# 6.2: Momentum Equation Application

## Momentum Equation Applied to Propellers

The propeller is a mechanical devise that is used to increase the fluid momentum. Many times it is used for propulsion purposes of airplanes, ships and other devices (thrust) as shown in Figure 6.4. The propeller can be stationary like in cooling tours, fan etc. The other common used of propeller is mostly to move fluids as a pump. The propeller analysis of unsteady is complicated due to the difficulty in understanding the velocity field. For a steady state the analysis is simpler and used here to provide an example of steady state. In the Figure 6.4 the fluid flows from the left to the right. Either it is assumed that some of the fluid enters into the container and fluid outside is not affected by the propeller. Or there is a line (or surface) in which the fluid outside changes only the flow direction. This surface is called slip surface. Of course it is only approximation but is provided a crude tool. Improvements can be made to this analysis. Here, this analysis is used for academic purposes. Fig. 6.4 Propeller schematic to explain the change of momentum due to velocity.

As first approximation, the pressure around control volume is the same. Thus, pressure drops from the calculation. The one dimensional momentum equation is reduced

$\label{mom:eq:propellerGov} F = \rho \left({U_2}^2 - {U_1}^2\right)$

Combining the control volume between points 1 and 3 with (note that there are no external forces) with points 4 and 2 results in

$\label{mom:eq:propellerGovR} \rho \left({U_2}^2 - {U_1}^2\right) = P_4 - P_3$ This analysis provide way to calculate the work needed to move this propeller. Note that in this analysis it was assumed that the flow is horizontal that $$z_1 = z_2$$ and/or the change is insignificant.

## Jet Propulsion

Jet propulsion is a mechanism in which the air planes and other devices are propelled. Essentially, the air is sucked into engine and with addition heating (burning fuel) the velocity is increased. Further increase of the exit area with the increased of the burned gases further increase the thrust. The analysis of such device in complicated and there is a whole class dedicated for such topic in many universities. Here, a very limited discussion related to the steady state is offered. The difference between the jets propulsion and propellers is based on the energy supplied. The propellers are moved by a mechanical work which is converted to thrust. In Jet propulsion, the thermal energy is converted to thrust. Hence, this direct conversion can be, and is, in many case more efficient. Furthermore, as it will be shown in the Chapter on compressible flow it allows to achieve velocity above speed of sound, a major obstacle in the past. The inlet area and exit area are different for most jets and if the mass of the fuel is neglected then

$\label{mom:eq:jetProGov} F = \rho\, \left(A_2\, {U_2}^2 - A_1\,{U_1}^2 \right)$

An academic example to demonstrate how a steady state calculations are done for a moving control volume. Notice that

Example 6.4

A sled toy shown in Figure 6.5 is pushed by liquid jet. Calculate the friction force on the toy when the toy is at steady state with velocity, $$U_0$$. Assume that the jet is horizontal and the reflecting jet is vertical. The Fig. 6.5 Toy Sled pushed by the liquid jet in a steady state for example .

velocity of the jet is uniform. Neglect the friction between the liquid (jet) and the toy and between the air and toy. Calculate the absolute velocity of the jet exit. Assume that the friction between the toy and surface (ground) is relative to the vertical force. The dynamics friction is $$\mu_d$$.

Solution 6.4

The chosen control volume is attached to the toy and thus steady state is obtained. The frame of reference is moving with the toy velocity, $$\pmb{U}_0$$. The applicable mass conservation equation for steady state is
\begin{align*}
A_{1} U_{1} = A_{2} U_{2}
\end{align*}
The momentum equation in the $$x$$ direction is

$\label{slendingS:a} \pmb{F}_{f} + \int_{c.v.} \pmb{g} \,\rho\, dV - \int_{c.v.}\pmb{P}\,dA + \int_{c.v.} \boldsymbol{\tau}\,dA = \int_{c.v.} \rho\, \pmb{U} \pmb{U}_{rn} dV$

The relative velocity into the control volume is
\begin{align*}
\pmb{U}_{1j} = \left(U_j - U_0\right)\,\hat{x}
\end{align*}
The relative velocity out the control volume is
\begin{align*}
\pmb{U}_{2j} = \left(U_j - U_0\right)\,\hat{y}
\end{align*}
The absolute exit velocity is
\begin{align*}
\pmb{U}_{2} = U_0 \hat{x} + \left(U_j - U_0\right)\,\hat{y}
\end{align*}
For small volume, the gravity can be neglected also because this term is small compared to other terms, thus
\begin{align*}
\int_{c.v.} \pmb{g} \,\rho\, dV \sim 0
\end{align*}
The same can be said for air friction as
\begin{align*}
\int_{c.v.} \boldsymbol{\tau}\,dA \sim 0
\end{align*}
The pressure is uniform around the control volume and thus the integral is
\begin{align*}
\int_{c.v.}\pmb{P}\,dA = 0
\end{align*}
The control volume was chosen so that the pressure calculation is minimized. The momentum flux is

$\label{slendingS:eq:mom} \int_{S_{c.v.}} \rho\, U_x\, U_i{rn}\, dA = A\,\rho\, {U_{1j}}^2$ The substituting (29) into equation (??) yields

$\label{slendingS:c} F_f = A\,\rho\, {U_{1j}}^2$ The friction can be obtained from the momentum equation in the $$y$$ direction
\begin{align*}
m_{toy} \,g + A\,\rho\, {U_{1j}}^2 = F_{earth}
\end{align*}
According to the statement of question the friction force is
\begin{align*}
F_f = \mu_d \left( m_{toy} \,g + A\,\rho\, {U_{1j}}^2\right)
\end{align*}
The momentum in the $$x$$ direction becomes
\begin{align*}
\mu_d \left( m_{toy} \,g + A\,\rho\, {U_{1j}}^2\right) = A\,\rho\, {U_{1j}}^2 = A\,\rho\,
\left( U_j- U_0\right)^2
\end{align*}
The toy velocity is then
\begin{align*}
U_0 = U_j - \sqrt{ \dfrac{\mu_d \,m_{toy} \,g}{ A\,\rho\, \left( 1 -\mu_d \right) } }
\end{align*}
Increase of the friction reduce the velocity. Additionally larger toy mass decrease the velocity.