Find. (a) Head loss $h_L$ (m); (b) energy loss per unit mass (J/kg); (c) power dissipated (kW).
Approach. Compute the specific energy $E=y+V^2/2g$ at each section from the given $Q$; the head loss is the drop in specific energy across the jump (the energy the turbulent roller dissipates as heat), which converts directly to a per-mass energy loss and, with the mass flow rate, to a dissipated power.
Velocities from continuity.
$$V_1=\frac{Q}{by_1}=\frac{9.4}{(2.4)(0.4)}=9.79\ \text{m/s}\qquad V_2=\frac{Q}{by_2}=\frac{9.4}{(2.4)(2.6)}=1.506\ \text{m/s}$$
Part (a) — specific energy and head loss.
$$E_1=y_1+\frac{V_1^2}{2g}=0.4+\frac{9.79^2}{2(9.81)}=5.287\ \text{m}\qquad E_2=y_2+\frac{V_2^2}{2g}=2.6+\frac{1.506^2}{2(9.81)}=2.716\ \text{m}$$
$$h_L=E_1-E_2=5.287-2.716=\boxed{2.571\ \text{m}}$$
Part (b) — energy loss per unit mass. Specific energy in metres is energy per unit weight (J/N); multiplying by $g$ converts to energy per unit mass,
$$\Delta e = gh_L=(9.81)(2.571)=\boxed{25.2\ \text{J/kg}}$$
Part (c) — power dissipated. With mass flow rate $\dot{m}=\rho Q=(1000)(9.4)=9400\ \text{kg/s}$,
$$P_{loss}=\dot{m}\,\Delta e=\rho gQh_L=(1000)(9.81)(9.4)(2.571)=\boxed{237\ \text{kW}}$$
