MOOG D661 Flow Sizing from Cylinder Area and Stroke Time
A machine must complete a 400 mm stroke in two seconds. Choosing a 40, 80 or 120 L/min D661 depends on cylinder area and the pressure drop available across the valve during movement. Calculate both before selecting the flow rating.
D661-G…A rated flows include 20, 40, 80, 90, 120, 160 and 200 L/min. The rating is measured at 35 bar per metering land, or 70 bar across two working lands. A 160 bar pump pressure does not mean that 160 bar is available across the valve.
Calculate the oil needed for the stroke
Consider a double-rod cylinder with equal effective areas of 50 cm². A 400 mm stroke in two seconds requires an average velocity of 200 mm/s, or 0.2 m/s.
Q (L/min) = 6 × effective area (cm²) × velocity (m/s).
Here, Q = 6 × 50 × 0.2 = 60 L/min. This is the steady-speed demand. The pump, piping and valve must deliver that flow under load.
| Stroke time | Average velocity | Required flow |
|---|---|---|
| 4 seconds | 0.10 m/s | 30 L/min |
| 2 seconds | 0.20 m/s | 60 L/min |
| 1 second | 0.40 m/s | 120 L/min |
Halving stroke time doubles flow demand for the same cylinder. If the specified time includes acceleration and deceleration, the intermediate speed exceeds the average. Calculate peak demand from the actual velocity profile.
For a single-rod cylinder, calculate cap-end and rod-end areas separately. Extension needs more inlet volume, while cap-end return flow during retraction can exceed inlet flow. Check both metering paths.
Why a 160 bar supply leaves only 50 bar for the valve
Continue with the equal-area cylinder. Assume 160 bar at valve port P during movement, a 100 bar pressure difference across the cylinder to overcome its load, and 10 bar at valve port T. Ignore line losses between valve and cylinder for this calculation.
The total pressure drop available across the two valve lands is:
160 − 100 − 10 = 50 bar.
With equal flow and symmetric metering edges, approximately 25 bar is available per land. This is below the 35 bar rated condition, so full-opening flow decreases.
Moog uses the square-root pressure-drop relation. Expressed with total valve pressure drop for this symmetric circuit:
Q = Qn × √(actual total valve pressure drop / 70).
| D661 rated flow | Estimated full-opening flow at 50 bar total drop | Comparison with 60 L/min demand |
|---|---|---|
| 40 L/min | 33.8 L/min | Insufficient |
| 80 L/min | 67.6 L/min | Covers steady motion with limited margin |
| 90 L/min | 76.1 L/min | More flow margin |
| 120 L/min | 101.4 L/min | Covers demand; assess low-speed operation too |
An 80 L/min rating is a candidate for the 60 L/min steady motion. If the two-second stroke includes acceleration that raises peak demand beyond 67.6 L/min, compare the 90 or 120 L/min options.
Why the same valve slows down at higher load
If the cylinder load requires 130 bar instead of 100 bar, the remaining valve drop becomes 160 − 130 − 10 = 20 bar.
The 80 L/min valve now delivers an estimated 42.8 L/min at full opening, below the required 60 L/min. A 120 L/min valve provides about 64.1 L/min with little margin, while a 160 L/min valve provides about 85.5 L/min.
A stroke that meets the two-second requirement unloaded but misses it at full load does not automatically indicate valve damage. Record P, T and both cylinder-chamber pressures during motion to establish the pressure budget. Standby pump pressure cannot replace these measurements.
If the pump itself continuously supplies only 50 L/min, a 160 L/min valve cannot create the missing flow. Valve capacity, supply flow and load pressure must all satisfy the movement requirement.
A larger flow rating must also suit slow motion
A mechanism with a 60 L/min rapid stage and a 1 L/min feed stage needs more than a rapid-flow check. A larger valve generally delivers the small flow at a smaller opening, making centre overlap, near-null flow gain and signal variations more relevant.
If the valve is already near full opening during a slow rapid stroke, investigate pressure drop and supply capacity. If the feed stage operates at very small commands, examine that working range. Increasing the flow rating does not solve every speed problem.
The D661 model and flow guide explains the flow codes. After selecting the main-stage rating of a MOOG D661, arrange suitable pilot supply, particularly for low-pressure startup or external-pilot circuits.
Conclusion
In this example, an 80 L/min rating covers the 60 L/min steady demand with 50 bar total valve drop. At 20 bar, the same rating is insufficient. Convert stroke time into flow, subtract load pressure and return backpressure from supply pressure, then compare valve sizes at the resulting operating condition.