The 180 ft Well Pump: Code-Minimum Conductor, Real-World Failure
A 5 hp, 230 V single-phase submersible well pump serves a rural house. The nameplate reads 28 A, matching the tabulated full-load current for a 5 hp, 230 V single-phase motor. The pump is 180 ft (one-way) from the house panel. The existing run is 10 AWG Cu THHN in buried PVC on a 70 A inverse-time breaker, with 75 C terminations at both ends, and the control box uses ambient-compensated overloads. On hot afternoons the pump hums, occasionally trips the 70 A breaker, and the motor shell gets hot enough to open its internal protector. A voltmeter reads 215 V at the panel and 202 V at the pump controller while the pump runs.
1. Calculate the running voltage drop on the existing 10 AWG run and interpret the field readings.
Read the worked answer
Single-phase voltage drop:
VD = 2 x K x I x D / CM, with K = 12.9 for copper, CM(10 AWG) = 10,380
VD = 2 x 12.9 x 28 x 180 / 10,380 = 130,032 / 10,380 = 12.5 V
12.5 V / 230 V = 5.4% on the branch circuit alone.
The field readings corroborate it: 215 V at the panel minus 202 V at the controller is 13 V, consistent with the calculation at running current. The motor is running about 6% below nominal voltage, which means it draws slightly more current for the same load - a compounding effect, and more heating in the windings.
Compare with the guidance in the 210.19(A) informational note: about 3% maximum on a branch circuit and 5% total from the service point. This circuit alone blows past both.
Starting is worse. Locked-rotor current for this motor is roughly 5 to 6 times FLC, about 150 to 170 A. At 160 A the drop is 2 x 12.9 x 160 x 180 / 10,380 = 71.6 V, about 31%. That is why the pump hums: the contactor and the motor see a deep sag on every start, and the internal protector trips on the resulting heat.
2. Is the 10 AWG conductor a Code violation? Evaluate the ampacity, the breaker, and the enforceable status of voltage drop.
Read the worked answer
Ampacity - it passes, barely.
- 430.22 requires branch-circuit conductors at not less than 125% of the full-load current: 1.25 x 28 A = 35 A.
- 10 AWG Cu THHN at 75 C is 35 A. With 75 C terminations (110.14(C)) and no adjustment factors, it meets the minimum exactly - zero margin.
- The 30 A small-conductor limit in 240.4(D) does not apply here; 240.4(G) exempts motor branch-circuit conductors, which is why a 10 AWG can sit behind a 70 A breaker at all.
Breaker - correctly sized.
430.52(C)(1) permits an inverse-time breaker at up to 250% of FLC: 2.50 x 28 A = 70 A, which is a standard rating in 240.6. The 70 A breaker is not the problem and must not be reduced to cure the tripping; a smaller breaker will simply nuisance-trip harder on the inrush.
Voltage drop - informational, but here it is the root cause.
The 3% and 5% figures are informational notes, not enforceable requirements in the general case. So the installation can be Code-minimum and still fail in service. That is the judgement call the exam is testing: Code compliance and functional adequacy are two separate questions. When the evidence is a hot motor, a sagging controller, and 13 V of measured drop, the answer is not 'it passes 430.22.' Note also that some equipment listings and the manufacturer's installation instructions establish a minimum supply voltage; if the pump nameplate or instructions specify one, that becomes enforceable through 110.3(b).
3. Size the replacement conductor for no more than 3% running drop, and state any associated requirements you must check.
Read the worked answer
Target: 3% of 230 V = 6.9 V.
CM required = 2 x K x I x D / VD_allowed = 2 x 12.9 x 28 x 180 / 6.9 = 130,032 / 6.9 = 18,845 circular mils
| Conductor | CM | Predicted VD | % of 230 V |
|---|---|---|---|
| 8 AWG | 16,510 | 7.88 V | 3.4% - still over |
| 6 AWG | 26,240 | 4.96 V | 2.2% - meets |
| 4 AWG | 41,740 | 3.12 V | 1.4% |
Use 6 AWG Cu. It also solves the starting sag: at 160 A locked rotor, VD falls to about 28 V (12%), which the controller and motor can ride through. Ampacity is not the constraint (6 AWG at 75 C = 65 A), so the run is being upsized for voltage drop only.
Check before you pull it:
- Equipment grounding conductor (250.122(B)). When ungrounded conductors are increased in size for voltage drop, the EGC must be increased proportionally by circular-mil area. Going from 10 AWG to 6 AWG is a ratio of 26,240 / 10,380 = 2.53, so the EGC sized from Table 250.122 for the branch-circuit rating must be multiplied by 2.53. This is widely overlooked, and it is a legitimate plan-review comment.
- Conduit fill and pull points. 6 AWG in the existing PVC may need a larger raceway and larger pull boxes; verify fill and bend radius rather than assuming the existing pipe works.
- Termination temperature and lugs. Confirm the panel and controller lugs accept 6 AWG and are rated 75 C.
- Overload settings. Leave the overloads per 430.32; only the conductor and the voltage drop are being changed.