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1. [Flow equalisation, effect of equalisation on BOD mass loading rate]. Consider the volumetric flowrates and BOD concentrations given in the following table: Time period midnight-1am 1am-2am 2am-3am 3am-4am 4am-5am 5am-6am 6am-7am 7am-8am 8am-9am 9am-10am 10am-11am 11am-12 12-1pm 1pm-2pm 2pm-3pm 3pm-4pm 4pm-5pm 5pm-6pm 6pm-7pm 7pm-8pm 8pm-9pm 9pm-10pm 10pm-11pm 11pm-midnight Average flowrate during time period (m³/s) 0.275 0.220 0.165 0.130 0.105 0.100 0.120 0.205 0.355 0.410 0.425 0.430 0.425 0.405 0.385 0.350 0.325 0.325 0.330 0.365 0.400 0.400 0.380 0.345 Average BOD concentration during time period (mg/L) 150 115 75 50 45 60 90 130 175 200 215 220 220 210 200 190 180 170 175 210 280 305 245 180 a) Calculate the equalised average flowrate for the day (flowrate fed to downstream processes). [Ans: 0.307 m³/s = 1105.2 m³/hr] b) Using the graphical method, (use spreadsheet and/or graph paper), calculate the inline storage volume required to equalise the flow rate (flowrate equalisation volume) [Ans: 4110 m³]

1. [Flow equalisation, effect of equalisation on BOD mass loading rate]. Consider the volumetric flowrates and BOD concentrations given in the following table: Time period midnight-1am 1am-2am 2am-3am 3am-4am 4am-5am 5am-6am 6am-7am 7am-8am 8am-9am 9am-10am 10am-11am 11am-12 12-1pm 1pm-2pm 2pm-3pm 3pm-4pm 4pm-5pm 5pm-6pm 6pm-7pm 7pm-8pm 8pm-9pm 9pm-10pm 10pm-11pm 11pm-midnight Average flowrate during time period (m³/s) 0.275 0.220 0.165 0.130 0.105 0.100 0.120 0.205 0.355 0.410 0.425 0.430 0.425 0.405 0.385 0.350 0.325 0.325 0.330 0.365 0.400 0.400 0.380 0.345 Average BOD concentration during time period (mg/L) 150 115 75 50 45 60 90 130 175 200 215 220 220 210 200 190 180 170 175 210 280 305 245 180 a) Calculate the equalised average flowrate for the day (flowrate fed to downstream processes). [Ans: 0.307 m³/s = 1105.2 m³/hr] b) Using the graphical method, (use spreadsheet and/or graph paper), calculate the inline storage volume required to equalise the flow rate (flowrate equalisation volume) [Ans: 4110 m³]

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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1. [Flow equalisation, effect of equalisation on BOD mass loading rate]. Consider the volumetric flowrates and BOD concentrations given in the following table: Time period midnight-1am 1am-2am 2am-3am 3am-4am 4am-5am 5am-6am 6am-7am 7am-8am 8am-9am 9am-10am 10am-11am 11am-12 12-1pm 1pm-2pm 2pm-3pm 3pm-4pm 4pm-5pm 5pm-6pm 6pm-7pm 7pm-8pm 8pm-9pm 9pm-10pm 10pm-11pm 11pm-midnight Average flowrate during time period (m³/s) 0.275 0.220 0.165 0.130 0.105 0.100 0.120 0.205 0.355 0.410 0.425 0.430 0.425 0.405 0.385 0.350 0.325 0.325 0.330 0.365 0.400 0.400 0.380 0.345 Average BOD concentration during time period (mg/L) 150 115 75 50 45 60 90 130 175 200 215 220 220 210 200 190 180 170 175 210 280 305 245 180 a) Calculate the equalised average flowrate for the day (flowrate fed to downstream processes). [Ans: 0.307 m³/s = 1105.2 m³/hr] b) Using the graphical method, (use spreadsheet and/or graph paper), calculate the inline storage volume required to equalise the flow rate (flowrate equalisation volume) [Ans: 4110 m³]
Transcribed Image Text:1. [Flow equalisation, effect of equalisation on BOD mass loading rate]. Consider the volumetric flowrates and BOD concentrations given in the following table: Time period midnight-1am 1am-2am 2am-3am 3am-4am 4am-5am 5am-6am 6am-7am 7am-8am 8am-9am 9am-10am 10am-11am 11am-12 12-1pm 1pm-2pm 2pm-3pm 3pm-4pm 4pm-5pm 5pm-6pm 6pm-7pm 7pm-8pm 8pm-9pm 9pm-10pm 10pm-11pm 11pm-midnight Average flowrate during time period (m³/s) 0.275 0.220 0.165 0.130 0.105 0.100 0.120 0.205 0.355 0.410 0.425 0.430 0.425 0.405 0.385 0.350 0.325 0.325 0.330 0.365 0.400 0.400 0.380 0.345 Average BOD concentration during time period (mg/L) 150 115 75 50 45 60 90 130 175 200 215 220 220 210 200 190 180 170 175 210 280 305 245 180 a) Calculate the equalised average flowrate for the day (flowrate fed to downstream processes). [Ans: 0.307 m³/s = 1105.2 m³/hr] b) Using the graphical method, (use spreadsheet and/or graph paper), calculate the inline storage volume required to equalise the flow rate (flowrate equalisation volume) [Ans: 4110 m³]
1.b) • Step 1 The first step is to develop a cumulative volume curve of the wastewater flowrate expressed in cubic meters. The cumulative curve is obtained by eumming the hourly values to obtain a cumulative volume curve. • Step 2 The second step is to prepare a plot of the cumulative volume. Examples of cumulative volume curves (also called inflow mass curves) are shown in the diagram below taken from Metcalf and Eddy. The slope of the line drawn from the origin till the end point of the cumulative volume curve represents the average flow rate (equalised flowrate) for the entire day! 2 M Average daily flowrate Step-by-step guidance Inflow mass diagram N Required equalization volume Inflow mass diagram M Time of day Average daily flowrate Coc= N (a) Flowrate pattern A (b) Flowrate pattern B Figure 1 - Schematic examples of cumulative volume plots • Step 3 The required storage volume is determined by drawing a line (or lines) that is (are) tangent to cumulative volume curve and parallel to the average flow rate line (see examples in figure 1). The required storage volume is given by the vertical distance from the tangency point to the average flow rate line (pattern A in Figure 1) or to the other tangent line (pattern B in Figure 1). Required equalization volume 1.c) There are different calculation methods to do this, but the simplest way to perform the necessary calculations is to start with the time period when the equalisation basin/tank is empty. In this particular problem the equalisation basin is empty at 8am (tangency point) hence the calculations should start at 8am. Vsc = Vsp + Vic - Voc With: Vsc = volume in basin at the end of current period; Vsp = volume in basin at the end of previous period; Vie Inflow volume during current period; Voc = Outflow volume during current period (= equalized flow rate time) • Step 1 The first step is to compute the volume in the equalisation basin at the end of each time period by using the following formula: Vic. Cic+Vsp Csp Vic + Vsp • Step 2 The second step is to compute the average concentration leaving the storage basin during each time period by using the following formula: /kgy Mass loading rate()= h M With: Coc = average BOD concentration in the outflow from storage basin during current time period; Cic = average BOD concentration in the inflow wastewater volume; Cap = average BOD concentration in the storage basin at the end of the previous time period; • Step 3 The third step is to compute the hourly mass loading using the following: (Coc. 9/m³). Equalised flowrate() 1000 2 • Step 4 The dampening effect of flow equalisation on BOD mass loading can be seen by plotting the unequalised and equalised hourly mass loadings.
Transcribed Image Text:1.b) • Step 1 The first step is to develop a cumulative volume curve of the wastewater flowrate expressed in cubic meters. The cumulative curve is obtained by eumming the hourly values to obtain a cumulative volume curve. • Step 2 The second step is to prepare a plot of the cumulative volume. Examples of cumulative volume curves (also called inflow mass curves) are shown in the diagram below taken from Metcalf and Eddy. The slope of the line drawn from the origin till the end point of the cumulative volume curve represents the average flow rate (equalised flowrate) for the entire day! 2 M Average daily flowrate Step-by-step guidance Inflow mass diagram N Required equalization volume Inflow mass diagram M Time of day Average daily flowrate Coc= N (a) Flowrate pattern A (b) Flowrate pattern B Figure 1 - Schematic examples of cumulative volume plots • Step 3 The required storage volume is determined by drawing a line (or lines) that is (are) tangent to cumulative volume curve and parallel to the average flow rate line (see examples in figure 1). The required storage volume is given by the vertical distance from the tangency point to the average flow rate line (pattern A in Figure 1) or to the other tangent line (pattern B in Figure 1). Required equalization volume 1.c) There are different calculation methods to do this, but the simplest way to perform the necessary calculations is to start with the time period when the equalisation basin/tank is empty. In this particular problem the equalisation basin is empty at 8am (tangency point) hence the calculations should start at 8am. Vsc = Vsp + Vic - Voc With: Vsc = volume in basin at the end of current period; Vsp = volume in basin at the end of previous period; Vie Inflow volume during current period; Voc = Outflow volume during current period (= equalized flow rate time) • Step 1 The first step is to compute the volume in the equalisation basin at the end of each time period by using the following formula: Vic. Cic+Vsp Csp Vic + Vsp • Step 2 The second step is to compute the average concentration leaving the storage basin during each time period by using the following formula: /kgy Mass loading rate()= h M With: Coc = average BOD concentration in the outflow from storage basin during current time period; Cic = average BOD concentration in the inflow wastewater volume; Cap = average BOD concentration in the storage basin at the end of the previous time period; • Step 3 The third step is to compute the hourly mass loading using the following: (Coc. 9/m³). Equalised flowrate() 1000 2 • Step 4 The dampening effect of flow equalisation on BOD mass loading can be seen by plotting the unequalised and equalised hourly mass loadings.
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