WATERWHEEL PROJECT

BILL OF MATERIALS

SKETCH BLADE DESIGN

UNIQUE FEATURES

Calculations for estimated power we get from the waterwheel

The waterwheel design possesses several unique features aimed at facilitating maximum power generation. These include: The shaft has been reinforced with several box sections. This is so as to ensure

that it handles the large force of the water flowing in the flume as well as the total weight of the blades and side-plates without collapsing or breaking.

A horizontally and vertically tapering channel has been placed in front of the waterwheel mouth. This increases the velocity of the water flowing within the flume such that it encounters the blades with a larger force hence generating more power.

The waterwheel is sunk rather than mounted or suspended. This is in order to ensure that the structure is stable and to reduce the chances of unwanted motion of the waterwheel structure. Furthermore this ensures that the floor clearance for water passing through the tapered channel is definite.

The following is a poster outlining the design of a micro-hydroelectric device to be used in a shallow flume to generate electricity. This design is aimed at achieving maximum device output, i.e. to generate the most electricity.

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Our design is based on an undershot water wheel, this type of wheel is the oldest type of wheel. The flow of water provides torque to turn the wheel, they gain no advantage from head and are suited best to shallow streams of water.

Group 12 BOM

Item no. Part no. Description Quantity Quantity Unit

1 919D111 High power 10:1 motor 1 -

2 919D10 Pulley 1 -

3 919D14 Pulley 1 -

4 919D21 Belt 1 -

5 919D22 Belt 1 -

6 EFOM-06 Flange Bearing 6 -

7 GFM-0608-04 Flange Bearing 4 -

8 691-8042 Clamp Collar 4 -

9 n/a Aluminium Box 4 m

10 n/a Aluminium Sheet (Ø=3mm) 2 m²

11 n/a Aluminium Sheet (Ø=2mm) 1 m²

12 n/a Aluminium Rod (Ø=6mm) 0.5 m

13 n/a M3 Steel Bolt (ℓ=30mm) 34 -

14 n/a M3 Steel Bolt (ℓ=10mm) 96 -

15 n/a M3 standard nut 130 -

16 n/a Flange Mount (Ø=64mm) 1 -

CALCULATIONS

The blade design consists of a curved end section which gradually flattens out. The reasoning behind this design is largely to reduce the loss of kinetic energy due to heat and sound; a considerable loss if the water were to strike an angled flat blade with no curvature. Ideally, the angle of the tip of the blade where the water first makes contact should be at the same angle as the water (horizontal) to minimize losses. By calculation it was worked out that the optimum angle of curvature would be 60 degrees as this would allow the blade to transfer maximum kinetic energy to the wheel by the time the water will have reached the flat section of the plate. Another reason for this design is that by having a curved plate, the total surface area exposed to the water stream is larger than an equivalent flat blade. Furthermore, the blade is made from a single sheet of metal so as to facilitate the manufacturing process involved in creating the part. The choice of metal is aluminium as it is a light and strong material hence simultaneously increasing angular velocity and optimizing strength of the wheel.

Blade Angle Efficiency = 2𝑐(1 − c)2 𝑑

𝑑𝑐 = 6𝑐2 − 8𝑐 + 2 = 0

3𝑐 − 1 2𝑐 − 2 = 0

𝑐 = 1 𝑜𝑟 1

3

If c=1 then no torque is produced, therefore

maximum efficiency occurs at c = 1

3

𝜃 = 𝑡𝑎𝑛−1 1

3 𝜽 = 𝟏𝟖. 𝟒𝟑o

Area of Blades Normal to flow, 𝐴𝑛 = 0.00976𝑚2 Area of water entering funnel, 𝐴1 = 0.0295 𝑚2 Area of water passing through end of funnel, 𝐴2 = 0.0122𝑚2 0.0122−0.00976

0.0122= 0.2 × 100% = 20% of water

passes blades Therefore the blades catch 80% of the water that flows normal to them Speed of water exiting funnel Speed of water entering funnel, 𝑣1= 1.67m/s To find speed of water exiting funnel, 𝑣2:

𝜌𝐴1𝑣1 = 𝜌𝐴2𝑣2 0.0295 × 1.67 = 0.0122𝑣2

𝑣2 = 0.0295 × 1.67

0.0122= 4.038 𝑚/𝑠

Speed of water exiting funnel, 𝒗𝟐 = 4.038m/s

Force on each Blades F = 𝜌𝐴𝑣2(1 − 𝑐)2

= 1000 × 0.0122 × 4.0382 (1 − 1

3 )2 × 𝟎. 𝟖

= 70.73N (multiplied by 0.8 to account for water that passes the blades)

Power 𝑃𝑎𝑐𝑡 = 𝜌𝐴𝑣1

3𝐶(1 − 𝐶)2 𝑃𝑎𝑐𝑡 = 1000 × 0.0122 × 4.03813 × (1 −1

3 )2 × 𝟎. 𝟖 = 95.21W

𝑃𝑚𝑎𝑥 = 1

2𝜌𝐴𝑣1

3

𝑃𝑚𝑎𝑥 =1

2 × 1000 × 0.0122 × 4.03813

= 401.7W

Efficiency, 𝜀 = 𝑃𝑎𝑐𝑡

𝑃𝑚𝑎𝑥 =

95.21

401.7 = 0.237 ×

100% = 23.7% Torque

Torque = 𝜌𝐴𝑣12(1 − 𝐶)2𝑅

𝑇 = 1000 × 0.0122 × 4.03812 × (1 − 1

3 )2 × 0.25

× 𝟎. 𝟖

= 17.68Nm Rotational Speed

𝜔 = 𝑣2

𝑅=

𝑣1

3×

1

𝑅=

4.038

3×0.25= 5.384 rad/s

= 5.384 ×60

2𝜋= 51.4 𝑅𝑃𝑀

The Pulley ratio is approximately 2:1, meaning the Power, Torque and Rotational Speed at the gearbox will all be double the value calculated from the Water Wheel Power at gearbox ~ 190 𝑊 / Torque at gearbox ~ 35 𝑁𝑚 / Rotational Speed at gearbox ~ 105 𝑅𝑃𝑀 The gearbox has a 10:1 ratio

Therefore rotational speed at gearbox output ~ 110𝑟𝑎𝑑/𝑠 ~ 1050 𝑅𝑃𝑀 If the mechanical efficiency of water wheel is ~ 50% Rotational speed at gearbox output ~ 1050 × 0.5 = 525 𝑅𝑃𝑀 525 𝑅𝑃𝑀 will supply approximately 2W across a 5 ohm load and produce 0.008Nm of torque at the motor.