Tank Vent Sizing Fundamentals | API 2000 Engineering Guide

1. Overview

Atmospheric and low-pressure storage tanks require properly sized vents to prevent dangerous pressure or vacuum build-up. API 2000 (harmonized with ISO 28300) defines the criteria for calculating venting requirements under various conditions.

Venting Scenarios

Scenario	Cause	Direction	Concern
Thermal Inbreathing	Cooling, vapor condensation	Air IN	Vacuum / tank collapse
Thermal Outbreathing	Heating, vapor expansion	Vapor OUT	Overpressure
Liquid Pump-Out	Liquid removal from tank	Air IN	Vacuum
Liquid Pump-In	Liquid filling tank	Vapor OUT	Overpressure
Blow-By Gas	Control valve fails open	Gas OUT	Overpressure
Fire Exposure	External fire heating tank	Vapor OUT	Overpressure / rupture

Design Basis: Per API 2000, consider the largest single contingency or any reasonable combination of contingencies. At minimum, combine liquid transfer effects with thermal effects for normal venting.

2. Blow-By Scenario (PHA Focus)

A critical scenario reviewed in Process Hazard Analyses (PHAs): a control valve on a vessel sump fails open, allowing upstream pressurized gas to "blow by" through the liquid drain path to a downstream atmospheric tank.

Typical Blow-By System Configuration

Blow-By Flow Path Diagram

Vessel Sump → Control Valve (Fails Open) → Restriction Orifice → Header → Condensate Tank → Vent

Key Components

Component	Function	Sizing Consideration
Vessel Sump	Collects liquids for drainage	Source pressure (typically 50-500 psig)
Dump Valve (LCV)	Controls liquid level	Cv determines max flow when failed open
Restriction Orifice	Limits blow-by gas rate	Critical for vent sizing
Collection Header	Combines multiple sources	Sum of all blow-by sources
Condensate Tank	Atmospheric storage	Pressure rating (typically 1 oz/in²)
Tank Vent	Relieves gas pressure	Must handle total blow-by flow

Blow-By Flow Calculation

Gas Flow Through Restriction Orifice (Choked Flow): W = Cd × Y × A × √(2ρ₁ × ΔP × gc) Where: W = Mass flow rate (lb/s) Cd = Discharge coefficient (0.61 for sharp-edge) Y = Expansion factor (0.65-0.67 for choked flow) A = Orifice area (ft²) ρ₁ = Upstream gas density (lb/ft³) ΔP = Pressure drop (lbf/ft²) gc = 32.174 lbm·ft/(lbf·s²)

Critical (Choked) Flow Check

Critical Pressure Ratio: (P₂/P₁)crit = (2/(k+1))^(k/(k-1)) For natural gas (k ≈ 1.28): Critical ratio ≈ 0.55 If P_tank / P_upstream < 0.55 → Flow is CHOKED

Practical rule: Most blow-by scenarios result in choked flow because upstream pressure (50-500 psig) is much higher than tank pressure (~0 psig). Flow is limited by sonic velocity at the orifice.

Multiple Blow-By Sources

When multiple dump valves feed a common tank, consider:

All valves fail simultaneously (conservative for PHA)
Only largest source fails (if independent failure modes)
Largest + 50% of others (engineering judgment)

PHA Action Items: Blow-by scenarios often result in LOPA requirements or mechanical safeguards like restriction orifices, check valves, or independent high-pressure shutdowns.

3. Thermal Breathing

Thermal breathing occurs due to temperature changes causing vapor expansion (outbreathing) or contraction (inbreathing). API 2000 Section 3.3.2 provides calculation methods.

Thermal Inbreathing

API 2000 Thermal Inbreathing: V_IT = C × (V_tk)^0.7 × R_i Where: V_IT = Inbreathing rate (SCFH) C = Factor based on vapor pressure and latitude V_tk = Tank volume (ft³) R_i = Insulation reduction factor C Factor Values: Low VP (<1.5 psia): C = 2.0 Mid VP (1.5-5 psia): C = 2.5 High VP (>5 psia): C = 3.0 Insulation Factors (R_i): No insulation: 1.0 1" insulation: 0.5 2" insulation: 0.3 4" insulation: 0.15

Thermal Outbreathing

Thermal outbreathing is typically 60% of inbreathing for atmospheric tanks:

V_OT = 0.6 × V_IT

Liquid Transfer Effects

Inbreathing from Pump-Out: V_ip = 8.02 × Q_out (SCFH) Outbreathing from Pump-In: V_op = 8.02 × Q_in (SCFH) For volatile liquids (VP > 5 kPa): V_op = 2 × 8.02 × Q_in Where: Q = Liquid flow rate (gpm)

Total Normal Venting

Total Inbreathing: V_in = V_IT + V_ip Total Outbreathing: V_out = V_OT + V_op Design Flow = max(V_in, V_out)

4. Emergency Venting (Fire Case)

Emergency venting provides relief capacity for external fire exposure per API 2000 Section 4 and API 521.

Heat Input Equations

Standard Case (adequate drainage): Q = 21,000 × F × A^0.82 (BTU/hr) Inadequate Drainage: Q = 34,500 × F × A^0.82 (BTU/hr) Confined Fire: Q = 21,000 × F × A^1.0 (BTU/hr) Where: Q = Heat input (BTU/hr) F = Environmental factor (1.0 for bare tank) A = Wetted surface area up to 25 ft above grade (ft²)

Emergency Vent Flow

Vapor Generation Rate: W = Q / λ (lb/hr) Where: λ = Latent heat of vaporization (BTU/lb) Convert to SCFH: V_emergency = W / ρ_std (SCFH)

Environmental Factors

Tank Condition	F Factor
Bare tank (no insulation)	1.0
Earth-covered storage	0.03
Insulated (1" approved)	0.3
Water spray (adequate)	0.3
Concrete-covered	0.1

Note: Emergency venting is typically provided by separate emergency vents, not the normal P/V vent. Emergency vents are larger and may use weighted pallets or rupture disks.

5. Vent Sizing

Vent Capacity Calculation

Required Vent Area: A = Q / (C × V_max) Where: A = Vent area (ft²) Q = Design flow rate (ft³/s actual) C = Discharge coefficient (0.62 for vent) V_max = Maximum velocity (120 ft/s) Vent Diameter: d = √(4A/π)

Standard Vent Pipe Sizes

NPS	ID (in)	Area (in²)	Capacity @ 120 ft/s (SCFH)
2"	2.067	3.36	~28,000
3"	3.068	7.39	~62,000
4"	4.026	12.73	~106,000
6"	6.065	28.89	~241,000
8"	7.981	50.03	~417,000
10"	10.02	78.85	~658,000
12"	11.938	111.9	~933,000

Pressure Drop Check

Verify that the tank pressure rise from vent flow does not exceed the tank design pressure:

ΔP = ρ × V² / (2 × gc × 144 × Cd²) Design Check: ΔP_calculated < P_design (tank)

6. Worked Example: Blow-By Scenario

Problem: Size the vent for a condensate tank that receives blow-by gas from two separator dump valves, each with a 1/2" restriction orifice. Upstream pressure is 100 psig.

Given

Number of sources	2
Restriction orifice diameter	0.5 in
Upstream pressure	100 psig (114.7 psia)
Tank pressure	~0 psig (14.7 psia)
Gas	Natural gas, SG = 0.65
Temperature	80°F = 540°R
Tank design pressure	1 oz/in² = 0.0625 psi

Solution

Step 1: Check for choked flow
Critical ratio = (2/2.28)^(1.28/0.28) = 0.55
Actual ratio = 14.7/114.7 = 0.128
0.128 < 0.55 → CHOKED FLOW ✓

Step 2: Calculate gas properties
k = 1.28 (natural gas)
MW = 28.96 × 0.65 = 18.8 lb/lbmol
ρ₁ = (114.7 × 18.8)/(0.95 × 10.73 × 540)
   = 0.392 lb/ft³

Step 3: Calculate orifice flow (choked)
Cd = 0.61
Y = √[1.28 × (2/2.28)^(2.28/0.28)] = 0.65
A = π × (0.5/12)² / 4 = 0.00136 ft²
ΔP = 114.7 × (1 - 0.55) = 51.6 psi

W = 0.61 × 0.65 × 0.00136 × √(2 × 0.392 × 51.6 × 144 / 32.17)
  = 0.00054 × √(1470.5)
  = 0.021 lb/s per orifice

Step 4: Convert to SCFH
ρ_std = 0.0473 lb/ft³ (@ 60°F, 14.7 psia)
SCFH per orifice = (0.021 × 3600) / 0.0473 = 1,600 SCFH

Step 5: Total blow-by (2 sources)
Total = 2 × 1,600 = 3,200 SCFH

Step 6: Size vent
Required area = 3,200 / (3600 × 120) = 0.0074 ft² = 1.07 in²
Required diameter = √(4 × 1.07 / π) = 1.17"

Result: Use 2" NPS vent (provides 3.36 in²)
Capacity = ~28,000 SCFH > 3,200 SCFH ✓
Margin = 775%
                        

Engineering Judgment: While a 2" vent provides ample margin for this example, always verify total venting requirements including thermal breathing and emergency cases. The controlling case determines final vent size.

References

API Standard 2000 – Venting Atmospheric and Low-Pressure Storage Tanks (7th Edition)
API Standard 650 – Welded Tanks for Oil Storage
API RP 521 – Pressure-relieving and Depressuring Systems
ISO 28300 – Petroleum and natural gas industries — Venting of atmospheric tanks
GPSA, Section 6

Tank Vent Sizing

API 2000

1 oz/in²

120 ft/s

Contents

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