CEE 481 Midterm Exam 1b, Aut 2008

Transcription

1 CEE 481 Midterm Exam 1b, Aut 8 Name: how all your work. If you can answer a question by inspection of the data, write a sentence stating your reasoning. A NUMERICAL ANWER WITHOUT ANY EXPLANATION WILL RECEIVE NO CREDIT. All data are given in metric units. Assume T = o C in all problems. g = 9.81 m/s 2. γ water@c = 9.79 kn/m kpa = 1. kn/m 2 =.2 m of head 1. The graph below characterizes the system described in the previous midterm, after the new factory is built. For each of the following questions, provide an answer and a brief explanation of the basis for that answer. If the community agrees to supply the factory s water needs: 1

2 9 Cumulative Volume ( m 3 ) upply Demand Time (h) (a) () At what time during the day will the reservoir be most full? Answer. The reservoir is most full when the difference between the cumulative inflow (the supply) and the cumulative outflow (the demand) is maximized. For the given system, this occurs at t = 6 h. (The value is almost the same at t = 7 h). (b) () At what time during the day will the reservoir level be declining at the fastest rate? Answer. The reservoir level is declining most rapidly when the difference between the rate of outflow (the slope of the demand curve) exceeds the rate of inflow (the slope of the supply curve) by the largest amount. Because the slope of the supply curve is constant, the time when the level is declining fastest is just the time when the slope of the demand curve is largest, which occurs around t = h. (c) () Approximately what is the rate of decline (m 3 /h) of the volume stored in the reservoir at the time you identified in part b? Answer. The rate at which the level is declining is the difference between the slopes of the demand and supply curves. The slope of the supply curve is always 81, m 3 /24 h, or 3,375 m 3 /h. The slope of the demand curve at t = h can be estimated by drawing a tangent at that time; I estimate this slope to be, m 3 /h, so the rate of decline is (, 3,375), or 6,625 m 3 /h. 2

3 Cumulative Volume ( m 3 ) upply Demand Time (h) 2. In the two-pipe network shown schematically below, point A is m lower than Point B. An equivalent pipe analysis indicates that the two-pipe network can be replaced by a single pipe with a.-m diameter and a 3.-km length. ome characteristics of the real and equivalent pipes are shown in the table. A B L (km) D (m) H-W C coefficient ? 5 Equivalent (a) () What is the diameter of pipe? For your calculations, assume that the total flow through the system is.1 m 3 /s. Answer. The headloss across the system can be found based on the conditions in the equivalent pipe. For D in meters and Q in m 3 /s, the friction slope in this pipe is: 3

4 Qeq f, eq Deq Ceq = = = 2.36x (.) 5 ( )( ) HLeq, = f, eqleq = 2.36x 27 m = 6.36 m The headloss through the real pipe system is also 6.36 m, so the flow through pipe can be found as follows: f, H 6.36 m = = = 3.18x L m LP, -1 Q 1/ 3 ( ) ( )( ) f, D / x. m = C = ( 5) = s The flow through pipe is therefore.1.552, or.448 m 3 /s. The headloss through is 6.36 m, so the friction slope and the diameter of that pipe can be found as follows: f, H 6.36 m = = = 2.12x L m LP, -2 D 1/4.87 1/ Q f, C 2.12x 5 = = =.1 The diameter of is thus 1 mm, or, rounding, mm. (b) () On the graphs below, sketch the HGL for the two pipes, for the conditions explored in part a. Assume the HGL at Point A is at 45 m for both pipes (note: a slight error is built into this assumption, since the velocities in the pipes are slightly different, but the error is small). In a sentence or two, explain the reasoning that led you to draw the lines the way did Length Along Pipe (km) Length Along Pipe (km) 4

5 Answer. The HGL declines linearly and by an amount equal to the headloss (6.36 m) through both pipes. The change in elevation has no effect on the HGL, since that change only converts pressure head to elevation head, or vice versa, and both types of head contribute to the HGL. The HGL therefore declines from 45. m to m over a distance of 2. m in and 3. m in Length Along Pipe (km) Length Along Pipe (km) 3. (a) () In the WaterGEM problem dealing with irrigation of the golf course, the problem statement specified an emitter coefficient for the sprinkler at each hole. Explain what an emitter coefficient is. Answer. The emitter coefficient is the coefficient relating the discharge from the sprinkler to the pressure at the sprinkler just upstream of the outlet. (The relationship is of the form.5 Q= cp, but that detail was not required for full credit.) As the water passes through the sprinkler head, it loses energy due to friction. This energy loss is manifested as loss of the pressure head and possibly a change (either increase or decrease) in the water s velocity; any change in velocity must be accompanied by a change in the cross-section of the flow to satisfy continuity (the mass balance). Thus, the emitter coefficient indirectly correlates the energy loss to the flow rate through the sprinkler head. (b) (15) In the WaterGEM problem dealing with water supply to a new sub-division, we represented the connection from the water main to the distribution pipes as a combination of a reservoir and a pump. Briefly explain the idea underlying this substitution. Answer. The main provides water to the sub-division at a fixed elevation and at a flow rate that depends on the pressure in the distribution pipe at the connection point the higher the pressure in the distribution pipe, the lower the flow from the main into that pipe. In this respect, the main operates just like a pump at the elevation of the connection, and with a pump curve (flow rate vs. discharge pressure) that corresponds to the flow-vs.-head relationship at the connection. 5