Question 4 (Part B, Question B1): Blowdown of an Insulated Air Tank through a Convergent–Divergent Valve: 22-Mec-B6 Advanced Fluid Mechanics, May 2013, Q4

Question 4 (Part B, Question B1): Blowdown of an Insulated Air Tank through a Convergent–Divergent Valve (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Tank volume	V	0.5 m³
Initial tank pressure / temperature	p_0i, T_0i	10 MPa abs, 30 °C = 303.15 K
Final tank pressure	p_0f	2 MPa abs
Throat area	A_t	1 mm² = 1 × 10⁻⁶ m²
Exit area	A_e	2 mm² = 2 × 10⁻⁶ m²
Back pressure	p_b	101.3 kPa abs
Gas properties	γ, R	1.4, 287 J/(kg·K)

Find. (a) The tank temperature at 2 MPa, and the exit speed and exit temperature at that instant; (b) the elapsed time from 10 MPa to 2 MPa.

Figure 4.1 — Insulated tank discharging through a convergent–divergent valve. The throat chokes as soon as the valve opens and stays choked throughout, so the exit runs at the supersonic solution of the area–Mach relation for A_e/A_t = 2.

Approach. The tank is insulated and frictionless, so the gas left inside expands isentropically — that fixes the tank temperature at any pressure. The area ratio fixes the exit Mach number provided the throat is choked, which is checked against the back pressure. The blowdown time then follows from a mass balance on the tank with the choked mass-flow formula, integrated in closed form.

Part (a) — tank temperature by isentropic expansion. The gas that remains in an insulated, frictionless tank undergoes a reversible adiabatic expansion, so $T_0 p_0^{(1-\gamma)/\gamma}$ is constant and $$T_{0f} = T_{0i}\left(\frac{p_{0f}}{p_{0i}}\right)^{(\gamma-1)/\gamma} = (303.15)\left(\frac{2}{10}\right)^{0.2857} = \boxed{191.4\ \text{K}}$$ that is -81.7 degrees Celsius. The tank chills dramatically as it empties, which is why blowdown lines frost over.
Part (a) — exit Mach number from the area ratio. If the throat is choked, the exit Mach number depends only on the area ratio through $$\frac{A_e}{A_t} = \frac{1}{M_e}\left[\frac{2}{\gamma+1} \left(1 + \frac{\gamma-1}{2}M_e^{2}\right)\right]^{(\gamma+1)/[2(\gamma-1)]} = 2$$ which has the supersonic root $M_e = 2.1972$ and the subsonic root $M_e = 0.3059$. Selecting between them requires the back pressure.
Part (a) — confirm the nozzle runs full and supersonic. The design (fully expanded supersonic) exit pressure at the closing condition is $$p_{e,\text{des}} = \frac{p_{0f}}{\left(1 + \frac{\gamma-1}{2}M_e^{2}\right)^{\gamma/(\gamma-1)}} = \frac{2{,}000}{10.646} = 187.9\ \text{kPa}$$ which is still above the atmospheric 101.3 kPa, so the nozzle is under-expanded and the exit plane carries the supersonic design solution. The nozzle would only unchoke if the back pressure rose above the first-critical value $1,874$ kPa, which never happens. At the start of the blowdown the design exit pressure is 939 kPa, even further above atmosphere, so the supersonic solution holds for the whole process.
Part (a) — exit temperature and speed. The nozzle flow is isentropic from the tank stagnation state, so $$T_e = \frac{T_{0f}}{1 + \frac{\gamma-1}{2}M_e^{2}} = \frac{191.4}{1.9655} = \boxed{97.4\ \text{K}}$$ and the speed follows from the local sound speed, $a_e = \sqrt{\gamma R T_e} = 197.8$ m/s, giving $$V_e = M_e a_e = (2.1972)(197.8) = \boxed{434.6\ \text{m/s}}$$ The jet leaves at about -176 degrees Celsius, cold enough to condense and freeze any moisture present.
Part (b) — the choked mass-flow law. Setting $M = 1$ at the throat in the aid-sheet mass-flow expression collapses it to a form proportional to the instantaneous tank pressure: $$\begin{aligned} \dot m &= \sqrt{\frac{\gamma}{R}}\left(\frac{2}{\gamma+1}\right)^{(\gamma+1)/[2(\gamma-1)]} \frac{p_0 A_t}{\sqrt{T_0}} \;=\; C\,\frac{p_0 A_t}{\sqrt{T_0}} \\ C &= 0.04042\ \text{(SI units)} \end{aligned}$$ At the start this is $\dot m = 23.214$ g/s and at the closing condition $5.843$ g/s, so the rate falls by a factor of four during the process and a constant-rate estimate would be badly wrong.
Part (b) — mass balance and closed-form integration. The tank contents obey $V\,\mathrm{d}\rho/\mathrm{d}t = -\dot m$. Writing the isentropic tank states as $p_0 = p_{0i}(\rho/\rho_i)^{\gamma}$ and $T_0 = p_0/(\rho R)$ turns the right-hand side into a pure power of density, because $p_0/\sqrt{T_0} = \sqrt{R\,\rho\,p_0} = K\rho^{(\gamma+1)/2}$ with $K = \sqrt{R\,p_{0i}/\rho_i^{\gamma}} = 1,934.7$. Separating variables and integrating from the initial to the final density gives $$t = \frac{V}{\tfrac{\gamma-1}{2}\,C A_t K} \left[\rho_f^{-(\gamma-1)/2} - \rho_i^{-(\gamma-1)/2}\right]$$ with the exponent $(\gamma-1)/2 = 0.2$ for air.
Part (b) — evaluate. The tank densities are $\rho_i = p_{0i}/(RT_{0i}) = 114.94$ and $\rho_f = p_{0f}/(RT_{0f}) = 36.41$ kg per cubic metre, corresponding to 57.47 kg and 18.20 kg of air, so 39.26 kg is vented. Substituting, $$t = \frac{0.5}{(0.2)(0.04042)(10^{-6})(1,934.7)} \left[0.48726 - 0.38718\right] = \boxed{3,200\ \text{s}}$$ that is 53.3 minutes. The answer sits sensibly between the two bounding estimates obtained by holding the mass flow at its initial value (1,691 s) and at its final value (6,720 s).

Final Results.

Quantity	Symbol	Value
Tank temperature at 2 MPa (a)	T_0f	191.4 K (-81.7 °C)
Exit Mach number	M_e	2.1972
Exit temperature (a)	T_e	97.4 K
Exit speed (a)	V_e	434.6 m/s
Design exit pressure at closing	p_e,des	187.9 kPa (under-expanded)
Initial / final tank mass	m	57.47 kg / 18.20 kg
Mass discharged	Δm	39.26 kg
Blowdown time (b)	t	3,200 s (53.3 min)

Concept & Theory

Topic. Quasi-steady blowdown of a finite reservoir through a choked convergent–divergent nozzle.

Key relations. Isentropic tank state $T_0 \propto p_0^{(\gamma-1)/\gamma}$; area–Mach relation $A/A^{*} = \tfrac{1}{M}\left[\tfrac{2}{\gamma+1}(1+\tfrac{\gamma-1}{2}M^{2}) \right]^{(\gamma+1)/[2(\gamma-1)]}$; choked mass flow $\dot m = C p_0 A_t/\sqrt{T_0}$ with $C = \sqrt{\gamma/R}\,[2/(\gamma+1)]^{(\gamma+1)/[2(\gamma-1)]}$; tank mass balance $V\,\mathrm{d}\rho/\mathrm{d}t = -\dot m$.

Why this works. Two separate idealisations do the work here. Inside the tank, the gas that has not yet left is compressed by nothing and heated by nothing, so it simply expands reversibly and adiabatically; that single fact converts the pressure history into a temperature history. In the nozzle, the residence time is microseconds against a blowdown lasting nearly an hour, so the nozzle can be treated as steady at each instant — the quasi-steady approximation. Once the throat is choked, the mass flow depends only on the stagnation state and the throat area, which makes the tank mass balance a separable first-order equation with an exact solution. The divergent section does not change the mass flow at all; it only sets how fast and how cold the gas is at the exit.

Common pitfall. The most damaging mistake is to assume the tank stays at 30 °C, which overstates the mass flow near the end and shortens the predicted time substantially. The second is to pick the subsonic root of the area–Mach relation without checking the back pressure, or conversely to assume the nozzle unchokes as the tank empties; here the first-critical pressure is about 1,874 kPa at the closing condition, far above atmosphere, so the throat stays choked to the end. A third is to use the exit area rather than the throat area in the choked mass-flow formula.

Source. Anderson, Modern Compressible Flow, 3rd ed., §5.4 (nozzle flow) and §7.2 (unsteady discharge); White, Fluid Mechanics, 8th ed., §9.4.

Question 4 of 6: Question 4 (Part B, Question B1): Blowdown of an Insulated Air Tank through a Convergent–Divergent Valve

Question 4 (Part B, Question B1): Blowdown of an Insulated Air Tank through a Convergent–Divergent Valve (30 marks)