Water turbine is a very common used throughout the world to generate power. Although the concept of the Impact of a Jet is essentially simple, and it can has a highly hydraulic efficiency, the factor of how can produce a powerful reaction force back is also has to be considerate truly. If we can manage well, we can determine the best situation and the optimum deflection angle out, for getting the optimum force that the impact can produce. But first of all, we have to understand hoe the deflection of the jet generates a force on the bucket, and how the force is related to the rate of momentum flow in the jet. The purpose of this experiment is exactly to measure out the reaction force generated by a jet of water striking on a solid surface with different degrees of flow deflection, and compare the results with the Theoretical equation.
This experimental apparatus consists of a water nozzle, a set of impact surfaces, a spring scale connected to a balance beam, a flow meter, and plumbing for recirculating the water. The pump draws water from the collection tank and provides sufficient head for the water to flow through the nozzle and the flow meter. The jet of water from the nozzle impinges on the impact surface. The balance beam attached to impact surface allows measurement of the force necessary to deflect the water jet. One thing is very important that the water jet hitting the impact surface is symmetric relative to the vertical axis of the surface.
First of all, use the Volume CollectedV and Time to Collectt to find Qt . By using equation, Qt=Vt ,Qt equal to flow rate. And, use v=QA to calculate velocity(v), where A equal to area of the nozzle, 5.0265×10-5. Then use Qm=ρQt=ρAv to find out Qm which is the mass flow rate and ρ=1000kgm-3. After knowing Qm, the Force exerted by deflector on water (Fy) is possible to be known by equation Fy=Qmvcosθ+1. For, static equilibrium, Fy is balanced by the applied load, W. It means Fy is equal to W. So, now we can use v2 and W to plot a graph to see the relative between them. The steeper slope means the larger force.
About the slope from the theory equation, just sub in the values of ρ,A and θ those provided above into the equation s=ρA(cosθ+1). Then, the slope can be solved.
Deflector Type| Volume Collected| Time to Collect| Mass Applied| | V (m3)| t (sec) | M (Kg)|
180.00| 0.0200| 44.000| 0.45|
180.00| 0.0100| 30.00| 0.40|
120.00| 0.0200| 44.000| 0.73|
120.00| 0.0100| 31.00| 0.30|
90.00| 0.0200| 44.000| 0.99|
90.00| 0.0060| 19.00| 0.20|
30.00| 0.0200| 44.000| 0.07|
30.00| 0.0100| 23.00| 0.05|
Flow rate, Qt=Vt
For every type of degree, Q1=0.0244=4.5455×10-4
For 180 degree, Q2=0.0130=3.3333×10-4
For 120 degree, Q2=0.0131=3.2258×10-4
For 90 degree, Q2=0.00619=3.1579×10-4
For 30 degree, Q2=0.0123=4.3478×10-4
For Q1, v=4.5455×10-45.0265×10-5.=9.0431 ms-1
For Q2 at 180 degree, v=3.3333×10-45.0265×10-5.=6.6315 ms-1 v2=43.9768 ms-12
For Q2 at 120 degree, v=3.2258×10-45.0265×10-5.=6.4176 ms-1 v2=41.1856 ms-12
For Q2 at 90 degree, v=3.1579×10-45.0265×10-5.=6.2825 ms-1 v2=39.4698 ms-12
For Q2 at 30 degree, v=4.3478×10-45.0265×10-5.=8.6498 ms-1 v2=74.819 ms-12
Mass flow rate (Qm), Qm=ρQt=ρAv
For 180 degree, Qm2=ρQt2=1000×3.3333×10-4=0.33333
For 120 degree, Qm2=ρQt2=1000×3.2258×10-4=0.32258
For 90 degree, Qm2=ρQt2=1000×3.1579×10-4=0.31579
For 30 degree, Qm2=ρQt2=1000×4.3478×10-4=0.43478
Force exerted by deflector on water(Fy), Fy=Qmvcosθ+1 where θ=180°-α, and α is the deflection angle. For θ=180 and Qm1, Fy1=0.454559.0431cos0+1=8.221
For θ=180 and Qm2,...