24 V Solenoid Circuit: Voltage Drop and Fuse Check
A cadet is checking a 24 V DC control circuit before commissioning. A solenoid valve rated 24 V, 43 W is installed 42 m of cable run away from the control panel and is wired in 1.0 mm^2 two-core copper cable, where 42 m is the one-way distance. Take copper resistivity as 0.0175 ohm mm^2/m. The manufacturer states that the coil must see at least 21 V to pull in reliably, and that inrush current is six times the holding current and lasts about 0.2 s. The panel is fitted with a 4 A slow-blow fuse, and the cable is clipped to a tray with three other cables, giving an installed current rating of 11 A for the 1.0 mm^2 conductor. On no load the cadet measures 24.1 V at the panel terminals and reports the circuit as 'fine, because the supply is 24 V'. The supervisor wants the actual coil voltage at the far end, the fuse choice checked, and a recommendation.
1. Calculate the steady-state current, the loop resistance and the voltage actually reaching the coil through the 1.0 mm^2 cable. Is the circuit acceptable?
Read the worked answer
Current: I = P / V = 43 / 24 = 1.79 A.
Loop resistance (current goes out and back, so use 2 x 42 = 84 m): R = rho L / A = 0.0175 x 84 / 1.0 = 1.47 ohm.
Voltage drop: V_drop = I R = 1.79 x 1.47 = 2.63 V.
At the coil: 24.0 - 2.63 = 21.37 V, against a 21 V minimum. The margin is 0.37 V, about 1.7%.
Judgement: this is not 'fine'. It only appears to pass because the calculation used the nominal 24 V. Supply tolerance, additional loads on the same supply, or a slightly higher coil current will take the coil below the 21 V pull-in threshold, and a solenoid held at marginal voltage chatters and overheats. The no-load reading of 24.1 V tells you nothing about the loaded condition - that is the reasoning error in the cadet's report. Note also that the minimum is usually a cold-coil figure.
2. Now check the inrush condition through the same cable, and say which of the two checks actually governs the cable size.
Read the worked answer
Inrush current: 6 x 1.79 = 10.7 A for about 0.2 s.
Inrush voltage drop along the same 1.47 ohm loop: 10.7 x 1.47 = 15.8 V.
Voltage at the coil during pull-in: 24.0 - 15.8 = about 8.2 V.
What that means: the steady-state check passed by a hair, but during the only moment that matters - the 0.2 s in which the valve must shift - the coil sees roughly a third of its rated voltage. The solenoid will produce very little pull-in force; expected symptoms are a valve that buzzes without shifting, a chattering armature, or intermittent operation that looks like a faulty coil.
Which check governs: the inrush check, because the cable resistance is fixed and the current is six times higher. A design that only passes the holding-current calculation is not designed at all.
Fix: go to 1.5 mm^2 (R = 0.0175 x 84 / 1.5 = 0.98 ohm): steady drop 1.76 V gives 22.2 V, and inrush drop 10.5 V gives about 13.5 V. That is better but still marginal, so the proper answer is 2.5 mm^2 (R = 0.59 ohm, inrush drop about 6.3 V) or a local interposing relay with a short coil loop.
3. Assess the 4 A slow-blow fuse and the 11 A cable rating against this duty, including the inrush, and state a conclusion for the commissioning report.
Read the worked answer
Steady state: 1.79 A on a 4 A fuse is 45% of rating - the fuse will not run hot or nuisance-blow.
Inrush: 10.7 A is 2.7 times the fuse rating for only 0.2 s. Slow-blow (time-lag) elements are designed for exactly this motor/solenoid starting duty and will hold. A fast-blow 4 A fuse in the same holder would very likely blow, and should not be substituted.
Cable protection: the installed rating of the 1.0 mm^2 conductor is 11 A, which is above the 4 A fuse, so the cable is protected against overload and the fuse is correctly sized as the protective device. Note that 10.7 A inrush is close to the 11 A conductor rating, but it is a 0.2 s transient, not a sustained load.
Conclusion for the report: the fuse and cable protection are correct; the cable size is not, because of voltage drop. Record: holding-current drop 2.63 V (coil 21.4 V, below a sensible margin), inrush drop 15.8 V (coil 8.2 V) - reject the circuit as installed, upsize the conductor or fit a local relay, keep the 4 A slow-blow fuse, and re-measure coil voltage at the far end under load, not at the panel on no load.