Question 10.8: Water flows in a wide channel at q = 10 m^3/(s · m) and y1 =...

Part (a)

The upstream velocity is

The upstream Froude number is therefore

From Fig. 10.12 this is a weak jump. The depth y_2 is obtained from Eq. (10.43):

\frac{2y_2}{y_1} = -1 + (1 + 8 Fr^2_1)^{1/2}                                        (10.43)

or                      y_2 = \frac{1}{2} y_1(5.54) = \frac{1}{2}(1.25)(5.54) = 3.46  m

Part (b)

From Eq. (10.44) the downstream velocity is

V_2 = \frac{V_1y_1}{y_2}                               (10.44)

Part (c)

The downstream Froude number is

Part (d)

As expected, Fr_2 is subcritical. From Eq. (10.45) the dissipation loss is

h_f = \frac{(y_2 – y_1)^3}{4y_1y_2}                                            (10.45)

Part (e)

The percentage dissipation relates h_f to upstream energy:

Hence                        Percentage loss = (100) \frac{h_f}{E_1} = \frac{100(0.625)}{4.51} = 14 percent

Part (f)

The power dissipated per unit width is

Part (g)

Finally, the mass flow rate is \dot{m} = \rho q = (1000  kg/m^3)[10  m^3/(s \cdot m)] = 10,000  kg/(s \cdot m), and the temperature rise from the steady flow energy equation is

Power dissipated = \dot{m} c_p \Delta T

or                          61,300 W/m = [10,000 kg/(s · m)][4200 J/(kg · K)]ΔT

from which

ΔT = 0.0015 K

The dissipation is large, but the temperature rise is negligible.