Why is the volume flow importantTo properly select an adsorber for use on a hydraulic tank, the air flow rate is an important value. The air flowing through the adsorber must be dehumidified by the desiccant. Here, the quantity and speed have an effect on the efficiency.

The figure shows the influence of air velocity on the adsorber index (efficiency). Very fast air does not have sufficient residence time to be dried effectively. In addition, a high amount of water is introduced into the adsorber, shortening the product life cycle. An adsorber filled with 1kg of silica gel should be used up to a volume flow of approx. 100 l/min to achieve a good adsorber index of 70%.

Figure 1: Adsorber index at 20 °C and an adsorber with 1 kg desiccant, as a function of volume flow rate.

## What are the possibilities to determine the volume flow?

There are three ways to determine the volume flow. On the one hand, a very strong simplification can be applied. The tank volume is used as a selection criterion. This number implies that, depending on the grouping of the tank volumes, a maximum air exchange is possible. This approach is sufficiently accurate in most cases. In practice, the following grouping has emerged.

• 0-50 ltr. Tank Volume = to 10 l/min / max. 30 l/min
• 50-100 ltr. Tank Volume = to 20 l/min / max. 100 l/min
• 100-400 ltr. Tank Volume = to 40 l/min / max. 260 l/min
• 400-800 ltr. Tank Volume = to 80 l/min / max. 490 l/min
• 800-1800 ltr. Tank Volume = to 110 l/min / max. 610 l/min
• 1800-3600 ltr. Tank Volume = to 160 l/min / max. 930 l/min
• 3600-5400 ltr. Tank Volume = to 210 l/min / max. 1250 l/min
• 5400-7200 ltr. Tank Volume = to 530 l/min / max. 2000 l/min

Furthermore, the pump capacity can be used as an indicator to determine the volume flow. This is based on the assumption that the pump transports the amount of oil from the tank according to its performance and that this volume must be replaced by ambient air. This approach ignores the fact that oil can be pumped in a circuit through a system or that counter-rotating cylinders are operated. In both cases, the air exchange volume would be approximately zero. The best and most accurate approach is to calculate or observe the tank. At the beginning, the base area of the tank is determined (length A x length B). Then the oil level movement (length C) is measured at a fixed time interval. As time unit 1 minute is recommended.

### Example 1

Hydraulic power unit operates cylinders at 20 strokes per hour

Tank size = A 0,5m x B 1,0m x C 0,5m = 0,25m³ = 250dm³ = 250Liter
Floor space = A 0,5m x B 1,0m = 0,5m²
Oil movement measured = C 0,1m / 1min.
Air pendulum volume = 0,5m² x 0,1m/min = 0,05m³/min = 50dm³/min= 50l/min

• based on tank volume = size … 3M
• based on air volume = size … 3L

### Example 2

Hydraulic unit operates test stand, oil circulates in circuit

Tank size = A 0,5m x B 1,0m x C 0,5m = 0,25m³ = 250dm³ = 250Liter
Floor space = A 0,5m x B 1,0m = 0,5m²
Oil movement measured = C 0,01m / 1min.
Air pendulum volume = 0,5m² x 0,01m/min = 0,01m³/min = 5dm³/min= 5l/min