Joukowsky surge pressure

Fragmentos de código

curl "https://api.oanor.com/waterhammer-api/v1/surge" \
  -H "x-oanor-key: oanor_test_..."

await fetch("https://api.oanor.com/waterhammer-api/v1/surge", {
  headers: { "x-oanor-key": "oanor_test_..." }
});

$ch = curl_init("https://api.oanor.com/waterhammer-api/v1/surge");
curl_setopt($ch, CURLOPT_HTTPHEADER, ["x-oanor-key: oanor_test_..."]);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$out = curl_exec($ch);

import requests
requests.get(
    "https://api.oanor.com/waterhammer-api/v1/surge",
    headers={"x-oanor-key": "oanor_test_..."}
)

Respuesta de ejemplo

Una respuesta real de este endpoint, capturada en la última comprobación de estado.

{
    "data": {
        "note": "The Joukowsky equation: a sudden stop of the flow spikes the pressure by Δp = ρ · a · Δv, where a is the pressure-wave speed — stopping 2 m/s of water (a≈1200 m/s) adds about 24 bar, far above the line pressure, which is what bangs the pipes and can split fittings. This is the worst case for an instantaneous closure; a slower valve (longer than the critical time) caps the surge well below it.",
        "inputs": {
            "wave_speed_m_s": 1200,
            "fluid_density_kg_m3": 998,
            "velocity_change_m_s": 2
        },
        "surge_head_m": 244.73,
        "surge_pressure_pa": 2395200,
        "surge_pressure_bar": 23.952,
        "surge_pressure_psi": 347.39
    },
    "meta": {
        "timestamp": "2026-06-07T08:18:01.770Z",
        "request_id": "7c54a78d-3fc2-4775-ba18-3c927e466629"
    },
    "status": "ok",
    "message": "Surge pressure",
    "success": true
}