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.(adsbygoogle = window.adsbygoogle || []).push({});

(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?

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# Homework Help: Water Wheel

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