fluid_mechanics
680 Chapter 12 ■ Turbomachines U W 1 V 1 U W 1 V 1 U W 1 V 1 U W 1 V 1 Inlet F I G U R E E12.7 If we consider the energy equation 1Eq. 5.842 for flow across the rotor we have (a) (b) (c) (d) p 1 g V 2 1 2g z 1 h S p 2 g V 2 2 W 2 = W 1 W 2 = W 1 U V 2 = 0 6 180° where h S is the shaft work head. This simplifies to 2g z 2 h L since p 1 p 2 and z 1 z 2 . Note that the impulse turbine obtains h S V 2 2 V1 2 h U V 2g L 2 (2) its energy from a reduction in the velocity head. The largest shaft work head possible 1and therefore the largest power2 occurs when all of the kinetic energy available is extracted by the turbine, giving U V 2 0 (Ans) V 2 W 2 = W 1 This is consistent with the maximum power condition represented by Fig. E12.7b. COMMENT As indicated by Eq. 1, if the exit absolute velocity is not in the plane of the rotor 1i.e., b 6 180° 2, there is a reduction V 2 in the power available 1by a factor of 1 cos b2. This is also U supported by the energy equation, Eq. 2, as follows. For b value of V 2 to zero—there is always a component in the axial direction. Outlet the inlet and exit velocity triangles are as shown in Fig. E12.7d. Regardless of the bucket speed, U, it is not possible to reduce the Thus, according to Eq. 2, the turbine cannot extract the entire velocity head; the exiting fluid has some kinetic energy left in it. W 2 = W 1 Dentist drill turbines are usually of the impulse class. A second type of impulse turbine that is widely used 1most often with air as the working fluid2 is indicated in Fig. 12.29. A circumferential series of fluid jets strikes the rotating blades which, as with the Pelton wheel, alter both the direction and magnitude of the absolute velocity. As with the Pelton wheel, the inlet and exit pressures 1i.e., on either side of the rotor2 are equal, and the magnitude of the relative velocity is unchanged as the fluid slides across the blades 1if frictional effects are negligible2. Typical inlet and exit velocity triangles 1absolute, relative, and blade velocities2 are shown in Fig. 12.30. As discussed in Section 12.2, in order for the absolute velocity of the fluid to be changed Nozzles Control volume T shaft ω Fluid jets (a) (b) F I G U R E 12.29 A multinozzle, non-Pelton wheel impulse turbine commonly used with air as the working fluid.
12.8 Turbines 681 U Section (1) W 2 W 1 V 2 U 2 Section (2) V 1 U 1 V θ1 F I G U R E 12.30 Inlet and exit velocity triangles for the impulse turbine shown in Fig. 12.29. as indicated during its passage across the blade, the blade must push on the fluid in the direction opposite of the blade motion. Hence, the fluid pushes on the blade in the direction of the blade’s motion—the fluid does work on the blade 1a turbine2. E XAMPLE 12.8 Non-Pelton Wheel Impulse Turbine (Dental Drill) GIVEN An air turbine used to drive the high-speed drill used by your dentist is shown in Figs. 12.29 and E12.8a. Air exiting from the upstream nozzle holes forces the turbine blades to move in the direction shown. The turbine rotor speed is 300,000 rpm, the tangential component of velocity out of the nozzle is twice the blade speed, and the tangential component of the absolute velocity out of the rotor is zero. FIND Estimate the shaft energy per unit mass of air flowing through the turbine. SOLUTION We use the fixed, nondeforming control volume that includes the turbine rotor and the fluid in the rotor blade passages at an instant of time 1see Fig. E12.8b2. The only torque acting on this control volume is the shaft torque. For simplicity we analyze this problem using an arithmetic mean radius, r m , where r m 1 2 1r 0 r i 2 A sketch of the velocity triangles at the rotor entrance and exit is shown in Fig. E12.8c. Application of Eq. 12.5 1a form of the moment-of-momentum equation2 gives w shaft U 1 V u1 U 2 V u2 where w shaft is shaft energy per unit of mass flowing through the turbine. From the problem statement, V u1 2U and V u2 0, where U vr m 1300,000 revmin211 min60 s212p radrev2 10.168 in. 0.133 in.22112 in.ft2 394 fts (1) (2) is the mean-radius blade velocity. Thus, Eq. 112 becomes w shaft U 1 V u1 2U 2 21394 fts2 2 310,000 ft 2 s 2 1310,000 ft 2 s 2 2132.1741ft # lbm21lb # s 2 22 9640 ft # lblbm (Ans) COMMENT For each lbm of air passing through the turbine there is 9640 ft # lb of energy available at the shaft to drive the drill. However, because of fluid friction, the actual amount of energy given up by each slug of air will be greater than the amount available at the shaft. How much greater depends on the efficiency of the fluid-mechanical energy transfer between the fluid and the turbine blades. Recall that the shaft power, W # shaft, is given by W # shaft m # w shaft Hence, to determine the power we need to know the mass flowrate, m # , which depends on the size and number of the nozzles. Although the energy per unit mass is large 1i.e., 9640 ft # lblbm2, the flowrate is small, so the power is not “large.”
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680 Chapter 12 ■ Turbomachines<br />
U W 1<br />
V 1<br />
U W 1<br />
V 1<br />
U W 1<br />
V 1<br />
U W 1<br />
V 1<br />
Inlet<br />
F I G U R E E12.7<br />
If we consider the energy equation 1Eq. 5.842 for flow across<br />
the rotor we have<br />
(a)<br />
(b)<br />
(c)<br />
(d)<br />
p 1<br />
g V 2 1<br />
2g z 1 h S p 2<br />
g V 2 2<br />
W 2 = W 1<br />
W 2 = W 1<br />
U V 2 = 0<br />
6 180°<br />
where h S is the shaft work head. This simplifies to<br />
2g z 2 h L<br />
since p 1 p 2 and z 1 z 2 . Note that the impulse turbine obtains<br />
h S V 2 2 V1<br />
2 h<br />
U V 2g<br />
L<br />
2<br />
(2)<br />
its energy from a reduction in the velocity head. The largest shaft<br />
work head possible 1and therefore the largest power2 occurs<br />
when all of the kinetic energy available is extracted by the turbine,<br />
giving<br />
U<br />
V 2 0<br />
(Ans)<br />
V 2 W 2 = W 1<br />
This is consistent with the maximum power condition represented<br />
by Fig. E12.7b.<br />
COMMENT As indicated by Eq. 1, if the exit absolute velocity<br />
is not in the plane of the rotor 1i.e., b 6 180° 2, there is a reduction<br />
V 2<br />
in the power available 1by a factor of 1 cos b2. This is also<br />
U<br />
supported by the energy equation, Eq. 2, as follows. For b<br />
value of V 2 to zero—there is always a component in the axial direction.<br />
Outlet<br />
the inlet and exit velocity triangles are as shown in Fig. E12.7d.<br />
Regardless of the bucket speed, U, it is not possible to reduce the<br />
Thus, according to Eq. 2, the turbine cannot extract the<br />
entire velocity head; the exiting <strong>fluid</strong> has some kinetic energy left<br />
in it.<br />
W 2 = W 1<br />
Dentist drill turbines<br />
are usually of<br />
the impulse class.<br />
A second type of impulse turbine that is widely used 1most often with air as the working<br />
<strong>fluid</strong>2 is indicated in Fig. 12.29. A circumferential series of <strong>fluid</strong> jets strikes the rotating blades<br />
which, as with the Pelton wheel, alter both the direction and magnitude of the absolute velocity.<br />
As with the Pelton wheel, the inlet and exit pressures 1i.e., on either side of the rotor2 are equal,<br />
and the magnitude of the relative velocity is unchanged as the <strong>fluid</strong> slides across the blades 1if frictional<br />
effects are negligible2.<br />
Typical inlet and exit velocity triangles 1absolute, relative, and blade velocities2 are shown in<br />
Fig. 12.30. As discussed in Section 12.2, in order for the absolute velocity of the <strong>fluid</strong> to be changed<br />
Nozzles<br />
Control<br />
volume<br />
T shaft<br />
ω<br />
Fluid jets<br />
(a)<br />
(b)<br />
F I G U R E 12.29 A multinozzle, non-Pelton wheel impulse turbine commonly used with air as the<br />
working <strong>fluid</strong>.