/v1/surge
Joukowsky surge pressure
Δοκιμάστε το ζωντανά
10 δωρεάν κλήσεις την ημέρα — χωρίς εγγραφή, χωρίς κλειδί API. Μέσω του gateway του oanor.
Δουλεύει. Πάρε ένα κλειδί API και χρησιμοποίησέ το στο έργο σου.
Λάβετε ένα κλειδί APIΑποσπάσματα κώδικα
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_..."}
)
Παράδειγμα απόκρισης
Πραγματική απόκριση αυτού του endpoint, από τον τελευταίο έλεγχο υγείας.
{
"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
}