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How much power does a passenger train take?

0ceanborn 15123 12
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Treść została przetłumaczona polish » english Zobacz oryginalną wersję tematu
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  • #1 17657031
    0ceanborn
    Level 25  
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    Hi :)

    I've been thinking about a technical solution recently, but I don't know enough. I mean what power is consumed by an average passenger train, e.g. 8 cars, not overcrowded and assuming that it has already collapsed to a speed of 110 km/h and maintains only this speed. I tried to calculate it more or less and I came up with strange things, so I am asking you for help. I mean the order of magnitude - how much is approximately kilowatts.
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  • #3 17657075
    nuszek
    Level 30  
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    You can read more here, although it's quite old.
    link
    I am curious myself how much electricity the Pendolino consumes, e.g. at 120 km/h /3000 V network/,
    there is probably a measurement in the cabin - maybe there are drivers here.
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  • #5 17658489
    buczkog
    Level 16  
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    Kol. Rezystor240, apart from the voltage of about 2800-3200V, nothing results. The previous posts show that the voltage is about 3000V, the current is about 600A (we do not count the heating of the cars because it adds a lot to the total power). It still gives, as the previous speakers wrote, ok. 8MW of energy.
  • #6 17658509
    karolark
    Level 42  
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    buczkog wrote:
    It still gives, as the previous speakers wrote, ok. 8MW of energy.


    On the fingers, other numbers come out
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  • #8 17658717
    vodiczka
    Level 43  
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    http://www.transportszynowy.pl/lokedane.php#ep07
    EP 07 electric locomotive - maximum continuous power = 2000kW; Vmax = 125 km/h
    EP 09 electric locomotive - maximum continuous power = 2900kW; Vmax = 160km/h

    It can be estimated that with a composition of 8 cars and a speed of 110 km / h, the power will be in line with what was given by atom 1477 in post # 2 and even lower. If I read the data from the links correctly, the current consumption was about 400A at a speed of 95km/h.
  • #11 17683005
    Anonymous
    Anonymous  
  • #12 17683025
    vodiczka
    Level 43  
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    stench wrote:
    the current consumption will of course depend on the power of the locomotive.
    It is correct to say "maximum current consumption depends on the power of the locomotive."
    Actual consumption depends on train weight and driving conditions (speed, gradient, etc.)

    The author's question is:
    0ceanborn wrote:
    ...what power is consumed by an average passenger train, e.g. 8 cars, not overcrowded and assuming that it has already limped to a speed of 110 km/h and only maintains this speed.
    it does not ask for the maximum power of the locomotive.
  • #13 17685324
    Anonymous
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Topic summary

✨ The discussion centers on the power consumption of an average passenger train, specifically one with eight cars traveling at a speed of 110 km/h. Responses indicate that the power consumption can vary, with estimates ranging from approximately 1.26 MW to 8 MW depending on factors such as locomotive power, efficiency, and additional systems like heating. Key figures mentioned include a current draw of about 600 A at 3000 V, resulting in around 1800 kW. The conversation also references specific electric locomotives, such as the EP 07 and EP 09, with maximum continuous powers of 2000 kW and 2900 kW, respectively. The importance of considering train weight, driving conditions, and efficiency factors in calculating actual power consumption is emphasized.
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FAQ

TL;DR: Typical 8‑car trains at 110 km/h draw about 2.23 MW—“current of approx 740 A” at 3 kV. This FAQ helps rail fans, engineers, and students estimate on‑wire power and current using real forum data. [Elektroda, Anonymous, post #17685324]

Why it matters: It helps you size traction needs, sanity‑check cab readings, and avoid order‑of‑magnitude errors.

Quick Facts

How much power does an 8‑car passenger train use at 110 km/h?

About 2.23 MW from a 3 kV DC catenary, including heating. Traction demand is about 1.26 MW at the wheel rim. Gear efficiency brings motor input near 1.29 MW. The calculated rolling resistance is 41.39 kN at that speed. Dividing 2.23 MW by 3 kV gives about 740 A. These figures model a typical modern locomotive with auxiliaries. They match forum calculations based on Polish railway formulas. [Elektroda, Anonymous, post #17685324]

How do I estimate current draw on a 3 kV DC line?

  1. Estimate resistance forces and multiply by speed to get traction power.
  2. Adjust for drivetrain efficiency (e.g., 98%) to get motor input.
  3. Add hotel loads (e.g., up to 800 kW), then divide by 3 kV for amperes. [Elektroda, Anonymous, post #17685324]

Why do some estimates say 1.8 MW at cruise?

Because many look at a typical ammeter reading near 600 A. On a 3 kV DC line, that equals 1.8 MW. That covers traction to hold speed on level track. It excludes coach heating and other hotel loads. It also excludes acceleration peaks. [Elektroda, Anonymous, post #17657059]

Does consumption depend on train weight, gradient, and speed?

Yes. The locomotive’s installed power only sets the maximum. Actual draw depends on mass, speed, gradients, and driving style. As one expert put it, “Actual consumption depends on train weight and driving conditions.” Use weight and route data for better estimates. [Elektroda, vodiczka, post #17683025]

What are typical continuous power ratings for Polish passenger locomotives?

EP07 has 2000 kW continuous power with 125 km/h top speed. EP09 has 2900 kW and reaches 160 km/h. These ratings define sustained capability, not the steady power at cruise. They indicate how much headroom exists for acceleration or grades at speed. [Elektroda, vodiczka, post #17658717]

How much do heating and auxiliaries add?

Coach heating can add up to about 800 kW. The traction chain includes other losses. The calculation used a 98% gear efficiency. Overall converter‑to‑pantograph efficiency sits around 0.88–0.9. Add these elements to the traction power to get on‑wire draw. [Elektroda, Anonymous, post #17685324]

What is the maximum permitted current draw in Poland?

EN 50388‑1 sets the limit referenced by operators. “In Poland it is 2500 A.” When near this cap, trains may limit traction to comply. Use this as a design ceiling for worst‑case calculations. [Elektroda, Anonymous, post #17683005]

Is 8 MW a realistic figure for an 8‑car train at 110 km/h?

No. A worked example gives about 2.23 MW at 110 km/h. That includes traction losses and passenger heating. The corresponding line current is about 740 A at 3 kV. The 8 MW claim results from a miscalculation, not measured operation. [Elektroda, Anonymous, post #17685324]

What current was observed around 95 km/h in practice?

A reported data read shows about 400 A at 95 km/h. That suggests lower demand at slightly lower speeds on level track. Use this as an order‑of‑magnitude check against your own readings. [Elektroda, vodiczka, post #17658717]

Where can I find driver‑level discussions or measurements?

See the long “encyclopedia” thread maintained by railway mechanics. It aggregates operational insights and instrument screenshots. Use it to cross‑check ammeter and voltmeter behavior under load. It is a practical complement to spec sheets and textbooks. [Elektroda, Parowy, post #17659739]

Do modern passenger locomotives really have about 6 MW continuous?

Yes. Modern passenger locomotives are around 6 MW continuous. That includes traction motors and auxiliary systems. Use the converter efficiency when turning this into line current. Then add heating to estimate total on‑wire demand. [Elektroda, Anonymous, post #17683005]

What DC supply voltage should I assume for “3 kV” lines?

Assume an operating range around 2.8–3.2 kV under normal load. On‑wire voltage varies with distance to substations and train draw. Use measured voltage where possible for precise current estimates. [Elektroda, buczkog, post #17658489]

How should I estimate Pendolino consumption at 120 km/h on 3 kV?

The thread provides no direct Pendolino reading. It suggests using cab instruments and the same method. Look for videos or driver posts with ammeter shots. Then apply the calculation approach for a consistent estimate. [Elektroda, nuszek, post #17657075]

What happens during winter or acceleration peaks?

Traction demand and auxiliary loads rise, raising line current. Operators account for converter efficiency and heating when computing limits. The current must remain within the 2500 A cap in Poland. Treat 2500 A as the worst‑case ceiling in planning. [Elektroda, Anonymous, post #17683005]
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