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Strategy: Use
vector plot at the exit to find the torque developed by the turbine.
Step 2: Apply conservation of mass at the exit
for the turbine:
Q =
2 π r2 b2 Vr2 So
Vr2 = Q / (2 π r2 b2
) =
17.68 ft/sec
and tan β2 =
Vr2 / Wt2
So Wt2 =
Vr2 / tan β2 =
30.63 ft/sec
By vector
addition: V2 = U2 + W2 and
Vt2 = ˗ r2 w
et +
Wt2 et
Now r2w =
20.1 ft/sec So
and Vt2 = ˗
20.1 + 30.63
= 10.52
ft/sec
Step 3: Apply conservation of angular momentum to
find the torque, T .
T
= r
Q (r1Vt1
˗ r2Vt2)
= r Q [ rout
Vtout ˗ rinVtin
]
T =
(1.94)(500)[ (3)( 10.5)
˗ (5)(˗ 39.63) ] = 222860 ft lb (result)
Step 4: For conservation of energy:
Power
generated by the turbine is P =
T w =
(222860)(6.7) = 1.5 x 106 ft lb/sec
Or P
= 2700 hp (result)
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