Mass balance equation right-hand side

The Right-Hand Side of the Mass Balance Equation... 

In Chapter 2 we developed models based on analyses of systems that had simple inputs. The right-hand side was either a constant or it was simple function of time. In those systems we did not consider the cause of the mass flow—that was literally external to both the control volume and the problem. The case of the flow was left implicit. The pump or driving device was upstream from the control volume, and all we needed to know were the magnitude of the flow the device caused and its time dependence. Given that information we could replace the right-hand side of the balance equation and integrate to the functional description of the system. 

The Navier-Stokes equations have a complex form due to the necessity of treating many of the terms as vector quantities. To understand these equations, however, one need only recognize that they are not mass balances but an elaboration of Newton s second law of motion for a flowing fluid. Recall that Newton s second law states that the vector sum of all the forces acting on an object ( F) will be equal to the product of the object s mass (m) and its acceleration (a), or XF = ma. Now consider the first of the three Navier-Stokes equations listed above, Eq. (10). The object in this case is a differential fluid element, that is, a small cube of fluid with volume dx dy dz and mass p(dx dy dz). The left-hand side of the equation is essentially the product of mass and acceleration for this fluid element (ma), while the right-hand side represents the sum of the forces... 

Control Volume Mass Balance. We can now combine equations (2.1), (2.5), (2.6), (2.11), and (2.13) into a mass balance on our box for Cartesian coordinates. After dividing hyV = dx dy dz and moving the diffusive flux terms to the right-hand side, this mass balance is... 

These equations must be supplemented by a kinetic equation for the time dependence of the degree of conversion P(t), and the dependence of the viscosity of a reactive mass on (3, temperature, and (perhaps) shear rate, if the reactive mass is a non-Newtonian liquid. The last two terms in the right-hand side of Eq. (2.89) are specific to a rheokinetic liquid. The first reflects the input of the enthalpy of polymerization into the energy balance, and the second represents heat dissipation due to shear deformation of a highly viscous liquid (reactive mass). 

The total balance would be F = W + 350 (in kg). Are the three component balances independent Because of the zero terms in the right-hand sides of Eqs. (b) and (c), no variation or combination of Eqs. (b) and (c) will lead to Eq. (a). Are Eqs. (b) and (c) redundant equations No constant exists that when multiplied into Eq. (b) gives Eq. (c), hence the three mass balances are independent. 

ATP]0 denote the initial concentration of bound hydrogen ions and ATP, respectively. The third term on the right-hand side of the mass balance equation computes the number of times the reference reaction has turned over, generating hydrogen ions. Given a set of initial concentrations, Equation (2.20) can be numerically solved for pH at any ATP concentration. 

In chemical reactions, there is always a mass balance, i.e., number of atoms of any species on the left-hand side of an equation is the same as that of the species on the right-hand side. In an electrochemical reaction, in addition to the mass balance, there is also a charge balance in which the total charge on the left-hand side of an equation is the same as that on the right-hand side. Thus, mass and charge balance are two basic requirements of an electrochemical equilibrium. 

The term on right hand side of Eq. (19) is the volume expansion due to liquid-vapor phase change. From the conditions of the mass continuity and energy balance at the vapor-liquid interface, the following equations are obtained. 

When we write down a chemical reaction in the form of an equation, we have to balance the equation to ensure that we don t make mass appear or disappear. That is to say, the number of atoms of any particular element that are present at the beginning of the reaction, on the left-hand side or LHS of the arrow, is equal to the number on the right-hand side (RHS). Looking at a simple equation such as the combustion or burning of hydrogen ... 

Where t is time, z are the axial position in the column, qt is the concentration of solute i in the stationary phase in equilibrium with Cu the mobile phase concentration of solute /, u is the mobile phase velocity, Da is the apparent dispersion coefficient, and F is the phase ratio (Vs/Vm). The equation describes that the difference between the amounts of component / that enters a slice of the column and the amount of the same component that leaves it is equal to the amount accumulated in the slice. The fist two terms on the left-hand side of Eq. 10 are the accumulation terms in the mobile and stationary phase, respectively [109], The third term is the convective term and the term on the right-hand side of Eq. 10 is the diffusion term. For a multi component system there are as many mass balance equation, as there are active components in the system [13],... 

A properly written chemical equation must contain properly written formulas and must be balanced. If the reaction represents a closed system, as is usually the case, then the law of mass conservation tells us that no mass can be gained or loss during the reaction. Furthermore, if no nuclear decay is occurring, the number of atoms of each element must remain constant. That is, there will be same number of each type of atom on the left and right hand side of the equation. 

Figure 2-2. An arbitrarily chosen control volume of fixed position and shape immersed in a fluid with velocity u. The velocity of the surface of the control volume is zero, and thus there is a net flux of fluid through its surface. Equation (2-3) represents a mass balance on the volume V, with the left-hand side giving the rate of mass accumulation and the right-hand side the net flux of mass into V that is due to the motion u.

Equation (10.1) is a statement of mass balance on a per volume of groundwater basis for a given component i. The first two terms on the right hand side of the equation refer to the net influxes of mass due to dispersive and advective transport in groundwater. 

Answer (a) The mass number and atomic number are 212 and 84, respectively, on the left-hand side and 208 and 82, respectively, on the right-hand side. Thus, X must have a mass number of 4 and an atomic number of 2, which means that it is an a particle. The balanced equation is... 

Equating the right-hand sides of equations (12.56) and (12.57) and using the component mass balances... 

In this last equation, the first term on the right-hand side is the entropy change of stream 1 from its initial temperature to the final temperature of stream 3, and the second term is the entropy change of the cold stream from its initial temperature to the exit temperature in stream 3. The entropy terms can be calculated since all temperatures are known. What is still missing is the mass flow rate of stream 2. This is obtained from the energy balance ... 

The component balance equation is expressed with mass fractions of the individual species, cOt and the density of the reaction mixture, Pg. In Eq. (5.1) Ft is the total volumetric flow rate, = zjL the dimensionless reactor length. The first term on the right-hand side corresponds to sources by chemical reactions the second term quantifies the local mass fluxes, jj(< ), entering through the membrane. In the FBR model the flux term is omitted. The catalyst density is denoted by p t and e is the averaged porosity of the catalyst bed. Vr is the reactor volume. Mi the molecular mass of species i, the surface area of the membrane wall. The reaction rate rj represents the jth rate of the overall Nr reactions taking place. 

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