p0778

# p0778 - 874 CHAPTER 7 CONSERVATION OF ENERGY 7.78 Chapter 7...

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874 CHAPTER7 . CONSERVATIONOFENERGY 7.78 Chapter 7, Problem 78 From the Ch´ezy-Manning equation, we have u = χ n S 1 / 2 o R 2 / 3 h The volume-flow rate is given by Q = uA ,whe re A is the channel cross-sectional area, so that Q = χ n AS 1 / 2 o R 2 / 3 h We are given Q =2 . 5 m 3 /sec, S o =0 . 0016 and n . 016 . Also, since we are using SI units, χ =1 . 00 m 1 / 3 /sec. Thus, for all cases, 2 . 5 m 3 sec = p 1 . 00 m 1 / 3 / sec Q 0 . 0016 0 . 016 AR 2 / 3 h = X 2 . 5 m 1 / 3 sec ~ AR 2 / 3 h Therefore, AR 2 / 3 h =1m 8 / 3 For a rectangle with b =3 y , A = by y 2 ,R h = by b +2 y = 3 y 2 5 y = 3 5 y Then, from the general equation developed above, AR 2 / 3 h y 2 w 3 5 y W 2 / 3 8 / 3 = y = w 1 3 W 3 / 8 w 5 3 W 1 / 4 m Therefore, the depth is y . 75 m Straightforward calculations show that area, A y 2 , perimeter, P =5 y ,h y d r a u l i c radius, R h y/ 5 and Froude number, Fr = Q/ ( A gy ) ,a A . 69 m 2 ,P . 75 m h . 45 m ,F r . 55 For a semi-circle with α = π , A = 1 8 ( π sin π ) D 2 = π 8 D 2 h = 1 4 w 1 sin π π W D = D 4

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7.78. CHAPTER 7, PROBLEM 78 875 Then, from the general equation developed above, AR 2 / 3 h = π 8 D 2 w D 4 W 2 / 3 =1m 8 / 3 = D = 2 w 8 π W 3 / 8 m Therefore, the diameter is D =2 . 01 m Straightforward calculations show that area, A = π D 2 / 8 , perimeter, P = π D/ 2 ,hydraulic radius, R h = 4 and Froude number, Fr = Q/ ( A ± gD/ 2) ,are A =1 . 59 m 2 ,P =3 . 16 m ,R h =0 . 50 m
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p0778 - 874 CHAPTER 7 CONSERVATION OF ENERGY 7.78 Chapter 7...

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