# The Power of Planning - Spaces of Control and Transformation by Oren Yiftachel, Jo Little, David Hedgcock, Ian Alexander PDF

By Oren Yiftachel, Jo Little, David Hedgcock, Ian Alexander

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Solution: 1. 135 MGD × 1,000,000 = 135,000 gpd 2. 6) where Q = Cubic feet per second (cfs). V = Velocity in feet per second (ft/sec). A = Area in square feet (ft2). 66 Problem: Find the flow in cubic feet per second in an 8-in. line if the velocity is 3 ft/sec. Determine the cross-sectional area of the line in square feet. Start by converting the diameter of the pipe to inches. ; therefore, the radius is 4 in. 33 ft. 3. 342 ft2 4. 67 Problem: Find the flow in gallons per minute when the total flow for the day is 75,000 gpd.

39 Problem: Convert 57°C to °F. 40 Problem: Convert 28°F to °C. ) is also involved. When a wastewater characterization study is required, pertinent data are often unavailable. When this is the case, population equivalent or unit per capita loading factors are used to estimate the total waste loadings to be treated. If we know the biochemical oxygen demand (BOD) contribution of a discharger, we can determine the loading placed on the wastewater treatment system in terms of equivalent number of people.

1500 gpm – 325 gpm = 1175 gpm Calculate how long until full, or detention time, and change minutes to hours. 2 hr Chemical Addition Conversions One of the most important water/wastewater operator functions is to make various chemical additions to unit processes. In this section, we demonstrate how to calculate the required amount of chemical (active ingredient and inactive ingredient), dry chemical feed rate, and liquid chemical feed rate. 1 mg/L of ferric chloride. 15 MGD. How many pounds of ferric chloride will be needed each day?