Dimensionless ideal lumped parameter pneumatic gun model

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This page will detail the development of a simplified model of the interior ballistics of a low-speed pneumatic gun (like a Nerf gun).

(What follows is largely a verbatim copy of some notes I made back in April to June 2010 - btrettel)

1 To-do
2 Assumptions
3 Labeled diagram
4 Variables
5 Derivation of equations
6 Non-dimensionalization of equations
7 Final non-dimensionalized equations
8 Conversion of ODEs into finite difference equations
9 Numerical solution of these equations
10 Regressions
11 Comparisons against empirical data
12 Footnotes

To-do

Add parts that distinguish between static and dynamic friction. As described, this model has no static friction.
Change names of sections such that there is no duplication.
Switch to $γ$ instead of $k$ .
Add more explanations and fill in missing parts.

Assumptions

Calorically perfect gas - Ideal gas model with constant specific heats. This has proved to be adequate.
Isentropic - All heat transfer and boundary layer effects (like friction on the gas) are ignored. The entropy remains constant as the gas moves through the valve. These effects have ignored with very acceptable results in simulations like GGDT and BAGS.
The gas has only thermal energy - Kinetic energy is assumed to be negligible (As is gravitational, etc., but kinetic is the only major component neglected). This may or may not be a good approximation, but it simplifies the equations substantially. This assumption also means that chambrage (i.e. difference in cross-sectional area of gas chamber and barrel) and chamber/barrel fineness ratio effects are ignored.
The gas has no momentum - Or at least it has no momentum when I want to make this simplifying assumption. This is necessary if we don't want to make any assumptions about the velocity distribution in the barrel. It's also reasonably accurate unless the projectile weight is approximately within an order of magnitude of the total gas weight.
The Mach number is low.
- Can treat some stagnation temperatures and pressures as the normal temperatures and pressures
- Pressures equalize instantaneously leading to the...
- Pressure, temperature, and density can be treated as lumped parameters - This is appropriate for low Mach numbers because the pressure, temperature, and density will not depend greatly on the location in each section of the gun.
Valve is assumed to not store any gas (has zero volume)
No leaks are present (including leaks around the projectile)
On the atmospheric side of the barrel, the pressure, density, and temperature are atmospheric
The gas never changes phase.

Labeled diagram

Variables

Derivation of equations

Chamber mass conservation (1)

$\text{rate of change of mass} = \text{rate of mass flowing in} - \text{rate of mass flowing out}\,\!$

$\dot{M} = \dot{M}_{in} - \dot{M}_{out}\,\!$

$M = V_c \rho_c\,\!$

$\dot{M}_{in} = 0\,\!$

$\dot{M}_{out} = \dot{m} \text{sqn}(P_c - P_b)\,\!$

$\tfrac{d}{dt} \left[V_c \rho_c\right] = 0 - \dot{m} \text{sqn}(P_c - P_b)\,\!$

$V_c \dot{\rho_c} = - \dot{m} \text{sqn}(P_c - P_b)\,\!$

Chamber energy conservation (2)

$\text{rate of change of energy} = \text{rate of energy flowing in} - \text{rate of energy flowing out}\,\!$

$\dot{E} = \dot{E}_{in} - \dot{E}_{out}\,\!$

Assuming the gas has no kinetic energy, the total energy of the chamber gas can be written very easily.

$E = V_c \rho_c c_v T_c\,\!$

$\dot{E}_{in} = 0$

$\dot{E}_{out} = \dot{m} c_p T_u \text{sgn}(P_c - P_b)\,\!$

$\dot{m} = \rho_u A_b u_{bnd}\,\!$

$V_c c_v \tfrac{d}{dt} \left[\rho_c T_c \right] + \rho_u A_b u_{bnd} c_p T_u \text{sgn}(P_c - P_b) = 0\,\!$

Barrel mass conservation (3)

$\text{rate of change of mass} = \text{rate of mass flowing in} - \text{rate of mass flowing out}\,\!$

$\dot{M} = \dot{M}_{in} - \dot{M}_{out}\,\!$

$M = \rho_b V_b = \rho_b \left(V_d + A_b x\right)\,\!$

$\dot{M}_{in} = \dot{m} \text{sqn}(P_c - P_b)\,\!$

$\dot{M}_{out} = 0\,\!$

$\tfrac{d}{dt} \left[\rho_b \left(V_d + A_b x \right) \right] = \dot{m} \text{sgn}(P_c - P_b)\,\!$

Barrel energy conservation (4)

$\text{rate of change of energy} = \text{rate of energy flowing in} - \text{rate of energy flowing out}\,\!$

$\dot{E} = \dot{E}_{in} - \dot{E}_{out}\,\!$

Assuming the gas has no kinetic energy, the total energy of the chamber gas can be written very easily.

$E = V_b \rho_c c_v T_c = \rho_b \left(A_b x + V_d \right) c_v T_b\,\!$

$\dot{E}_{in} = \dot{m} c_p T_u \text{sgn}(P_c - P_b)\,\!$

$\dot{E}_{out} = -F_d \dot{x}\,\!$

$\dot{m} = \rho_u A_b u_{bnd}\,\!$

$\tfrac{d}{dt} \left[\rho_b \left(A_b x + V_d \right) c_v T_b \right] - \rho_u u_{bnd} A_b c_p T_u \text{sgn}(P_c - P_b) = - F_d \dot{x}\,\!$

Projectile equation of motion (5)

If the mass of the projectile is far higher than the mass of the air ( $\tfrac{m_{air}}{m_d} < 10^{-2}\,\!$ or so) then the mass of the air can be fully neglected from the analysis.

Simple dynamic pressure terms from the Bernoulli equation are used. This can be justified by using the Reynolds transport theorem applied to momentum conservation and dropping the integral terms (as the gas has negligible mass).

The equation below can be derived directly with Newton's second law.

$m_d \ddot{x} = A_b \left[P_b - P_{atm} - P_f \text{sgn} \left(\dot{x} \right) + \tfrac{1}{2} \left(\rho_b - \rho_{atm} \right) \dot{x}^2 \right]\,\!$

Valve mass flow rate (6)

The valve mass flow rate equations from Fluid Power Control are used^[1] with a few modifications. These equations treat the valve as an orifice with a discharge coefficient to account for inefficiencies.

$\dot{m} = \frac{C_d A_{ref} P_u}{\sqrt{R T_u}} \text{f} \left(\tfrac{P_d}{P_u}, k\right)\,\!$

$\left(\frac{P_d}{P_u}\right)_{crit} = \left(\frac{2}{k + 1}\right) ^ {\tfrac{k}{k - 1}}\,\!$

If greater accuracy is desired, empirical values of $\left(\tfrac{P_d}{P_u}\right)_{crit}\,\!$ can be used, along with modified equations to account for a throat Mach number other than 1.

$C d$ - discharge coefficient (dimensionless)

$A r e f$ - reference area [ $L 2$ ] - Ideally, this is the minimum cross sectional area of the flow path through the valve, however, that area is not always easy to find. For this model $A r e f = A b$ , where $A b$ is the cross sectional area of the barrel.

$P u$ - upstream stagnation pressure [ $F L - 2$ ]

$P d$ - downstream pressure [ $F L - 2$ ]

$T u$ - upstream stagnation temperature [ $T$ ]

$P u$ and $T u$ are assumed to be equal to the "normal" pressures and temperatures, respectively. This is accurate at low Mach numbers.

$k$ - ratio of specific heats - 1.4 for air

$\text{f} \left(\tfrac{P_d}{P_u}, k\right) = \begin{cases} \displaystyle \sqrt{\frac{2 k}{k - 1}} \left(\frac{P_d}{P_u}\right)^{\tfrac{1}{k}} \sqrt{1 - \left(\frac{P_d}{P_u}\right)^{\tfrac{k - 1}{k}}} & \displaystyle \text{if } \left(\frac{P_d}{P_u}\right) > \left(\frac{P_d}{P_u}\right)_{crit} \\ \displaystyle \sqrt{\frac{k}{\left(\tfrac{k + 1}{2}\right)^{\tfrac{k + 1}{k - 1}}}} & \displaystyle \text{if } \left(\frac{P_d}{P_u}\right) < \left(\frac{P_d}{P_u}\right)_{crit} \end{cases}\,\!$

The function $\text{f}_0\,\!$ determines whether the velocity in the valve is choked. Choked flow occurs when the downstream pressure drops below a certain critical value. Below this value the velocity of the flow through the valve is limited to the speed of sound. This means that the upstream side of the valve is not aware of the pressure on the downstream side; the information can only travel at the speed of sound up the flow, and the flow is at that speed, so at best the information remains stationary. At pressure ratios (across the valve) that are below the critical pressure ratio $\left(\tfrac{P_d}{P_u}\right)_{crit}\,\!$ , the velocity is independent of the pressure ratio.

A simple approximation for the valve opening time is used.

$C_d = \begin{cases} \displaystyle \frac{t}{t_{open}} C_{d0} & \displaystyle \text{if } t < t_{open} \\ \displaystyle C_{d0} & \displaystyle \text{if } t \geq t_{open} \end{cases}\,\!$

This makes the flow coefficient linearly increase until the valve is completely open.

More complicated equations would model the dynamics of the valve; this is not necessary for this relatively simple model.

Non-dimensionalization of equations

Defining some initial dimensionless parameters

$\rho^* = \frac{\rho}{\rho_{atm}}\,\!$

$T^* = \frac{T}{T_{atm}}\,\!$

$P^* = \frac{P}{P_{atm}}\,\!$

$x^* = \frac{x}{L_b}\,\!$

Ideal gas law

$P = \rho R T\,\!$

Substitute in $ρ$ , $T$ , $P$ ...

$P^* P_{atm} = \rho^* \rho_{atm} R T^* T_{atm}\,\!$

Note that $P_{atm} = \rho_{atm} R T_{atm}\,\!$ . Substitute this in for $P a t m$ .

$P^* (\cancel{\rho_{atm} R T_{atm}}) = \rho^* \cancel{\rho_{atm} R} T^* \cancel{T_{atm}}\,\!$

$P^* = \rho^* T^*\,\!$

Non-dimensional time

We need to develop a non-dimensional time that simplifies the other equations. A few different possibilities were tested. The one below works, and why will be demonstrated as it is derived from a variation of (1).

$\tau = \frac{A_b \sqrt{R T_{atm}}}{V_c} t\,\!$

$\frac{dt}{d \tau} = \frac{Vc}{A_b \sqrt{R T_{atm}}}\,\!$

$V_c \dot{\rho_c} = - \dot{m} \underbrace{sgn(P_c - P_b)}_{\begin{smallmatrix} \text{Assumption:} \\ \text{Always +1} \end{smallmatrix}}\,\!$

$\dot{m} = \frac{C_d A_b P_u}{\sqrt{R T_u}} f \left(\frac{P_d}{P_u}, k \right)\,\!$

Note that the u subscript refers to upstream conditions. Here upstream refers to the gas chamber.

$V_c \dot{\rho_c} = - \frac{C_d A_b P_c f}{\sqrt{R T_c}}\,\!$

$V_c \dot{\rho_c}^* \rho_{atm} = - \frac{C_d A_b P_c^* P_{atm} f}{\sqrt{R T_c^* T_{atm}}}\,\!$

$\dot{\rho_c}^* = - C_d f \frac{P_c^*}{\sqrt{T_c^*}} \frac{A_b}{V_c} \frac{P_{atm}}{\rho_{atm} \sqrt{R T_{atm}}}\,\!$

$\dot{\rho_c}^* = - C_d f \frac{P_c^*}{\sqrt{T_c^*}} \underbrace{\frac{A_b \sqrt{R T_{atm}}}{V_c}}_{\begin{smallmatrix} \text{Has dimensions} \\ \text{of time}^{-1} \end{smallmatrix}}\,\!$

Physically, this time is proportional to the time constant or rise time of the gas chamber when the flow is choked.

Valve mass flow rate (6)

A reference mass is defined as the total mass in the gas chamber before pressurization.

$m^* = \frac{m}{\rho_{atm} V_c}$

$\frac{d m^*}{d \tau} = \frac{d m}{dt} \frac{dt}{d \tau} \left(\frac{1}{\rho_{atm} V_c}\right) = \dot{m} \left(\frac{\cancel{V_c}}{A_b \sqrt{R T_{atm}}}\right) \left(\frac{1}{\rho_{atm} \cancel{V_c}}\right)\,\!$

$\frac{d m^*}{d \tau} = \frac{\dot{m}}{A_b \rho_{atm} \sqrt{R T_{atm}}}\,\!$

$\frac{d m^*}{d \tau} = \left(\frac{C_d \cancel{A_b} P_u^* P_{atm} \text{f}}{\sqrt{R T_u^* T_{atm}}}\right)\left(\frac{1}{\cancel{A_b} \rho_{atm} \sqrt{R T_{atm}}}\right) = \text{f} C_d \frac{P_u^*}{\sqrt{T_u^*}} \cancel{\frac{P_{atm}}{\rho_{atm} R T_{atm}}}\,\!$

$\frac{d m^*}{d \tau} = C_d \frac{P_u^*}{\sqrt{T_u^*}} \text{f} \left(\tfrac{P_u^*}{P_d^*}, k\right)\,\!$

The following will be useful elsewhere.

$\dot{m} = A_b \rho_{atm} \sqrt{R T_{atm}} \frac{d m^*}{d \tau}\,\!$