fluid_mechanics

12.8 Turbines 679
or
Thus, the nozzle diameter for maximum power output is
(Ans)
(b) The corresponding maximum power can be determined
from Eq. 6 as
or
W shaft 1.04 106 10.2392 2
31 15210.2392 4 4 3 2 3.25 104 ft lb/s
W shaft 3.25 10 4 ft lb/s
59.0 hp
304 D 4 1 1
D 1 0.239 ft
(Ans)
The rotor speed at the maximum power condition can be obtained
from
U R V 1
2
where V 1 is given by Eq. 4. Thus,
113.5
V 1
2R 21 15210.2392 4 ft/s
2a 3 2 ftb
1 hp
550 ftlb/s
COMMENT The reason that an optimum diameter nozzle
exists can be explained as follows. A larger diameter nozzle will
allow a larger flowrate, but will produce a smaller jet velocity because
of the head loss within the supply side. A smaller diameter
nozzle will reduce the flowrate but will produce a larger jet velocity.
Since the power depends on a product combination of flowrate
and jet velocity (see Eq. 1), there is an optimum-diameter nozzle
that gives the maximum power.
These results can be generalized (i.e., without regard to the
specific parameter values of this problem) by considering Eqs. 1
and 3 and the condition that U V 1 /2 to obtain
W shaft UV 12 p r 11 cos b2
16
12gz 0 2 3/2 D 2 1^a1 f / 3/2
D 5 D4 1b
By setting dW shaft/dD 1 0, it can be shown (see Problem 12.67)
that the maximum power occurs when
D 1 D^ a2f / D b 1/4
which gives the same results obtained earlier for the specific parameters
of the example problem. Note that the optimum condition
depends only on the friction factor and the length-to-diameter
ratio of the supply pipe. What happens if the supply pipe is frictionless
or of essentially zero length?
30.9 rad/s 1 rev/2 rad 60 s/min
295 rpm
(Ans)
In previous chapters we mainly treated turbines 1and pumps2 as “black boxes” in the flow that
removed 1or added2 energy to the fluid. We treated these devices as objects that removed a certain
shaft work head from or added a certain shaft work head to the fluid. The relationship between the
shaft work head and the power output as described by the moment-of-momentum considerations is
illustrated in Example 12.7.
E XAMPLE 12.7
Maximum Power Output for a Pelton Wheel Turbine
GIVEN Water flows through the Pelton wheel turbine shown
in Fig. 12.24. For simplicity we assume that the water is turned
180° by the blade.
FIND Show, based on the energy equation 1Eq. 5.842, that the
maximum power output occurs when the absolute velocity of the
fluid exiting the turbine is zero.
SOLUTION
As indicated by Eq. 12.51, the shaft power of the turbine is given by
W # shaft rQU1U V 1 211 cos b2
2rQ1U 2 V 1 U2
For this impulse turbine with b 180°, the velocity triangles
simplify into the diagram types shown in Fig. E12.7. Three possibilities
are indicated:
(a) the exit absolute velocity, V 2 , is directed back toward the
nozzle,
(1)
(b) the absolute velocity at the exit is zero, or
(c) the exiting stream flows in the direction of the incoming
stream.
According to Eq. 12.52, the maximum power occurs when
U V 12, which corresponds to the situation shown in Fig.
E12.7b, that is, U V 12 W 1 . If viscous effects are negligible,
then W 1 W 2 and we have U W 2 , which gives
V 2 0
(Ans)