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Q1: 6. Plot the same graph for the orifice meter. Determine the flow discharge coefficient for orifice from the graph.See Answer
Q2: Given the following data for 3 water reservoirs shown in the schematic below, determine all flow rates in concrete pipes using the Hazen Williams equation. See Answer
Q3: Show at least 3 iterations to find the flow rates in all pipes under given the inflows and outflows.All pipes are cast iron with a diameter of 25 cm.Use f= 0.02 for the Darcy-Weisbach equation.Note that initial Q and its direction in each pipe are given in the figure. Check if your AQ and sum of head loss are close to zero. You may directly fill out the following table for easy hand-calculation. If you have used Excel, copy and paste your worksheet to the answer sheet neatly. See Answer
Q4: 7. Calculate water velocities using the pitot tube measurements, u=\sqrt{2 g \Delta h} Calculate flow rates based on velocities (Q = v.A). Compare the calculated flow rates with actual flow rates in a scatter plot.See Answer
Q5: 7.12. If a channel with the same cross-sectional and flow properties as the channel of Problem 7.11 is laid on a slope of 0.01 ft/ft, determine whether the flow is supercritical or sub critical. Find the depth of flow at a point 1000 ft downstream from the point where y = 1.5 ft. (A trial-and-error solution may be necessary.)See Answer
Q6: 7.13. Classify the water surface profiles according to Table 7–2 of (a) Problem 7.11and (b) Problem 7.12.See Answer
Q7: 7.10. Determine the local change in water surface elevation caused by a 0.2-ft-highobstruction in the bottom of a 10-ft-wide rectangular channel on a slope of0.0005 ft/ft. The rate of flow is 20 cfs and the unobstructed flow depth is 0.9 ft.(See Fig. P7-10). Assume no head loss. See Answer
Q8: 7.11. A rectangular channel with n = 0.012 is 5 ft wide and is built on a slope of0.0006 f/ft. At point a, the flow rate is 60 cfs and y, = 3 ft. Using one reach,find the distance to point b where y, = 2.5 ft and determine whether this point is upstream or downstream of point a.See Answer
Q9: 7.8. Determine the critical depth and the critical velocity for the Colorado River System Aqueduct (Problem 7.1) if Q = 1500 cfs.See Answer
Q10: 7.7. Find the normal depth y, for the triangular channel shown in Figure P7–7 if So = 0.0005 m/m, Q = 40 m³/s, and n = 0.030. See Answer
Q11: 1. Fit a Horton infiltration formula to the following measurements: See Answer
Q12: Problem 1. Gate AB in Figure 2 (shown below), is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric pressure, compute the force F on the gate and its center-of-pressure position X. (ywater=9810 N/m³, and pwater=1000 kg/m³). See Answer
Q13: 3.3.6. Water flowing in a positive x-direction passes through a 90° elbowin a 6-in.-diameter pipeline and heads in a positive y-direction (FigureP3.3.6). If the flow rate is 3.05 ft³/s, compute the magnitude and directionof the reaction force (F). The pressure upstream of the elbow is 15.1 psi;just downstream it is 14.8 psi. See Answer
Q14: 5. Now, plot log(Qactual) vs. log(Ah), for venturi meter. Determine the flow dischargecoefficient for venturi from the graph.See Answer
Q15: Assume welded steel pipe and use Hardy Cross method to solve for all pipe network flow rates with Hazen Williams head loss equationSee Answer
Q16: Assume welded steel pipe and use Hardy Cross method to solve for all pipe network flow rates with Darcy Weisbach head loss equationSee Answer
Q17: Assume flow rates (Q) and directions in each pipe that satisfy continuitySee Answer
Q18: Calculate net head loss (Ehf clockwise Ehf counterclockwise) around loopSee Answer
Q19: \text { Equations } \text { 1. } \quad \Delta \mathrm{S}=\mathrm{P}-(\mathrm{E}+\mathrm{T}+\mathrm{I}+\mathrm{Q}) \text { 2. Average precipitation }=\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{A}_{\mathrm{i}} / \Sigma \mathrm{A}_{\mathrm{i}}\right) \text { 3. } \quad Q_{p}=C I A \text { 3a. } \quad \Delta \mathrm{S}=\mathbf{P}-\mathrm{R}-\mathrm{G}-\mathrm{E}-\mathrm{T} \text { 4. } f=f_{\mathrm{c}}+\left(f_{0}-f_{\mathrm{c}}\right) \mathrm{e}^{-\mathrm{k}^{\mathrm{t}}} \text { 5. } \quad F(t)=\int_{0}^{t} f d t=f_{c} t+\left[\frac{f_{0}-f_{c}}{k}\right]\left(1-e^{-k t}\right) \text { 24. } \quad I-Q=\frac{\Delta S}{\Delta t} \text { 25. } \quad \frac{I_{1}}{2}+\frac{I_{2}}{2}-\frac{Q_{1}}{2}-\frac{Q_{2}}{2}=\frac{S_{2}-S_{1}}{\Delta t} \text { 28. } \quad Q_{2}=C_{0} I_{2}+C_{1} I_{1}+C_{2} Q \text { 29. } \quad C_{0}=\frac{-K x+0.5 \Delta t}{D} \text { 30. } \quad C_{1}=\frac{K x+0.5 \Delta t}{D} C_{2}=\frac{K-K x-0.5 \Delta t}{D} \text { 32. } \quad D=K-K x+0.5 \Delta tSee Answer
Q20: A steel pipeline feeding water from a river to a large storage tank as shown has a flow rate 60,000 GPD(24 hr/day). The pipe system has following specifications: 1. Water temperature in the pipe system is 85°F. For the purpose of initial pipe size selection, the velocity of water in the pipe is assumed at 7ft/s. 2. The pump is the centrifugal type. 3. Pipe material is steel, schedule 40. All connections are flanged. 4. Length of the pipe as shown with six 90° regular elbows, three gate insolation valves, one swing check valve, one strainer (K = 3.0). 5. In the storage tank, the gauge pressure 28 inches of mercury. 6. Pressure drop from the tank connection to the storage is 3.2 psig.See Answer

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