Question 15.10: Water flows over a 0.25 m bump in a constant width, horizonta...

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, y_C = [Q²/gw²]^1/3, to obtain y_C = [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 h_B = 0.25 m, thus E_B = E₁ − h_B = 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 h_C, the bump height needed to make the flow critical. If we write (A) as h_C = E₁ − E_min, we find h_C = E₁ − E_min = 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 y_B 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 E_B = 0.252 m = (5.09 × 10⁻⁴ m³)/y^2 _B + y_B. 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 y_B = − 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 y_C = 0.1 m. Thus the solution of interest is y_B = 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 h_C = 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.