The inlet line to the pump should be at the same end of the reservoir as the return line, with the baffle between them forcing returning fluid to travel to the opposite end of the tank and back. The inlet must be below the fluid level and may include a strainer. If the inlet line is just a straight piece of pipe installed vertically, it is best to cut the line’s open end at 45° so it is impossible to butt it against the floor of the tank, which could block flow. Outside the tank, this line should lead as directly to the pump as possible, with no unnecessary bends or connections. Never use a pipe union in the inlet line; unions are almost impossible to seal against air leaks. Even a minute leak in the inlet line can cause pump cavitation and all of its problems.
The return line should be located in the same end of the tank as the inlet line and on the opposite side of the baffle (as shown in Figure 6-5). The return line must terminate below fluid level to reduce turbulence and aeration. The open end of the return line also should be cut at 45° to eliminate the chance of stopping flow if it gets pushed to the bottom. Another good practice is to point the opening toward the side wall to get the most heat-transfer surface contact possible.
When a hydraulic reservoir is part of the machine base or body, it may not be possible to incorporate some of the features discussed in this chapter. However, keep in mind the different functions mentioned to try to eliminate ongoing problems.
Non-pressurized and pressurized reservoirs
Reservoirs are seldom pressurized because that feature is not required under most circumstances. One reason for using a pressurized reservoir is to provide the positive inlet pressure required by some pumps -- usually in line piston types. Another reason is to force fluid into a cylinder through an undersized prefill valve. Both of these reasons may require pressures between 5 and 25 psi and could not use a conventional rectangular reservoir design.
Figure 6-6. Formula for estimating how much heat a tank of a given size can dissipate.
Another reason for pressurizing a tank is to keep out contaminates. If the reservoir always has a positive pressure in it there is no way for atmospheric air with its contaminants to enter. Pressure for this application is very low – 0.1 to 1.0 psi -- and may be all right even in a rectangular design tank.
A pressurized reservoir would be built like any pressure vessel, but with the baffling and other features shown in Figure 6-5. Note however that the reservoir pictured in Figure 6-5 is non-pressurized. The symbol for this type tank is shown at the left. The symbol also indicates how lines that terminate above and below fluid level are shown. If a drain line comes from an area that might have suction part of the time, it might not be best to terminate it below fluid level. If this type line terminated below fluid level, it could suck oil into the unit and possibly cause sluggish action. The drain line from the case drain of a pressure-compensated piston pump and an air bleed valve should terminate below fluid level at all times. This keeps air from being sucked in and causing problems.
Heat in hydraulic systems
All heat in a hydraulic circuit comes from wasted energy. Any horsepower put into the circuit that does not do useful work wastes energy.
Any circuit has inefficiencies that can be up to 15% of input power. This is bypass fluid in pumps, valves or other components and pressure drop through these components and the flow lines. These losses can be reduced, but never completely eliminated in a typical hydraulic circuit. Some ways to reduce inefficiency losses is to correctly size piping and valves, keep working pressure at or only slightly higher than required for all operations, and never allow fluid to relieve to tank. Flow controls also generate heat because they restrict flow. Reducing valves, counterbalance valves, and sequence valves also waste energy... especially if they are not set correctly. A pressure sequence or a counterbalance valve will do its job even when set too high, but will waste more energy at an unnecessarily high setting.
Each lost horsepower generates 2,545 BTU/hour of heat. To put this in perspective, 10 hp would heat a three-bedroom home when the outside air temperature is 30°F. Thus it is obvious what that much heat would do to the temperature of a 20-gallon tank of hydraulic fluid.
In a hydraulic circuit, you must calculate wasted horsepower to determine heat generation. In a highly efficient circuit this figure could be low enough to use the reservoir’s cooling capacity to keep maximum operating temperature below 130°F. If heat generation is slightly higher than a standard tank will handle, it may be best to oversize the tank rather than adding a heat exchanger. An oversize tank is less expensive than a heat exchanger; and avoids the cost of installing water lines.
It is easy to figure heat generation by figuring horsepower input and subtracting horsepower output. With the gauge at a fixed-volume pump’s outlet reading 150 psi and a gauge at the working cylinder reading 125 psi, there is a 25-psi pressure difference between energy in and work out. To figure horsepower loss, multiply (0.000583)(gpm)(psi). For this example, assume a 40-gpm pump. Then, lost horsepower = (0.000583)(40)(25) = .58. To determine actual heat loss, this figure must be divided by the percentage of the cycle that it occurs. If this figure is from the cylinder extending and the extend time was four seconds during a total cycle of 12 seconds, then figure 1/3 of the .58 hp or .19 hp as waste. Do this operation for each actuator in both directions of travel to determine total wasted horsepower. After all actuator losses are calculated, add them together to determine total wasted horsepower. (Note that when this horsepower total is less than the answer from the formula in Figure 6-6, no heat exchanger is required.)
The example just cited would be a straightforward circuit without any flow controls or other added restrictions. With a flow control in the circuit, pressure drop would be much higher and energy waste would increase drastically. Most circuits using a pressure-compensated pump would have flow controls so the pump would be at compensator setting while the actuator would be at whatever pressure is required. A circuit without flow controls or other restrictive plumbing usually has low energy losses. This type system may get by without a heat exchanger when ambient temperature stays below 110°F.
Tank cooling capacity
Use the formula in Figure 6-6 to estimate how much heat a given tank can dissipate. This formula assumes the tank is open on all sides with free air movement around it. (Remember, pipes, cylinders, and valves also have surface areas that can dissipate heat but those areas are usually not included when calculating cooling capacity.) For a 100-gal tank and a 30°F. temperature difference, this amounts to about 1.4 hp. A single cylinder circuit wasting only 0.7 hp would not require a heat exchanger on a 100°F. day while maintaining a maximum fluid temperature of 130°F.
Heating and cooling devices
Tank heaters: Most industrial hydraulic units operate in a warm indoor atmosphere so low temperature is not a problem. For circuits that see temperatures below 65° to 70°F., some sort of fluid heater is recommended. The most common tank heater is an electric-powered immersion type. These units consist of resistive wire in a steel housing with a mounting option. They often have an integral thermostatic control. The heating element on most units contacts the fluid directly, similar to a hot water heating system. Some immersion heaters in a sealed housing heat the air in the enclosure. The air then transfers heat to the fluid. This heater-in-tube arrangement allows the heating unit to be changed without draining the reservoir.
Figure 6-7. Typical electric-powered in-tank heater.
Figure 6-7 shows an electric immersion heater and its ISO symbol for schematic drawings. (This symbol may be replaced by some sort of pictorial rendition of the heating unit on many schematics.)
Electric heating units should not have concentrated heat like those used to heat water. Oil viscosity at low temperature is thick enough to retard movement. This could allow fluid next to the heating rods to overheat and possibly breakdown. The usual recommendation is for the heating rods not to have over 8 to 10 watts per square inch density. This limit may require multiple heating rods to meet the heat requirement of some systems.
Another way to electrically heat a tank is with a mat that has heating elements similar to those in an electric blanket. This mat is attached to the outside of the bottom of the reservoir and adds heat during low temperature conditions. This type heater requires no ports in the tank for insertion. It also evenly heats the fluid even during times of low or no fluid circulation.
Figure 6-8. Formula for estimating heater capacity to increase fluid to a minimum temperature.
Heat can be introduced through a heat exchanger by using hot water or steam in place of the cooling water. The exchanger becomes a temperature controller when it also uses cooling water to take away heat at other times. Figure 6-10 depicts the symbol for this type heat exchanger. (Chapter 7 shows an alternate way to add heat while filtering system fluid.) Temperature controllers are not a common option in most climates because the majority of industrial applications operate in controlled environments.
The formula for estimating how many kilowatts are needed to heat a certain size tank from an expected minimum ambient to a nominal working temperature of 50° to 70°F. is shown in Figure 6-8. Use it to size a heater when the tank is exposed to low temperatures.