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ME 363 - Fluid Mechanics
Homework #7 solutions
Due Wednesday, March 12, 2008
Spring Semester 2008
1] Pelton wheel turbines are commonly used in hydroelectric power plants to generate electric power. In
these turbines, a high-speed jet at a velocity of Vj impinges on buckets, forcing the wheel to rotate. The
buckets reverse the direction of the jet, and the jet leaves the bucket making an angle  with the
direction of the jet, as shown in Fig. 1. Use the following relationship to obtain the power produced by
a Pelton wheel: P = Tshaft, where Tshaft is the total torque produced by the wheel and  is an angular
velocity of the wheel. Show that the power P = rQ(Vj - r)(1 – cos ), where  is the density and Q
is the volume flow rate of the fluid. Obtain the numerical value for  = 1000 kg m-3, r = 2 m, Q = 10
m3s-1,  = 5 s-1,  = 160°, and Vj = 50 m/s.
2] A liquid of density ρ flows through a 90° bend as and issues vertically from a uniformly porous section
of length L. Neglecting weight, find an expression for the support torque M required at point O.
3] A power plant on a river, as in Fig. 3, must deliver 55 MW of waste heat to the river. The river
conditions upstream are Qi = 2.5 m3/s and Ti = 18°C. The river is 45 m wide and 2.7 m deep. If heat
losses to the atmosphere and ground are negligible, estimate the downstream river conditions (Q0, T0).
4] A simple turbomachine is constructed from a disk with two internal ducts which exit tangentially
through square holes, as in the figure. Water at 20C enters the disk at the center, as shown. The disk
must drive, at 250 rev/min, a small device whose retarding torque is 1.5 Nm. What is the proper mass
flow of water, in kg/s?
5] Multnomah Falls in the Columbia River Gorge has a sheer drop of 543 ft. Estimate the water
temperature change in degrees F caused by this drop.
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Problem 1 A Pelton wheel is considered for power generation in a hydroelectric power
plant. A relation is to be obtained for power generation, and its numerical value is to be
obtained.
Mshaft

r
Vj - r
Shaft

Nozzle
Vj - r
Vj
r
Assumptions 1 The flow is uniform and cyclically steady. 2 The water is discharged to
the atmosphere, and thus the gage pressure at the nozzle outlet is zero. 3 Friction and
losses due to air drag of rotating components are neglected. 4 The nozzle diameter is
small compared to the moment arm, and thus we use average values of radius and
velocity at the outlet.
Properties
We take the density of water to be 1000 kg/m3 = 1 kg/L.
Analysis
The tangential velocity of buckets corresponding to an angular velocity of
  2n is Vbucket  r . Then the relative velocity of the jet (relative to the bucket)
becomes
Vr  V j  V bucket  V j  r
We take the imaginary disk that contains the Pelton wheel and rotates with the wheel as the control volume.
The inlet velocity of the fluid into this control volume is Vr, and the component of outlet velocity normal to
the moment arm is Vrcos . The angular momentum equation can be expressed as
M
rm V 
rm V where all moments in the counterclockwise direction are positive, and all in



out
in
the clockwise direction are negative. Then the angular momentum equation about the axis of rotation
becomes
 Vr cos   rm
 Vr
M shaft  rm
or M shaft  rm V r (1  cos  )  rm (V j  r )(1  cos  )
  V , the shaft power output of a Pelton turbine becomes
Noting that W shaft  2nM shaft  M shaft and m
W shaft  Vr (V j  r )(1  cos  )
which is the desired relation. For given values, the shaft power output is determined to be
 1 MW 
W shaft  (1000 kg/m 3 )(10 m 3 /s)( 2 m) (15.71 rad/s)(50 - 2 15.71 m/s)(1 - cos160) 6
  11.3 MW
 10 N  m/s 
where
  2n  2 (150 rev/min) 
1 min 
  15 .71 rad/s
 60 s 
Discussion The actual power will be somewhat lower than this due to air drag and
friction. Note that this is the shaft power; the electrical power generated by the generator
connected to the shaft is be lower due to generator inefficiencies.
Problem 2
Problem 3
Problem 4
Problem 5