Pelton Wheel Lab Report 3e6x4v

FACULTY OF ENGINEERING & TECHNOLOGY

LAB REPORT FLUID DYNAMICS EME 3026 TRIMESTER 1 (2018/2019)

Experiment 1: PELTON WHEEL

Name

ID

Chew Way Sern

1132700167

Ng Hsien Siong

1131121638

Ng Jin Han

1142701320

Tan Chuin Chiet

1132701813

OBJECTIVE 1. To familiarize with the working principle of Pelton Wheel. 2. To determine the characteristics of Pelton Wheel. 3. To determine the optimum efficiency point. ABSTRACT In this experiment, we are dealing with the Pelton Wheel. In this experiment, we observed the load on the Pelton Wheel with every decrement of the speed and observing the flow rate of the water. We also observing whether the nozzle setting effect the performance of the Pelton Wheel. By those parameters, we can determine the characteristics of the Pelton Wheel and it optimum efficiency point. All he data and the related graphs are plotted for further analysis.

INTRODUCTION Working Principle of Turbine Wheel Pelton Wheel is a type of impulse turbine. These turbines work on the principle whereby the direction of high velocity fluid in motion is changed to generate an impulse force causing the blades to spin the turbine. The kinetic energy of the fluid leaving the Pelton Wheel will be reduced. The fluid’s pressure head is converted into velocity head prior to entering the wheel with a nozzle. The water jet is directed to the rotor and hits the sharp edge of the blades so that it divides into two symmetrical parts. Each blade is composed of two symmetrical spoons. The water jet (circular with diameter 𝑑) comes out of the distributor at velocity 𝑣, 𝑣 = √2𝑔ℎ where ℎ is the height of water fall and 𝑔 is the gravitational acceleration, and hits the blade with this velocity. The wheel diameter 𝐷 and the rotational speed 𝑛 are selected so that the runner peripheral velocity 𝑣ro is 𝑣ro = 𝜋𝑛𝐷 = 𝑣⁄2 Therefore the relative velocity (with respect to the blades) of the water flow at the blade inlet is 𝑣 − 𝑣ro = 𝑣⁄2 The water flow will be expelled at a velocity reversed to the inlet velocity.

Figure 1: Structure of Pelton Turbine. Mounted on the periphery is a series of double spoonshaped blades (P). The water flows through the forced duct (C) and is expelled as a jet (E) by a distributor. The distributor has a circular section nozzle (B) that can be adjusted by a taper shutter pin S which closes the nozzle as it moves forward. Baffle (T) is used to stop the wheel quickly.

Figure 2: speed indicator

Figure 3: load applied

Figure 4: water level indicator

Apparatus Setting: 1. The water tank is filled to 4/5 of its full capacity. 2. The apparatus is connected to its power supply. 3. The water hose is connected to the inlet of the Pelton Wheel. 4. The belt is put on to the pulley of the wheel and is tightened as necessary. 5. The speed sensor cable is connected to the table top control . 6. The power supply of the Pelton Wheel and the hydraulic bench is switched on. 7. The speed sensor is adjusted to ensure that the screw of the pulley can be sensed by the speed sensor head. 8. The control valve and by valve of the hydraulic bench are ensured to be fully opened. 9. The pump is switched on, and the by valve is closed slowly. 10. The frequency at the frequency inverter is set to zero.

Procedure: 1. The nozzle jet is preset at fully opened position. The wheel from the nozzle will cause the wheel to rotate. 2. The frequency is switched to 30Hz. The wheel is allowed to run for about 3 minutes to let the flow rate reading become stable. 3. The belt is attached to the pulley at the back of the Pelton Wheel. The wheel is loaded by tightening the loading nut. The belt is ensured to be seated properly on the groove of the pulley. 4. The values of load on the spring scale is recorded every 100 rpm decrement. 5. The load on the wheel is increased until it reaches 500 rpm. Result

Load 1 (N)

Load 2 (N)

Speed (rpm)

Pressure (bar)

1.50 2.50 5.50 7.50 9.00 10.50 11.75 13.00 14.25

1.5 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75

1300 1200 1100 1000 900 800 700 600 500

1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3

Water Collection Time (s) 14.10 14.38 14.79 14.56 14.53 14.89 14.77 14.35 14.73

Average Speed (rad/s)

Torque (N.m)

Output Power (W)

136.14 125.66 115.19 104.72 94.25 83.78 73.30 62.83 52.36 Table 2

0.000 0.020 0.098 0.150 0.189 0.228 0.260 0.293 0.325

0.000 2.450 11.231 15.656 17.766 19.059 19.059 18.378 17.017

Table 1

Average Speed (rpm)

Load (N)

1300 1200 1100 1000 900 800 700 600 500

0.00 0.75 3.75 5.75 7.25 8.75 10.00 11.25 12.50

Pressure (bar)

Flow Rate (Lpm)

Flow Rate (m³/s)

Input Power (W)

Output Power (W)

Efficiency (%)

1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3

42.55 41.72 40.57 41.21 41.29 40.30 40.62 41.81 40.73

0.00070922 0.00069541 0.000676133 0.000686813 0.000688231 0.000671592 0.000677048 0.000696864 0.000678887

92.20 90.40 87.90 89.29 89.47 87.31 88.02 90.59 88.26

0.00 2.45 11.23 15.66 17.77 19.06 19.06 18.38 17.02

0.00 2.71 12.78 17.53 19.86 21.83 21.65 20.29 19.28

Table 3

Discussion 1. From the graph of output power against speed, we can notice that when the speed is at 52.36 rad/s – 73.30 rad/s, the output power is increasing while after the wheel achieves 83.78 rad/s, the output power started to drop until the speed reached 136.14 rad/s the output power is zero. This is because initially there is no force applied, there is no friction between the belt and the wheel, so that the output power is zero. When there is load applied, there is friction between the belt and the wheel, it creates a torque and cause there is the output power. 2. Theoretically, we know that when the load is applied on the wheel by the belt, there is some friction cause the wheel hard to turn. Due to this, the speed of the wheel will decrease. When the speed of the wheel decreases, it will cause the output power to decrease as well. 3. From the graph of torque against speed, we can notice that the torque is inversely proportional to the speed. This means that when the speed increase, the torque decrease. This is due to the load that applied directly and perpendicular to the Pelton Wheel. As we all know, torque is the force applied multiply by the perpendicular distance. When the load applied increase, torque increase, friction increase, the only is the speed decrease.

4. Based on the result from the experiment, the speed that gives the best performance is 800rpm. It produces the highest output power (19.06W) and highest efficiency (21.83%). 5. The maximum power generated by Pelton Wheel is 19.06W at speed 800rpm. 6. The nozzle will affect the performance of the Pelton Wheel by the diameter of the water come out from the nozzle. When the diameter of the water flow out from the nozzle is small, the water will flow in high velocity and hit the blades of the wheel. The area of the water on the blades is small and cause high pressure on it. This is because pressure is force per unit area, when the area is small there will be a high pressure. When pressure is high, the impulse force applied on the blade will increase as well, therefore it will produce high average of the velocity on the Pelton Wheel. On the other hand, if the diameter of the water flow out from the nozzle is big, the lower the velocity that the water flow out, the impulsive force act on the blades is small and cause the Pelton Wheel turn with low speed. 7. In this experiment, there is some error occur when we conduct this experiment. The leaking of water at the water meter will lead us to a wrong data for the volume of water collected. There is also the parallax error when taking the reading from the water meter and the pressure gauge because the level of the eye is not perpendicular to the level of the reading.

Conclusion From this experiment, we can conclude that when the diameter of the water flow out from the nozzle is small, then will generate high velocity of the water flow and cause the Pelton Wheel turn in high speed, The load is applied increase, the torque increase. This will cause the output power will increase and will generate higher efficiency as well. The maximum power generated from this experiment is 19.06W at speed 800rpm.