeper total pressure line through the waist, because a 175 mm duct at 8.2 m/s rubs far harder than a 250 mm duct at 4.0 m/s.
3. The open end of a duct
Consider a plain duct discharging into a room. By the time the air reaches the end, the static pressure has been spent on friction. The air still carries its velocity pressure, because it is still moving.
Once it leaves the duct, that kinetic energy disperses into the room as turbulence and eventually heat. Nothing downstream recovers it, so it is a genuine loss and it is called the exit loss. Its value is the velocity pressure of the air at the opening, 0.6v².
That loss forms part of the system total pressure, so it belongs in the FTP calculation. It does not belong in FSP. The fan produces velocity pressure at its own discharge as an unavoidable consequence of moving the air. Where the fan discharge area and the duct area are the same, the exit loss and the fan’s discharge velocity pressure are equal and cancel.
Example. A 10 m straight duct with two elbows has a friction loss of 24.7 Pa. Air travels at 3.5 m/s, giving a velocity pressure of 7.35 Pa. System total pressure is 24.7 + 7.35 = 32.05 Pa. The fan static pressure to specify is 24.7 Pa, because the 7.35 Pa comes with the fan.
The open end, and the loss we cannot get back
A plain duct discharging into a room. The static pressure has been spent on friction, so what leaves the opening is pure velocity pressure.
The exit loss is 7.35 Pa, which is 23 per cent of the 32.05 Pa the system really needs. It still belongs in the fan total pressure. It does not belong in the fan static pressure, because the fan produces the same velocity pressure at its own discharge: where the discharge area and the duct area match, the two cancel each other.
4. Reducing the exit loss: the evasé
Section 3 points to using the velocity pressure at the exit. The exit loss is 0.6v² at the opening, so if we slow the air down before it leaves, we throw away less energy. The change in velocity pressure is converted to static pressure and so the overall static pressure requirement can be reduced.
The fitting that does this is an evasé, a smoothly flared expansion at the discharge. A bellmouth is the same shape used at an inlet; at a discharge it is an evasé, and the two are often confused in specifications.
As the section expands, the air slows and part of its velocity pressure converts back into static pressure. That regained static pressure acts against the system resistance, so the fan has less work to do. But all of the velocity pressure is not converted to static pressure. Expansion generates turbulence, and the fraction actually recovered is called regain coefficient which is typically 0.7 to 0.8 for an included angle of 15° or less. If the angle is greater than this there are chances that the flow separates from the walls and the regain coefficient reduces again.
Example. Fit a 2:1 evasé at 15° to the exit of the same 3.5 m/s duct. Velocity halves to 1.75 m/s and the exit loss falls from 7.35 Pa to 1.84 Pa. The velocity pressure difference is 5.51 Pa, and at a regain coefficient of 0.75 about 4.13 Pa of that returns as static pressure working against the duct friction.
The evasé, and the angle where it stops paying
Increase the discharge area and the air slows before it leaves, so less velocity pressure is thrown away and part of the difference comes back as static pressure. Check also by increasing the angle beyond 15 degrees.
4.13 Pa of the 5.51 Pa returns as static pressure and works against the duct friction, so the fan static pressure drops from 24.70 Pa to 20.57 Pa. This helps reduce the shaft power and lower the noise. But an additional fitting with 315 mm of length is required, which is usually why it does not get built.
5. Velocity changes inside the run
The same conversion works in reverse. Where a duct narrows part way along its run, the air must speed up to carry the same volume flow, and its velocity pressure rises. That energy is converted out of the static pressure available at that point, and the fan has to have supplied it upstream.
So a mid-run contraction increases loss in two ways, first the static to dynamic conversion and also the narrow passage will have higher friction loss for the same flow.
Example. A fan discharges at 3 m/s, giving Pv = 5.4 Pa. Mid-run the duct narrows and the air accelerates to 4 m/s, giving Pv = 9.6 Pa. The fan must supply an extra 4.2 Pa of static pressure purely to create that acceleration, before the higher friction of the smaller duct is counted.
A narrowed mid-run charges the fan twice
Same flow, two ducts.
The conversion costs 4.20 Pa and the fan has to have supplied it upstream. Then the narrow duct needs 2.05 times harder per metre, so 6.0 m of it adds another 5.69 Pa. Together that is 9.89 Pa.
A mid-run expansion does the opposite and gives back static pressure, subject to the same regain coefficient as the evasé. The general point is that the fan pays for every acceleration and recovers part of every deceleration.
6. Terminals and catalogue data
Taking a diffuser or grille pressure drop at face value without reading the column heading is one of the most common causes of an over-pressurised system, because manufacturers publish the figure both ways.
Where the catalogue quotes total pressure drop, that figure is the full energy the air needs at the spigot to get through the terminal and out into the room. The exit loss is already inside it. To get the static pressure the fan has to provide, subtract the velocity pressure of the air arriving at the spigot. Otherwise the exit loss gets counted twice.
Where the catalogue quotes static pressure drop, the manufacturer has already separated the velocity component out. Add that number straight into the duct friction summation.
Example. A diffuser is listed at 25 Pa total pressure drop, with air arriving at the spigot at 3 m/s, so Pv = 5.4 Pa. The static pressure required at that node is 25 − 5.4 = 19.6 Pa. Adding the full 25 Pa to the friction losses over-specifies the fan.
Read the column heading before we use the number
The same diffuser is published both ways.
The 25 Pa is the full energy the air needs at the spigot, so the 5.40 Pa it already carries is inside the figure. Subtract it and the fan has to find 19.60 Pa at that node. Add the whole 25 Pa instead and the fan is scheduled 5.40 Pa high, because the exit loss has been paid for twice.
7. Putting it together
Where the terminal catalogue gives static pressure, the full energy balance for the fan is:
The bracketed term is the section 5 conversion applied between the fan discharge and the terminal spigot. If the spigot velocity pressure is higher than the fan discharge velocity pressure, add the difference, because the fan has to assist the acceleration. If it is lower, subtract it, because the air has slowed and converted kinetic energy back into static pressure. If the two are equal they cancel, and the fan static pressure is duct friction plus the terminal static pressure drop, which is the case in section 3.
Example. Duct friction 50 Pa, diffuser ΔP_s 25 Pa, spigot velocity 4 m/s (Pv = 9.6 Pa), fan discharge velocity 3 m/s (Pv = 5.4 Pa).
The whole fan static pressure, assembled
Duct friction, the terminal's static drop, and the conversion between the fan discharge and the spigot.
The spigot is faster than the fan discharge, so the fan has to assist the acceleration and the 4.20 Pa gets added. If we leave it out, the fan is scheduled 4.20 Pa short of what is required by the system.