# Integral momentum equation

(Difference between revisions)
 Revision as of 08:55, 5 November 2009 (view source)← Older edit Revision as of 20:52, 5 November 2009 (view source)Newer edit → Line 1: Line 1: - The integral form of Newton’s second law for a control volume that includes only one phase is expressed by eq. (2.10). The surface integral terms in eq. (2.10) can be rewritten using the divergence theorem: + Newton’s second law states that, in an inertial reference frame, the time rate of momentum change of a fixed mass system is equal to the net force acting on the system, and it takes place in the direction of the net force. Mathematically, Newton’s second law of motion for fixed-mass in a reference frame that moves at a constant velocity ${V_{ref}}$ is written as - (2.62) + - (2.63) +
$\sum {{\mathbf{F}} = \frac{{d{{(m{\mathbf{V}})}_{rel}}}}{{dt}}} \qquad \qquad(1)$
- Substituting eq. (2.62) and (2.63) into eq. (2.10), and considering eq. (2.48), the entire equation can be rewritten as a volume integral: + - (2.64) + where the left-hand side is the net force vector acting on the fixed-mass system, and the right-hand side is the rate of momentum change. - As was the case for the continuity equation, the integrand must equal zero to assure the general validity of eq. (2.64); so, one obtains the desired differential form of the momentum equation: + - (2.65) + For control volumes that contain only one phase, the integral form of Newton’s second law can be obtained by using eq. (2.3). With the applicable value of $\Phi$ and $\phi$ in eq. (2.3) defined as $\Phi = m{{\mathbf{V}}_{rel}}$ and $\phi = {{\mathbf{V}}_{rel}}$, one obtains - The derivatives on the left-hand side of eq. (2.65) may be expanded to yield + - (2.66) +
$\sum {\mathbf{F}} = \frac{\partial }{{\partial t}}\int_V {\rho {{\mathbf{V}}_{rel}}dV} + \int_A {\rho ({{\mathbf{V}}_{rel}} \cdot {\mathbf{n}}){{\mathbf{V}}_{rel}}dA} \qquad \qquad(2)$
- The first bracketed term on the left vanishes, as required by the continuity eq. (2.51). The second term may be written more simply in substantial derivative form, and the entire equation becomes + - (2.67) + which is in the vector form and is valid in all three directions. - The stress tensor,  is the sum of an isotropic thermodynamic stress, , and the viscous stress tensor,   [defined in eq. (1.56)], i.e., + - (2.68) + Forces acting on the control volume include body forces and contact forces that act on its surface. For example, for a multicomponent system that contains $N$ components, if the body force per unit volume acting on the $i$th species is ${X_i}$, the total body force acting on the control volume is $\int_V {\left[ {\sum\limits_{i = 1}^N {{\rho _i}{{\mathbf{X}}_i}} } \right]} dV$, where ${\rho _i}$ is the mass concentration (kg/m3) of the $i$th species. If the body force per unit mass is the same for different species (as is the case with gravity), the body force term is reduced to $\int_V {\rho {\mathbf{X}}} dV,$ where $\rho$ is the density of the mixture. The stress tensor acts on the surface of a fluid control volume, and includes both normal and shear stresses. The net force may be written as - Substituting eq. (2.68) into eq. (2.67), the momentum equation becomes + - (2.69) +
$\sum {{\mathbf{F}} = \int_V {\left[ {\sum\limits_{i = 1}^N {{\rho _i}{{\mathbf{X}}_i}} } \right]} dV + \int_A {{{{\mathbf{\tau '}}}_{rel}} \cdot {\mathbf{n}}dA} } \qquad \qquad(3)$
- The viscous stress tensor measured in the reference frame, , can be determined by using Newton’s law of viscosity [see eq. (1.59)]: + - (2.70) + where ${\mathbf{\tau '}}$ is the total stress tensor and $n$ is the local normal unit vector on surface $A$. The dot product of a tensor of rank two, ${\mathbf{\tau '}}$, and a vector, $n$, is a vector that represents the force acting on the surface of the control volume per unit area. - where Drel is the rate of strain tensor, i.e., + Combining eqs. (2) and (3), we obtain the momentum equation for the control volume of a single-phase system: - (2.71) + - and I in eq. (2.70) is the unit tensor that satisfies  for any tensor a. The diagonal components of I are equal to one and all other components are zero: +
$\begin{array}{l} \int_V {\left[ {\sum\limits_{i = 1}^N {{\rho _i}{{\mathbf{X}}_i}} } \right]} dV + \int_A {{{{\mathbf{\tau '}}}_{rel}} \cdot {\mathbf{n}}dA} \\ = \frac{\partial }{{\partial t}}\int_V {\rho {{\mathbf{V}}_{rel}}dV} + \int_A {\rho ({{\mathbf{V}}_{rel}} \cdot {\mathbf{n}}){{\mathbf{V}}_{rel}}dA} \\ \end{array} \qquad \qquad(4)$
- (2.72) + - If the fluid is incompressible ( ), the second term on the right-hand side of eq. (2.70) will be zero according to eq. (2.55). The momentum equation (2.67) then becomes + where the two terms on the left-hand side represent, respectively, the body force and stress on the control volume, and the two terms on the right-hand side represent, respectively, the rate of momentum change in the control volume and the rate of the momentum flow into or out of the control volume. ${V_{rel}}$ is the bulk velocity of the mixture that contains $N$ components. - (2.73) + - where the left-hand side is the inertial term (mass per unit volume  times acceleration, DVrel/Dt). The three terms on the right-hand side represent body force per unit volume, pressure force per unit volume, and viscous force per unit volume, respectively. + When the control volume includes multiple phases, integrations must be performed for each subvolume. In that case, the momentum equation becomes (Faghri and Zhang, 2006) - For  , we have Stokes’ flow or creep flow, and eq. (2.73) becomes elliptic and is similar to the steady-state conduction equation. + - In a Cartesian coordinate system, the vector form of the momentum equation, eq. (2.73), for incompressible and Newtonian fluid with constant viscosity can be written as three equations in the x-, y-, and z-directions: +
$\begin{array}{l} \sum\limits_{k = 1}^\Pi {\left[ {\int_{{V_k}(t)} {\left[ {\sum\limits_{i = 1}^N {{\rho _{k,i}}{{\mathbf{X}}_{k,i}}} } \right]} dV + \int_{{A_k}(t)} {{{{\mathbf{\tau '}}}_{k,rel}} \cdot {{\mathbf{n}}_k}dA} } \right]} \\ = \sum\limits_{k = 1}^\Pi {\left[ {\frac{\partial }{{\partial t}}\int_{{V_k}(t)} {{\rho _k}{{\mathbf{V}}_{k,rel}}dV} + \int_{{A_k}(t)} {{\rho _k}({{\mathbf{V}}_{k,rel}} \cdot {{\mathbf{n}}_k}){{\mathbf{V}}_{k,rel}}dA} } \right]} \\ \end{array} \qquad \qquad(5)$
- (2.74) + - (2.75) + Equations (4) and (5) are momentum equations in a coordinate system that is attached to and moves with an inertial reference frame. For a fixed coordinate system that does not move with the reference frame (while the control volume still moves with the reference frame at velocity ${V_{ref}}$), one can substitute the general variables $\Phi = m{\mathbf{V}}$ and $\phi = {\mathbf{V}}$ to obtain the momentum equation. - (2.76) + - where  are the components of body force per unit volume acting on the ith species in the x-, y-, and z- directions, respectively. + ==References== - For the case that the only body force is gravity,  , eq. (2.73) becomes + - (2.77) + Faghri, A., and Zhang, Y., 2006, ''Transport Phenomena in Multiphase Systems'', Burlington, MA. - For natural convection problem, it is often assumed that the fluid is incompressible except in the first term on the right-hand side of eq. (2.77); this is referred to as the Boussinesq assumption. The density of a mixture is a function of temperature and mass fractions of species. It can be expanded using a Taylor’s series near the vicinity of a reference point ( ): + - (2.78) + ==Further Reading== - where  is density at the reference point,   is the coefficient of thermal expansion, and  is the composition coefficient of volume expansion. Substituting eq. (2.78) into eq. (2.77), the momentum equation for natural convection is obtained + - (2.79) + ==External Links== - where the second and third terms on the right-hand side of eq. (2.79) describe the effect of buoyancy force due to temperature and composition variation within the system, respectively. +

## Revision as of 20:52, 5 November 2009

Newton’s second law states that, in an inertial reference frame, the time rate of momentum change of a fixed mass system is equal to the net force acting on the system, and it takes place in the direction of the net force. Mathematically, Newton’s second law of motion for fixed-mass in a reference frame that moves at a constant velocity Vref is written as $\sum {{\mathbf{F}} = \frac{{d{{(m{\mathbf{V}})}_{rel}}}}{{dt}}} \qquad \qquad(1)$

where the left-hand side is the net force vector acting on the fixed-mass system, and the right-hand side is the rate of momentum change.

For control volumes that contain only one phase, the integral form of Newton’s second law can be obtained by using eq. (2.3). With the applicable value of Φ and φ in eq. (2.3) defined as $\Phi = m{{\mathbf{V}}_{rel}}$ and $\phi = {{\mathbf{V}}_{rel}}$, one obtains $\sum {\mathbf{F}} = \frac{\partial }{{\partial t}}\int_V {\rho {{\mathbf{V}}_{rel}}dV} + \int_A {\rho ({{\mathbf{V}}_{rel}} \cdot {\mathbf{n}}){{\mathbf{V}}_{rel}}dA} \qquad \qquad(2)$

which is in the vector form and is valid in all three directions.

Forces acting on the control volume include body forces and contact forces that act on its surface. For example, for a multicomponent system that contains N components, if the body force per unit volume acting on the ith species is Xi, the total body force acting on the control volume is $\int_V {\left[ {\sum\limits_{i = 1}^N {{\rho _i}{{\mathbf{X}}_i}} } \right]} dV$, where ρi is the mass concentration (kg/m3) of the ith species. If the body force per unit mass is the same for different species (as is the case with gravity), the body force term is reduced to $\int_V {\rho {\mathbf{X}}} dV,$ where ρ is the density of the mixture. The stress tensor acts on the surface of a fluid control volume, and includes both normal and shear stresses. The net force may be written as $\sum {{\mathbf{F}} = \int_V {\left[ {\sum\limits_{i = 1}^N {{\rho _i}{{\mathbf{X}}_i}} } \right]} dV + \int_A {{{{\mathbf{\tau '}}}_{rel}} \cdot {\mathbf{n}}dA} } \qquad \qquad(3)$

where ${\mathbf{\tau '}}$ is the total stress tensor and n is the local normal unit vector on surface A. The dot product of a tensor of rank two, ${\mathbf{\tau '}}$, and a vector, n, is a vector that represents the force acting on the surface of the control volume per unit area. Combining eqs. (2) and (3), we obtain the momentum equation for the control volume of a single-phase system: $\begin{array}{l} \int_V {\left[ {\sum\limits_{i = 1}^N {{\rho _i}{{\mathbf{X}}_i}} } \right]} dV + \int_A {{{{\mathbf{\tau '}}}_{rel}} \cdot {\mathbf{n}}dA} \\ = \frac{\partial }{{\partial t}}\int_V {\rho {{\mathbf{V}}_{rel}}dV} + \int_A {\rho ({{\mathbf{V}}_{rel}} \cdot {\mathbf{n}}){{\mathbf{V}}_{rel}}dA} \\ \end{array} \qquad \qquad(4)$

where the two terms on the left-hand side represent, respectively, the body force and stress on the control volume, and the two terms on the right-hand side represent, respectively, the rate of momentum change in the control volume and the rate of the momentum flow into or out of the control volume. Vrel is the bulk velocity of the mixture that contains N components.

When the control volume includes multiple phases, integrations must be performed for each subvolume. In that case, the momentum equation becomes (Faghri and Zhang, 2006) $\begin{array}{l} \sum\limits_{k = 1}^\Pi {\left[ {\int_{{V_k}(t)} {\left[ {\sum\limits_{i = 1}^N {{\rho _{k,i}}{{\mathbf{X}}_{k,i}}} } \right]} dV + \int_{{A_k}(t)} {{{{\mathbf{\tau '}}}_{k,rel}} \cdot {{\mathbf{n}}_k}dA} } \right]} \\ = \sum\limits_{k = 1}^\Pi {\left[ {\frac{\partial }{{\partial t}}\int_{{V_k}(t)} {{\rho _k}{{\mathbf{V}}_{k,rel}}dV} + \int_{{A_k}(t)} {{\rho _k}({{\mathbf{V}}_{k,rel}} \cdot {{\mathbf{n}}_k}){{\mathbf{V}}_{k,rel}}dA} } \right]} \\ \end{array} \qquad \qquad(5)$

Equations (4) and (5) are momentum equations in a coordinate system that is attached to and moves with an inertial reference frame. For a fixed coordinate system that does not move with the reference frame (while the control volume still moves with the reference frame at velocity Vref), one can substitute the general variables $\Phi = m{\mathbf{V}}$ and $\phi = {\mathbf{V}}$ to obtain the momentum equation.

## References

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Burlington, MA.