### The water rocket: Thrust from water

Once the rocket has left the launch tube, the water thrust phase of the flight begins. But before we begin those calculations, we need to have equations for ballistic flight, which is the *last* step described in the introduction. Why? Because during ballistic flight (coasting through the air), the only forces acting on the rocket are gravity and wind drag — but these are acting on the rocket during the thrust phase too. Ballistic flight is exactly the same, just without the thrust.

### Ballistic flight

We need to know how gravity and drag affect acceleration, velocity, and ultimately altitude during all thrust phases as well as the ballistic trajectory afterward, so we'll start with the ballistic flight equations.

#### Air resistance

First we need to get the ambient air density. Denser air results in higher drag, and dry air is more dense than moist air. To get the density, we first need to know the partial pressure of water vapor in the air. To do this, we first get the saturation vapor pressure of water from the Arden Buck equation: $$P_\text{sat} = 0.61121 \exp \left[\left( 18.678 - \frac{T+273.15} {234.5}\right)\left( \frac{T+273.15} {257.14 + T+273.15} \right)\right] \tag{B0}$$ where \(T\) is the temperature in Kelvin (you'll notice the value 273.15 added to each \(T\) above because the original Arden Buck equation uses Celsius).

The air density is given by $$\rho_a = \frac{P_\text{barometer} - \phi P_\text{sat}}{R_\text{dry} T} + \frac{\phi P_\text{sat}}{R_v T} \tag{B1}$$ where- \(P_\text{barometer}\) = Current barometric pressure (Pa)
- \(R_\text{dry}\) = Specific gas constant for dry air, 287.058 J/(kg·K)
- \(T\) = Temperature in Kelvin
- \(\phi\) = relative humidity (0.0 to 1.0)
- \(R_v\) = Specific gas constant for water vapor, 461.495 J/(kg·K)
- \(T\) = ambient air temperature (K)

The air resistance, or drag, is given by: $$F_D = \frac{1}{2}\rho_a v_r^2 C_D A_f \tag{B2}$$ where

- \(F_D\) = drag force against motion due to air resistance (N)
- \(\rho_a\) = the density of the air (kg/m
^{3}) from (B1) above - \(v_r\) = velocity of rocket (m/s)
- \(C_D\) = coefficient of drag, 0.295 for a bullet-shaped object
- \(A_f\) = frontal area of the rocket (m
^{2}) calculated from largest radius. The radius of a standard US 2-liter bottle is about 0.54 m (108 mm diameter) depending on the pressure.

It's reasonable to conclude that a water rocket might travel slightly higher on hot, humid days when the air density is lowest. And indeed it does! The altitude achieved by a 2 liter bottle rocket can be different by as much as 15 meters on a cool dry day versus a warm humid day. Atmospheric conditions have a *significant* impact on apogee altitude. But if you don't want to calculate this, you can use the standard air density value of 1.225 kg/m^{3}.

We are ignoring the effects of altitude on temperature, because this change is insignificant for the altitudes typically achieved by soda bottle rockets.

#### Acceleration of the rocket

During ballistic flight, the rocket will slow down as it coasts upward to apogee, then speed up as it falls downward until it reaches a terminal velocity determined by the air drag. We use the convention that upward force is positive and downward force is negative. The drag changes sign depending on the direction of the rocket; the drag is a negative force when the rocket is ascending and a positive force when descending. The sign is the same as the sign of velocity. $$F=F_T - m g - \text{sign}(v_r)F_D \tag{B3}$$ where

- \(F_T\) = force of thrust = \(F_w\) or \(F_a\) depending if the thrust is from water or air, respectively
- \(m\) = total mass of the rocket at the time of the calculation
- \(g\) = acceleration due to gravity (9.81 m/s
^{2})

The rocket's acceleration is given by Newton’s Second Law \(F = ma\), that is: $$a=\frac{F}{m}\tag{B4}$$

This is an instantaneous value of acceleration for a snapshot of mass and force. The mass of the rocket changes rapidly as water is ejected, the thrust decreases as internal pressure decreases, and drag increases with the square of velocity. All of these changes are nonlinear. We can simulate the flight of the rocket by breaking it up into such small time intervals that all values change approximately linearly within each interval. Then we calculate values for pressure, mass, force, acceleration, velocity, and distance for the next time increment. An increment of 0.5 or 1 milliseconds is sufficient to approximate the thrust phase of the flight. After all the water and air is exhausted and no more thrust is being produced, we can use a time increment much larger, like 20 to 50 milliseconds.

The next velocity \(v_\text{next}\) and altitude \(d_\text{next}\) at each iteration over time increment \(\Delta t\) are calculated from the previous values: $$v_\text{next}= v_\text{prev} + a_\text{prev} \Delta t \tag{B5}$$ $$d_\text{next}= d_\text{prev} + v_\text{prev}t + \frac{1}{2}a_\text{prev}(\Delta t)^2$$

### Nozzle velocity of water

Finally we can work out the water thrust equations. We also use the calculations here to get the water loss from leakage around the launch tube, giving us the initial water volume and pressure available for the water thrust phase.

From Bernoulli's principle: \(\frac{v^2}{2}+GH+\frac{p}{\rho} = \text{constant} \)

Solving for velocity: $$v = C_v\sqrt{2\left(GH+\frac{p}{\rho}\right)}\tag{W1}$$ where

- \(v\) = velocity of the watter from the nozzle
- \(C_v\) = the velocity coefficient (0.97 for water)
- \(G\) = the total acceleration \(g+a\), where
- \(g = 9.81\) m/s
^{2}due to gravity - \(a\) = the acceleration of the rocket (m/s
^{2})

- \(g = 9.81\) m/s
- \(p\) = the excess pressure in the tank above ambient pressure (Pa)
- \(\rho\) = the fluid density (1000 kg/m
^{3}for water)

**Important note:** The Bernoulli equation is a steady-state equation, but we aren't going to be using it this way. The values of \(G\) and \(p\) are instantaneous values at a snapshot in time. They change rapidly with time. We will calculate the velocity of water from the nozzle over extremely short time intervals during which we assume a steady state, and from there we get thrust and acceleration at the beginning of that time interval. Then at the end of the time interval, we calculate a new \(G\) based on the acceleration, and a new \(p\) based on the adiabatic expansion of air in the bottle.

### Volume flow of water from pressurized container

The value of \(C_v\) above comes from an Engineering Toolbox page about mass flow from a pressurized container. From the formula found there, we can combine the terms to get the mass rate of water flowing out of the nozzle under pressure: $$m' = C_c \rho A v \tag{W2}$$ where, in addition to the values defined above:

- \(m'\) = mass flow rate (kg/s)
- \(C_c\) = contraction coefficient (0.62 for a sharp edge, 0.97 for a rounded edge; some soda bottles have a nice rounded transition to the neck, some have more of a conical transition, so we can use 0.9 as a conservative value)
- \(A\) = aperture area, in this case the area of the bottle neck opening (m
^{2})

During the launch tube phase, the nozzle aperture area \(A\) is blocked by the cylinder cross section of area \(A_\text{tube}\), so we would use the difference \(A-A_\text{tube}\) in (W2) to calculate the mass flow of the water leak. The mass of the water lost \(\Delta m\) while traversing the tube is simply
$$\Delta m = m't_\text{tube}$$
and we subtract this from the initial mass of water to get the remaining mass of water available for thrust. We convert it to volume easily, since water has a density of 1000 kg/m^{3}.

Once the nozzle is no longer blocked by the launch tube, we use the full area of the nozzle opening for a 21.75 mm diameter nozzle, converted to 0.00037154 m^{2} of area.

### Thrust from a water jet

Again from the Engineering Toolbox, the jet of watter from the nozzle imparts a force \(F_w\) equal to the mass flow rate times velocity: $$F_w = m'v \tag{W3}$$

### Pressure change as a function of water flow

To calculate \(F_w\) at subsequent time steps, we must first determine the air pressure remaining after each time step. We *could* use the relationship from the ideal gas law: $$\text{pressure}\times\text{volume}=\text{constant}$$
but this assumes that pressure changes inversely with volume as an isothermic process (i.e. temperature remains constant). That's a reasonable assumption if volume changes slowly. However, when air expands rapidly, as it does in a water rocket, it experiences a dramatic reduction in temperature. This temperature reduction results in a lower pressure than if the temperature were constant throughout the expansion. Therefore, the isothermic ideal gas law results in an over-optimistic estimate of thrust.

If a tank of volume \(V\) filled with water of initial volume \(W_0\) is emptied quickly to water volume \(W_1\) by the tank's internal pressure, then the pressure resulting from the change in volume is: $$P_1=P_0 \left(\frac{V-W_0}{V-W_1}\right)^\gamma \tag{W4}$$ where, in addition to the values defined in the launch tube formulas:

- \(W_0\) = volume of water in the bottle at the beginning of the time interval
- \(W_1\) = volume of water at the end of the time interval

The instant when all of the water is gone from the bottle, \(W_1 = 0\) and the remaining internal tank pressure will provide additional short burst of thrust as the remaining air escapes. This remaining empty-tank pressure is given by a formula for adiabatic gas expansion: $$P_e = P_0 \left(\frac{V-W_0}{V}\right)^\gamma \tag{W5}$$

The temperature of the air inside, if there were no water vapor, would be $$T_e = T_0\left(\frac{P_e}{P_0}\right)^\frac{\gamma-1}{\gamma} \tag{W6}$$ where, in addition to the definitions above:

- \(T_e\) = temperature of air in the tank (in degrees Kelvin) at the time the tank becomes empty of water

This gives an unrealistic result for the temperature, however. A 2-liter bottle filled one-third full of water and pressurized to 100 psi (689476 Pa) at 20°C will end up at −44°C according to this calculation. This may be true for an ideal dry gas in a reversible process, but we have humid air to modulate the temperature, and the process isn't reversible. With such a low calculated ending temperature, the calculated final pressure in (W5) will also be too low. Calculating the heat contributed by cooling (or freezing) water vapor is possible but needlessly complicated. We can, however, fudge the constant \(\gamma\) to account for effects of water vapor. For the typical conditions inside a bottle rocket, a value \(\gamma = 1.34\) accounts for a few degrees of heat imparted to the air by water condensation, resulting in a final temperature still quite cold, but not as much.

Fudging \(\gamma\) in this way is an example of a polytropic process, somewhere in between an isothermal and adiabatic process, in which we use an exponent between 1.0 (isothermal) and 1.4 (isentropic or adiabatic).

### References

- ↵ Shape Effects on Drag, NASA Glenn Research Center. There’s a bit of a contradiction here. The page says a bullet-shape has a drag coefficient of 0.295, but a rocket, which should be a more streamlined shape but still with a blunt tail end, has a higher drag coefficient of 0.75.
- ↵ Isentropic Compression (or expansion), NASA Glenn Research Center.
- ↵ Physics Forum discussion I had with another scientist to reach a conclusion about the value of \(\gamma\) to account for condensation of humidity.

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