# 6.2.2: Momentum Application to Unsteady State

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## Rocket Mechanics

A rocket is a devise similar to jet propulsion. The difference is the fact that the oxidant is on board with the fuel. The two components are burned and the gases are ejected through a nozzle. This mechanism is useful for specific locations because it is independent of the medium though which it travels. In contrast to other mechanisms such as jet propulsion which obtain the oxygen from the medium which they travel the rockets carry the oxygen with it. The rocket is accelerating and thus the frame for reference is moving the with the rocket. The velocity of the rocket in the rocket frame of reference is zero. However, the derivative with respect to time, $$dU/dteq0$$ is not zero. The resistance of the medium is Denote as $$F_R$$. The momentum equation is

$\overbrace{\int_{c.v.} \boldsymbol{\tau} dA}^{F_R} + \int_{c.v.} \pmb{g}\,\rho\,dV + \overbrace{\int_{c.v.} \pmb{P} dA}^{0} - \int \rho\,a_0 \,dV =\\ \dfrac{d}{dt} \int_{V_{c.v.}} \rho U_y dV + \int_{c.v.} \rho U_y \, U_{rn} dA \label{mom:eq:govRocket}$

There are no external forces in this control volume thus, the first term $$F_R$$, vanishes. The pressure term vanish because the pressure essentially is the same and the difference can be neglected. The gravity term is an instantaneous mass times the gravity times the constant and the same can be said for the acceleration term. Yet, the acceleration is the derivative of the velocity and thus

$\label{mom:eq:rocketAcc} \int \rho\,a_0 \,dV = \dfrac{dU}{dt}\,\left( m_R + m_f \right)$ The first term on the right hand side is the change of the momentum in the rocket volume. This change is due to the change in the volume of the oxidant and the fuel.

$\label{mom:eq:zeroChangeRocket} \dfrac{d}{dt} \int_{V_{c.v.}} \rho U_y dV = \dfrac{d}{dt} \left[\left( m_R + m_f \right)\, U\right]$ Clearly, the change of the rocket mass can be considered minimal or even neglected. The oxidant and fuel flow outside. However, inside the rocket the change in the velocity is due to change in the reduction of the volume of the oxidant and fuel. This change is minimal and for this analysis, it can be neglected. The last term is

$\label{mom:eq:lastTerm} \int_{c.v.} \rho U_y \, U_{rn} dA = \dot{m} \, \left( U_g - U_R \right)$ Combining all the above term results in

$\label{mom:eq:resGovRoc} - F_R - \left( m_R + m_f\right)\,g + \dfrac{dU}{dt} \left( m_R + m_f\right) = \dot{m}\, \left( U_g - U_R \right)$
Denoting $$\mathcal{M}_T = m_R + m_f$$ and thus $$d\mathfrak{M}/dt = \dot{m}$$ and $$U_e = U_g - U_R$$. As first approximation, for constant fuel consumption (and almost oxidant), gas flow out is constant as well. Thus, for constant constant gas consumption equation (35) transformed to

$\label{mom:eq:diffRocket} - F_R - \mathcal{M}_T\, g + \dfrac{dU}{dt} \, \mathcal{M}_T = \dot{\mathcal{M}_T}\, U_e$ Separating the variables equation (36) yields

$\label{mom:eq:kineticN} dU = \left( \dfrac{- \dot{\mathcal{M}_T}\, U_e}{\mathcal{M}_T} - \dfrac{F_R}{\mathcal{M}_T} - g \right) dt$ Before integrating equation (37), it can be noticed that the friction resistance $$F_R$$, is a function of the several parameters such the duration, the speed (the Reynolds number), material that surface made and the medium it flow in altitude. For simplicity here the part close to Earth (to the atmosphere) is assumed to be small compared to the distance in space. Thus it is assume that $$F_R = 0$$. Integrating equation (37) with limits of $$U(t=0) =0$$ provides

$\label{mom:eq:solutionR} \int_0^U dU = - \dot{\mathcal{M}_T}\, U_e \int_0^t \dfrac{dt}{\mathcal{M}_T} - \int_0^t g\, dt$ the results of the integration is (notice $$\mathcal{M} = \mathcal{M}_0 - t \, \dot{\mathcal{M}}$$)

$\label{mom:eq:solutionRf} U = U_e \ln \left( \dfrac{\mathcal{M}_0}{\mathcal{M}_0 - t \, \dot{\mathcal{M}} } \right) -g\,t$ The following is an elaborated example which deals with an unsteady two dimensional problem. This problem demonstrates the used of control volume to find method of approximation for not given velocity profiles

Example 6.5 Fig. 6.7 Schematic of a tank seating on wheel for unsteady state discussion

A tank with wheels is filled with liquid is depicted in Figure . The tank upper part is opened to the atmosphere. At initial time the valve on the tank is opened and the liquid flows out with an uniform velocity profile. The tank mass with the wheels (the solid parts) is known, $$m_t$$. Calculate the tank velocity for two cases. One the wheels have a constant resistance with the ground and two the resistance linear function of the weight. Assume that the exit velocity is a linear function of the height.

Solution 6.5

This problem is similar to the rocket mechanics with a twist, the source of the propulsion is the potential energy. Furthermore, the fluid has two velocity components verse one component in the rocket mechanics. The control volume is shown in Figure 6.7. The frame of reference is moving with the tank. This situation is unsteady state thus equation (12) for two dimensions is used. The mass conservation equation is

$\label{tankWheel:a} \dfrac{d}{dt} \int_{V_{c.v.}} \rho\, dV + \int_{S_{c.v.}} \rho dA = 0$

Equation (40) can be transferred to

$\label{tankWheel:b} \dfrac{dm_{c.v.}}{dt} = - \rho\,U_0\,A_0 = - m_0$ Where $$m_0$$ is mass flow rate out. Equation (41) can be further reduced due to constant density to

$\label{tankWheel:massE} \dfrac{d\,\left( A\,h \right) }{dt} + U_0\, A_0 = 0$ It can be noticed that the area of the tank is almost constant ($$A=constant$$) thus

$\label{tankWheel:mass} A\dfrac{dh}{dt} + U_0\, A_0 = 0 \Longrightarrow \dfrac{dh}{dt} = - \dfrac{U_0\, A_0}{A}$

The relationship between the height and the flow now can be used.

$\label{tankWheel:hU} U_0 = \mathcal{B} \, h$

Where $$\mathcal{B}$$ is the coefficient that has the right units to mach equation (44) that represent the resistance in the system and substitute the energy equation. Substituting equation (44) into equation (??) results in

$\label{tankWheel:dff_h} \dfrac{dh}{dt} + \dfrac{\mathcal{B} \, h \, A_0}{A} =0$

Equation (45) is a first order differential equation which can be solved with the initial condition $$h(t=0) =h_0$$. The solution (see for details in the Appendix ) is

$\label{tankWheel:soDff} h(t) = h_0\, \text{e}^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}$ Fig. 6.8 A new control volume to find the velocity in discharge tank for example.

To find the average velocity in the $$x$$ direction a new control volume is used. The boundary of this control volume are the tank boundary on the left with the straight surface as depicted in Figure 6.8. The last boundary is variable surface in a distance x from the tank left part. The tank depth, is not relevant. The mass conservation for this control volume is

$\label{tankWheel:xCV} \cancel{w}\, x \dfrac{dh}{dt} = - \cancel{w}\, h \, \overline{U_x}$

Where here $$w$$ is the depth or width of the tank. Substituting (46) into (??) results

$\label{tankWheel:xCVs} \overline{U_x}(x) = \dfrac{x\,A_0\,\cancel{h_0}\mathcal{B} }{A\,\cancel{h}} \cancel{\text{e}^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}} = \dfrac{x\,A_0\mathcal{B} }{A}$ The average $$x$$ component of the velocity is a linear function of $$x$$. Perhaps surprising, it also can be noticed that $$\overline{U_x}(x)$$ is a not function of the time. Using this function, the average velocity in the tank is

$\label{tankWheel:Uyaverge} \overline{U_y} = \dfrac{dh}{dt} = - \dfrac{h_0\,A_0\,\mathcal{B}}{A} \text{e}^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}$ It can be noticed that the velocity in the $$y$$ is a function of time as oppose to the $$x$$ direction. The applicable momentum equation (in the tank frame of reference) is (11) which is reduced to

$\label{tankWheel:governing} -\pmb{F}_R - \left( m_t+ m_f\right) \pmb{g} - \overbrace{\pmb{a} \left( m_t+ m_f\right)}^{acceleration} = \dfrac{d}{dt} \left[ \left( m_t+ m_f\right)\, \pmb{U}_r \right] + U_0 \, m_o$ Where $$\pmb{U}_r$$ is the relative fluid velocity to the tank (if there was no tank movement). $$m_f$$ and $$m_t$$ are the mass of the fluid and the mass of tank respectively. The acceleration of the tank is $$\pmb{a} = -\hat{i} a_0$$ or $$\hat{i}\cdot\pmb{a} = -a$$. And the additional force for accelerated system is

\begin{align*}
- \hat{i} \cdot \int_{V_{c.v.}} \pmb{a} \rho dV = m_{c.v.} \, a
\end{align*}
The mass in the control volume include the mass of the liquid with mass of the solid part (including the wheels).
\begin{align*}
m_{c.v.} = m_f + m_{T}
\end{align*}
because the density of the air is very small the change of the air mass is very small as well ($$\rho_a << \rho$$). The pressure around the control volume is uniform thus
\begin{align*}
\int_{S_{c.v.}} P \cos\theta_x dA \sim 0
\end{align*}
and the resistance due to air is negligible, hence
\begin{align*}
\int_{S_{c.v.}} \boldsymbol{\tau} dA \sim 0
\end{align*}
The momentum flow rate out of the tank is

$\label{tankWheel:rho} \int_{S_{c.v.}} \rho \, U_x \, U_{rn} dA = \rho \, {U_{o}} ^2 A_{o} = m_o\,U_{o}$

In the $$x$$ coordinate the momentum equation is

$\label{tankWheel:governingX} -{F}_x + \left( m_t+ m_f\right) a = \dfrac{d}{dt} \left[ \left( m_t+ m_f\right)\, {U}_x \right] + U_0 \, \dot{m}_f$ Where $${F}_x$$ is the $$x$$ component of the reaction which is opposite to the movement direction. The momentum equation in the $$y$$ coordinate it is

$\label{tankWheel:governingY} {F}_y - \left( m_t+ m_f\right) g = \dfrac{d}{dt} \left[\strut \left( m_t+ m_f\right)\, U_y \right]$ There is no mass flow in the $$y$$ direction and $${U}_y$$ is component of the velocity in the $$y$$ direction. The tank movement cause movement of the air which cause momentum change. This momentum is function of the tank volume times the air density times tank velocity ($$h_0\times A\times \rho_{a}\times U$$). This effect is known as the add mass/momentum and will be discussed in the Dimensional Analysis and Ideal Flow Chapters. Here this effect is neglected. The main problem of integral analysis approach is that it does not provide a way to analysis the time derivative since the velocity profile is not given inside the control volume. This limitation can be partially overcome by assuming some kind of average. It can be noticed that the velocity in the tank has two components.
The first component is downward ($$y$$) direction and the second in the exit direction ($$x$$). The velocity in the $$y$$ direction does not contribute to the momentum in the $$x$$ direction. The average velocity in the tank (because constant density and more about it later section) is
\begin{align*}
\overline{U_x} = \dfrac{1}{V_t} \int_{V_f} U_x dV
\end{align*}
Because the integral is replaced by the average it is transferred to
\begin{align*}
\int_{V_f} \rho\, U_x dV \sim m_{c.v.} \, \overline{U_x}
\end{align*}
Thus, if the difference between the actual and averaged momentum is neglected then

$\label{tankWheel:dmomdt} \dfrac{d}{dt} \int_{V_f} \rho\, U_x \, dV \sim \dfrac{d}{dt} \left( m_{c.v.} \, \overline{U_x} \right) = \dfrac{d \, m_{c.v.}}{dt} \overline{U_x} + \overbrace{\dfrac{d \, \overline{U_x}}{dt}}^{\sim 0} m_{c.v.}$ Noticing that the derivative with time of control volume mass is the flow out in equation (??) becomes

$\label{mom:eq:d} \dfrac{d \, m_{c.v.}}{dt} \overline{U_x} + \dfrac{d \, \overline{U_x}}{dt} m_{c.v.} = - \overbrace{\dot{m}_0}^{\text{mass rate out}}\,\overline{U_x} = - m_0 \, \dfrac{L\, A_0 \, \mathcal{B} }{2\,A}$ Combining all the terms results in

$\label{tankWheel:combined} -{F}_x + a\, \left( m_f + m_t \right) = - m_0 \, \dfrac{L\, A_0 \, \mathcal{B} }{2\,A} - U_0 \, {m}_0$ Rearranging and noticing that $$a = dU_T/dt$$ transformed equation (56) into

$\label{tankWheel:combinedR} a = \dfrac{ {F}_x}{ m_f + m_t } - m_0 \left( \dfrac{L\, A_0 \, \mathcal{B} + 2\,A\,{U}_0 \left( m_f + m_t \right) } {2\,A \left( m_f + m_t \right) \dfrac{}{} } \right)$ If the $${F}_x \geq m_0 \left( \dfrac{L\, A_0 \, \mathcal{B} }{2\,A\dfrac{}{}} + U_0 \right)$$ the toy will not move. However, if it is the opposite the toy start to move. From equation (44) the mass flow out is

$\label{tankWheel:m0} m_0(t) = \overbrace{\mathcal{B}\,\overbrace{h_0\, e^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}}^{h}} ^{U_0} A_0 \, \rho$ The mass in the control volume is

$\label{tankWheel:DmcvDt} m_f = \rho \, \overbrace{A \, h_0\, e^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}}^{V}$ The initial condition is that $$U_T(t=0) = 0$$. Substituting equations (58) and (??) into equation (??) transforms it to a differential equation which is integrated if $$R_x$$ is constant. For the second case where $$R_x$$ is a function of the $$R_y$$

$\label{tankWheel:RxRy} R_x = \mu\,R_y$ The $$y$$ component of the average velocity is function of the time. The change in the accumulative momentum is

$\label{tankWheel:dMOM} \dfrac{d}{dt} \left[\strut \left( m_f\right)\, \overline{U_y} \right] = \dfrac{d m_{f}}{dt} \overline{U_y} + \dfrac{d \overline{U_y} }{dt} m_{f}$ The reason that $$m_f$$ is used because the solid parts do not have velocity in the $$y$$ direction. Rearranging the momentum equation in the $$y$$ direction transformed

$\label{tankWheel:governingYR} {F}_y = \left( m_t+ \overbrace{\rho\,A\,h_0 \,e^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}}^{m_f} \right)\,g + 2 \left( \dfrac{\rho\,h_0 {A_0\dfrac{}{}}^2\, \mathcal{B}^2} {A} \right) ^2 \, e ^{-\dfrac{t\,A_0\,\mathcal{B} }{A}}$ The actual results of the integrations are not provided since the main purpose of this exercise to learn how to use the integral analysis.

## Averaged Velocity! Estimates

In example 6.1 relationship between momentum of maximum velocity to average velocity was presented. Here, relationship between momentum for the average velocity to the actual velocity is presented. There are situations where actual velocity profile is not known but is function can be approximated. For example, the velocity profile can be estimated using the ideal fluid theory but the actual values are not known. For example, the flow profile in example 6.5 can be estimated even by hand sketching. For these cases a correction factor can be used. This correction factor can be calculated by finding the relation between the two cases. The momentum for average velocity is

$\label{mom:eq:UmomAve} M_a = m_{c.v} \overline{U} = \rho\,V \int_{c.v} U dV$

The actual momentum for control volume is

$\label{mom:eq:UmomAct} M_c = \int_{c.v.} \rho\, U_x dV$ These two have to equal thus,

$\label{mom:eq:UmEQ} \mathcal{C}\, \rho\,V \int_{c.v} U dV = \int_{c.v.} \rho\, U_x dV$ If the density is constant then the coefficient is one ($$\mathcal{C} quiv 1$$). However, if the density is not constant, the coefficient is not equal to one.