1. Modelling & Simulation of Chemical Engineering Systems Department of Chemical Engineering King Saud University 501 هعم : تمثيل الأنظمة الهندسية على الحاسب الآلى

2. LECTURE #5 Equations of Change

3. Total Mass balance Mass in: The mass entering in the x-direction at the cross sectional area (  y  z) is : (  v x )| x  y  z  t The mass entering in the y-direction at the cross sectional area (  x  z) is   )| y  x  z  t The mass entering in the z-direction at the cross sectional area (  x  y) is (  v z )| z  x  y  t Mass out: The mass exiting in the x-direction is: (  v x )| x+  X  y  z  t The mass exiting in the y-direction is:   )| y+  y  x  z  t The mass exiting in the z-direction is: (  v z )| z+  z  x  y  t Rate of accumulation: The rate of accumulation of mass in the elementary volume is: (  )|   x  y  z - (  )| t  x  y  z

4. Total Mass balance Continuity equation If the fluid is incompressible

5. Component Balance Equation (  A | t+  t −  A | t )  x  y  z = (n Ax | x+  x − n Ax | x )  y  z  t + (n Ay | y+  y − n Ay | y )  x  z  t + (n Az | z+  z − n Az | z )  x  y  t +r A  x  y  z  t n A =  A v + j A

6. Component Balance Equation If the binary mixture is a dilute liquid and can be considered incompressible, then density  and diffusivity D AB are constant In molar units

7. Momentum Balance Navier-Stock’s equation (Newtonian, incompressible fluid)

8. Energy balance for incompressible fluids

9. For solids, the density is constant and with no velocity, i.e. v = 0, the equation is reduced to:

10. Balance Equations in Cartesian Coordinates

11. Balance Equations in Cylindrical Coordinates x = rcos(  ),y = rsin(  ),z = z

12. Examples of Application of Equations of change 1. Liquid flow in a Pipe One dimensional flow through the pipe of an incompressible fluid Cylindrical coordinate, Continuity Equation The plug flow assumptions imply that v r = v  = 0

13. Examples of Application of Equations of change 2 Diffusion with Chemical Reaction in a Slab Catalyst To model the steady state diffusion with chemical reaction of species A in a slab catalyst, we use the component balance equation of change. The fluid properties are assumed constant One dimensional, Cartesian coordinate Steady State No bulk flow: v x = v y = v z = 0 One dimension, z

14. Examples of Application of Equations of change- Plug Flow Reactor The isothermal plug flow reactor can be modeled using the component balance equation The one dimensional assumption and plug flow conditions imply that vr = v  = 0 and. :

15. Non-isothermal PFR For fluid with constant properties and at constant pressure, we have: Using the plug-flow assumptions, v r = v  = 0 and, and neglecting the viscous forces, the term  H includes the heat generation by reaction rate R A and heat exchanged with the cooling jacket, h t A(T – Tw). :

16. Energy Transport with Heat Generation A solid cylinder of radius R in which heat is being generated due to some reaction at a uniform rate of F H ( J/m2s ). A cooling system is used to remove heat from the system and maintain its surface temperature at the constant value Tw The solid is of constant density, thermal conductivity and heat capacity. The system is at steady state i.e. The variation of temperature is only allowed in radial directions Derive the temperature variations in the cylinder.

17. The energy balance is reduced to: Boundary conditions: The temperature at the wall is constant: The maximum temperature will be reached at the center ( r = 0):