Lab 2: Tank blowdown

A rigid tank of air vents to atmosphere through a small orifice. Students derive the closed-form pressure and temperature history for choked adiabatic blowdown, compare it with the shipped Tank blowdown example, measure the error against time step, and then make the tank isothermal to see how much the answer depends on the energy equation.

Case 1 of the tank transient validation report uses the same orifice and tank volume at a higher starting pressure. The tank blowdown tutorial is a gentler walk-through of the same example.

Learning objectives

By the end of the lab, students should be able to:

1. Decide whether an orifice is choked, and compute the choked mass flow of an ideal gas.
2. Derive the closed-form choked adiabatic blowdown solution and the cooling law T/T₀ = (m/m₀)^(γ−1) from the open-system mass and energy balances.
3. Explain why the gas cools, using the difference between the enthalpy that leaves and the internal energy that is stored.
4. Show that a backward-Euler integrator converges at first order in the time step, and choose a time step for a required accuracy.
5. Compare adiabatic and isothermal blowdown and explain which assumption a real tank is closer to.

Prerequisites

• Thermodynamics: first law for open systems, ideal-gas relations, isentropic flow, choking.
• Basic ODEs: separable first-order equations.
• Lab 1 or the Getting started page.

Time

About 45 minutes of pre-lab work and 1.5 hours in the lab.

Theory

Model. The example has three elements (user manual §7.3):

Element	Type	Values
Tank	internal node with volume	V = 0.1 m³, initially P₀ = 500 kPa, T₀ = 300 K
Ambient	boundary node	P_b = 101 325 Pa, 300 K
Orifice	orifice branch	A = 1×10⁻⁴ m², Cd = 0.6

The fluid is the ideal-gas air preset: R = 287 J/(kg·K), γ = 1.4, cp = 1005 J/(kg·K) (src/core/fluids/index.ts). The solver uses cv = cp − R = 718 J/(kg·K) in the energy equation, so cp/cv = 1.3997. The resulting difference of about 0.01 % shows up only in the time-step study.

Balances. The tank stores mass and internal energy; the outflow carries enthalpy (user manual §4.1.3; report "Tank ODEs"):

dm/dt = −ṁ
d(m cv T)/dt = −ṁ cp T
P = m R T / V

Eliminating time gives T/T₀ = (m/m₀)^(γ−1) and P/P₀ = (m/m₀)^γ for any adiabatic blowdown, whatever the orifice law.

Choked orifice. The gas orifice chokes when P_b/P falls below the critical ratio r* = (2/(γ+1))^(γ/(γ−1)) = 0.5283 (manual §4.3.2), and the mass flow then no longer depends on the back pressure:

ṁ = Cd A P sqrt(γ/(R T)) (2/(γ+1))^((γ+1)/(2(γ−1)))

Closed form. Substituting the choked law and the cooling law into the mass balance gives a separable ODE whose solution is (report, "Adiabatic Blowdown"):

m/m₀ = [1 + ((γ−1)/2) (t/τ)]^(−2/(γ−1))
P/P₀ = [1 + ((γ−1)/2) (t/τ)]^(−2γ/(γ−1))
τ = V / ( Cd A sqrt(γ R T₀) (2/(γ+1))^((γ+1)/(2(γ−1))) )

The solution holds only while the orifice stays choked.

Isothermal limit. If heat exchange holds the gas at T₀, the choked mass balance becomes linear in P, and P/P₀ = exp(−t/τ) with the same τ.

Time integration. OpenFLUME integrates transients with backward (implicit) Euler (manual §4.6.2), which is first order: halving dt should halve the deviation from the exact solution.

Pre-lab hand calculation

1. Compute the initial gas mass m₀ and the initial choked mass flow ṁ₀.
2. Find the tank pressure below which the orifice unchokes, P_b / r*. Confirm that the orifice is choked at t = 0.
3. Compute τ.
4. Tabulate P, T, and m from the closed form at t = 1, 2, 3, 4, 5 s.
5. Find the time at which the tank pressure reaches the unchoking pressure. Is the closed form valid for the whole 5 s run?
6. Repeat steps 4 and 5 for an isothermal tank.
7. Backward Euler evaluates the outflow at the end of each step, where the tank pressure is lower. Predict whether the computed pressure will lie above or below the exact curve, and whether the gap grows or shrinks during the run.

Procedure in OpenFLUME

Set Units ▾ to SI.

Part A: Baseline run

1. Open the example: Tank blowdown. (Or Examples ▾ → Applications → Tank blowdown.)
2. Click the Tank node and check Pressure, Temperature, and Volume. Click the Orifice and check its area and discharge coefficient. Open the Setup tab: under Solver, Mode is Transient, with Fixed dt, Time Step 0.01 s, and End Time 5 s.
3. Return to the Model tab and click Run. While it runs, the toolbar shows t = … s / 5 s. The status pill should end at Converged.
4. Open the Results tab. In the empty plot, use or plot a whole set… to choose Node pressure. Read the tank pressure at t = 1, 2, 3, 4, 5 s. Click the chart and use the arrow keys to step the cursor, or use Export CSV and read the values from the file.
5. Click + Plot and choose the Node & solid temperature set. Read the tank temperature at the same times.
6. Add a third plot with the orifice mass flow. Note the value at the first time step.

Part B: Time-step convergence

1. In Setup → Solver, change Time Step to 1 s and click Run. Record the tank pressure at t = 5 s.
2. Repeat with Time Step = 0.5, 0.25, 0.1, and 0.05 s. Every run is kept in the run history. On the pressure plot, + Compare run… overlays up to four earlier runs on the same axes.
3. For each time step, compute the relative error in P(5 s) against your closed-form value. Plot error against dt on log-log axes and measure the slope.
4. Set Time Step back to 0.01 s before Part C.

Part C: Isothermal tank

To make a gas tank nearly isothermal, tie it to a fixed-temperature reservoir with a convection conductor whose hA is much larger than 10⁴ W/K (user manual §3.6).

1. On the Model tab, place an Ambient node from the left rail (Thermal nodes). Set its Temperature to 300 K.
2. Drag from the Tank node's handle to the ambient node. In Connect with, under Thermal ties, choose Convection.
3. In the conductor's property panel, leave Heat-transfer model on Specified h (constant or equation), and set h to 10000 W/(m²·K) and Area to 10 m², so that hA = 10⁵ W/K.
4. Click Run. Compare tank pressure and temperature with the adiabatic run, using + Compare run….

Questions

1. How close is the dt = 0.01 s run to your closed-form pressure at 5 s? Is the solver above or below the exact curve? Does that match your prediction in pre-lab step 7?
2. What convergence order did you measure? The validation report measured 0.99–1.00 on its own blowdown case. If your slope differs noticeably, what else limits the comparison at very small dt?
3. Which time step would you choose to keep the 5 s pressure within 0.5 % of the exact value? Within 0.1 %?
4. The tank temperature falls by about 60 K in 5 s even though no heat leaves the tank. Where does the internal energy go?
5. Why does the isothermal tank hold its pressure longer, even though its mass flow at t = 0 is the same?
6. A real tank has metal walls. Is a real blowdown closer to the adiabatic or the isothermal result? What would you need to add to the model to answer this properly?
7. What value does the orifice mass flow show at t = 0, and why does it not match ṁ₀?

Extensions

• Past unchoking. Set End Time to 15 s and rerun the adiabatic case. Find the time at which the flow unchokes and compare it with your pre-lab estimate. Once the flow stops, the tank has expanded isentropically to ambient pressure. Predict the final temperature from T = T₀ (P_b/P₀)^((γ−1)/γ).
• Validation case. Rebuild case 1 of the validation report (10 bar tank, 1 bar boundary, 8 s, dt = 0.05 s) and compare your final pressure and temperature with the report's 2.91 bar and 210.8 K.
• Orifice size. Use the Sweep tab to vary the orifice area. How does τ scale with area, and does the solver agree? The tutorial covers this sweep.
• Fill heating. Reverse the problem: a tank at 1 bar filled from a 5 bar, 300 K supply heats above 300 K. Case 4 of the validation report gives the closed form, T_f = γ T₀ P_f / (P_f + (γ−1) P₀) = 388.9 K.

Answer key

Closed-form values were computed independently in Python from the formulas above, with γ = 1.4 and R = 287 J/(kg·K). A Runge–Kutta (RK4) integration of the tank ODEs that uses the preset's actual cp = 1005 and cv = 718 gives P(5 s) = 225.425 kPa and T(5 s) = 238.96 K, within 0.013 % of the closed form. Solver values are OpenFLUME 0.3.1, run locally with the shipped example.

Pre-lab

• m₀ = P₀V/(RT₀) = 0.5807 kg. ṁ₀ = 0.07001 kg/s.
• r* = 0.5283, so the unchoking pressure is 101 325/0.5283 = 191.8 kPa. At t = 0, P_b/P₀ = 0.203, so the orifice is choked.
• τ = 8.295 s (the same V, Cd A, and T₀ as the validation report's case 1, which also gives 8.295 s).
• Adiabatic unchoking time: 6.08 s. The 5 s run stays choked throughout, so the closed form holds for the whole run. Isothermal unchoking time: τ ln(P₀/191.8 kPa) = 7.95 s.

t [s]	P adiabatic [kPa]	T adiabatic [K]	m [kg]	P isothermal [kPa]
0	500.00	300.00	0.5807	500.00
1	423.20	286.04	0.5155	443.22
2	359.58	273.03	0.4589	392.88
3	306.67	260.89	0.4096	348.26
4	262.47	249.55	0.3665	308.71
5	225.40	238.92	0.3287	273.65

At 5 s the tank has vented 0.252 kg, which is 43.4 % of its charge.

• Step 7: the implicit step discharges slightly too little mass in each step, so the computed pressure lies above the exact curve. Within a fixed-step run, the gap accumulates and grows with time. The validation report describes the same one-sided lag.

Part A (solver, dt = 0.01 s)

t [s]	P [kPa]	error vs closed form	T [K]
1	423.27	+0.018 %	286.07
2	359.71	+0.034 %	273.09
3	306.82	+0.048 %	260.97
4	262.63	+0.061 %	249.64
5	225.56	+0.073 %	239.03

The run takes 500 steps. The orifice mass flow at the first step is 0.0699 kg/s, against 0.0700 kg/s exact at t = 0.

Part B: time-step study (solver, P at 5 s; closed form 225.396 kPa)

dt [s]	P(5 s) [kPa]	error	T(5 s) [K]
1	239.22	+6.13 %	245.70
0.5	232.59	+3.19 %	242.53
0.25	229.07	+1.63 %	240.80
0.1	226.90	+0.67 %	239.71
0.05	226.11	+0.32 %	239.34
0.01	225.56	+0.073 %	239.03

Successive slopes on a log-log plot are 0.94, 0.97, 0.98, 1.08, and 0.91. The order is about 1. The last slope is pulled down by the 0.013 % offset between the closed form and the preset's cp/cv (see above). Measured against the RK4 reference, the slopes are 0.95, 0.97, 0.99, 1.11, and 1.00. For 0.5 % accuracy at 5 s, dt ≈ 0.075 s is enough; for 0.1 %, about dt ≈ 0.015 s. The example's 0.01 s meets both.

Part C: isothermal (solver, hA = 10⁵ W/K, dt = 0.01 s)

P(5 s) = 273.71 kPa (+0.02 % against 500 exp(−5/8.295) = 273.65 kPa), and the tank temperature stays between 299.95 and 300 K. With hA = 10⁴ W/K the tank runs 0.3 to 0.5 K cold and P(5 s) = 273.34 kPa.

Questions

• Q4: Each kilogram that leaves carries enthalpy cp T, but each kilogram stored only holds cv T. The difference, R T per kilogram, is the flow work the remaining gas does pushing the outflow out. It comes from the remaining gas's internal energy, so the gas cools.
• Q5: The choked mass flow is proportional to P/sqrt(T). In the adiabatic tank, T falls, which on its own would raise ṁ, but P falls much faster (P ∝ m^1.4, compared with P ∝ m isothermally). For the same mass vented, the adiabatic tank loses more pressure.
• Q6: For a short, fast blowdown the wall cannot exchange much heat, so the answer is closer to adiabatic. For slow venting, it moves toward isothermal. Answering properly needs a solid wall node with mass and cp, tied to the gas by a convection conductor with a realistic h and to the ambient (see Lab 3).
• Q7: In OpenFLUME 0.3.1 the t = 0 sample of a branch mass flow is the solver's starting value, not a solved flow. Here it is the default initial flow guess, 0.1 kg/s. Read the flow from the first step onward.

Extensions

• Past unchoking (solver, dt = 0.01 s): P(6 s) = 194.4 kPa, so the flow unchokes just after 6 s. The pressure reaches 101.3 kPa by about 15 s with T = 190.2 K, against the isentropic value 300 × (101 325/500 000)^(0.4/1.4) = 190.1 K.