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The "two-thirds" Rule for Locatiing Sensors to Control Variable Flow Systems
November 4, 2007
Recently, I was out in the field scoping out an existing facility for retrocommissioning opportunities. One of the operating engineers pointed out that the differential pressure sensor controlling the chilled water distribution pumps was located so that it sensed the pressure across the mains leaving the plant. He had heard that you could save energy by locating the sensor two thirds of the way between the pumps and the most remote load, so he thought that relocating the sensor might represent an opportunity. But, he wasn't exactly sure why. For one thing, what did "two thirds of the way to the most remote load" really mean; physical distance or feet of pipe or something else? He also was wondering why "two thirds" instead of "three quarters" or "seven eights" or "fifteen sixteenths"? Finally, he was curious about exactly how the point where the sensor measured pressure impacted the energy consumption of the system. He knew that both the pressure and flow a pump produced impacted the power it required, all as stated by the equation for pump power.
But he wasn't sure how the relationships applied in a working system. Finally, he wondered if there were other things to consider. For instance, if the sensor was located at remote point in the system, what would happen if someone isolated that portion of the system for service? And from a practical standpoint, it seemed like it would be a lot harder to remember the location of a sensor at some remote point in a 500,000 square foot facility versus a location that was in the immediate vicinity of the pumps it served.
These are all good questions and come up a lot in the field when I start talking about using a remote pressure sensor to optimize the energy consumption of a variable flow system, be it an air system or a water system like we are discussing here. So, I thought the subject might make good fodder for a few blog posts.
As a starting point, lets consider a simple system like the one illustrated in the system diagram below.
Our discussion will focus on the distribution piping network served by Pump P1; i.e. the flow path from A through B, C, D, E, F, and back to A. For the purposes of our discussion, I've made a few simplyfying assumptions.
The reference pressure is the pressure established by the expansion tank at the suction of pump P1 and is assumed to be 15 psi (6.5 ft.w.c.).
The loads served by the system are identical; specifically at full load, both AHU1 and AHU2 require 400 gpm and a differential pressure of 20 ft.w.c. at the point where they connect.
The pumps, piping network, and air handling units are all at about the same elevation; thus the effects of elevation on the pressure readings can be ignored.
Piping lenghts will be discussed in terms of equivalent feet. In other words, when I say that the distance from point A to point B is 200 equivalent feet of pipe, I'm saying that if I were to convert the resistance due to flow of all of the fittings between point A and B to an equivalent length of straight pipe, and add it to the actual length of straight pipe, it would be the same as 200 feet of straight pipe with no fittings in it. If there were no fittings, then the distance would literally be 200 feet, but as the number of fittings increased, the physical distance associated with 200 equivalent feet of straight pipe would be reduced.
One of the key concepts that you need to understand with regard to this topic is that the pressure required to move water through a system is a function of the flow in the system. This means that if the control valves on AHU1 and 2 are both closed, the pressure reading at the discharge of pump P1 would be identical to the reading at the tee where the supply piping splits to serve the two AHUs, even though the tee is three hundred equivalent feet away from the pump. (Remember, we have assumed that both points are at the same elevation, so the impact of elevation on pressure can be ignored). If the control valves in the air handling units start to open, the pressure at the tee will start to drop relative to the pressure at the pump. The magnitude of the drop will be a function of the flow rate. Common wisdom is that this relationship is a "square law".
In other words, if you reduce the flow by 50%, the pressure drop associated with the flow drops to 25% of what it was (50% of 50%), assuming nothing changed in the system (no valves changed position, the piping was not modified, etc.). Research has shown that for most real piping systems the exponent is more like 1.89 versus 2, but out in the field, we probably couldn't measure the difference as illustrated by the following graph.
So for our purposes, we can still think of it as the "square law" instead of the "one point eight nine law".
Now, lets look at the pressures required by our system in two different operating modes. In the first operating mode, the system is at full load. Both AHUs are using 400 gpm and the pump needs to deliver 20 ft.w.c. at the tee serving the AHUs while moving 800 gpm through the mains to and from the air handling units. The total length of the piping circuit is 600 equivalent feet; 300 equivalent feet from the pump discharge to the tee that serve the units and 300 equivalent feet back to the pump suction.
In the second operating mode, one AHU has shut down but the other is still operating at full load. This means that the pump still needs to deliver 20 ft.w.c. at the tee serving the unit, but only needs to move 400 gpm through the piping network. The two operating modes are compared in the following table.
Note that the friction rate for the second mode is significantly different from that associated with the first mode. In the next post, we will look at the results in more detail.
Posted by David Sellers on November 4, 2007 | Comments (0)
