Sizing PSV for Flashing Water: API 520 Direct Integration

Subcooled condensate at P1 = 94.45 psia and T1 = 248 °F relieving through a nozzle to a 45 ft atmospheric tailpipe sits squarely in the high-subcooling regime of two-phase relief. The homogeneous direct integration (HDI) method in API 520 Part I is the right tool, but the answer it gives is anchored by a closed-form result that takes thirty seconds to check. Run both. If they disagree by more than a few percent, the spreadsheet is wrong, not the physics.

Flow Regime Determination Before Any Sizing Equation

Before anything else, confirm three numbers from the same steam-table set used for the flash: the saturation pressure at the relieving temperature, the saturated-liquid entropy at that temperature, and the back pressure the nozzle actually sees.

1. Read Psat(248 °F). Interpolating logarithmically between the 240 °F and 250 °F entries gives approximately 28.8 psia. This is the pressure at which the isentrope leaves the compressed-liquid region.
2. Read sf(248 °F). It falls at roughly 0.3648 Btu/lbm-°F, which matches the reported inlet value s1 = 0.364867 to five decimals. That agreement is not a coincidence — compressed-liquid entropy at moderate pressure is essentially the saturated-liquid value at the same temperature, and it confirms the inlet state was read correctly.
3. Compare Psat to the back pressure. At 28.8 psia versus 14.7 psia atmospheric, the fluid flashes inside the nozzle and two-phase sizing applies. Had the built-up back pressure exceeded 28.8 psia, no flashing would occur in the valve at all and the device would be sized with the API 520 liquid equation.

The subcooling ratio Psat/P1 = 28.8/94.45 = 0.305 puts this case deep in the high-subcooling region. In that region the critical flow condition occurs at flashing inception, so the maximum mass flux landing exactly at the flash pressure is the expected physical result, not a numerical artifact. Do not move on until Psat(T1) and the back pressure are both fixed numbers.

The Constant-Temperature Flash Shortcut and Its Error Bound

Holding temperature constant while stepping pressure down, instead of solving for the temperature that holds entropy constant, is acceptable here. The magnitude of the error is bounded by a Maxwell relation:

(∂s/∂P)_T = -(∂v/∂T)_P = -vβ

v  ≈ 0.0170 ft³/lbm at 248 °F
β  ≈ 4.8e-4 /°F (water at 248 °F)

ds = -(0.0170)(4.8e-4)(144)/778.17 = -1.5e-6 Btu/lbm-°F per psi

That predicted drift of about 1.5 to 2 microunits per psi is exactly what the tabulated steps show (0.364867 → 0.364869 → 0.364869 → 0.364872). The residual is at steam-table round-off level, not a modelling error.

The true isentropic temperature change over the same interval is:

(∂T/∂P)_s = T·v·β/cp = (707.7)(0.0170)(4.8e-4)(144)/(778.17)(1.01)
                = 0.0011 °F per psi  →  <0.08 °F over 65.6 psi

An 0.08 °F temperature error changes liquid density by well under 0.01 percent. Cumulative entropy drift to the flash point is roughly 1e-4 Btu/lbm-°F, which at sfg ≈ 1.33 Btu/lbm-°F corresponds to a quality error near 8e-5 — and it lands at the point where quality is zero anyway. The shortcut does not move the answer.

Method Comparison for a Deeply Subcooled Inlet

The enthalpy form G = ρ·223.8·√(h1 - h), with enthalpy in Btu/lbm, is mathematically identical to the integral along an isentrope and removes the numerical integration entirely. It is a precision trap in the subcooled leg: at the flash point h1 - h is only about 0.21 Btu/lbm, a difference between two numbers near 216 Btu/lbm. Four-significant-figure tables cannot resolve it. Use the enthalpy form only once quality is established.

Recommended Path: Refined HDI Anchored to the Closed Form

Keep the direct integration as the primary calculation and use the closed form as the acceptance criterion.

1. Replace the entire subcooled leg with the incompressible result. Liquid density varies less than half a percent over 65 psi, so ∫dP/ρ = (P1 - P)/ρL is exact to the precision of the table.
2. Compute the anchor value at the flash point:

   G = √(2 · gc · ρL · (P1 - Psat))
     = √(2 × 32.174 × 58.8 × 65.6 × 144)
     ≈ 5,980 lbm/(s·ft²)

   with pressures converted to lbf/ft² and ρL = 1/vf(248 °F) ≈ 58.8 lbm/ft³.
3. Place an integration node exactly at Psat. The specific volume is continuous there but its slope is not; a trapezoid straddling the kink smears the peak and shifts it by half a step.
4. Refine the grid to 0.1 psi or finer within ±2 psi of Psat. The 1 psi grid is more than adequate in the liquid leg and too coarse where quality first appears.
5. Continue the integration into the two-phase region far enough to prove G is falling, typically 5 to 10 psi below Psat. Once G has dropped 10 percent from the peak it will not recover.
6. Halve the step size and re-run. If G_max moves more than 1 percent, the grid is still too coarse.
7. Confirm the refined HDI peak lands within a few percent of the 5,980 lbm/(s·ft²) anchor. A large discrepancy points to a density read at the wrong state or an integrand of ρ rather than 1/ρ.

Evaluate G at every pressure as G(P) = ρ(P)·√(2·gc·∫dP/ρ), using the local mixture density at that pressure. Computing the integral correctly and then multiplying by the inlet density is the most common spreadsheet error in this method.

Back Pressure and the 45 ft Tailpipe

Integrating to 14.7 psia without knowing the built-up back pressure worked here only because the choke sits at 28.8 psia and everything below it is discarded. That does not make the tailpipe calculation optional — it moves it to the valve-selection stage.

1. Size the tailpipe for the flashing two-phase discharge over its full 45 ft run using a homogeneous or omega-based pipe method. Flashing water accelerates along the line and the exit will usually choke.
2. Work back from the choked exit pressure to the valve outlet to obtain the built-up back pressure Pb.
3. Test Pb against 28.8 psia. Below it, the nozzle choke and G_max stand unchanged. At or above it, the fluid never flashes in the nozzle and the sizing basis switches to the liquid equation.
4. Test Pb against 10 percent of set pressure. Assuming 10 percent overpressure, P1 = 94.45 psia implies a set pressure near 72.5 psig, so a conventional spring valve tolerates roughly 7.2 psi of built-up back pressure — about 22 psia at the outlet flange — before lift degrades and Kb = 1 stops being defensible.
5. If Pb exceeds that limit, enlarge the tailpipe or specify a balanced-bellows or pilot-operated valve and take the correction factor from the manufacturer's certified curve.

Compute the reaction force on the 45 ft line per API 520 Part II and check the discharge orientation at grade. A flashing 248 °F water jet is a personnel hazard at the outlet regardless of how the nozzle was sized.

Coefficients and the Area Equation Units Audit

The constant in A = 0.04 W / (Kd·Kb·Kc·Kv·G)It is correct only when W is in lb/h and G is in lbm/(s·ft²). If G is carried in lb/(h·ft²), the constant becomes 144; if in lb/(s·in²), it becomes 1/3600. Audit the units of G in the spreadsheet before trusting the area.

At G = 5,980 lbm/(s·ft²) and Kd = 0.65, the area works out to roughly 0.0103 in² per 1,000 lb/h of required relief rate. Raising Kd to 0.85 would drop that to 0.0079 in² per 1,000 lb/h — a 24 percent area swing that rests entirely on the certification basis of the valve, which is why the coefficient is a documented input and not an assumption.

Verification Before Issuing the Datasheet

1. Check that s1 matches sf(T1) within table round-off. It does here, at 0.364867 against roughly 0.3648.
2. Check that the computed flash pressure equals Psat(T1) read independently from the saturation table. A mismatch means the quality-solving loop has an entropy datum error.
3. Check that G_max agrees with √(2·gc·ρL·(P1 - Psat)) ≈ 5,980 lbm/(s·ft²).
4. Check that quality at the step immediately below Psat is small and positive, on the order of 1e-3 or less, and that mixture density there falls steeply. Quality jumping straight to several percent in one step means the grid is too coarse.
5. Check inlet-line pressure loss against the 3 percent non-recoverable limit at the rated capacity of the selected orifice, not at the required rate.
6. Round the required area up to the next API 526 letter orifice. Below about 10,700 lb/h the calculation calls for less than a D orifice (0.110 in²) — at that point the oversizing ratio invites chatter, and a smaller valve body or a pilot-operated design should be evaluated before the datasheet is released.
7. Recognize the direction of the HEM bias: delayed nucleation in a short nozzle makes real critical flow at high subcooling exceed the equilibrium prediction. The computed area is therefore conservative for capacity but optimistic for tailpipe pressure drop and reaction force. Size the discharge line for the higher flow.

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