California State University, Northridge



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|College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 390

Fluid Mechanics | |

| |Spring 2008 Number: 11971 Instructor: Larry Caretto |

May 6 Open-Channel-Flow-Homework Solutions

10.60 Two canals join to form a lager canal as shown in Figure P10.60 copied at the right. Each of three rectangular canals is lined wit the same material and has the same bottom slope. The water depth in each is to be 2 m. Determine the width of the merged canal, b3. Explain physically (i. e., without using any equations) why it is expected that the width of the merged canal is less than the combined widths of the two original canals (i.e., b3 < 4 m + 8 m = 12 m).

The merged flow is the sum of the two original flows: Q3 = Q1 + Q2. Each of these flows is given by the Manning equation so that we can write.

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Since each canal has the same bottom slope and is lined with the same material the values of S0 and n are the same in all three canals. Thus we can divide the equation above by the common values of S01/2/n and substitute A/P for Rh to obtain the following equation.

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If y is the depth and b is the width then the area is yb and the wetted perimeter is 2y + b. Making these substitutions in the last equation gives.

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Since y1 = y2 = y3 = 2 m we can cancel the common y values from the numerator of the central equation above and insert the values of b1 = 4 m and b2 = 8 m obtain the following equation.

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We see that the equations will be dimensionally consistent if b3 is in meters. Omitting the dimensions, we can obtain the following computational equation.

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Using a computer root finder to solve this equation gives the new width, b3 = 10.66 m.

The open channel flow has an equilibrium between the gravity driving force and the resistance of the viscous forces. The single channel has less wall space to generate viscous forces, so the viscous forces are less, but the gravity force is the same, because the depth is the same. To maintain the same depth, with the same slope, we have to have a final width that is less than the original width to maintain a balance between the gravity driving force of depth and the viscous forces retarding the flow.

10.63 The cross section of a long tunnel carrying water through a mountain is shown in Figure P10.63, copied at the right. Plot a graph of flowrate as a function of water depth, y, for 0 ( y( 18 ft. The slope is 2 ft/mi and the surface of the tunnel is rough rock (equivalent to rubble masonry). At what depth is the flowrate maximum? Explain.

We can use the Manning equation to find the flow rate.

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For BG units k = 1.49. For rubble masonry, n = 0.025 from Table 10.9. The slope of 2 ft/mi is equivalent to (2 ft) / (5280 ft) = 0.000379 = S0. For depths less than 12 ft, the area is the depth, y, times the width, 12 ft and the wetted perimeter is the 12-ft width plus twice the depth. This gives the hydraulic radius and resulting flow rate as follows.

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The flow is in ft3/s when the depth, y, is in feet.

Above a depth of 12 ft, the circular top gives a different area and wetted perimeter. The filled area below 12 ft is a square whose area is (12 ft)2 = 144 ft2. The total area for a depth above this 12-ft height is the area of a semicircle whose radius is 6 ft; this area is π(6 ft)2/2. The flow area is the area of this semicircle minus the empty area above the depth, y. The calculation of this unfilled area was done in Example 10.5 and the relation between this unfilled area, A(1) and the depth, y, is linked through the angle θ.

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Thus, the flow area is given by the following equation.

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The wetted perimeter when y > 12 ft is the wetted perimeter of the filled lower area which is 3(12 ft) = 36 ft plus the wetted part of the perimeter of the semicircular area. The entire perimeter of the semicircle is 2πR/2 = π(6 ft). The perimeter of the semicircle that is not wetted is θR = θ(6 ft). So, the total wetted perimeter when the depth, y, is greater than 12 ft is 36 ft + (6 ft)(π – θ) = 54.85 – 6θ. We can now construct the following set of equations to compute the flow for depths, y, above 12 ft.

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The flow rates at various depths were computed on a spreadsheet using the equations derived above. The appropriate equations were used for depths below 12 ft and depths above 12 ft. The maximum flow rate was determined using the Excel Solver and found to occur at a depth of 17.1 m. This maximum flow rate is Qmax = 581 ft3/s.

For depths greater than 17.1 ft, the wall perimeter retarding the flow increases much more rapidly than the flow area and the flow rate decreases.

10.72 Water flows in a rectangular channel with a bottom slope of 4.2 ft/mi and a head loss of 2.3 ft/mi. At a section where the depth is 5.8 ft and the average velocity is 5.9 ft/s, does the flow depth increase of decrease in the direction of flow? Explain.

The change in depth with distance, dy/dx, is given in terms of the friction slope for head loss, Sf, the bottom slope, S0 and the Froude number, Fr, by the following equation.

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For the data of this problem S0 = 4.2 ft/mi, Sf = 2.3 ft/mi, y = 5.8 ft, and V = 5.9 ft/s. We can then compute dy/dx as follows.

[pic] –2.34 ft/mi

The head loss retarding the flow is less than the slope giving the gravity force. This force imbalance accelerates the fluid so the level becomes shallower to maintain the same flow rate.

10.76 A 2.0-ft standing wave is produced at the bottom of the rectangular channel in an amusement park ride. If the water depth upstream of the wave is estimated to be 1.5 ft, determine how fast the boat is traveling when it passes through this standing wave (hydraulic jump) for its final “splash.”

We can use solve the equation for the depth ratio (y2 /y1) across a hydraulic jump in terms of the upstream Froude number, Fr1 = V1/(gy1)1/2, which can be used to find the upstream velocity, V1. The boat will be moving at this water velocity when it passes through this standing wave.

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From the data given for this problem y1 = 1.5 ft and y2 = y1 + 2 ft so y2/y1 = (3.5 ft) /(1.5 ft) = 7/3. We can use the equation above to find V1.

[pic] 13.7 ft/s

10.44 Under appropriate conditions, water flowing from a faucet, onto a flat place, and over the edge of the place can produce a circular hydraulic jump as shown in Figure P10.78 copied at the right. Consider a situation where a jump forms 3.0 in from the center of the plate with depths upstream and downstream of the jump of 0.05 in and 0.20 in, respectively. Determine the flowrate from the faucet.

We can use solve the equation for the depth ratio (y2 /y1) across a hydraulic jump in terms of the Froude number for the upstream Froude number, Fr1 = V1/(gy1)1/2, which can be used to find the upstream velocity. This water velocity can then be used to find the flow rate.

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From the data given for this problem y1 = 0.05 in = 0.004167 ft and y2 = 0.20.

[pic] 1.158 ft/s

The area of the flow before the jump will be πDy1 = π(6 in)(0.05 in)(1 ft2/144 in2) = 0.006545 ft2. The flow rate, Q = V1A1.

[pic] 0.00758 ft3/s

10.90 Water flows from a storage tank, over two triangular weirs, and into two irrigation channels as shown in Figure P10.90, copied at the right. The head for each weir is 0.4 ft and the flowrate in the channel fed by the 90-degree V-notch weir is to be twice the flowrate in the other channel. Determine the angle θ for the second weir.

The flow rate over a triangular weir is given by equation 10.32.

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In this problem we have two channels flows such that Q2 = 0.5Q1, where H2 = H1 = 0.4 ft and θ1 = 90o. Using the weir equation for both flows and setting Q2 = 0.5Q1 gives

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In the last step above we used the result that tan(45o) = 1. The weir coefficients depend on both H and θ. For the 90o weir with θ1 = 90o and H = 0.4 ft we see that Cwt,1` = 0.59 from Figure 10.25 on page 599, copied at the left. Using this value in the equation above and canceling common factors in the two streams gives the following result.

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|Input θ2 |Cwt,2 |Computed θ2 |

|20 |0.626 |50.47 |

|45 |0.607 |51.84 |

|60 |0.597 |52.59 |

|90 |0.590 |53.13 |

We cannot solve this equation explicitly since Cwt,2 depends on θ2. To solve this problem with the sparse data relating θ and Cwt,2, we can find Cwr,2 for the four values of θ shown on the figure and determine the relationship between the value of θ on the figure versus the computed value of θ = 2tan-1(0.295/Cwt,2). The results are shown in the table at the right. Here the first column represents the angle for which there is a curve on Figure 10.25 and the value of Cwt is read from that figure. Because there is such a small range of Cwt values the computed values of θ2 (from the tan-1 equation above) are in a narrow range from 50.47o to 53.13o. The plot of the computed values versus the input values shown below uses the narrow range of the computed angle results. The intersection of the input versus computed q line with a 1:1 line marks the correct answer: θ2 = 52.3o.

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