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A concrete channel to bring water to Crystal Lake is being designed. It will have vertical walls and be 15 feet wide. It will be 10 feet deep, have a slope of .0015 feet/foot, and a roughness coefficient of .014. How deep will the water be when 1000 cubic feet per second is flowing through the channel? To solve this problem, we can use Manning's equation 1.486 AR/3s\/2 N where Q is the flow of water (cubic feet per second), N is the roughness coefficient (unitless), A is the area (square feet), S is the slope (feet/foot), and R is the hydraulic radius (feet).

A concrete channel to bring water to Crystal Lake is being designed. It will have vertical walls and be 15 feet wide. It will be 10 feet deep, have a slope of .0015 feet/foot, and a roughness coefficient of .014. How deep will the water be when 1000 cubic feet per second is flowing through the channel? To solve this problem, we can use Manning's equation 1.486 AR/3s\/2 N where Q is the flow of water (cubic feet per second), N is the roughness coefficient (unitless), A is the area (square feet), S is the slope (feet/foot), and R is the hydraulic radius (feet).

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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A concrete channel to bring water to Crystal Lake is being designed. It will have vertical walls and be 15 feet wide. It will be 10 feet deep, have a slope of .0015 feet/foot, and a roughness coefficient of .014. How deep will the water be when 1000 cubic feet per second is flowing through the channel? To solve this problem, we can use Manning's equation 1.486 "AR²/³g\/2 N where Q is the flow of water (cubic feet per second), N is the roughness coefficient (unitless), A is the area (square feet), S is the slope (feet/foot), and R is the hydraulic radius (feet). The hydraulic radius is the cross-sectional area divided by the wetted perimeter. For square channels like the one in this example, Hydraulic radius = depth x width /(2.0 x depth + width) To solve this problem, design a program that allows the user to guess a depth and then calculates the corresponding flow. If the flow is too little, the user should guess a depth a little higher; if the flow is too high, the user should guess a depth a little lower. The guessing is repeated until the computed flow is within 0.1% of the flow desired. To help the user make an initial guess, the program should display the flow for half the channel depth. Note the example run: At a depth of 5.0000 feet, the flow is 641.3255 cubic feet per second. Enter your initial guess for the channel depth when the flow is 1000.0000 cubic feet per second Enter guess> 6.0 Depth: 6.0000 Flow: 825.5906 cfs Target: 1000.0000 cfs Difference: 174.4094 Error: 17.4409 percent Enter guess> 7.0 Depth: 7.0000 Flow: 1017.7784 cfs Target: 1000.0000 cfs Difference: -17.7784 Error: -1.7778 percent Enter guess> 6.8
Transcribed Image Text:A concrete channel to bring water to Crystal Lake is being designed. It will have vertical walls and be 15 feet wide. It will be 10 feet deep, have a slope of .0015 feet/foot, and a roughness coefficient of .014. How deep will the water be when 1000 cubic feet per second is flowing through the channel? To solve this problem, we can use Manning's equation 1.486 "AR²/³g\/2 N where Q is the flow of water (cubic feet per second), N is the roughness coefficient (unitless), A is the area (square feet), S is the slope (feet/foot), and R is the hydraulic radius (feet). The hydraulic radius is the cross-sectional area divided by the wetted perimeter. For square channels like the one in this example, Hydraulic radius = depth x width /(2.0 x depth + width) To solve this problem, design a program that allows the user to guess a depth and then calculates the corresponding flow. If the flow is too little, the user should guess a depth a little higher; if the flow is too high, the user should guess a depth a little lower. The guessing is repeated until the computed flow is within 0.1% of the flow desired. To help the user make an initial guess, the program should display the flow for half the channel depth. Note the example run: At a depth of 5.0000 feet, the flow is 641.3255 cubic feet per second. Enter your initial guess for the channel depth when the flow is 1000.0000 cubic feet per second Enter guess> 6.0 Depth: 6.0000 Flow: 825.5906 cfs Target: 1000.0000 cfs Difference: 174.4094 Error: 17.4409 percent Enter guess> 7.0 Depth: 7.0000 Flow: 1017.7784 cfs Target: 1000.0000 cfs Difference: -17.7784 Error: -1.7778 percent Enter guess> 6.8
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