Diagnosing Feeder Overload from Melted Insulation and Duty Cycles

Someone asked me to inspect the electrical system in a service building. The reported symptom was that a technician had found insulation melting at scattered points along the main cable, but the laundry machines still ran every day. No breakers tripped, and nobody complained about voltage drops. I went to collect the full set of data myself, from each machine's nameplate to timing its actual operating cycle, because in cases like this, the figures on paper do not always match behavior on site. In this article, I walk through the whole diagnosis, from the first number that caught my attention to the explanation for why such a heavily overloaded system could "seem to work."

On-site symptoms and numbers that did not add up

What I found was a single 95 mm² copper cable running from the building's main panel, supplying six zones in sequence: laundry, central sterile supply, pharmacy, inpatient ward, physiotherapy, and finally administration. The total cable run was nearly 290 meters, with a splice at every pole. The melted insulation was scattered along the early part of the run.

The damage pattern matters a great deal to the diagnosis. The melted insulation was scattered at points along the run, rather than burned at a single location. This points to accumulated heat from repeatedly carrying excessive current over a long period, rather than damage from a single short circuit. A short circuit would cause severe damage at the fault location and trip the breaker. Accumulated heat, by contrast, gradually eats away at the insulation along the entire cable without warning.

Collect every nameplate first; do not guess from machine size

For this kind of diagnosis, I always start with the nameplates, not a meter, because a meter only tells me "how much it is drawing now," while the nameplate tells me "how much it could draw at maximum." This room had six machines. For any machine whose plate did not state the current, I calculated it from the listed power using the standard three-phase formula.

I = P / (sqrt(3) x V x pf)          ระบบ 3 เฟส 380 V

SPE-50 ฮีตเตอร์ 24 kW  ->  24000 / (1.732 x 380) = 36.5 A
IM50   ฮีตเตอร์ 30 kW  ->  30000 / (1.732 x 380) = 45.6 A

หมายเหตุ: โหลดฮีตเตอร์เป็นความต้านทานล้วน ตัวประกอบกำลังราว 1.0
ถ้าเป็นมอเตอร์ต้องหารด้วย pf จริงของเครื่อง ไม่ใช่ 1.0

371.5 A for the laundry and 482 A for the entire run

Adding the 371.5 A laundry load to the other loads connected to the same cable—the central sterile supply unit's sterilizers and the air conditioning in the ward, physiotherapy department, pharmacy, and offices—brought the theoretical total for this cable to 482 A. Meanwhile, that 95 mm² copper cable could carry 180 A under the actual installation method in use.

At 2.7 times the rating, anyone hearing that number would immediately ask why it had not burned up yet. This is the interesting part of diagnostic work, because the answer is not in the figures on paper. It is in the connection locations and the timing of the process. I narrowed it down to three points.

First, the heavy load was near the start of the cable

This is the point people most often miss. They see a 289-meter cable and assume the laundry machines are at the far end. In fact, the laundry branches off the main cable roughly 40 meters from the start. That means the machines most sensitive to voltage drop in this system happened to be in the safest position along the cable. The accumulated voltage drop at that point was only about 1.4%, while the larger drop of 2.9% appeared at the far end, in the office zone.

# ประมาณแรงดันตกด้วยความต้านทานล้วน เพื่อดูอันดับของตัวเลข
RHO_CU = 0.0175        # โอห์ม x ตร.มม. / เมตร  ค่ามาตรฐานทองแดงที่ 20 C
V_LL   = 380.0

def vd_percent(i_amp, length_m, area_mm2):
    r_per_m = RHO_CU / area_mm2          # โอห์มต่อเมตร ต่อเส้น
    v_drop  = 1.732 * i_amp * length_m * r_per_m
    return round(v_drop, 1), round(v_drop / V_LL * 100, 2)

print(vd_percent(482, 40,  95))     # จุดต่อห้องซักรีด อยู่ต้นสาย
print(vd_percent(120, 289, 95))     # ปลายสาย เหลือเฉพาะโหลดสำนักงาน

The values from this script are only rough estimates. In the actual report, I used R and X values at operating temperature from the manufacturer's tables, then accounted for the current decreasing along successive cable sections. The results were 1.4% at the laundry connection and 2.9% at the far end. The script above is simply there to show that the variables governing the result are current times distance divided by cross-sectional area.

Second, the process kept the heaters from overlapping

This is the core of the case. Almost all the power consumed by the laundry machines goes to heaters, not motors. The washers use their heaters only during the water-heating stage, roughly the first 15 minutes of the cycle, while the dryers use theirs for almost the entire drying cycle. The laundry must finish washing and spin-drying before it can move into the dryers. This sequence naturally keeps the two groups of heaters from operating at the same time.

Third, the voltage-drop symptoms appeared at the far end

The only loads at the far end were air conditioners and office computers. The air conditioners could tolerate a brief voltage drop; at worst, the compressors would cut out and restart on their own. Every computer already had a UPS. On top of that, the system's full phase voltage ran close to 250 V, so even after a 2.9% drop, it remained comfortably within what the equipment could handle. The actual symptoms were therefore masked, and nobody ever reported them.

Put simply, this system survived because of three stacked coincidences. The heavy load happened to be near the start of the cable, the process happened to keep the heaters from overlapping, and the loads exposed to the full voltage drop happened to tolerate it. None of these was the result of design.

Model the current minute by minute for a full day, then count the time above the rating

I did not want to argue based on impressions, so I wrote a minute-by-minute model of the entire working day. I entered the wash cycles I had actually timed, added the dryer thermostats' behavior of cutting out for roughly two minutes every ten minutes, and then added the base loads of the other zones by time of day. After that, I counted the total number of minutes the graph rose above the 180 A line.

# แบบจำลองกระแสรวมบนสายเดิม รายนาที ตลอดวันทำงาน
WASH_HEAT = 170.4     # ฮีตเตอร์เครื่องซัก 3 ตัวรวม
DRY_HEAT  = 201.1     # ฮีตเตอร์เครื่องอบ 3 ตัวรวม
LIMIT     = 180       # พิกัดสายเดิม 95 ตร.มม.

def base_load(minute):                 # แอร์ จ่ายกลาง สำนักงาน ตามช่วงเวลา
    hour = minute // 60
    return 105 if 9 <= hour <= 15 else 60

def laundry(minute, start=8*60):       # รอบเดียวต่อวัน เริ่ม 08:00
    t = minute - start
    if 0 <= t < 15:                    # ต้มน้ำ
        return WASH_HEAT
    if 15 <= t < 40:                   # ซัก ล้าง ปั่นหมาด เหลือแต่มอเตอร์
        return 24.7
    if 40 <= t < 80:                   # อบ เทอร์โมสตัทตัด 2 นาทีทุก 10 นาที
        return 0 if (t - 40) % 10 >= 8 else DRY_HEAT
    return 0

series = [base_load(m) + laundry(m) for m in range(24 * 60)]
over   = [m for m, i in enumerate(series) if i > LIMIT]
print('peak      ', max(series))
print('average   ', round(sum(series) / len(series)))
print('over limit', len(over), 'minutes')

ผลที่บันทึกไว้จากแบบจำลองของไซต์นี้

  พีคสูงสุด                301 A   (ช่วงอบ ฮีตเตอร์ 3 เครื่องพร้อมกัน)
  กระแสเฉลี่ยทั้งวัน        115 A
  เวลารวมที่เกิน 180 A      1.0 ชม./วัน  (11% ของวัน)
  ช่วงเกินต่อเนื่องยาวสุด    15 นาที
  ค่าทางทฤษฎีที่ไม่เคยเกิด   482 A

These figures explained everything. The average current for the whole day was just 115 A against a rating of 180 A. The cable was not hot all the time, but it was subjected to overload pulses totaling one hour a day, each lasting no more than 15 minutes. That was too short to cause an immediately visible failure, but long enough for the insulation to heat up and deteriorate a little at a time, day after day.

755 units of energy per day, with half going to the laundry

Another way to help the owner understand the situation was to look at the energy share. I estimated it from the power of each zone multiplied by its actual operating hours. For air-conditioning loads, I used a duty factor of 0.7; for the laundry, I used the operating cycles I had timed.

A single load consumed half the electricity in the building and still shared one conductor with five other zones. At that point, I did not think any further calculation was needed to say it should have its own feeder and meter, both for safety and for allocating costs by department.

1. Collect the nameplate data for every machine, and photograph each plate. For machines without a current value, calculate it from kW and note that it is a calculated value, not a nameplate value.
2. Write out the actual process sequence as a timeline. Do not accept "everything runs at once" until you have stood there and watched a full cycle yourself.
3. Measure the actual connection locations along the cable, including the distance from the main panel to each zone. These figures change the entire explanation of the symptoms.
4. Build a minute-by-minute model and count the "time above the rating," rather than looking only at the peak, because the damage comes from accumulated time.
5. Photograph the damage clearly enough to show whether it is scattered at multiple points or concentrated at one location. This can distinguish accumulated heating from a short circuit.

Diagnostic conclusion

The main cause in this case was a single conductor carrying a load above its rating, and the evidence was the insulation that had already melted. Phase imbalance and loose connections made it worse, but they were not the root cause. These must not be treated as interchangeable explanations. Retightening the connections along the entire run did restore operation, but as long as the overload remained, those connections would loosen again. It bought time; it did not solve the problem.

The solution I proposed was to move the laundry load onto a new feeder of its own. The existing cable would then carry around 98 A, or 54% of its rating, returning it to a safe operating condition with room for expansion. I cover how to size the cable and breaker for the new feeder separately in the article on ampacity and installation methods, because that is a substantial topic on its own. The upstream problem that prevented phase balancing in the first place is covered in the article on a neutral shared across transformers.

If you encounter a similar case, I recommend starting with two questions. First, where along the cable is the heaviest load connected? Second, which loads does the production process keep from operating at the same time? Once those two questions are answered, the apparently conflicting figures will fall into place. If I have missed anything in the explanation, or if a case comes up with a different load pattern, I will come back and add to this later.