In a hydroelectric power plant power is generated by allowing water to flow from

Question

In a hydroelectric power plant power is generated by allowing water to flow from a reservoir through tunnels and penstocks in the dam to turn turbines located in a powerhouse. The large difference in elevation between the intake (reservoir) and the powerhouse causes high-energy flow through the turbine to generate power. To dissipate the high-energy flow of water downstream of the turbine, abrupt expansions are commonly used. Suppose an abrupt expansion is to be used to dissipate the high-energy flow of water in a 5 ft diameter penstock as shown in the figure below. The flow velocity at section 1 is 25 ft/s. What power (in horsepower) is lost through the expansion? Assume the head loss due to a sudden expansion can be expressed as: Hloss = (Vupstream - Vdownstream)2/2g; where Vupstream is the velocity of flow upstream of the expansion, Vdownstream is the velocity downstream of the expansion, and g is gravitational acceleration. If the pressure at section 1 is 5 psi, what is the pressure at section 2? What force is needed to hold the expansion in place?

Explanation / Answer

a) Given V1=25ft/s            D1=5ft D2=10ft Now A1=19.625ft2 A2=78.5ft2 A2=78.5ft2 From continuity equation A1V1=A2V2 Siubstituting the values V2=6.25ft/s Discharge Q=A1V1=490.625ft3/s Accelleration due to gravity g=32.174ft/s2 Siubstituting the values V2=6.25ft/s Discharge Q=A1V1=490.625ft3/s Accelleration due to gravity g=32.174ft/s2 a)HLoss = (Vupstream-Vdownstream)2/2g               =(V1-V2)2/2g               =(25-6.25)2/2x32               =5.46ft Power lost =gQHLoss=(62.4)(490.625)(5.46)=167157.9ft-lbf/s=304HP b) Given P1=5 Psi=720lb/ft2 Assuming Z1=Z2 Apply benoullis equation at 1 and 2 (P1/w)+ [(V1)2/2g ] = P2/w +[(V2)2/2g] P2=1290.9lb/ft2=8.96Psi c) Density of water =w/g Force in x direction F=Q(V1-V2)+P1A1-P2A2 =-69.196lbf =-69.196lbf

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