Exact Solution of the Continuity and Conservation of Linear Momentum Equations

An introduction to hydraulic structures

Amir Hossein Azimi , in Water Engineering Modeling and Mathematic Tools, 2021

16.2.5 Momentum equation

The momentum equation is used in open channel flow problems to determine unknown forces ( F) acting on the walls or bed in a control volume. In comparison to the energy equation that deals with scalar quantities such as mass (m), pressure (P), and velocity magnitude (V), the momentum equation deals with vector quantities such as velocity vector and forces (F). Therefore it is critical to write a momentum equation in a known direction and use the component of the forces within the defined direction. Using Newton's second law of motion, which equates the rate of change of momentum (M=mV) with the algebraic sum of all external forces, the momentum equation can be written as:

(16.18) $\frac{dM}{dt}=\sum F$

Further, the rate of change of momentum can be divided into two terms as:

(16.19) $\frac{dM}{dt}=m\frac{dV}{dt}+V\frac{dm}{dt}$

Assuming homogeneous fluid in open channel flow problems, the variations of fluid mass with time are zero and Eq. (16.19) reduces to dM/dt=m(dV/dt). By replacing the mass of fluid with a product of density and volume, Newton's second law of motion (Eq. 16.18) can be simplified to:

(16.20) $\rho Q(V_2-V_1)=\sum F$

Consider a control volume in a one-dimensional straight channel with a general cross section as shown in Fig. 16.7, where the momentum equation along the bed can be written as:

Figure 16.7. Schematic of the momentum principle and force balance in a prismatic open channel with rectangular cross section.

(16.21) $\rho A_2{V}_2^2-\rho A_1{V}_1^2=F_{P1}-F_{P2}+F_g-F_{\tau }$

where (F _P ) is the pressure force (F _P =ρgy′A), (y′) is the depth from water surface to the center of pressure, (F _g ) is the gravitational force along the direction of the channel (F _g =Wsin θ), (W) is the weight of water, and (F _τ ) is the friction force between water and channel boundaries. It should be noted that forces along the direction of flow are considered positive and forces against the flow direction are considered negative in Eq. (16.21).

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GENERALIZED GOVERNING EQUATIONS FOR MULTIPHASE SYSTEMS: AVERAGING FORMULATIONS

Amir Faghri , Yuwen Zhang , in Transport Phenomena in Multiphase Systems, 2006

4.4.2 Momentum Equation

The momentum equation for phase k in the multifluid model is expressed in eq. (4.34). By adding together the momentum equations for all Π phases, one obtains

(4.94) $\begin{array}{l}\frac{\partial }{\partial t}\left({\displaystyle \sum _{k=1}^{\Pi }{\epsilon }_k{〈{\rho }_k〉}^k{〈{\text{V}}_k〉}^k}\right)+\nabla \cdot \left({\displaystyle \sum _{k=1}^{\Pi }{\epsilon }_k{〈{\rho }_k〉}^k{〈{\text{V}}_k{\text{V}}_k〉}^k}\right)\\ =\nabla \cdot {\displaystyle \sum _{k=1}^{\Pi }{\epsilon }_k{〈{{\tau }^{\prime }}_k〉}^k+{\displaystyle \sum _{k=1}^{\Pi }\left({\epsilon }_k{〈{\rho }_k〉}^k\right)}\text{X}+{\displaystyle \sum _{k=1}^{\Pi }{\displaystyle \sum _{j=1(j\ne k)}^{\Pi }\left(〈{\text{F}}_{jk}〉+〈{{{\dot{m}}^{″}}^{\prime }}_{jk}〉{〈{\text{V}}_{k,I}〉}^k\right)}}}\end{array}$

The stress tensor of the multiphase mixture is

(4.95) $〈{\tau }^{\prime }〉={\displaystyle \sum _{k=1}^{\Pi }{\epsilon }_k{〈{\tau }_k〉}^k=-〈p〉I+\mu \left[\nabla \tilde{\text{V}}+\nabla {\tilde{\text{V}}}^T\right]}-\frac{2}{3}\mu (\nabla \cdot \tilde{\text{V}})\text{I}$

The summation of all interphase forces must be zero since 〈F _jk 〉 = -〈F _kj 〉, i.e.,

(4.96) ${\displaystyle \sum _{k=1}^{\Pi }{\displaystyle \sum _{j=1(j\ne k)}^{\Pi }〈{\text{F}}_{jk}〉=0}}$

Considering eqs. (4.92), (4.95) and (4.96), the momentum equation becomes

(4.97) $\frac{\partial }{\partial t}\left(〈\rho 〉\tilde{\text{V}}\right)+\nabla \cdot \left({\displaystyle \sum _{k=1}^{\Pi }{\epsilon }_k{〈{\rho }_k〉}^k{〈{\text{V}}_k{\text{V}}_k〉}^k}\right)=\nabla \cdot 〈{\tau }^{\prime }〉+〈\rho 〉\text{X}+{{{\dot{\text{M}}}^{″}}^{\prime }}_I$

where

(4.98) ${{{\dot{\text{M}}}^{″}}^{\prime }}_I={\displaystyle \sum _{k=1}^{\Pi }{{\displaystyle \sum _{j=1(j\ne k)}^{\Pi }〈{{{\dot{m}}^{″}}^{\prime }}_{jk}〉〈{\text{V}}_{k,I}〉}}^k}$

Equation (4.98) represents the momentum production rate due to interaction between different phases along their separating interfaces. It must be specified according to the combination of phases in the multiphase system that is under consideration.

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Real-Time Transient Model–Based Leak Detection

Morgan Henrie PhD, PMP, PEM , ... R. Edward Nicholas , in Pipeline Leak Detection Handbook, 2016

Momentum Equation (Newton's Second Law of Motion)

The momentum equation ( Eq. 4.2), an expression of Newton's second law of motion, represents the transient force balance on the fluid within a slice of the pipeline cross-section.

•

The left side, $\rho \left(\partial v/\partial t+v\partial v/\partial x\right)$ , is mass times acceleration per unit volume of fluid (there is a velocity change in time, $t$ , as well as a change as it moves in distance, $x$ ).

•

The right side (RHS) represents the forces acting on a unit mass of fluid.

•

The first RHS term, $-\partial p/\partial x$ , is the net force imposed by the pressure gradient.

•

The second RHS term, $-\rho g\partial z/\partial x$ , is the force of gravity on the element as it moves in the vertical direction (due to the slope of the pipeline).

•

The final term $-\rho fv\left|v\right|/2D$ is the frictional force that acts in a direction opposite to the velocity.

Equation 4.2. Momentum Equation (Momentum Balance)

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Description of single-point wind time series along railways

Hui Liu , in Wind Forecasting in Railway Engineering, 2021

3.2.2.1.2 Momentum equation

The momentum equation expresses the law of conservation of momentum for moving fluid. The rate of change of total momentum of any micro unit in flow field is equal to the resultant force of all external forces acting on the micro unit. The expression of fluid momentum equation is as follows [ 14]:

(3.2) $\{\begin{array}{c}\frac{\partial \left(\rho u\right)}{\partial t}+\nabla \cdot \left(\rho uU\right)=-\frac{\partial \rho }{\partial x}+\frac{\partial {\tau }_{xx}}{\partial x}+\frac{\partial {\tau }_{yx}}{\partial y}+\frac{\partial {\tau }_{zx}}{\partial z}+\rho f_x\\ \frac{\partial \left(\rho v\right)}{\partial t}+\nabla \cdot \left(\rho vU\right)=-\frac{\partial \rho }{\partial y}+\frac{\partial {\tau }_{xy}}{\partial x}+\frac{\partial {\tau }_{yy}}{\partial y}+\frac{\partial {\tau }_{zy}}{\partial z}+\rho f_y\\ \frac{\partial \left(\rho w\right)}{\partial t}+\nabla \cdot \left(\rho wU\right)=-\frac{\partial \rho }{\partial z}+\frac{\partial {\tau }_{xz}}{\partial x}+\frac{\partial {\tau }_{yz}}{\partial y}+\frac{\partial {\tau }_{zz}}{\partial z}+\rho f_z\end{array}$

where $U=ui+vj+wk$ is velocity of micro unit, ${\tau }_{ij}$ is viscosity stress component of micro unit, and $f_i$ is body force of micro unit.

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Operation examples of emission control systems

Masaaki Okubo , Takuya Kuwahara , in New Technologies for Emission Control in Marine Diesel Engines, 2020

Particle momentum equation

A momentum equation for simulating the motion of one test particle may be written as follows, according to the Lagrange method.

(4.23) $m_{\text{p}}\frac{\mathrm{d}{\mathit{u}}_{\mathrm{p}}}{\mathrm{d}t}=\frac{C_{\text{D}}}{C_{\text{C}}}\rho \left(\mathit{u}-{\mathit{u}}_{\mathrm{p}}\right)|\mathit{u}-{\mathit{u}}_{\mathrm{p}}|\frac{A_{\text{p}}}{2}+m_{\text{p}}\mathit{g}+q\mathit{E}+{\mathit{F}}_{\text{Ba}}+{\mathit{F}}_{\text{Saff}}+{\mathit{F}}_{\text{p}}+{\mathit{F}}_{\text{Br}}$

where m _p is the mass of a single particle, u _p is the particle velocity vector, C _D is the drag coefficient, C _C is Cunningham's correction factor, ρ is the surrounding fluid density, A _p (for spherical particle, ${\mathit{A}}_{\text{p}}={\mathit{\pi d}}_{\text{p}}^2/4$ ) is the particle projection area, and g is the gravity vector. On the right side of Eq. (4.23) are given the drag force, gravity force, electrostatic force, and the following unsteady drag forces: F _Ba, the Basset force, F _Saff, the Saffman lifting force, F _p, the force due to pressure gradient including buoyancy, and F _Br, the force due to Brownian diffusion. Furthermore, the inertial force due to virtual mass is small in solid-gas two-phase flows; thus, it generally may be ignored.

On the right side of Eq. (4.23), the influence of drag force, gravity, and electrostatic force are generally significant. The drag force is influenced not only by particle shape and direction but also by particle Reynolds number, turbulence level, and other factors. For nanoparticles having sizes near the mean free path of gas fluid molecules, particle slip occurs. As a result, the drag force is needed to be corrected. As shown in Eq. (4.23), corrections may be applied by dividing C _D by C _C (the value of which is larger than one). Furthermore, F _Ba, F _Saff, F _p, and F _Br ought to be properly considered in the precise calculation.

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Tribology for Energy Conservation

Chao Zhang , ... Zugan Qiu , in Tribology Series, 1998

2.2 A viscoelastic fluid

The momentum equations and the constitutive equations of the Maxwell liquid for a dynamically loaded finite journal bearings are derived by Rastogi et al. [19] and Paranjpe [22]:

(15) $\frac{\partial {\mathrm{\tau }}_{\mathrm{\chi }\mathrm{y}}}{\partial \mathrm{y}}=\frac{\partial \mathrm{p}}{\partial \mathrm{\chi }},\frac{\partial {\mathrm{\tau }}_{\mathrm{y}\mathrm{z}}}{\partial \mathrm{y}}=\frac{\partial \mathrm{p}}{\partial \mathrm{z}}$

(16) ${\mathrm{\tau }}_{\mathrm{\chi }\mathrm{y}}+\mathrm{\kappa }\frac{\partial {\mathrm{\tau }}_{\mathrm{x}\mathrm{y}}}{\partial \mathrm{t}}=\mathrm{\mu }\frac{\partial \mathrm{u}}{\partial \mathrm{y}},{\mathrm{\tau }}_{\mathrm{y}\mathrm{z}}+\mathrm{\kappa }\frac{\partial {\mathrm{\tau }}_{\mathrm{y}\mathrm{z}}}{\partial \mathrm{t}}=\mathrm{\mu }\frac{\partial \mathrm{w}}{\partial \mathrm{y}}$

Where κ is the relaxation time and viscosity is treated as a function of the second invariant of strain rate.

(17) $\mathrm{Defining}:\mathrm{p}\text{'}=\mathrm{p}+\mathrm{\kappa }\frac{\partial \mathrm{p}}{\partial \mathrm{t}}$

Eq. (16) can be reduced to, respectively:

(18) $\frac{\partial \mathrm{p}\text{'}}{\partial \mathrm{\chi }}=\frac{\partial }{\partial \mathrm{y}}\mathrm{\mu }\frac{\partial \mathrm{u}}{\partial \mathrm{y}},\frac{\partial \mathrm{p}\text{'}}{\partial \mathrm{z}}=\frac{\partial }{\partial \mathrm{y}}\mathrm{\mu }\frac{\partial \mathrm{w}}{\partial \mathrm{y}}$

Following the usual procedure, from Eq. (18) we can derive the Reynolds equation of the rough journal bearings for a Maxwell fluid. It has the same form as that with the same viscosity and without elasticity, except the variable is $\overline{\mathrm{p}\text{'}}$ instead of $\overline{\mathrm{p}.}$ For the Maxwell fluid with the viscosity of the power law fluid, its Reynolds equation is in the same form as Eq. (2), except that the variable is $\overline{\mathrm{p}\text{'}}$ instead of $\overline{\mathrm{p}.}$

At the boundaries $\partial \mathrm{p}/\partial \mathrm{t}=0\text{,}$ so $\overline{\mathrm{p}\text{'}}=\overline{\mathrm{p}}\text{.}$ Solving Eq. (2) with the same boundary conditions for $\overline{\mathrm{p}\text{'}}\text{,}$ we can get the correct pressure distribution as follows:

(19) $\overline{\mathrm{p}}={\mathrm{e}}^{-\mathrm{t}/\mathrm{\kappa }}\left[{\int }_0^{\mathrm{t}}{\mathrm{e}}^{\mathrm{t}\text{'}/\mathrm{\kappa }}\frac{\overline{\mathrm{p}\text{'}}}{\mathrm{\kappa }}\mathrm{d}\mathrm{t}\text{'}+{\mathrm{e}}^{-\mathrm{T}/\mathrm{\kappa }}{\int }_0^{\mathrm{T}}{\mathrm{e}}^{\mathrm{t}/\mathrm{\kappa }}\frac{\overline{\mathrm{p}}\text{'}}{\mathrm{\kappa }}\mathrm{d}\mathrm{t}\right]$

The inverse problem (i.e., the load is specified) is treated in the way similar to that used by Paranjpe[22]. Defining F'_ex as follows

(20) $\mathrm{F}{\text{'}}_{\mathrm{ex}}=-\mathrm{F}{\text{'}}_{\mathrm{oilx}}-{\mathrm{F}}_{\mathrm{c}\mathrm{x}}+\mathrm{M}{\ddot{\mathrm{e}}}_{\mathrm{x}}$

where

(21) $\mathrm{F}{\text{'}}_{\mathrm{oilx}}=-\int \overline{\mathrm{p}\text{'}}cos\mathrm{\theta }dA$

(22) $\mathrm{F}{\text{'}}_{\mathrm{oilx}}={\mathrm{F}}_{\mathrm{oilx}}+\mathrm{\kappa }\frac{\partial {\mathrm{F}}_{\mathrm{oilx}}}{\partial \mathrm{t}}$

Substituting for F'_oilx from Eq. (13) into Eq. (22) and substituting it into Eq. (20) we obtain F'_exto be

(23) $\mathrm{F}{\text{'}}_{\mathrm{ex}}={\mathrm{F}}_{\mathrm{ex}}+{\mathrm{F}}_{\mathrm{c}\mathrm{x}}+\mathrm{\kappa }\frac{\partial }{\partial \mathrm{t}}\left({\mathrm{F}}_{\mathrm{ex}}+{\mathrm{F}}_{\mathrm{c}\mathrm{x}}-\mathrm{M}{\ddot{\mathrm{e}}}_{\mathrm{x}}\right)$

Similarly F'_ey is given by

(24) $\mathrm{F}{\text{'}}_{\mathrm{e}\mathrm{y}}={\mathrm{F}}_{\mathrm{e}\mathrm{y}}+{\mathrm{F}}_{\mathrm{c}\mathrm{y}}+\mathrm{\kappa }\frac{\partial }{\partial \mathrm{t}}\left({\mathrm{F}}_{\mathrm{e}\mathrm{y}}+{\mathrm{F}}_{\mathrm{c}\mathrm{y}}-\mathrm{M}{\ddot{\mathrm{e}}}_{\mathrm{y}}\right)$

These expressions are greatly simplified if the contact load and the inertia of the journal in Eqs. (23) and (24) are neglected. In that case

(25) $\mathrm{F}{\text{'}}_{\mathrm{ex}}={\mathrm{F}}_{\mathrm{ex}}+\mathrm{\kappa }\frac{\partial {\mathrm{F}}_{\mathrm{ex}}}{\partial \mathrm{t}},\mathrm{F}{\text{'}}_{\mathrm{e}\mathrm{y}}={\mathrm{F}}_{\mathrm{e}\mathrm{y}}+\mathrm{\kappa }\frac{\partial {\mathrm{F}}_{\mathrm{e}\mathrm{y}}}{\partial \mathrm{t}}$

Therefore, if we transform the real loads F_ex and F_ey to F'_ex and F'_ey and and solve the dynamic bearing problem as before we can calculate the journal orbit for the viscoelastic fluid as well.

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Basic Equations of Fluid Mechanics

Singiresu S. Rao , in The Finite Element Method in Engineering (Sixth Edition), 2018

17.3 Methods of Describing the Motion of a Fluid

The motion of a group of particles in a fluid can be described by either the Lagrangian method or the Eulerian method. In the Lagrangian method, the coordinates of the moving particles are represented as functions of time. This means that at some arbitrary time t ₀, the coordinates of a particle (x ₀, y ₀, z ₀) are identified and that thereafter we follow that particle through the fluid flow. Thus, the position of the particle at any other instant is given by a set of equations of the form

$x=f_1\left(x_0,y_0,z_0,t\right),y=f_2\left(x_0,y_0,z_0,t\right),z=f_3\left(x_0,y_0,z_0,t\right)$

The Lagrangian approach is not generally used in fluid mechanics because it leads to more cumbersome equations. In the Eulerian method, we observe the flow characteristics in the vicinity of a fixed point as the particles pass by. Thus, in this approach the velocities at various points are expressed as functions of time as

$u=f_1\left(x,y,z,t\right),v=f_2\left(x,y,z,t\right),w=f_3\left(x,y,z,t\right)$

where u, v, and w are the components of velocity in x, y, and z directions, respectively.

The velocity change in the vicinity of a point in the x direction is given by

(17.1) $\text{d}u=\frac{\partial u}{\partial t}\text{d}t+\frac{\partial u}{\partial x}\text{d}x+\frac{\partial u}{\partial y}\text{d}y+\frac{\partial u}{\partial z}\text{d}z$

(total derivative expressed in terms of partial derivatives).

The small distances moved by a particle in time dt can be expressed as

(17.2) $\text{d}x=u\text{d}t,\text{d}y=v\text{d}t,\text{d}z=w\text{d}t$

Thus, dividing Eq. (17.1) by dt and using Eq. (17.2) leads to the total or substantial derivative of the velocity u (x component of acceleration) as

(17.3a) $a_x=\frac{\text{d}u}{\text{d}t}\equiv \frac{\text{D}u}{\text{D}t}=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}$

The other components of acceleration can be expressed in a similar manner as

(17.3b) $a_y=\frac{\text{d}v}{\text{d}t}\equiv \frac{\text{D}v}{\text{D}t}=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}$

(17.3c) $a_z=\frac{\text{d}w}{\text{d}t}\equiv \frac{\text{D}w}{\text{D}t}=\frac{\partial w}{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}$

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Aircraft Drag Analysis

Snorri Gudmundsson BScAE, MScAE, Ph.D., FAA DER (ret.) , in General Aviation Aircraft Design (Second Edition), 2022

16.2.5 Drag of a 2D Body in a Wind-Tunnel

Drag force of a 2D body (e.g., an airfoil) is estimated in wind tunnel testing by two means: (1) Using a force balance and (2) based on velocity profile in front and aft of model using the momentum theorem (see Section 14.2.1, Conservation Laws). The former is outside the scope of this book. The latter warrants a short introduction.

The momentum equation of Equation (14-26) can be simplified to allow the extraction of drag of a body in a wind tunnel, by assuming steady flow, absence of body forces, and uniformity of pressure in front and aft. Thus, only the momentum flux term remains. For instance, as shown in references such as Refs. [7–9], this leads to the following expression for drag force per unit length:

(16-21) $D^{\prime }={\int }_{-h/2}^{h/2}\mathrm{\rho }{u}_e\left(u_i-u_e\right)\mathit{dy}$

Where h is the height of the test section, u _i and u _e are the velocity profiles at the inlet and exit of the test section, respectively. It is assumed the body is placed near the vertical center of the test section (where y  =   0). See Example 16-3 for details of its use.

Example 16-3

The wind tunnel setup shown in Figure 16-12 is used to measure drag of an airfoil. The distribution of velocity is measured at the inlet and exit of the test section. The airspeed profile at the inlet is uniform, but at the exit is $u_e\left(y\right)=\frac{1}{2}{u}_i\left(1+4ysgn\left(y\right)/h\right)$ between −  h/4   ≤ y  ≤ h/4 and sgn() is the sign function. Elsewhere, u _e   = u _i . Assume equal and uniform static pressures at entry and exit. Determine the drag force per unit length (D′) acting on the airfoil.

Figure 16-12. A 2D body in a wind tunnel.

SOLUTION:

Let us break the integral of Equation (16-21) into four smaller integrals

$D^{\prime }={\int }_{-h/2}^{h/2}\mathrm{\rho }{u}_e\left(u_i-u_e\right)\mathit{dy}=\stackrel{=0}{\overbrace{{\int }_{-h/2}^{-h/4}\mathrm{\rho }{u}_i\left(u_i-u_i\right)\mathit{dy}}}+{\int }_{-h/4}^0\mathrm{\rho }{u}_e\left(u_i-u_e\right)\mathit{dy}+{\int }_0^{h/4}\mathrm{\rho }{u}_e\left(u_i-u_e\right)\mathit{dy}+\stackrel{=0}{\overbrace{{\int }_{h/4}^{h/2}\mathrm{\rho }{u}_i\left(u_i-u_i\right)\mathit{dy}}}$

As shown, the upper and lower fourths of the region return zero (because u _e   = u _i ). Due to the symmetry of this problem, the center two integrals are equal. Therefore, we can drop the sign function and integrate from 0 to h/4 and multiply the result by two, which reduces this to

$D^{\prime }=2\mathrm{\rho }{\int }_0^{h/4}\left(\frac{u_i}{2}\left(1+\frac{4y}{h}\right)\right)\left(u_i-\frac{u_i}{2}\left(1+\frac{4y}{h}\right)\right)\mathit{dy}$

Substitute the given velocity profiles into Equation (16-21) and evaluate the integrals to give

$D^{\prime }=\mathrm{\rho }{u}_i^2{\int }_0^{h/4}\left(1+\frac{4y}{h}\right)\left(2-\left(1+\frac{4y}{h}\right)\right)\mathit{dy}=\mathrm{\rho }{u}_i^2{\left[y-\frac{16y^3}{3h^2}\right]}_0^{h/4}=\frac{\mathrm{\rho }{u}_i^{2}h}{6}$

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Equations of Motion

E.L. Houghton , ... Daniel T. Valentine , in Aerodynamics for Engineering Students (Seventh Edition), 2017

Momentum Equation

The momentum equation requires that the time rate of momentum change in a given direction be equal to the sum of the forces acting in that direction. This is known as Newton's second law of motion and in the model used here the forces concerned are gravitational (body) and surface.

Consider a fluid in steady flow, and take any small stream tube as in Fig. 2.4. The distance s is measured along the tube's axis from some arbitrary origin. A is the cross-sectional area of the stream tube at distance s from the arbitrary origin.

Figure 2.4. Stream tube and element with definitions related to the momentum equation.

p, ρ, and v represent pressure, density, and flow speed, respectively. A, p, ρ, and v vary with s (i.e., with position along the stream tube) but not with time since the motion is steady. Now consider the small element of fluid shown in Fig. 2.5, which is immersed in fluid of varying pressure. The element is the right frustrum of a cone of length δs, area A at the upstream section, area $A+\delta A$ on the downstream section. The pressure acting on one face of the element is p, and on the other face is $p+(dp/ds)\delta s$ . Around the curved surface the pressure may be taken to be the mean value $p+(dp/ds)\delta s/2$ . In addition, the weight W of the fluid in the element acts vertically as shown. Shear forces on the surface due to viscosity would add another force, which is ignored here.

Figure 2.5. Forces on the element.

As a result of these pressures and the weight, there is a resultant force F acting along the axis of the cylinder where F is given by

(2.6) $F=pA-(p+\frac{dp}{ds}\delta s)(A+\delta A)+(p+\frac{dp}{ds}\frac{\delta s}{2})\delta A-W\cos \alpha$

where α is the angle between the axis of the stream tube and the vertical.

From Eq.(2.6) it is seen that on neglecting quantities of small order such as $(dp/ds)\delta s\delta A$ and with like terms summing to zero,

(2.7) $F=-\frac{dp}{ds}A\delta s-\rho gA(\delta s)\cos \alpha$

since the gravitational force on the fluid in the element is ρgA δs, volume × density × g.

Now, Newton's second law of motion ( $\text{force}=\text{mass}\times \text{acceleration}$ ) applied to the element of Fig. 2.5 in the direction of flow gives a scalar equation:

(2.8) $-\rho gA\delta s\cos \alpha -\frac{dp}{ds}A\delta s=\rho A\delta s\frac{dv}{dt}$

where t represents time. Dividing by Aδs this becomes

$-\rho g\cos \alpha -\frac{dp}{ds}=\rho \frac{dv}{dt}$

But the acceleration can be written as

$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}$

and therefore

$\rho v\frac{dv}{ds}+\frac{dp}{ds}+\rho g\cos \alpha =0$

or

$v\frac{dv}{ds}+\frac{1}{\rho }\frac{dp}{ds}+g\cos \alpha =0$

Integrating along the stream tube, this becomes

$\int \frac{dp}{\rho }+\int vdv+g\int \cos \alpha ds=\text{constant}$

but since

$\int \cos \alpha ds=\text{increase in vertical coordinate}z$

and

$\int vdv=\frac{1}{2}{v}^2$

then

(2.9) $\int \frac{dp}{\rho }+\frac{1}{2}{v}^2+gz=\text{constant}$

This result is known as Bernoulli's equation and is discussed momentarily.

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Compressor and Turbine Performance Characteristics

Meherwan P. Boyce , in Gas Turbine Engineering Handbook (Fourth Edition), 2012

Momentum Equation

The momentum equation is a mathematical formulation of the law of conservation of momentum. It states that the rate of change in linear momentum of a volume moving with a fluid is equal to the surface forces and the body forces acting on a fluid. Figure 3-8 shows the velocity components in a generalized turbomachine. The velocity vectors as shown are resolved into three mutually perpendicular components: the axial component (V_a ), the tangential component (V_θ ), and the radial component (V_m ).

Figure 3-8. Velocity vectors in compressor rotor flow.

By examining each of these velocities, the following characteristics can be noted: the change in the magnitude of the axial velocity gives rise to an axial force, which is taken up by a thrust bearing; the change in radial velocity gives rise to a radial force, which is taken up by the journal bearing. The tangential component is the only component that causes a force, which corresponds to a change in angular momentum; the other two velocity components have no effect on this force, except for what bearing friction may arise.

By applying the conservation of momentum principle, the change in angular momentum obtained by the change in the tangential velocity is equal to the summation of all the forces applied on the rotor. This summation is the net torque of the rotor. A certain mass of fluid enters the turbomachine with an initial velocity $V_{{\theta }_1}$ at a radius r ₁ and leaves with a tangential velocity $V_{{\theta }_2}$ at a radius r ₂. Assuming that the mass flow rate through the turbomachine remains unchanged, the torque exerted by the changes in angular velocity can be written as follows:

(3-23) $\tau =\frac{\dot{m}}{g_c}\left(r_1{V}_{{\theta }_1}-r_2{V}_{{\theta }_2}\right)$

The rate of change of energy transfer (ft-lb _f /s) is the product of the torque and the angular velocity (ω):

(3-24) $\tau \omega =\frac{\dot{m}}{g_c}\left(r_1\omega V_{{\theta }_1}-r_2\omega V_{{\theta }_2}\right)$

Thus, the total energy transfer can be written as follows:

(3-25) $E=\frac{\dot{m}}{g_c}\left(U_1{V}_{{\theta }_1}-U_2{V}_{{\theta }_2}\right)$

where U ₁ and U ₂ are the linear velocity of the rotor at the respective radii. The previous relation per unit mass flow can be written in terms of the total enthalpy (H):

(3-26) $H=\frac{1}{g_c}\left(U_1{V}_{{\theta }_1}-U_2{V}_{{\theta }_2}\right)$

where H is the energy transfer per unit mass flow (ft-lb _f /lb _m ) or fluid pressure. Equation (3-26) is known as the Euler turbine equation.

The equation of motion as given in terms of angular momentum can be transformed into other forms that are more convenient to understand some of the basic design components. To understand the flow in a turbomachine, the concepts of absolute and relative velocities must be grasped. Absolute velocity (V) is the gas velocity with respect to a stationary coordinate system. Relative velocity (W) is the velocity relative to the rotor. In turbomachinery, the air entering the rotor will have a relative velocity component parallel to the rotor blade and an absolute velocity component parallel to the stationary blades. Mathematically, this relationship is written as follows:

(3-27) $\stackrel{\rightharpoonup }{V}=\stackrel{\rightharpoonup }{W}\to \stackrel{\rightharpoonup }{U}$

where the absolute velocity (V) is the algebraic addition of the relative velocity (W) and the linear rotor velocity (U). The absolute velocity can be resolved into its components, the radial or meridional velocity (V_m ) and the tangential component V_θ . From Figure 3-9, the following relationships are obtained:

(3-28) $\begin{array}{ll}{V}_1^2\hfill & =V_{{\theta }_1}^2+V_{m_1}^2\hfill \\ V_2^2\hfill & =V_{{\theta }_2}^2+V_{m_2}^2\hfill \\ W_1^2\hfill & ={\left(U_1-V_{{\theta }_1}\right)}^2+V_{m_1}^2\hfill \\ W_2^2\hfill & ={\left(U_2-V_{{\theta }_2}\right)}^2+V_{m_2}^2\hfill \end{array}$

Figure 3-9. Velocity triangles for an axial-flow compressor.

By placing these relationships into Equation (3-26), the Euler turbine equation, the following relationship is obtained for the total enthalpy:

(3-29) $H_o=\frac{1}{2g_c}\left[\left(V_1^2-V_2^2\right)+\left(U_1^2-U_2^2\right)+\left(W_2^2-W_1^2\right)\right]$

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Source: https://www.sciencedirect.com/topics/engineering/momentum-equation

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