Fundamentals of Open Channel Flow 1st Edition Moglen Solutions Manual Chapter 1: Introductory Material - Solutions 1-1. What slope would lead to a 1.0 percent difference between depth in the vertical plane rather than depth measured perpendicular to the channel bottom? Compare this slope to the observation that a channel slope of S0=0.01 m/m is generally considered quite steep for open channel flow. Solution: If θ is the angle between the horizontal plane and the plane of the channel then, θ cos x
Thus, 1.01x
or, in terms of rise/run , θ = 8.1o S = tan(8.1o) = 0.14 m/m v v Comparing this number to a channel slope of S0=0.01 m/m we see that the slope corresponding to a 1.0 percent difference between depths is more than an order of magnitude larger. 1-2. Using Bernoulli’s equation, write the energy balance in general terms for flow in an open channel from location 1 to 2 where hL is the head loss between these two locations. Simplify the equation by taking the perspective of a point on the water surface at both locations. Note: your solution should show that the pressure term from Bernoulli’s equation is not relevant for open channel flow. Solution: p v 2 p v 2 1 1 z 2 2 z h γ 2g 1 γ 2g 2 L If we take a point on the water surface at both locations, the p1 equals p2 equals atmospheric pressure, and thus these terms may be cancelled from both sides of the equality, 2 1 z 2 2 z h 2g 1 2g 2 L The remaining equation if y is substituted for z and if hL is set to zero, forms the basis for the specific energy equation which is the focus for Chapter 2. 1-2 1-3. Parts a), b), and c) require simple multiplication/division and/or addition/subtraction to solve. The reader is cautioned to be report the final answer very carefully in terms of significant digits. a) If the density of water is 1000 kg/m3 and gravitational acceleration is 9.81 m/s2, what is the unit weight of water? b) If the density of water is 1.0x103 kg/m3 and gravitational acceleration is 9.81 m/s2, what is the unit weight of water? c) The cross -sectional area of a channel is broken into three separate sub -areas with the following sizes: 1.3 m2, 0.92 m2, and 15 m2. What is the total cross -sectional area of the channel? Solution: a) The unit weight of water is the product of density and gravitational acceleration so, γ ρg 1000 9.81 9810 N Since density is given with one significant figure. The answer has one significant figure resulting in: 10,000 N. b) The new statement gives density with two significant figures, so the answer becomes: 9800 N. c) The calculator -based sum of the three provided numbers is 17.22. However, the number “15” indicates uncertainty in the “ones” place of the number. This same uncertainty needs to be conveyed in the answer, so the correct answer is 17 m2. 1-3
Thus, 1.01x
or, in terms of rise/run , θ = 8.1o S = tan(8.1o) = 0.14 m/m v v Comparing this number to a channel slope of S0=0.01 m/m we see that the slope corresponding to a 1.0 percent difference between depths is more than an order of magnitude larger. 1-2. Using Bernoulli’s equation, write the energy balance in general terms for flow in an open channel from location 1 to 2 where hL is the head loss between these two locations. Simplify the equation by taking the perspective of a point on the water surface at both locations. Note: your solution should show that the pressure term from Bernoulli’s equation is not relevant for open channel flow. Solution: p v 2 p v 2 1 1 z 2 2 z h γ 2g 1 γ 2g 2 L If we take a point on the water surface at both locations, the p1 equals p2 equals atmospheric pressure, and thus these terms may be cancelled from both sides of the equality, 2 1 z 2 2 z h 2g 1 2g 2 L The remaining equation if y is substituted for z and if hL is set to zero, forms the basis for the specific energy equation which is the focus for Chapter 2. 1-2 1-3. Parts a), b), and c) require simple multiplication/division and/or addition/subtraction to solve. The reader is cautioned to be report the final answer very carefully in terms of significant digits. a) If the density of water is 1000 kg/m3 and gravitational acceleration is 9.81 m/s2, what is the unit weight of water? b) If the density of water is 1.0x103 kg/m3 and gravitational acceleration is 9.81 m/s2, what is the unit weight of water? c) The cross -sectional area of a channel is broken into three separate sub -areas with the following sizes: 1.3 m2, 0.92 m2, and 15 m2. What is the total cross -sectional area of the channel? Solution: a) The unit weight of water is the product of density and gravitational acceleration so, γ ρg 1000 9.81 9810 N Since density is given with one significant figure. The answer has one significant figure resulting in: 10,000 N. b) The new statement gives density with two significant figures, so the answer becomes: 9800 N. c) The calculator -based sum of the three provided numbers is 17.22. However, the number “15” indicates uncertainty in the “ones” place of the number. This same uncertainty needs to be conveyed in the answer, so the correct answer is 17 m2. 1-3
