CHAPTER 1. Problem 1: (a) Trapezoidal channel with side slopes m 1 and m 2 y y

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1 CHPTER Problem : (a) Trapeoidal channel with side slopes m and m ( b + m + m ) T b + ( m ) + m + P b + + m + m D / T R / P (b) Trapeoidal channel with one vertical side Set m and m m in the equations given in part (a). (c) Right triangular channel Set m, b, and m m in the equations given in part (a). Problem : (a) From the problem statement: γ 6.4 lbf/ft Y c depth to centroid ft d ( ft/ft) 4 ft ( ft/ft) 4 ft /ft B using Equation.9 Fp γ Yc 6.4()(4) 499 lbf / foot The e hdrostatic pressure force is normal to the vertical sidewall. (b) From the problem statement: γ 6.4 lbf/ft Y c depth to centroid ft (d +(md) ) / ( ft/ft) (4 +(x4) ) / ft ( ft/ft) 8.94 ft /ft B using Equation.9 Fp γ Yc 6.4()(4) 6 lbf / foot The hdrostatic pressure force is normal to the inclined side. Problem :

2 v.5v s * Let s /. Then v.5v * Define q discharge per unit width. B definition (see Equation.) q vd.5v * d [.5v ] d [ v ] d *.5 * q.5v*.5v*.5v* ( ) Problem 4: [ ] [ ] Noting that q discharge per unit width and s / as in Problem, b definition (see Equation.) ( ) q.5v *.5v * Because >>, we have (-). Therefore.5v* Problem 5: For the velocit distribution given, obviousl, the velocit is imum at the free surface. Substituting in v.5v* s we obtain v.5v*.5v s with s /. * 4

3 Problem 6: Using the results of Problems 4 and 5.5v* v.5v * with s /. Problem 7: We found an expression for in Problem 4. Here we will determine the value of for which v. In other words with s /.5v*.5v * and / (.78).78 The point velocit at distance.7 from the bottom or.6 from the free surface will be equal to the cross sectional average velocit. Therefore, the velocit measured at.6 from the surface will be a good approximation to the average velocit. Problem 8: With s /, we have v.5v * B definition β v d Using the velocit distribution given, for unit width, we can write (.5v *) β d ( ) 5

4 Let us first evaluate the integration d ( )( ) d ( d + [ ] d d + [ ( ) ( ) ] d + d ) d d ( ) + ( ) ( ) (*) lso from Problem 4.5v* (.5v* ) (.5v*) + With >> and (-) ( ) (.5v* ) + (**) Substituting Equations (*) and (**) into the expression for β β [(.5v ) ] * ( ) + ( ) ( ) (.5v* ) + Noting (-) and simplifing + + ( ) ( ) β + From Problem 6 + 6

5 v Substituting this into the expression for β and rearranging β v + Problem 9: / / τ.7 v *.6m / s ρ From Problem 4 (.94).5v.5v (.5)(.6) s (.) s q (.94)(.4).m / s * * From Problem 5 v. (.94) (.) v β + s + (.)..4m / s g m m gm Rate of momentum transfer βρ q (.)( )(. )(.4 ) 894 / m m s s s v α + v + (.) (.). ρ g m m Nm Rate of inetic transfer α q (.) ( )(. )(.4 ) 6 / m m s s s Problem : m x m m 4 m x 4 m 6 m.5( m x m) m 7

6 Q + +.5() +.5(6) +.() 7. m / s Q Q m / s Problem : From Equation () +.5 (6) +. β.79 (47) Then b using Equation. with ρ g/m ().4 Rate of momentum transfer βρ Q.4()(7.)(.79) 4,8 g m / s Liewise, from Equation. α () +.5 (6) +. ().79 (47).45 Then b using Equation.9 ρ Rate of inetic energ transfer α Q Problem :.45 (7.)(.79) 8,57 g m / s Q Q. fps ( b + m).5(5 +.5) ( b + m).5(5 +.5) 5.6 R. 86 P b + + m 5 + (.5) + 9. ( b + m).5(5 +.5) 5.6 D. ft b + m ft (a) Using Equation. 4R 4R 4(.8)(.86) R e 5 ν ν.7 The flow is turbulent.,95, (b) Using Equation.4 F r gd

7 The flow is subcritical. Problem : (a) Nonuniform (b) Nonuniform (c) Uniform (d) Nonuniform Problem 4: (a) Unstead (b) Stead (c) Stead Problem 5: erif that ( Y C ) Consider a horiontal strip of flow area having a length of b and thicness of d and extending from one side of a flow section to the other side. Suppose the vertical distance between the centroid of this strip and the channel bottom is. Then b definition ( Y C ) ( ) bd Using the Leibnit rule, we can write this as ( Y C ) ( ) bd + ( ) bd ( ) bd The last two terms on the right hand side are ero. Then ( Y ) C ( ) bd ( bd) ( bd) x x Since and b are not functions of x at a given section, the last term on the right hand side is ero. lso, since is not a function of and b and d are not functions of x ( Y C ) ( bd) bd x 9