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Chapter 15

Q. 15.10

Water flows over a 0.25 m bump in a constant width, horizontal rectangular channel as shown in Figure 15.34A. The upstream depth and velocity are 0.5 m and 0.2 m/s, respectively. Draw the specific energy diagram for this flow and find the depth and velocity over the bump at point B. Then find the depth and velocity downstream of the bump at point 2. Neglect the effect of friction.

15.34

Step-by-Step

Verified Solution

We first calculate the upstream Froude number for the specified conditions as
Fr1 = V1/\sqrt{gy_1}=0.2 m/s/\sqrt{9.81  m/s^2(0.5  m)} = 0.09 Thus the flow is subcritical and we expect the depth to decrease over the bump (see Figure 15.14A). The specific energy for flow in a rectangular channel is given by Eq. 15.38 as E = V²/2g + y. Substituting V = V1y1/y, we find E = V^2 _1 y^2_1/2gy2 + y, inserting the data yields E = (5.09 × 10−4 m3)/y2 + y. This equation forms the basis for the specific energy diagram as shown in Figure 15.34B. Various points on this curve can now be identified. Upstream at point 1 we have V1 = 0.2 m/s and y1 = 0.5 m, thus we can calculate E1 = V^2 _1 /2g + y1 = (0.2 m/s )2/2(9.81 m/s2) + (0.5 m ) = 0.502 m. The critical point C can be located by substituting Q = V1y1w into Eq. 15.41a, yC = [Q2/gw2]1/3, to obtain yC = [V^2 _1 y^2_ 1/g]1/3. Inserting the data, we find

y_{C}=\left[\frac{(0.2\ m/s )^2(0.5\ m )^2 }{9.81\ m/s^2 }\right]^{1/3}=0.1006\ m.

Next we use Eq. 15.41c to compute the velocity at the critical point:

V_{C}=\sqrt{gy_C}=\sqrt{(9.81\ m/s^2)(0.1006\ m)}=0.993\ m/s

Finally we use Eq. 15.41b to write E_{min}=\frac{3}{2} y_C =\frac{3}{2} (0.1006 m) = 0.151 m.

To calculate the depth and velocity over the bump and downstream, recall that in Section 15.3.1 we analyzed frictionless flow over a bump and obtained Eq. 15.6:

\frac{V^2_1}{2g}+y_1+h(x_1)=\frac{V(x)^2}{2g}+y(x)+h(x)

Applying this equation between point 1 and any point downstream, we have V^2 _1 /2g + y1 + h1 = V2/2g + y + h. Introducing the specific energy and noting that with the choice of horizontal datum in Figure 15.34A h1 = 0, this equation becomes E1 = E + h. For calculations we can write this as

E = E1 − h                                          (A)

For the bump we know hB = 0.25 m, thus EB = E1 − hB = 0.502 m − 0.25 m = 0.252 m. Drawing a vertical line at this value of specific energy gives us two possible depths over the bump, as expected. Before we find these two depths, note that we can also use (A) to calculate the critical bump height hC, the bump height needed to make the flow critical. If we write (A) as hC = E1 − Emin, we find hC = E1 − Emin = 0.502 m − 0.151 m = 0.35 m. We see immediately that the flow must remain subcritical, since the bump is only 0.25 m high, a value less than the 0.35 m needed for the flow to become critical.

To find the depth yB over our 0.25 m bump, we make use of the equation of the specific energy diagram E = (5.09 × 10−4 m3)/y2 + y and write EB = 0.252 m = (5.09 × 10−4 m3)/y^2 _B + yB. This yields the cubic equation y^3 _B − 0.252 m y^2 _B + 5.09 × 10−4 m3 = 0. The resulting three solutions are yB = − 0.04 m, + 0.05 m, and + 0.24 m. However, we already know from the specific energy diagram that the only one of interest to us must lie between y1 = 0.5 m and yC = 0.1 m. Thus the solution of interest is yB = 0.24 m. The velocity over the bump can now be calculated as

V_B=\frac{V_1y_1 }{y_B}=\frac{(0.2\ m/s)(0.5\ m)}{(0.24\ m)}=0.42\ m/s

To find the depth and velocity at point 2 downstream, we use (A) to write E2 = E1 − h2. Since h2 = 0 we have E2 = E1 = 0.502 m. The specific energy at point 2 is given by E2 = 0.502 m = (5.09 × 10−4 m3)/y^2 _2 + y2, which we can write as the cubic equation y^3 _2 − 0.502 m y^2 _2 + 5.09 × 10−4 m3 = 0. One solution is known immediately:  y2 = y1 = 0.5 m, corresponding to a subcritical flow identical to that upstream of the bump. The other two solutions are + 0.032 m, corresponding to a supercritical flow, and − 0.030 m, which is meaningless. The flow cannot become supercritical after the bump unless it has gone through the critical point. Since the bump height of 0.25 m is less than the critical bump height of 0.35 m calculated earlier, this could not have happened. Thus we conclude that at point 2, the depth and velocity are the same as that upstream.

Before leaving this example, we could ask ourselves what we would observe if the upstream conditions were fixed and the bump height was increased to exactly the critical height hC = 0.35 m. The answer is found by looking at the specific energy diagram of Figure 15.34B and realizing that we would move along the upper branch of the specific energy curve from point 1 to point C. The flow would speed up going up the bump and become critical at the crest of the bump. What would happen as the flow went down the bump? The specific energy diagram reveals we could move back along the upper branch to the initial subcritical state (points 1 and 2) or move along the supercritical branch to point 2′. This point corresponds to a supercritical flow. Conditions far downstream determine which of these two possibilities occurs.

We could also ask what would happen if the upstream conditions were fixed and the bump height increased beyond the critical height of 0.35 m. The answer to this question is interesting, since now point B would not be on the specific energy curve corresponding to the upstream conditions. Thus this flow is impossible with the upstream depth and flowrate fixed. If the flowrate is fixed, and the flow at the crest is critical when the bump height is slowly increased, then the flow at the crest will remain critical and the upstream depth will increase as needed.