# 6.1.5.1: Momentum for For Constant Pressure and Frictionless Flow

- Page ID
- 1225

Another important sub category of simplification deals with flow under approximation of the frictionless flow and uniform pressure. This kind of situations arise when friction (forces) is small compared to kinetic momentum change. Additionally, in these situations, flow is exposed to the atmosphere and thus (almost) uniform pressure surrounding the control volume. In this situation, the mass flow rate in and out are equal. Thus, equation (15) is further reduced to

\[

\label{mom:eq:mom:eq:egRTTststS}

\pmb{F} = \int_{out} \rho \pmb{U} \overbrace{\left( \pmb{U} \cdot \hat{n} \right)}^{U_{rn}} dA

- \int_{in} \rho \pmb{U} \overbrace{\left( \pmb{U} \cdot \hat{n} \right)}^{U_{rn}} dA \tag{16}

\]

In situations where the velocity is provided and known (remember that density is constant) the integral can be replaced by

\[

\label{mom:eq:egRTTststSU}

\pmb{F} = \dot{m} \overline{\pmb{U}_o} - \dot{m} \overline{\pmb{U}_i} \tag{17}

\]

The average velocity is related to the velocity profile by the following integral

\[

\label{mom:eq:UmaxUaveI}

{\overline{U}}^2 = \dfrac{1}{A}\int_A \left[U(r)\right]^2\,dA \tag{18}

\]

Equation (18) is applicable to any velocity profile and any geometrical shape.

Example 6.1

Calculate the average velocity for the given parabolic velocity profile for a circular pipe.

Solution 6.1

The velocity profile is

\[

\label{mom:eq:Uprofilec}

U\left(\dfrac{r}{R}\right) = U_{max} \left[ 1 - \left( \dfrac{r}{R} \right)^2 \right] \tag{19}

\]

Substituting equation (19) into equation (??)

\[

\label{mom:eq:UmaxUaveIC}

{\overline{U}}^2 = \dfrac{1}{2\,\pi\,R^2}\int_0^R \left[U(r)\right]^2\,2\,\pi\,r\, dr \tag{20}

\]

results in

\[

\label{mom:eq:Uave}

{\overline{U}}^2 = \left(U_{max}\right)^2\, \int_0^1\left(1-\bar{r}^2\right)^2 \bar{r} d\bar{r} = \dfrac{1}{6}

\left(U_{max}\right)^2 \tag{21}

\]

Thus,

\begin{align*}

\overline{U} = \dfrac{U_{max}}{\sqrt{6}}

\end{align*}

Example 6.2

A jet is impinging on a stationary surface by changing only the jet direction (see Figure 6.2). Neglect the friction, calculate the force and the angle which the support has to apply to keep the system in equilibrium. What is the angle for which maximum force will be created?

Solution 6.2

Equation (11) can be reduced, because it is a steady state, to

\[

\label{areaJet:gov}

\pmb{F} = \int_{out} \rho \pmb{U} \overbrace{\left( \pmb{U} \cdot \hat{n} \right)}^{U_{rn}} dA

- \int_{in} \rho \pmb{U} \overbrace{\left( \pmb{U} \cdot \hat{n} \right)}^{U_{rn}} dA

= \dot{m} \pmb{U_o} - \dot{m} \pmb{U_i} \tag{22}

\]

It can be noticed that even though the velocity change direction, the mass flow rate remains constant. Equation (22) can be explicitly written for the two coordinates. The equation for the \(x\) coordinate is

\begin{align*}

F_x = \dot{m}\,\left(\cos\theta\, {U_o} - {U_i} \right)

\end{align*}

or since $U_i = U_o$

\begin{align*}

F_x = \dot{m}\,{U_i}\,\left(\cos\theta - 1 \right)

\end{align*}

It can be observed that the maximum force, \(F_x\) occurs when \(\cos\theta=\pi\). It can be proved by setting \(dF_x/d\theta=0\) which yields \(\theta=0\) a minimum and the previous solution. Hence,

\begin{align*}

\left. F_x\right|_{max} = - 2\,\dot{m}\,{U_i}

\end{align*}

and the force in the \(y\) direction is

\begin{align*}

F_y = \dot{m} \, {U_i} \,\sin\theta

\end{align*}

the combined forces are

\begin{align*}

F_{total} = \sqrt{{F_x}^2 + {F_y}^2} = \dot{m}\,{U_i}

\sqrt{\left( \cos\theta - 1\right)^2 + \sin^2\theta}

\end{align*}

Which results in

\begin{align*}

F_{total} = \dot{m}\,{U_i} \sin \left( \theta/2\right)

\end{align*}

with the force angle of

\begin{align*}

\tan \phi = \pi - \dfrac{F_y}{F_x} = \dfrac{\pi}{2} - \dfrac{\theta}{2}

\end{align*}

For angle between \( 0 <\theta < \pi\) the maximum occur at \(\theta = \pi\) and the minimum at \(\theta \sim 0\). For small angle analysis is important in the calculations of flow around thin wings.

Example 6.3

Liquid flows through a symmetrical nozzle as shown in the Figure 6.3 with a mass

*Fig. 6.3 Nozzle schematic for the discussion on the forces and for example .
flow rate of 0.01 [\(gk/sec\)]. The entrance pressure is 3[\(Bar\)] and the entrance velocity is 5 [\(m/sec\)]. The exit velocity is uniform but unknown. The exit pressure is 1[Bar]. The entrance area is 0.0005[\(m^2\)] and the exit area is 0.0001[\(cm^2\)]. What is the exit velocity? What is the force acting the nozzle? Assume that the density is constant \(\rho = 1000 [kg/m^3]\) and the volume in the nozzle is 0.0015 [\(m^3\)].*

Solution 6.3

The chosen control volume is shown in Figure 6.3. First, the velocity has to be found. This situation is a steady state for constant density. Then

\begin{align*}

A_{1}\,U_{1} = A_{2}\,U_{2}

\end{align*}

and after rearrangement, the exit velocity is

\begin{align*}

U_{2} = \dfrac{A_{1} }{A_{2} } \, U_{1} = \dfrac{0.0005}{0.0001} \times 5 = 25 [m/sec]

\end{align*}

Equation (12) is applicable but should be transformed into the z direction which is

\begin{multline*}

\sum F_z + \int_{c.v.} \pmb{g}\cdot \hat{k} \,\rho\, dV + \int_{c.v.} \pmb{P}\cos\theta_z\, dA +

\int _{c.v.} \boldsymbol{\tau}_z \,dA = \\

\overbrace{\dfrac{t}{dt} \int_{c.v.} \rho\,\pmb{U}_z\,dV }^{=0} +

\int_{c.v.} \rho\,\pmb{U}_z\cdot\pmb{U}_{rn} dA

\end{multline*}

The control volume does not cross any solid body (or surface) there is no external forces. Hence,

\begin{multline*}

\overbrace{\sum F_z}^{=0} + \int_{c.v.} \pmb{g}\cdot \hat{k} \,\rho\, dV +

\overbrace{\int_{c.v.} \pmb{P}\cos\theta_z\, dA}^{\text{ liquid surface}} + \\

\overbrace{\overbrace{\int_{c.v.} \pmb{P}\cos\theta_z\, dA}^{\text{solid surface}} + \int _{c.v.} \boldsymbol{\tau}_z \,dA}^{\text{forces on the nozzle $F_{nozzle}$}} =

\int_{c.v.} \rho\,\pmb{U}_z\cdot\pmb{U}_{rn} dA

\end{multline*}

All the forces that act on the nozzle are combined as

\[

\label{mom:eq:egRTTcompbined}

\sum F_{nozzle} + \int_{c.v.} \pmb{g}\cdot \hat{k} \,\rho\, dV +

\int_{c.v.} \pmb{P}\cos\theta_z\, dA

= \int_{c.v.} \rho\,\pmb{U}_z\cdot\pmb{U}_{rn} dA \tag{23}

\]

The second term or the body force which acts through the center of the nozzle is

\begin{align*}

\pmb{F}_b = - \int_{c.v.} \pmb{g}\cdot\hat{n}\,\rho \, dV = - g\,\rho V_{nozzle}

\end{align*}

Notice that in the results the gravity is not bold since only the magnitude is used. The part of the pressure which act on the nozzle in the \(z\) direction is

\begin{align*}

- \int_{c.v.} P dA = \int_{1} P dA - \int_{2} P dA

= \left.PA\right|_{1} - \left.PA\right|_{2}

\end{align*}

The last term in equation (23) is

\begin{align*}

\int_{c.v.} \rho\,\pmb{U}_z\cdot\pmb{U}_{rn} dA =

\int_{A_{2}} U_{2} \left( U_{2} \right) dA -

\int_{A_{1}} U_{1} \left( U_{1} \right) dA

\end{align*}

which results in

\begin{align*}

\int_{c.v.} \rho\,\pmb{U}_z\cdot\pmb{U}_{rn} dA =

\rho\, \left( {U_{2}}^2 A_{2} - {U_{1}}^2 A_{1} \right)

\end{align*}

Combining all transform equation (23) into

\[

\nonumber

F_z = - g\,\rho V_{nozzle} + \left.PA\right|_{2} - \left.PA\right|_{1} +

\rho\, \left( {U_{2}}^2 A_{2} - {U_{1}}^2 A_{1} \right) \tag{24}

\]

\begin{align*}

F_z = 9.8 \times 1000 \times

\end{align*}

### 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.