The water rocket: Thrust from air

After the water is expended, the tank (bottle) still contains substantial pressure that releases quickly through the nozzle. This final burst of air can impart a significant boost to velocity (at least 30% depending on the mass of the rocket), so we shouldn’t ignore this contribution to thrust. For convenience, however, we will ignore any further affects on air temperature due to water vapor although we will still calculate pressure changes adiabatically. Any water vapor will likely have condensed by the time airflow begins.

Thrust from choked airflow

When the ratio of ambient pressure to total absolute tank pressure is less than the "choke ratio" $$\alpha_c = \frac{P_a}{P_e} = \left(\frac{2}{\lambda+1}\right)^\frac{\lambda}{\lambda-1} \tag{A1}$$ then the outflow is choked, or limited, to the speed of sound: $$c = \sqrt{\lambda RT} \tag{A2}$$ where

  • \(c\) = speed of sound in air, approximately 343 m/s at 20°C, 331 m/s at 0°C
  • \(R\) = ideal gas constant divided by gas molecular weight (8.3145 / 0.02896 for air)
  • \(T\) = constant gas temperature in Kelvin (e.g. 0°C = 273 K)

The choke ratio \(\alpha_c = 0.455\) for moist air, using \(\lambda=1.34\); the traditional value is 0.528 using \(\lambda=1.4\) for dry air.

We can use a value of \(T\) from equation (W6), resulting from the expansion of the air that ejected all of the water. The presence of water vapor would have a moderating effect on the final temperature and pressure, so we use the modified value of \(\lambda\) (1.34 instead of 1.4) to calculate the speed of sound. The range of possible values of \(c\) turns out to make just a small difference to apogee altitude, however, typically less than a meter.

To figure out the mass flow rate of air during choked flow, we first need to find the density of compressed humid air in the tank. To do this, we need to know the saturation vapor pressure \(P_\text{sat}\). The Arden Buck equation provides a good approximation: $$P_\text{sat} = 0.61121 \exp \left[\left( 18.678 - \frac{T} {234.5}\right)\left( \frac{T} {257.14 + T} \right)\right]\tag{A3}$$ where in this case \(T\) is expressed in °C rather than Kelvin.

We assume that the air in the bottle is saturated to 100% relative humidity. The density of humid air is given by $$\rho_\text{ humid~air} = \frac{P_\text{d}}{R_\text{d} T} + \frac{P_\text{v}}{R_\text{v} T} \tag{A4}$$ where \(\rho_\text{ humid~air}\) has units of kg/m3, and all of the terms on the right are known:

  • \(P_\text{d} = P_0 - P_\text{v}\) = partial pressure of dry air, or the difference between normal dry air pressure and water vapor pressure (Pa)
  • \(R_\text{d} = 287.058\) = specific gas constant for dry air in J/(kg·K)
  • \(T\) = air temperature (K)
  • \(P_\text{v} = P_\text{sat}\) = pressure of water vapor (Pa)
  • \(R_{v} = 461.495\) = specific gas constant for water vapor in J/(kg·K)

From this, we can calculate the total initial mass of air and water vapor available for thrust: $$m_{a0} = \frac{P_e}{P_a}\rho_\text{ humid~air}V\tag{A5}$$ where \(V\) is the total volume of the bottle, and \(P_e\) from equation (W5), the absolute pressure of the air in the tank emptied of water, will already be adjusted for adiabatic cooling during the expansion that provided the water thrust.

The density of the air in the bottle is simply the mass/volume ratio: $$\rho_0 = \frac{m_a}{V}\tag{A6}$$ where \(m_a = m_{a0}\) initially. This is the mass of air remaining in the bottle at each time step.

The mass flow rate for choked flow is: $$m'=C_dA\sqrt{s \rho_0 P_e} \tag{A7}$$ where in addition to the previous definitions, \(s = \text{constant} = \lambda \left( \frac{2}{\lambda+1} \right)^\frac{\lambda+1}{\lambda-1} = 0.4548\) for \(\lambda = 1.34.\)

The choked airflow thrust is then the same formula as (W3) given previously for water, the mass flow rate multiplied by the velocity: $$F_a = m'c \tag{A8}$$ where the speed of sound \(c\) from (A2) is a function of temperature that changes with each time increment.

Thrust from non-choked airflow

As choked airflow thrust is calculated for new time increments, the internal pressure will decrease until the choke condition is no longer met. Then we can use similar formulas as (W1), (W2), and (W3) for compressible fluid flow: $$v=C_v \sqrt{2\frac{\lambda}{\lambda-1}\frac{p_e}{\rho_0}} \tag{A9}$$ $$m'=C_c \rho_0 A v \tag{A10}$$ $$F_a=m'v \tag{A11}$$

Remember, here \(p_e\) is the excess (measured) pressure, not absolute pressure \(P_e\). Also we'll use the conservative value \(C_c = 0.9\) as we did in (W2).

The value \(C_v\) for air can be found by setting \(v=c\) and using (A1) to set \(p_e=P_a/\alpha_c − P_a\), then solving for \(C_v\) in (A9). This value yields the speed of sound when the internal pressure is right at the choke threshold. \(C_v\) will fall in a range between 0.45 and 0.6 depending on the initial launch conditions of temperature, pressure, and fill volume. \(C_v\) needs to be recalculated at each time step during choked flow to determine when non-choked flow begins.

Final pressure at the end of the time interval

We need to calculate the pressure at the end of the time interval, which would be the pressure to use for calculations at the beginning of the next time interval. To do this, we need to get the new air density for the next interval \(i\): $$\rho_{a~i} = \frac{m_a - m'\Delta t}{V} \tag{A12}$$ then from this relation $$\frac{P_1}{\rho_1^\gamma}=\frac{P_2}{\rho_2^\gamma}=\text{constant}$$ we get the new pressure for the next time step: $$P_{e~i} = P_e \left(\frac{\rho_{a~i}}{\rho_a}\right)^\gamma \tag{A13}$$ where the subscript \(i\) denotes the start of the next time interval, and values without the subscript \(i\) are for the current time step.

To finish up initializing the next time step:

  • Use (A13) to update \(C_v\) in (A9),
  • Use (W6) to calculate the temperature in the bottle from the new pressure,
  • Get a new speed of sound value in (A2) using this new temperature.

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