14.50: PROBLEM DEFINITION
Situation: A jet of water strikes the buckets of an impulse wheel and turns water by
180 degrees with bucket speed 1/2 of jet speed.
Find: (a) Jet force on the bucket.
(b) Resolve the discrepancy with Eq. 14.24 (EFM 10e).
PLAN
Apply the momentum principle.
SOLUTION
Consider the power developed from the force on a single bucket. Referencing velocities
tothebucketgives
Momentum principle
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14.51: PROBLEM DEFINITION
Part (a)
Situation: Compare reaction turbine and centrifugal pump.
Find:Difference between reaction turbine and centrifugal pump.
SOLUTION
The power is provided to the impeller of a centrifugal pump to increase pressure of
fluid. The pressure on the runner of a reaction turbine causes rotating and power
production.
Part (b)
Situation: Runner in reaction turbine.
Find: Meaning of “runner” in reaction turbine.
SOLUTION
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14.52: PROBLEM DEFINITION
Situation: A Francis turbine has an outside diameter of 5 m and inside diameter of 3
m, an inlet blade angle of 60oand an outlet angle of 90o. Runner width is 1 m and
discharge is 126 m3/s.
Find:(a)α1for non-separating flow conditions .
(b) Maximum attainable power.
(c) Changes to increase power production.
Properties: From Table A.5, ρ=1000kg/m3at 10oC
SOLUTION
Flow rate equation
a)
b) From Eq. (14.27 (EFM 10e))
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14.53: PROBLEM DEFINITION
Situation: A Francis turbine with a discharge of 3.3 m3/s operates at 60 rpm, has
outside and inside diameters of 1.5 and 1.2 m respectively, inlet and outlet blade
angles of 85 and 165 degrees and a runner width of 33 cm.
Find:(a)α1for non-separating flow conditions.
(b) Power.
(c) Torque.
Properties: From Table A.5, ρ=1000kg/m3at 10oC
SOLUTION
V=Q/(2π∗r∗B)
Vtan1=r1ω+Vr1cot β1=1.5m×2πrad/s +1.061 m/s ×0.0875
=9.518 m/s
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14.54: PROBLEM DEFINITION
Situation: A Francis turbine operates at 120 rpm with discharge of 200 m3/s. The
outer radius, vane angle and runner width are 3 m, 45oand 0.9 m respectively.
Find:α1for non-separating flow conditions.
SOLUTION
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14.55: PROBLEM DEFINITION
Situation: A small hydroelectric project is fed by an elevation difference of 400 ft by
1000 ft of 12 in diameter steel pipe. Design calls for 8 cfs and 80% efficiency.
Find: (a) Power output.
(b) Draw the HGL and EGL.
Assumptions:Ke=0.50; KE=1.0; Kb=0.2; ρ=62.4lbm/ft3,ν=1.22 ×
10−5ft2/s..Assume two bends in system.
PLAN
To get power apply the energy equation. Apply the flow rate equation to get Vfor
the head loss. Then apply the power equation.
SOLUTION
Assume there is an entry loss, sudden expansion loss and loss due to a bend.
Energy equation
2g(fL
D+KE+Kee+2Kb)
–
Reynolds number.
Power equation
Power output from the turbine
Plot of HGL & EGL
EGL
HGL
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14.56: PROBLEM DEFINITION
Situation: Performance of wind turbines
Find: Factors determining maximum and minimum wind speed for turbine.
SOLUTION
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14.57: PROBLEM DEFINITION
Situation: Using the internet and other resources, identify at least four different
types of wind turbines. For each type, discribe its distinguishing characteristics, and
its relative advantages and disadvantages.
SOLUTION
No solution provided for this problem, answers will vary.
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14.58: PROBLEM DEFINITION
Situation: Sizing a wind turbine given a power requirement and wind data.
Find: Minimum capture area
Properties:ρ=1.2kg/m3,V =10mph
PLAN
Use equations for maximum power of a windmill.
A=Pmax
54
16
1
ρV 2
SOLUTION
Convert windspeed to m/s.
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14.59: PROBLEM DEFINITION
Situation: A conventional horizontal-axis wind turbine with 2.3 m diameter propeller
ina47km/hwind.
Find: Maximum deliverable power.
Properties:ρ=1.2kg/m3.
PLAN
Use equation for theoretical maximum power.
SOLUTION
The wind speed in m/s
Maximum power
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14.60: PROBLEM DEFINITION
Situation: Wind farm with 20 Darrieus wind turbines 15 m tall in 20 m/s wind to
produce 2 MW power. Turbine has shape of circular arc.
Find: Width of wind turbine.
Properties:ρ=1.2kg/m3.
PLAN
Apply the wind turbine maximum power equation and find capture area of each
turbine..
SOLUTION
Each windmill must produce 2 MW/20 = 100,000 W.
Wind turbine maximum power
Consider the figure for the section of a circle.
R
The area of a sector is given by
so
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Solving graphically gives θ=52
o. The width of the windmill is
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14.61: PROBLEM DEFINITION
Situation: A 10 ft diameter windmill used to pump water from 10 ft deep well in
30 mph wind. Pump efficiency is 80%.There is 20 ft of 2 in galvcanized iron pipe in
system.
Find: Discharge of pump.
Properties:ρ=0.07 lbm/ft3.
Assumptions: Assume one 90 degree elbow in system and a fully rough galvanized
pipe.
PLAN
Apply energy equation between well and exit to find system head in terms of Q.
Apply the wind turbine maximum power equation to get Pforthepowerequation
and solve for Q.
SOLUTION
Wind speed
Energy equation between well surface and outlet.
Theroughnessforgalvanizedironis0.006insorelativeroughnessis0.006in/2
in=0.003. From Fig 10.14 (EFM 10e), the fully rough Darcy Weisback friction factor
is 0.027. The system head is
The system head is
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Wind turbine maximum power
Power equation
Equating the heads
Solving for discharge gives Q=0.645 cfs
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