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HAIPHONG VIETNAM  PENSTOCKS
Sizing Calculation for Recommendation of Actuator
I have calculated for the largest unit, as a worse case.
The maximum stem thrust to open the penstock door must firstly be calculated:
Water Force on the Door
The water force on the door acting in the onseating case:
Water Force = Average Head x Door Area x Density of Fluid Head x Force due to Gravity
Water head to centre of door = 3.9 – 1.2 = 2.7m
Area of Door = 1600mm x 2400mm = 3.84m^{2}
Density of Water (clean) = 1000 kg/m^{3}
Gravitational Constant = 9.81
Water Force on Door = 2.7 x 3.84 x 1000 x 9.81 = 101.71kN
Max stem thrust (rising stem) = Door Mass (inclusive of pump) + spindle mass + (Water Force x Coefficient of friction) + hydraulic down force (due to bottom rib and head ratio)
Coefficient of friction for rollers on frame (conservative) = 0.4
Door mass (inclusive of pump) = 3810kg = 37.37kN
Friction due to water force = 101.71 kN x 0.4 = 40.68 kN
Mass of spindles (2 off) = 510kg = 5kN
Hydraulic down force= (1.6 x 0.25 x 9.81 x 1000 x 3.9) x (3.9/15) = 3.98 kN
Max. stem thrust to open door = 37.37 + 40.68 + 5 + 3.98 = 87.03 kN
Therefore, the force required to open for each spindle = 82.03kN / 2 = 43,515 N
Stem Torque
For an Acme type thread, the stem torque to open the door is given by the formula:
Where;
F = Force per spindle to open door as calculated above (43,515N)
d_{m} = mean diameter of thread = nominal diameter  thread depth
for 3” rolled stub acme thread 0.5” pitch, thread depth = 4.2 mm
d_{m} = 77.1 – 4.2 = 72.9 mm
l = lead; for a single start thread lead = pitch = 12.7 mm
= coefficient of friction between the spindle and the nut
for a rolled stainless steel rising spindle and bronze nut this is 0.14
= angle of inclination of thread form to the thread axis = 1/2 thread angle
for stub acme threads, the thread angle is so
Therefore;
= = 319.73 Nm (per spindle)
Selection
Utilising Rotork Gears IB9 3:1 ratio bevel gearboxes.
Actual gearbox ratio = 3:1
Efficiency of gearbox = 85%
Efficiency due to cross shaft losses = 95%
Therefore to find the effective ratio of each gearbox:
3 x 0.85 x 0.95 = 2.42 so the effective ratio is 2.42:1
So the required input torque per spindle can be calculated:
= 132.12 Nm (per spindle)
Total torque required = 132.12 x 2 = 264.24 Nm
Selecting an actuator that uses maximum 80% of its rated torque to achieve the above figure (we do this as a conservative measure):
= 330.3 Nm
Therefore, an actuator with a rated torque of approximately 331 Nm or greater would be suitable to operate the penstock under the maximum head condition.
Conclusion
Looking at the Auma actuator performance data:

SA14.6 unit running at 63 rpm (for 267 mm/min door opening speed – This is within the British Standard BS7775:2005 suggestion for penstocks which states 250300 mm/minute).

This unit would have a 30 minute rated motor due to the opening and closing time being in excess of the standard 15 minute rating.

This has a max rated torque of 360Nm.
Based on the above, this is a suitable unit and would be perfectly adequate for the operation of the penstock.
James Jones – Group Engineering Manager (12/02/15)
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