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## 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. ## 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.