# 6.4: More Examples on Momentum Conservation

- Page ID
- 721

Example 6.7

A design of a rocket is based on the idea that density increase of the leaving jet increases the acceleration of the rocket see Figure

*Fig. 6.10 Nozzle schematics water rocket for the discussion on the forces for example.*

Assume that this idea has a good engineering logic. Liquid fills the lower part of the rocket tank. The upper part of the rocket tank is filled with compressed gas. Select the control volume in such a way that provides the ability to find the rocket acceleration. What is the instantaneous velocity of the rocket at time zero? Develop the expression for the pressure (assuming no friction with the walls). Develop expression for rocket velocity. Assume that the gas is obeying the perfect gas model. What are the parameters that effect the problem.}

Solution 6.7

Under construction for time being only hintsootnotemark

ootnotetext{This problem appeared in the previous version (0.2.3) without a solution. Several people ask to provide a solution or some hints for the solution. The following is not the solution but rather the approach how to treat this problem.} In the solution of this problem several assumptions must be made so that the integral system can be employed.

The surface remained straight at the times and no liquid residue remains behind.

The gas obeys the ideal gas law.

The process is isothermal (can be isentropic process).

No gas leaves the rocket.

The mixing between the liquid and gas is negligible.

The gas mass is negligible in comparison to the liquid mass and/or the rocket.

No resistance to the rocket (can be added).

The cross section of the liquid is constant.

In this problem the energy source is the pressure of the gas which propels the rocket. Once the gas pressure reduced to be equal or below the outside pressure the rocket have no power for propulsion. Additionally, the initial take off is requires a larger pressure. The mass conservation is similar to the rocket hence it is

\[

\label{waterRocket:mass}

\dfrac{dm}{dt} = - U_{e}\,A_{e} \tag{73}

\]

The mass conservation on the gas zone is a byproduct of the mass conservation of the liquid. Furthermore, it can be observed that the gas pressure is a direct function of the mass flow out. The gas pressure at the initial point is

\[

\label{waterRocket:iniP}

P_0 = \rho_0 \, R\, T \tag{74}

\]

Per the assumption the gas mass remain constant and is denoted as \(m_g\). Using the above definition, equation (74) becomes

\[

\label{waterRocket:iniPm}

P_0 = \dfrac{m_g\,R\,T}{V_{0g}} \tag{75}

\]

The relationship between the gas volume

\[

\label{waterRocket:h_lV}

V_g = \overline{h_g}\, A \tag{76}

\]

The gas geometry is replaced by a virtual constant cross section which cross section of the liquid (probably the same as the base of the gas phase). The change of the gas volume is

\]

\label{waterRocket:DgasH}

\dfrac{dV_g}{dt} = {A}\, \dfrac{dh_g}{dt} = - {A}\,\dfrac{dh_{ll}}{dt} \tag{77}

\]

The last identify in the above equation is based on the idea what ever height concede by the liquid is taken by the gas. The minus sign is to account for change of "direction'' of the liquid height. The total change of the gas volume can be obtained by integration as

\[

\label{waterRocket:gasH}

V_g = A \, \left(h_{g0} - \Delta h_{ll} \right) \tag{78}

\]

It must be point out that integral is not function of time since the height as function of time is known at this stage. The initial pressure now can be expressed as

\[

\label{waterRocket:gP1}

P_0 = \dfrac{m_g\,R\,T}{h_{g0}\,A} \tag{79}

\]

The pressure at any time is

\[

\label{waterRocket:gP}

P = \dfrac{m_g\,R\,T}{h_{g}\,A} \tag{80}

\]

Thus the pressure ratio is

\[

\label{waterRocket:PP0}

\dfrac{P}{P_0} = \dfrac{h_{g0} } {h_g} = \dfrac{ h_{g0} }{ h_{g0} - \Delta h_{ll} }

= h_{g0} \dfrac {1} { 1 - \dfrac{\Delta h_{ll} } {h_{g0} } } \tag{81}

\]

Equation (73) can be written as

\[

\label{waterRocket:massG}

m_{ll} (t) = m_{ll 0} - \int_0 ^t U_e\, A_e dt \tag{82}

\]

From equation (73) it also can be written that

\[

\label{waterRocket:massH}

\dfrac{d h_{ll}} { dt} = \dfrac{U_e\,A_e}{\rho_e\,A} \tag{83}

\]

According to the assumption the flow out is linear function of the pressure inside thus,

\[

\label{waterRocket:P-u}

U_e = f(P) + g\,h_{ll}\,rho \backsimeq f(P) = \zeta \, P \tag{84}

\]

Where \(\zeta\) here is a constant which the right units. The liquid momentum balance is

\[

\label{waterRocket:momentum}

- g \left( m_R+ m_{ll} \right) - a \left( m_R+ m_{ll} \right) =

\overbrace{\dfrac{d}{dt} \left( m_R+ m_{ll} \right) U}^{=0}

+ bc + \left( U_R + U_{ll} \right) m_{ll} \tag{85}

\]

Where \(bc\) is the change of the liquid mass due the boundary movement.

Example 6.8

A rocket is filled with only compressed gas. At a specific moment the valve is opened and the rocket is allowed to fly. What is the minimum pressure which make the rocket fly. What are the parameters that effect the rocket velocity. Develop an expression for the rocket velocity.

Example 6.9

In Example 6.5 it was mentioned that there are only two velocity components. What was the assumption that the third velocity component was neglected.

### Contributors

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.