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G121 SERIES
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G121 SERIES
TRANSCRIPT
Relationship of Elevation & Pressure
INTRODUCTION This is the second segment in Hydrostatics. You can use the link below to get to the first segment. In our first segment on Hydrostatics, we reviewed five (5) basic principles that are applied to fluids. In this segment we will focus on how these hydraulic principles are applied with regard to fire protection systems. Of the five principles that we previously discussed; the third principal was that “pressure created by a liquid in an open container is directly proportional to the depth of the liquid.” Let’s explore that with an example. PRESSURE & ELEVATION In our example, we used water as the incompressible fluid. The specific weight of water is 62.4 lbf/cu ft; this translates to an important constant 0.433 psi / ft. This constant means that as we go deeper into a column of water, the pressure will increase 0.433 psi for every foot we go down. This is an important one to remember. PRESSURE OF ELEVATED TANK As an example, what’s the pressure developed at the ground level if the water level in an elevated tank is 85 feet above a ground level outlet? The 3rd principle is applied; Pressure (psi) = w (psi/ft) x h (ft) Where w = specific weight / unit height h = height from a datum to the level of the liquid In IP units; Pressure = 0.433 psi/ft x 85 ft = 36.8 psi In SI units; = 0.098 bar/m x 26 m = 2.55 bar If the liquid outlet is not at the base of the tower, then that change in elevation is added (or deducted) from the reference datum. Let’s expand the example, If the liquid outlet is further below the reference datum, add the additional height. If the outlet is an additional 36 ft below the base of the tower … [FADE 05A INTO 05B] In IP units; Pressure = 0.433 psi/ft x (85+36) ft = 52.39 psi STANDPIPE PRESSURE EXAMPLE What if we need to supply 100 psi (6.9 bar) at the roof of a high rise? But... NFPA 14 Edition 2016 Section 7.2.3.2 limits connections to 175 psi (12.1 bar). So here, we need to supply 100 psi (6.9 bar) at the roof of a high-rise, but the pressure at the bottom of the high-rise connection cannot be more than 175 psi (12.1 bar). We know the pressure increases, we go deeper in the water column. How tall can that high-rise be then? If each floor is about 10-feet (about 3.0m) in height, how many stories can the highrise be before we’re over our maximum limit? You can only lose 75 psi, that’s the 175 psi limit less 100 psi between the standpipe connection on the ground floor and the roof connection. In SI, that becomes 5.2 bar. 75 psi / (0.433 psi/ft) = 173 ft or about 17 stories In SI, that’s 5.2 bar / (0.098 bar/m) = 53.1 m, or about 17 floors Wait, that’s not all. That’is the static condition. Once the water is flowing, friction loss occur. By applying Hazen-Williams equation (the subject of another talk), an equivalent length of 180 ft of 4” standpipe (including all of the fittings) could easily account for ~14 Psi in additional loss if 500 gpm is required at the rooftop. In SI, we’re talking of a loss around 0.97 bar. So, including the friction loss through the pipe, we really only have 61 psi (or 4.2 bar) to work with. Now, that’s 61 psi / (0.433 psi/ft) = 140 ft or about 14 floors. In SI, that’s 4.2 bar / (0.098 bar/m) = 42.9 m, or about 14 floors. In this segment, we covered how height and pressure are related. We looked specifically at a standpipe system as an example to determine how tall a building could be and still accommodate no more than 75 psi (or 5.2 bar) difference between the top and bottom of the building. In our next segment, we’ll talk about the difference between pressure and head. I’m Ed Henderson; this is MeyerFire University.
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