Closed Loop Chilled Water System Pressure
Chilled water systems circulate chilled water in a closed loop from chillers to air handling units and fan coil units. I’ve always wondered how to decide the pressure rating of valves and fittings. So, I decided to find out the closed loop chilled water system pressure.
Generally, most chilled water systems operate at a pressure of around 10-12 bar. Some chilled water systems work at more than 16 bar of pressure but they rarely exceed 20 bar. High pressure chilled water systems usually use heat exchangers to reduce the pressure below 20 bar.
Components such as valves, fittings and cooling coils have a certain pressure rating. Typically, they are rated at either PN10, PN16, PN20 or PN25 which represent 10 bar, 16 bar, 20 bar and 25 bar respectively. The higher the pressure rating, the higher the initial cost.
| Pressure Rating | Maximum Working Pressure |
|---|---|
| PN10 | 10 Bar |
| PN16 | 16 Bar |
| PN20 | 20 Bar |
| PN25 | 25 Bar |
To know what pressure rating should be used, HVAC engineers often add the static pressure and the pump head to get the total pressure in a closed loop chilled water system.
But, is that the right way to calculate it?
To find out how to calculate the total pressure in a closed loop chilled water system, I spent many hours researching Hydronics and pipes and the relationship between flow and pressure. Furthermore, I’ve spoken to a few HVAC seniors to understand the principle behind flow and pressure.
What is Pressure?
Pressure is the amount of force exerted on a surface. In a closed loop chilled water system, there are a few types of pressures that need to be accounted for when deciding the pressure rating of the equipment.
Static Pressure
Static pressure is the pressure exerted on pipes, valves and fittings due to the weight of the column of water. Imagine a valve being pressed by the water above it. This pressure increases with the water height.
The formula used to calculate static pressure is:
Ps = ρgh
where,
Ps = Static pressure, Pa
ρ = Density of water, 997 kg/m3
g = Gravity, 9.81 m/s2
h = Height, m
Dynamic Pressure
When a valve is opened halfway, its shaft blocks some of the chilled water. This additional pressure exerted on the shaft is called dynamic pressure.
The formula used to calculate dynamic pressure is:
Pd = 0.5ρv2
where,
Pd = Dynamic pressure, Pa
ρ = Density of water, 997 kg/m3
v = Water velocity, m/s
Since water velocity is involved, dynamic pressure only occurs when the chilled water is moving. However, if the shaft of the valve gives away and flows together with the chilled water at the same velocity, the dynamic pressure is zero.
For instance, chilled water is typically designed at 3 m/s. Using the above formula, the dynamic pressure is 4486.5 Pa (0.65 psi). Due to the low water velocity in chilled water systems, this dynamic pressure is negligible when compared to other pressures.
Stagnation Pressure
The sum of static pressure and dynamic pressure is stagnation pressure. Stagnation pressure is also known as total pressure and the formula used to calculate total pressure is:
Pt = Ps + Pd
Pt = ρgh + 0.5ρv2
Again, if the dynamic pressure is negligible, static pressure is equivalent to stagnation pressure. For high water velocity applications, the dynamic pressure could be significant and therefore, need to be accounted for.
Fill Pressure
Fill pressure is the additional pressure added to the chilled water system in order to maintain a minimum pressure at the highest point.
When a chilled water system is first started, the chilled water pipes have to be filled with water. Upon the entire chilled water pipeline is filled with water, there is no pressure at the highest point. This can cause air pockets to be left there.
Hence, to eliminate air pockets, the system must be slightly pressurized and this additional pressure is known as fill pressure or minimum pressure.
Pump Pressure
Chilled water pumps exert pressure when running and this pressure is known as pump pressure or head pressure. Pipes, valves and fittings nearer to the pump experience a higher pump pressure than those that are further away from the pump.
Valves that are further away from the pump experience less pump pressure because there is a head pressure loss along the way. The head pressure loss is due to the friction present in the pipeline.
As the chilled water travels to the farthest air conditioning unit, the water pressure drops accordingly. This is illustrated in the below diagram:
Dead Head Pressure
Dead head pressure is the maximum pump pressure that can occur when all valves are shut. This is a phenomenon known as pump dead head; the pump is running but there is no flow.
Regardless of the pump impeller size, as the flow rate drops, the pump head increases and thus, the pump pressure increases.
Based on the above pump curve, if the flow rate drops from the design operating point of 1050 GPM to 600 GPM, the pump head increases from 4.5 bar to 4.7 bar.
In some cases, the maximum pump pressure could be double its design pump pressure. Hence, dead head pressure must be accounted for when deciding the pressure rating of the water-side equipment in a chilled water system.
In variable flow systems, the minimum flow of the pump is always maintained either via a bypass pipe or a 3-way valve. Hence, dead head will not occur. Nonetheless, it is wise to design for the worst condition.
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Chilled Water System Working Pressure Calculation
In a closed loop chilled water system, pressure is always higher on lower floors than on upper floors due to static pressure. However, when the chilled water pump is running, should the system pressure be higher?
1. Calculate the Static Pressure
To determine the total pressure or working pressure of a chilled water system, the first pressure to calculate is the static pressure. This pressure is calculated when the chilled water pump is not running.
Let’s use the following diagram as an example for the calculation:
From the above diagram, the column of water above AHU3 and its associated valve is as high as 150 m (492 ft). Using the formula mentioned earlier, the static pressure at AHU3 can be calculated as follows:
Ps = ρgh
Ps = (997)(9.81)(150)
Ps = 1467 kPa (213 psi)
Therefore, the static pressure exerted on AHU3 and its associated valves, pipes and fittings is about 213 psi which is equivalent to 14.7 bar. For AHU2, the static pressure is 142 psi and AHU1 experiences almost no static pressure because it is the highest point in the system.
2. Add the Fill Pressure
As mentioned earlier, in order to eliminate air pockets, the highest point in the system should see a minimum pressure of 20 psi. This additional fill pressure is exerted on all equipment. Hence, all AHUs need to add 20 psi of fill pressure on top of their static pressure.
With the fill pressure, the pressure exerted on AHU3 becomes 233 psi which is equivalent to 16.1 bar. Before adding the fill pressure, the valve could be rated at PN16 but now, the valve needs to be rated at PN20 in order to withstand the pressure.
But, we’re not done yet.
3. Add the Pump Pressure
The amount of pump pressure (head pressure) experienced by each AHU is different and it is based on the pump head. Recalling that pump pressure is pump head minus head loss, both parameters must be determined in order to calculate the resulting pump pressure.
Assuming that the chilled water pump has a pump head of 40 meters and the head loss from the pump to the valve of the AHU1 is 10 meters, the pump pressure can be calculated as follows:
Pump Pressure = Pump Head – Head Loss
Pump Pressure = 40 – 10
Pump Pressure = 30 m
Using the hydrostatic pressure formula, the pump pressure can be converted as follows:
P = ρgh
P = (997)(9.81)(30)
P = 293.4 kPa (42.6 psi)
With the pump pressure, static pressure and fill pressure, the total pressure can be determined.
4. Calculate the Total Pressure
From the above diagram, AHU3 and its associated pipes, valves and fittings experience the greatest pressure. There is a static pressure of 213 psi, a fill pressure of 20 psi and a pump pressure of 55.3 psi.
As a result, the total pressure is 288.3 psi which is equivalent to 19.9 bar.
Therefore, the pressure rating of the valves, flanges, pipes and cooling coils at that level must be at least PN25. The calculated total pressure shows that it is too risky to use PN20. A safety factor must be included when deciding the pressure rating of the equipment.
5. Check the Dead Head Pressure
To check the dead head pressure, the pump curve must be examined. Assume that the pump head of the chilled water pump in our example changes from 40 meters to 45 meters during low-flow conditions, the pump pressure can be recalculated and the result is as follows:
As shown in the above diagram, when the pump head increases to 45 meters, the pump pressure at each AHU increases as well. This pressure increase must be examined to make sure the decided pressure rating can accommodate it.
Conclusion
Experienced engineers can calculate the total pressure of a chilled water system very quickly. Basically, 150 meters of water height is equivalent to approximately 15 bar.
Then, a pump head of 40 meters is equivalent to approximately 4 bar. With a fill pressure of about 1-1.5 bar, the pressure rating is around 20-21 bar.
Hence, a pressure rating of PN25 for the equipment is required.
However, this is applied to water only since the rule of thumb is based on water density. If you’re dealing with other types of fluids, for example, glycol water mixture in a chilled water system with thermal energy storage tanks, then the result would be different.
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Hello, Yu. HOpe all is well.
That is great content, keep that up.
I work with critical systems, specially datacenters.
CHW is a closed loop system and we can often apply closed loop in the distribution for redundancy & availability purposes allowing for manitenance or helping to cope for failures.
Suppose I have two supply and two return mains/headers coming down on opposite sides of the building (supply return ||========|| supply return), each floor I attend by creating a closed loop that connects both mains (opposite raisers).
Does the static parcel of the head pressure remains the same through out the loop?
The loop behave like a plenum, right? Where I understand static remains about the same in every point, would the static be equal to the head I need for the most critical consuming point in that floor loop? FInding hard to get reliable info in this scenario, would you suggest any reference?
Hi Alex,
I don’t quite understand. Can you elaborate more or email me a drawing or something? I’m interested to have a look. Frankly, information on closed loop pressure is scarce.