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Test 3 with Answers - Civil and Environmental Engineering Fluid Mechanics | CEE 3500, Exams of Civil Engineering

Material Type: Exam; Professor: Urroz; Class: Civil and Environmental Engineering Fluid Mechanics; Subject: Civil & Environmental Engr; University: Utah State University; Term: Fall 2005;

Typology: Exams

Pre 2010

Uploaded on 07/30/2009

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CEE 3500 - Fall 2005 - TEST 3 - Friday, November 11, 2005 - Key [1]. A spillway model is to be built to a scale of 1:25 across a flume that is 2 ft wide. The prototype is 37.5 ft high, and the maximum head expected is 5.0 ft. If the flow over the model at 0.20 ft head is 0.70 cfs, what is the flow per unit width (in féls) in the prototype? Solution. A spillway model follows Froude similarity since gravity is the dominant force in the flow, & thus, Fr,= = = 1. Since the value of the acceleration of gravity cannot be controlled, g, = — = 8," L, 8m 1, and the velocity ratio is V,=./L, . The discharge ratio, in terms of the length ratio, is given by 1 5 Q, = V+L2=L,2 +L2= L2. (1,7) Lr :=25: Or =evalf Lr? 3125.000000 (1) The data given is the following: P, =375 fi, A, =5.0 ft,Q,, = 0.70 cfs, H,,=0.20ft. Using Maple: Pp := 37.5: Hp := 5.0: Om := 0.70 : Hm := 0.20: From Q,= 0, then 0,= 9, - O,, ie, Qp = Or - Om 2187.500000 (2) The result is Q, = 2187.5 cfs. [2]. The discharge per unit length g = Q/B over a rectangular weir depends on the approach depth H and on the acceleration of gravity g. (Here, Q is the discharge and B is the width of the weir). First, using dimensional analysis, form a II (Pi) term involving the parameters q, B, and g. Then, using the definition of the unit discharge, namely, g = Q/B, and making the IT term equal to a constant C,obtain an expression for the discharge Q. Solution. The problem is represented by the function {(g, H, g)=0. The dimensions involved are [gq] = foveal = (22-7~'\r= 22-7, (H] =L, and [g] =L-T. Thus, we have m = 3 and n = 2, requiring m-n =] TI term. This term can be written as II = g*-B”-g. In terms of the dimensions: M.. 19. = (2-r-1). (LY Le T72 =e tt! Ten? The resulting equations are: Eql :=0=2-x+y+1:EqT =0=-x—-2: The solution is: solve( {EqL, EqT }, {x,y}) {x= —2, y=3} (3) Thus, with xi=—2:y:53: the I term is: Tl:=q°: +g (4)