How vacuum ejectors work
A vacuum ejector (also called a venturi generator) passes compressed air through a nozzle at high velocity. The jet entrains surrounding air through a suction port, creating vacuum at the gripper. Ejectors have no moving parts, are light enough to mount on the robot or tool, and create vacuum instantly. Their trade-off is that they consume compressed air for every second they operate.
There are two broad designs:
- Single-stage ejectors use one nozzle and diffuser. They can reach a high vacuum level but with relatively low suction flow, suiting sealed, non-porous parts.
- Multi-stage ejectors use several nozzles in series. They give higher suction flow at low and medium vacuum for the same motive air, so they evacuate larger volumes faster and cope better with small leaks.
Most ejectors have an optimum motive-air pressure stated by the maker, commonly in the range of about 4–6 bar (58–87 psig). Running above it often does not raise vacuum and simply uses more air; running below it reduces both vacuum and suction flow.
Holding force, suction flow and air consumption are three separate things
Catalogues print these values side by side, but they answer different questions:
| Quantity | What it means | What decides it |
|---|---|---|
| Holding force | Force keeping the part on the cup | Vacuum level (pressure difference) × effective cup area |
| Suction flow | How fast air is removed from cups and lines | Ejector design, vacuum level, motive pressure |
| Evacuation time | Time to reach working vacuum after contact | Volume of cups and tubing, suction flow, leakage |
| Motive-air consumption | Compressed air drawn from the plant | Nozzle size, supply pressure, on-time |
| Release time | Time to drop the part | Blow-off pulse, vacuum line volume |
A strong ejector with a high vacuum rating cannot hold a part if the cup leaks badly; a gentle ejector can hold a sealed part securely with very little air. Keeping these quantities apart is the key to a good selection.
Calculating the holding force
Theoretical holding force is the pressure difference multiplied by the effective area of the cup. For example, at a vacuum of −0.6 bar (60 kPa below atmosphere), a 50 mm cup with an effective area of about 19.6 cm² gives roughly 60,000 Pa × 0.00196 m², about 118 N.
That force is then reduced by a safety factor that depends on how the load acts:
- Horizontal cup, vertical lift (load pulls straight off the cup): a factor of at least 2 is common.
- Vertical cup, load in shear (part hanging on its face): friction carries the load, so the usable force is much lower and a factor of 4 or more is common, depending on the friction between cup and surface.
- Fast robot motion adds acceleration forces; add the mass multiplied by the highest acceleration, including emergency stops, to the gravity load.
The cup maker’s data for the specific cup material and surface should be used where available. Oily, dusty, textured or curved surfaces reduce effective area and friction. Hot parts and certain plastics can deform cups.
Very high vacuum is not always better. Many applications work best around −0.6 to −0.7 bar; higher vacuum increases cup wear, can mark soft parts, and needs more time to evacuate for a small gain in force.
Porous and leaky parts
Cardboard cartons, sacks, wood, textiles and some castings let air leak through the cup. Holding then depends on the ejector removing air as fast as it leaks in, so suction flow at the working vacuum becomes the main selection criterion. A multi-stage ejector or an electrically driven vacuum pump is often a better fit. The practical method is to test the actual part: measure the vacuum level reached with a given ejector and check that the resulting force, with the safety factor, still holds the load.
Leaks also arise from worn cup lips, cracked tubing, loose fittings and cups that do not lie flat on the part. A gripper that worked when new may drop parts after months of wear, so periodic checks of reached vacuum are part of maintenance.
Calculating compressed-air consumption
An ejector’s motive-air consumption is published in normal litres per minute at a stated supply pressure. Total consumption is that figure multiplied by the fraction of time the ejector is on, plus blow-off air.
Worked example:
- An ejector uses 60 normal litres per minute at 5 bar (72.5 psig).
- The handling cycle is 10 seconds, with the ejector on for 4 seconds (approach, pick, transfer, place).
- Average motive air = 60 × 4 ÷ 10 = 24 normal litres per minute.
- Blow-off: a 0.2 second pulse at 100 normal litres per minute each cycle adds about 0.33 litre per cycle, or 2 normal litres per minute at 6 cycles a minute.
- Total per gripper is about 26 normal litres per minute, roughly 0.9 cfm.
- Twenty such grippers on a packaging line need about 520 normal litres per minute, roughly 18 cfm, before allowing for leaks and simultaneous peaks.
The peak demand matters as well as the average. If all grippers can switch on together, the instantaneous flow is the full motive-air figure for every ejector, which must be supported by the supply pipework and a local receiver without the pressure falling below the ejector’s optimum.
Reducing air consumption
Ejectors are among the most visible compressed-air loads on packaging and automation lines. Practical ways to cut their demand:
- Air-saving circuits. Many ejectors include a non-return valve and vacuum switch. Once the set vacuum is reached, motive air switches off and the check valve holds vacuum; air restarts only if vacuum falls. On sealed parts this can reduce consumption dramatically.
- Correct supply pressure. Set the regulator to the ejector’s optimum, not to the plant header pressure.
- Short on-times. Switch the ejector on as close to contact as the evacuation time allows, and off as soon as the part is released.
- Short, correctly sized vacuum lines. Less volume to evacuate means shorter on-time.
- Maintain cups and tubing. Leaks keep air-saving circuits running continuously.
- Consider an electric vacuum pump for large central systems with many cups running continuously, especially on porous products.
Remember that each extra bar on the compressor system costs about 7% more power. Raising plant pressure to make ejectors work harder usually costs more than fixing the ejector supply.
Ejector or vacuum pump: how to choose
| Factor | Compressed-air ejector | Electric vacuum pump or blower |
|---|---|---|
| Response | Very fast, mounted at the cup | Slower if far from the cups; reservoir helps |
| Weight and size | Light, compact, no moving parts | Heavier, usually fixed on the machine frame |
| Running cost | Uses compressed air whenever on | Direct electrical drive, often lower for continuous duty |
| Porous loads | Multi-stage types cope with moderate leakage | High flow types handle heavy leakage well |
| Maintenance | Clean silencers and nozzles; replace cups | Pump servicing, filters, vanes or bearings |
| Best fit | Intermittent picks, robots, many separate grippers | Continuous duty, large area grippers, palletising |
Air quality and supply for ejectors
Ejector nozzles are small and silencers clog easily. Water and oil aerosol in the motive air, together with dust drawn in through the vacuum side, gradually reduce performance. A typical arrangement is a refrigerated dryer at about +3 °C pressure dew point and coalescing filtration at the compressor house, with a point-of-use filter-regulator at the machine. A vacuum-side filter between cup and ejector protects against dust from cartons, paper and powders.
Supply pressure must be stable at the ejector inlet. Long, small-bore tubing between the valve and ejector causes pressure drop during operation, so the ejector may receive well below its rated pressure while the gauge at the regulator reads correctly. Measure at the ejector during operation when troubleshooting.
Troubleshooting common problems
| Symptom | Likely cause | Check |
|---|---|---|
| Part dropped during fast moves | Insufficient safety factor for acceleration; worn cups | Holding force calculation including acceleration; cup lips |
| Slow pickup | Large vacuum line volume, low suction flow, low motive pressure | Tubing length, ejector type, pressure at ejector inlet |
| Vacuum never reaches set point | Leakage through part or cup, cracked tubing | Leak test with part replaced by a sealed plate |
| Air-saving circuit cycles constantly | Leak in cup, check valve or tubing | Cup seating, check valve condition |
| Part sticks on release | No or weak blow-off; vacuum line volume | Blow-off pulse settings, release valve |
| Gradual loss of performance | Clogged silencer or nozzle; contaminated air | Silencer, air filtration, dew point |
How Reddy Compressors helps
Packaging, pharmaceutical, food and automotive lines across Hyderabad, Telangana and Andhra Pradesh often have dozens of ejectors, and together they make a significant compressed-air load. Reddy Compressors, at Tarnaka, Secunderabad since 2002, is an independent multi-brand specialist. We sell (new and used/refurbished), rent and service compressors from ELGi, Atlas Copco, Ingersoll Rand, KAESER, Chicago Pneumatic and others, and we sell, rent and service vacuum pumps and blowers, including side-channel types used for vacuum duties. We are not an authorised dealer of any brand; gripper and ejector selection itself is usually finalised with the automation supplier. We have tie-ups with all major and reputed brands.
- Air audits that measure line-by-line demand, so ejector consumption and leaks can be quantified in litres per minute and kWh.
- Compressor, dryer, filter and receiver sizing for lines with large or coincident ejector demand.
- Service, annual maintenance contracts (AMC) and breakdown support with more than 20 service engineers and technicians, and rental compressors from our own fleet, supplied across India.
What to send us
- Number of grippers and ejectors per line, with ejector models or motive-air ratings.
- Cycle time and on-time for each gripper.
- Supply pressure at the regulator and, if possible, at the ejector during operation.
- Compressor and dryer nameplate photos and the plant header pressure.
- Products handled, especially porous or dusty items, and the problem you see.
Frequently asked questions
How much compressed air does a vacuum ejector use?
An ejector uses its rated motive-air flow, published in normal litres per minute at a stated pressure, for every second it is switched on. Multiply that flow by the fraction of the cycle it runs and add any blow-off air. For example, a 60 litres per minute ejector on for 4 seconds in a 10-second cycle averages about 24 litres per minute.
How do I calculate the holding force of a suction cup?
Multiply the vacuum level as a pressure difference by the effective area of the cup. At −0.6 bar, a 50 mm cup gives about 118 N in theory. Then divide by a safety factor: at least 2 for a straight vertical lift and 4 or more where the load acts in shear, and add acceleration forces from fast robot moves.
Is a vacuum pump cheaper to run than an ejector?
For continuous duty with many cups, especially on porous products, an electric vacuum pump or blower usually costs less to run because it avoids the losses of making compressed air first. For intermittent picks, robots and separate grippers, ejectors with air-saving circuits can be very economical. Compare kWh per shift using measured on-times rather than catalogue figures.
Why does my vacuum gripper drop parts?
The usual causes are worn or damaged cups, leaks in tubing or fittings, porous or dusty parts, low motive-air pressure at the ejector, or a safety factor that ignored acceleration and emergency stops. Check the vacuum level actually reached on the part, measure pressure at the ejector inlet during operation, and recalculate force including the highest acceleration.
What pressure should a vacuum ejector run at?
Use the optimum motive pressure stated by the ejector maker, commonly in the region of 4–6 bar (58–87 psig). Running higher rarely improves vacuum and wastes air; running lower reduces both vacuum level and suction flow. Set a regulator for the ejectors rather than running them at full header pressure, and check the pressure at the ejector itself.
What is an air-saving function on a vacuum ejector?
It is a built-in arrangement of a non-return valve and vacuum switch. When the cup reaches the set vacuum, the motive air shuts off and the check valve holds the vacuum. If vacuum falls below a lower setting, air switches back on. On sealed, non-porous parts this greatly cuts compressed-air use, provided cups and tubing are leak-free.
Related pages
Calculators and worked examples
These tools give indicative figures for screening, comparison and for preparing an enquiry. They run in your browser and send nothing. No prices or tariffs are assumed: enter your own to see costs. Efficiencies and other figures shown are example values: replace them with your own. Results are not a performance guarantee or a design certificate; final selection is confirmed by a service engineer against the machine data sheet and the site.
Vacuum holding force
Holding force of suction cups or a vacuum table from the vacuum at the part and the sealed area, reduced by a safety factor, and checked against the part weight and acceleration.
The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.
Worked example
Inputs
- Vacuum at the part: pressure difference below atmosphere (bar): 0.6
- Sealed area entered as: Suction cups: diameter and number
- Cup effective diameter (mm): 50
- Number of cups sharing the load: 1
- Table or pad area under vacuum (m²): 1
- Share of that area covered and sealed by the workpiece (%): 100
- Load case: Horizontal cup, vertical lift (safety factor 2)
- Own safety factor, if chosen (times): 2
- Part mass (kg; 0 = skip the check): 0
- Highest acceleration incl. emergency stop (m/s²): 0
Result
- Pressure difference: 600 mbar = 60 kPa (−0.6 bar g)
- Effective sealed area: 19.6 cm² (1 × 19.63 cm²)
- Theoretical holding force: 118 N = 0.118 kN = 12 kgf
- Force per square metre of sealed area: 6.118 tonnes-force per m²
- Usable force after the safety factor: 58.9 N (safety factor 2)
Method. Force F = Δp × A, with Δp the pressure difference below atmosphere in Pa (1 bar = 100,000 Pa; 1 mbar = 100 Pa) and A the effective sealed area in m² (cup: π/4 × d² × number of cups; table: area × share covered). 0.2 bar acts as about 2 tonnes-force per m². Usable force = F ÷ safety factor: at least 2 is common for a straight vertical lift, 4 or more for a cup loaded in shear on a vertical face (friction carries the load). Required force = m × (g + a), with a the highest acceleration including emergency stops. Use the vacuum measured at the part, not the pump rating; the maximum possible Δp is the local atmospheric pressure (about 1.013 bar at sea level, about 0.95 bar at Hyderabad). Use the cup maker’s data for oily, dusty, textured or curved surfaces.
What to take from it. More vacuum gives little extra force and more cup wear: add cup area or cups first, and measure the vacuum at the part on the worst workpiece.
Technical content reviewed by D. Srinivas Reddy, Managing Partner and service engineer, Reddy Compressors. Last updated 17 October 2026. How our engineering content is reviewed.
Vacuum ejector compressed-air consumption
Average and peak compressed air used by vacuum ejectors from their rated motive-air flow, the on-time per cycle and blow-off pulses, for one gripper and for the whole line.
The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.
Worked example
Inputs
- Ejector motive-air consumption at its feed pressure (Nl/min): 60
- Cycle time (s): 10
- Ejector on-time per cycle (s): 4
- Blow-off flow (Nl/min; 0 = none): 100
- Blow-off pulse per cycle (s): 0.2
- Number of ejectors on the line: 20
- Number of ejectors that can be on at the same moment: 20
- Compressor specific power (kW per m³/min; 0 = skip energy): 0
- Running hours per year of the line: 6000
Result
- Ejector duty: 40 % of the cycle (6 cycles a minute)
- Average motive air per gripper: 24 Nl/min
- Blow-off per gripper: 2 Nl/min (0.333 litres per cycle)
- Total per gripper: 26 Nl/min = 0.918 cfm
- Average for all 20 ejectors: 520 Nl/min = 18.36 cfm = 8.67 L/s = 0.52 m³/min
- Peak motive flow with 20 on together: 1,200 Nl/min = 42.4 cfm (2.3 × the average)
Method. Average motive air = rated consumption (Nl/min at the stated feed pressure) × on-time ÷ cycle time. Blow-off per cycle = blow-off flow × pulse ÷ 60 (litres); per minute = that × 60 ÷ cycle time. Total per gripper = motive + blow-off; line total = × number of ejectors. Peak = rated consumption × the number that can be on at once: it sizes the supply pipework, regulator and local receiver. 1 cfm = 28.317 l/min; normal litres are at reference conditions. Energy: line flow (m³/min) × specific power × hours. Air-saving (vacuum-hold) circuits reduce the on-time on sealed parts; use the measured on-time. Leaks are not included.
What to take from it. Average demand sizes the compressor, peak concurrency sizes the pipework and local receiver; on porous, high-duty stations an electric vacuum pump or blower usually costs less to run.
Compressed air demand estimate
Add up the air your tools and processes use, apply duty and simultaneity, then add leakage and growth allowances to reach a capacity to enquire about.
The worked example below shows the method with fixed numbers. Turn on JavaScript to change the inputs.
Worked example
Inputs
- Unit for the air each tool uses: L/s
- Group A: number of tools: 6
- Group A: free air each (in the unit above): 4
- Group A: duty (%): 40
- Group B: number of tools: 2
- Group B: free air each (in the unit above): 15
- Group B: duty (%): 60
- Group C: number of tools: 1
- Group C: free air each (in the unit above): 40
- Group C: duty (%): 80
- Simultaneity of all tools (%): 65
- Leakage allowance (%): 15
- Growth allowance (%): 10
Result
- Connected demand: 94 L/s = 199 cfm
- Average demand (duty-weighted): 59.6 L/s = 126 cfm
- Simultaneous peak: 61.1 L/s = 129 cfm
- Design basis (higher of average and peak): 61.1 L/s = 129 cfm
- Design capacity with allowances: 77.29 L/s = 4.64 m³/min = 163.8 cfm
Method. Average demand = Σ(number × air per tool × duty). Peak = Σ(number × air per tool) × simultaneity. The peak can never be lower than the average, so the design basis is the higher of the two. Design capacity = design basis × (1 + leakage) × (1 + growth).
What to take from it. Leakage and growth allowances move the answer as much as any single tool. Where you can, replace catalogue figures with measured ones.