II)Gas/liquid separation

It is often based on the principal of gravity settling, when liquid droplets sus- pended in rising gas vapors settle down at the bottom of the separation vessel and are eventually taken out through the bottom. Gas stream separated from liquid is taken out from the top of the separation vessel.

Gas/Liquid separation is usually accomplished in three stages :

The first stage, primary separation, uses an inlet diverter to cause the largest droplets to impinge by momentum and then drop by gravity. The next stage; secondary separation, is gravity separation of smaller droplets as the vapor flows through the disengagement area. Gravity separation can be aided by utilizing distribution baffles that create an even velocity distribution in the fluid, this allowing enhanced separation. The final stage; is mist elimination, where the smallest droplets are coalesced on an impingement device, such as a mist pad or vane pack, followed by gravity settling of the larger formed droplets

The separation of liquid droplets from vapor phase can be explained with the help of equation for terminal velocity of liquid droplets. In the gravity settling section of a separator, liquid droplets are removed using the force of gravity. Liquid droplets will settle out of a gas phase if the gravitational force acting on the droplet is greater than the drag force of the gas following around droplet. These forces can be described mathematically using the terminal or free settling velocity 

a)Equation of motion

Gravity separation is conceptually simple. The droplets of any liquid in a vapor flow are acted on by three forces: gravity, buoyancy, and drag. The resultant of these forces causes motion in the direction of the net force. A primary design goal is to size the separator such that the drag and buoyancy forces succumb to the gravity force causing the droplet to disengage, i.e. separate, from the vapor flow. The force balance on a typical liquid droplet can be established by application of Newton’s Law: where the forces, Fi, and acceleration, a, are functions of time, t, and md is the mass of the droplet. The magnitudes of the gravity, buoyancy and drag forces, 

 

                   

respectively, are defined as follows:

FG ρL Vd  g                                  (1)

FB ρv  Vd  g                               (2)

FD ρv  U2  C Ad/2                   (3)

 

The gravity force is always directed downward, the buoyancy force is opposite the gravity force, and the drag force is opposite the direction of droplet velocity. The droplet Reynolds number is defined as the ratio of inertia and viscous forces and the characteristic length is the droplet diameter. The droplet Reynolds number is defined as follows: 

$Re_d=\frac{\rho_vUD_d}{\mu}$

where rhov and μv are the vapor density and absolute viscosity, respectively, and U is the velocity of the vapor past the droplet relative to the droplet’s velocity. The drag coefficient, CD, for a smooth sphere can be numerically estimated using the following (Bird, 1960) 

$C_D=\frac{24}{Re_d}$      (Re_d<1)

$C_D=\frac{18.5}{Re_d}$   (1<Re_d<500)

$C_D\sim 0.44$                      ($Re_d<2e10^5$)

While both of the estimates are equally valid, Equation (8) will be used for the present model development because it is defined over the entire Reynolds number range of interest.

The gravity force Fg acts downward and thus incites a velocity in that di- rection while the buoyancy force Fb and the drag force acts FD on the opposite direction of movement and thus upward. This can be expressed as follow : 

$\sum F=m_pa=F_g-F_b-F_D$

$\sum F=m_pa=m_pg-\rho_fV_pg-\frac{1}{2}C_DA_p\rho_f(u-v)^2$

b)Terminal Velocity

Terminal Velocity Most separators will have a given dimension stated by the terminal or maximum velocity achieved by particles. So in that manner it is the first expression to be obtained. Furthermore, for the case of maximum velocity , there will be no acceleration( a=0). Hence the equation becomes :

$m_pg-\rho_fV_pg-\frac{1}{2}C_DA_P\rho_f(u_{max})^2$

Given the volume and area of a sphere and the density expressed via Vp= corresponding volume of the particles, rp=volumetric radius 

$d=2r_p=\left(\frac{6V_p}{\pi})^\frac{1}{3}\right)$

$A_p=A_{sphere}=4\pi r_{p}^{2}$

$A_p>A_{sphere}(non-sphereical)$

$\rho_p=\frac{m_p}{V_p}$

To consider the shape of the particle, surface area is used. An increase in it may cause an increase in drag. 

$\frac{1}{2}C_D\pi r^2\rho_fu_{max}^2=V_fg(\rho_p-\rho_f)$

$\frac{1}{2}C_D\pi r^2\rho_fu_{max}^2=\frac{4}{3}\pi r^3g(\rho_p-\rho_f)$

$u_{max}^{2}=\frac{8rg(\rho_p-\rho_f)}{3\rho_fC_D}$

Leading to the expression for maximum velocity expressed as 

$u_{max}=\sqrt{\frac{8rg(\rho_p-\rho_f)}{3\mu_f}}$

Which can also be expressed as : But due to the drag coeddiecent used ( based on stokes ), only applicable at low Re 

$u_{max}=\frac{2r^2g(\rho_p-\rho_f)}{9\mu_f}$

Terminal velocity using the Stokes correction factor : Another way to express the terminal velocity is by the use of a correction factor . This cor- rection factor is a coefficient by which the flow in question would be assimilated to the one assumed by stokes following 

$F_D=-3\pi d\mu_fu_f=-\frac{1}{8}\pi d^2\rho_f\vert u \vert C_D$

Thus yielding 

$u_{max}\frac{2}{9}\frac{r^2(\rho_p-\rho_f)}{\mu_ff}$

And would be applicable for every interval in which a set of correction f exist being calculated experimentally for every case of flow and factors involved. 

 

 

Stokes regime

Transient Velocity for Stokes Regime.

The transient velocity of such particle is time dependent. It will increase over the time until it reaches the final maximum velocity umax .What follows is the deriving of such expression : 

$\sum F=m_pa=m_p\frac{du}{dt}=F_g-F_b-F_D$

$\sum F=m_pa=m_p\frac{du}{dt}=m_pg-\rho_fV_pg-\frac{1}{2}C_DA_P\rho_fu^2$

Using the same expressions as for solving for umax  leads to 

$\rho_pV_p\frac{du}{dt}=V_pg(\rho_p-\rho_f)-\frac{1}{2}C_D\pi r^2\rho_fu^2$ 

By substituting CD with the Reynolds dependent equation derived by Stokes expression and substituting this Reynolds with its definition yields 

$\rho_pV_p\frac{du}{dt}=V_pg(\rho_p-\rho_f)-\frac{1}{2}\frac{24\mu_f}{2r\rho_fu}\pi r^2\rho_fu^2$

$\frac{2}{9}\frac{r^2\rho_p}{\mu_f}\frac{du}{dt}=\frac{2}{9}\frac{r^2g(\rho_p-\rho_f)}{mu_f}-u$

Which by introducing $\lambda$ , defined by someone as a particle relaxation time, which is assuming $\rho_f>>\rho_p  ---> \delta\rho=\rho_p yeilds$

$\lambda\frac{2}{9}\frac{\rho_pr^2}{\mu_f}$

$\lambda\frac{du}{dt}=\lambda g-u$

Noticing that g = umax and integrating from zero to a time t in which a velocity u is present yields 

$\lambda\frac{du}{dt}=u_{max} -u$

$u=u_{max}(1-exp\frac{-t}{\lambda})+Constant$

Assuming the particle starts at rest, in other words that u0 = 0 at t = 0 then the Transient velocity can be expressed as follow 

$u_{max}\frac{2}{9}\frac{r^2g(\rho_p-\rho_f)}{\mu_f}(1-exp(-f\frac{t}{\frac{2}{9}\frac{\rho_pr^2}{\mu_f}}))$ 

Notice that $\lambda$ is the coefficeint determining how fast terminal velocity is obtained.

Terminal Velocity for Intermediate Value of Re_p

Solving for the transient velocity profile for intermediate values of Re, equa- tions proposed by Putnam, Schiller-Naumann and similar have to be incor- porated and solved accordingly and incorporating them yields 

$\rho_pV_p\frac{du}{dt}=V_pg(\rho_p-\rho_f)-/frac{1}{2}\pi r^2\rho_f(C_1u+C2u^k)$

which by introducing the constants a,b and c yields

$a=\frac{V_pg(\rho_p-\rho_f)}{\rho_pV_p}$

$b=\frac{\frac{1}{2}\pi r^2\rho_fC_1}{\rho_pV_p}$

$c=\frac{\frac{1}{2}\pi r^2 \rho_f C_2}{\rho_pV_p}$

$\frac{du}{dt}=a-bu-cu^k$

to solve with respect to a varying velocity , a solution for the integral is required. Hence a method for generally approximating the integral with respect to u is proposed.

$\int\frac{du}{a-bu-cu^k}$

This integral is to be solved in the open interval for which u varies , where u=z>0 . 

Solving & approximation of the integral.Having proved that the approximation is reasoable, and processding with calculating the integral by susbituing the approximation into the integral, yields the following

$\int \frac{du}{a-bu-cu^k}$

This integral is trivial to solve and when computed yields

$\frac{A}{2(u_{top}-z}ln(\frac{u-2u_{top}+z}{w-z})+Bu+C$

​where C is a constant of integration.