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Solenoid valves are available in a variety of DC and AC voltages. The common DC voltages are 6, 12, 24, 120, and 240 volts. The common AC voltages at 60 Hz are 24, 120, 240, and 480 volts. The common AC voltages for 50 Hz are 110 and 220 volts.
The power consumption for each valve will be slightly different due to the size of the valve stem that's being moved open and closed and the size of the coil needed to build the magnetic field. Ill. 1 shows a data sheet for a typical solenoid valve. This data sheet shows that the wattage for a normally closed solenoid with 8-inch openings will range from 6.1-17.1 W for a coil that's powered with AC voltage, and from 10.6-11.6 W for a coil that's powered with DC voltage. So if the coil is powered with 120 volts and it consumes 17.1 W it would draw 0.14 A for holding current. Multiplying this value by 3, one obtains inrush current = 0.42 A.
Some data sheets provide the rating for solenoid valves as the amount of volt-amperes (VA) that the coil consumes. The apparent power is determined by measuring the current and the voltage and multiplying them to provide the VA value. True power, which is the actual wattage the circuit consumes, is calculated by the formula P = I^2R. Wattage is determined by measurement with a wattmeter. The true power (W) and the apparent power (VA) will only be equal values if the circuit load is purely resistive. If the load is magnetic (inductive) or if it has capacitance, the apparent power value will always be larger. The ratio of the true power divided by the apparent power is called the power factor. The data sheet shown in Ill. 1 also shows typical ratings for the solenoid valve. Notice that these ratings include the pipe size, flow factor, pressure differential, and maximum temperature. The wattage rating for each coil is also provided in the column at the far right.
Above: Ill. 1: Technical data sheet for a typical solenoid valve. CLICK IMAGE TO ENLARGE.
When one is providing fuse protection for a coil in a circuit, one can size the fuse at the amount of holding current that one measures with an ammeter. The fuse will need to be a dual-element time-delay fuse so that the inrush current can occur without blowing the fuse, while still providing protection at the holding current level.
Exercise
Select a solenoid valve from the data sheets in Ill. 1, or via this calculator, that will be used to replace the normally closed solenoid valve that controls the coolant flow to the milling machine that was delivered to a factory. The coolant for cutting is a mixture of light oils. This valve needs to have 3/8-inch opening and a flow-factor rating of 0.72. The minimum pressure for this system is 0 psi and the maximum pressure is 100 psi. The control voltage for this machine is 110 V AC.
Solution
Since this valve is used to control the flow of light oil that's used as a coolant and it needs an opening of 8 in., one should start his/her search in the last four rows of normally closed part of the table. Since the control voltage for the system is 110 V AC, one can ignore the DC-powered coils. The pressure for the system ranges from 0-100 psi and the fluid is light oil, so one will have to select a 8263G206 Brass Body Valve listed (for normally closed solenoid).
