Page 460 - Mechatronics with Experiments

JWST499-Cetinkunt
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were zero leakage, that is 100% volumetric efficiency. The overall efficiency takes into
account both the leakage and other mechanical efficiency.
One of the main operational concerns is the noise level of the pump. The noise due to
cavitation can be very serious. Low pressure levels at the pump input may result in allowing
air bubbles to enter the pump inlet port. Under high pressure, the bubbles collapse which
leads to high noise levels. Therefore, some pumps may require higher than normal levels of
inlet port pressure, that is so called boost inlet pressure using another small pump (called
charge pump) between the main pump and tank. Cavitation may also be caused by too high
hydraulic fluid viscosity, as a result, the pump is not able to suck in enough fluid. This may
happen especially in cold start-up conditions. Heaters may be included in the pump inlet to
limit the fluid viscosity.

3
Example Consider a pump with fixed displacement D = 100 cm ∕rev and following
p
nominal operating conditions: input shaft speed w = 1200 rpm, torque at the input shaft
shaft
3
T = 250 Nt ⋅ m,theoutputpressurep = 12 MPa, andoutputflowrate Q = 1750 cm ∕s.
out out
Determine the volumetric, mechanical, and overall efficiencies of the pump at this operating
condition.
Q out
= (7.88)
v
D ⋅ w shaft
p
3
1750 cm ∕s
= = 87.5% (7.89)
3
100 cm ⋅ 20 rev∕s
p out ⋅ D p
= (7.90)
m
T
2
6
3
12 × 10 N∕m ⋅ 100 × 10 −6 m ∕rev
= = 76.4% (7.91)
2 ⋅ ⋅ 250
= ⋅ m (7.92)
v
o
= 0.875 ⋅ 0.764 ⋅ 100%= 66.85% (7.93)
Notice that if any of the efficiency calculations results in being larger than 100%, it is an
indication that there is an error in the given information or measured variables since the
efficiencies cannot be larger than 100%. In order to confirm the results, let us calculate the
input mechanical power to the pump and output hydraulic power from the pump.
Power = T ⋅ w (7.94)
in shaft
= 250 N ⋅ m ⋅ 20 rev∕s ⋅ 2 rad∕rev (7.95)
= 31.41 N ⋅ m∕s = 31.41 kW (7.96)
Power = p ⋅ Q (7.97)
out out out
3
2
3
6
3
6
= 12 ⋅ 10 N∕m ⋅ (1750 cm ∕s) ⋅ (1 m ∕10 cm ) (7.98)
= 12 ⋅ 1750 N ⋅ m∕s (7.99)
3
= 21 ⋅ 10 N ⋅ m∕s = 21 kW (7.100)
Power out 21
= (7.101)
Power in 31.41
= 66.85% (7.102)
= ⋅ (7.103)
v m
= (7.104)
o
Example Consider the hydrostatic transmission shown in Figure 7.35. Assume that
the pressure drop across the connecting hydraulic line is 200 psi. The pump variables and