The inventor of a water wheel claims that his design is better than a conventional overshot or breast wheel at extracting energy from water flows at low heads. He states that the design shown can generate up to 260 W of power from a head of 0.8 m. average household energy use over a 24-hour period is about 28 kW h. (a) The claimed efficiency of the device is 70%. With a head of 0.8 m, calculate the flow rate of water that would be required to deliver 260 W of power. Efficiency = 70% Gravity = 9.81 m/s Head size = 0.8 m Efficiency = Flow Rate x Gravity x Head Size E = Q x g x h So… Q = E gh Q = 70 9.81 x 0.8 Flow rate = 8.9 kg/s (b) When the wheel is operating, there are twelve full troughs of water on the downward-travelling side of the belt. The diameter of the toothed sprocket is given as 0.25 m. If any shaft rotates at 41 rpm when delivering 260 W of power, calculate how many litres of water each trough must be able to contain. Flow rate = 8.9 kg/s Power = 260 W Sprocket = 0.25 m 41 rpm into angular speed = 41 x 2π 60 ω = 4.3 rad/s v = rω v = 0125 x 4.3 v = 0.5375 rad/s Power = Torque x Angular speed Torque = Power Angular speed Torque = 260 4.3 Torque = 60.47 Nm Power = Gravity x Flow rate x Head size (P = G x Q x H) H = P GQ H = 260 9.81 x 8.9 H = 2.98 m unable to get the answer to part b)......any suggestions?