An illustrative case shaped like an F-16 external wing tank. It is not F-16 data from General Dynamics or Lockheed Martin: the tank’s real pressure limits and vent valves are not public. These are the page’s default inputs. Load this example into the calculator.
| Input | Value | Source |
|---|
| Tank volume | 370 US gal (1,401 L), empty | The commonly quoted size of the F-16 wing tank. The USAF fact sheet lists “two 2400-pound external fuel tanks”: about 360 gal of jet fuel at 6.7 lb/gal. |
| Ceiling | 50,000 ft | USAF F-16 fact sheet: “above 50,000 feet”. |
| Climb rate | 50,000 ft/min, approximate | Widely quoted in secondary references. Not given by the USAF fact sheet or Lockheed Martin. |
| Descent rate | 30,000 ft/min | Assumed design case for a steep descent or dive. |
| Limits | 1.5 psid (10.34 kPa) burst, 0.5 psid (3.45 kPa) collapse | Assumed. |
| Vent valves | Cv = 4.0, xT = 0.70 | Assumed. |
| Tank gas | 59 °F (15 °C), isothermal, dry air | Conservative choices (cards 04 and 08). |
Step 1: pressure change at sea level
$$\dot H = 50{,}000\ \text{ft/min} = 833\ \text{ft/s} = 254.0\ \text{m/s}$$
$$\frac{dP_a}{dt} = -\rho_a g_0 \dot H = -(1.2250)(9.80665)(254.0)$$
$$\frac{dP_a}{dt} = -3{,}051\ \text{Pa/s} = -0.443\ \text{psi/s}$$
Step 2: air that must leave
$$\dot m = \frac{V}{R T}\left|\frac{dP_a}{dt}\right| = \frac{1.4006}{(287.05)(288.15)}(3{,}051)$$
$$\dot m = 0.0517\ \text{kg/s} = 6.83\ \text{lb/min}$$
Step 3: one valve at the 1.5 psid limit
$$p_1 = 101.325 + 10.342 = 111.67\ \text{kPa}$$
$$x = \frac{10.342}{111.67} = 0.0926 < F_k x_T = 0.70\ \text{(not choked)}$$
$$Y = 1 - \frac{0.0926}{3(0.70)} = 0.956$$
$$\rho_1 = \frac{111{,}667}{(287.05)(288.15)} = 1.350\ \text{kg/m}^3$$
$$w = 2.73\,(4.0)(0.956)\sqrt{(0.0926)(111.67)(1.350)}$$
$$w = 39.0\ \text{kg/h} = 1.43\ \text{lb/min}$$
Step 4: valves for climb
$$N = \frac{6.83}{1.43} = 4.77 \;\Rightarrow\; 5\ \text{valves}$$
$$\text{total } C_v = 19.1\ \ (C_d A \approx 0.50\ \text{in}^2)$$
Step 5: the rest of the climb
$$P_a(50{,}000\ \text{ft}) = 1.682\ \text{psia}, \quad \frac{dP_a}{dt} = -0.0674\ \text{psi/s}$$
$$\dot m = 7.87\ \text{g/s} = 1.04\ \text{lb/min}$$
At the ceiling the flow is only 15 % of the sea-level value. Sea level governs. If the tank did reach 1.5 psid up there, the pressure ratio would be 1.682/3.182 = 0.529, right at the 0.528 critical ratio where an orifice chokes. The Cv valve would be at x = 0.471, under its 0.70 limit.
Step 6: descent at 30,000 ft/min, arriving at sea level
$$\frac{dP_a}{dt} = +0.266\ \text{psi/s}, \quad \dot m = 31.0\ \text{g/s} = 4.10\ \text{lb/min}$$
$$w_{valve}(0.5\ \text{psid}) = 0.811\ \text{lb/min}\quad (x = 0.034,\ Y = 0.984)$$
$$N = \frac{4.10}{0.811} = 5.06 \;\Rightarrow\; 6\ \text{valves}, \quad \text{total } C_v = 20.2$$
Step 7: result
$$\text{6 two-way vents:}\quad \Delta P_{climb} = 0.950\ \text{psid}\ \ (\text{limit } 1.5)$$
$$\Delta P_{descent} = 0.351\ \text{psid}\ \ (\text{limit } 0.5)$$
$$\dot H_{climb,\max} = 62{,}910\ \text{ft/min}, \qquad \dot H_{descent,\max} = 35{,}610\ \text{ft/min}$$
Descent governs, even though the climb moves more air (6.83 against 4.10 lb/min). The collapse limit is a third of the burst limit, and flow rises only with the square root of ΔP. Thin external tanks are often weakest in collapse, so check the inflow side first. Separate valves could be sized instead: 5 for outflow and 6 for inflow (tick “different valve for inflow”).
The flight simulation with 6 vents peaks at 0.773 psid on climb near 8,950 ft (a 60 s climb) and 0.338 psid on descent at sea level. The tank lags the outside pressure, so the quasi-steady values are never quite reached. Size on the quasi-steady values anyway (card 07).