Question

Question 1 (5 Points) Determining the dispersed oil content in the effluent water from a CPI plate separator Given: water flow rate25,000 bbl/day 125°F, Feed water specific gravity 1.06 125°F, Feed water viscosity Dispersed oil concentration 650 mg/l, Dissolved oil concentration= 10 mg/l. Total oil & grease -0.65 cp @125°F =660 mg/l. The dispersed oil droplet size distribution in feed water is as follows: 40-60 60-80 80-100 100-120 120 100% Microns <40 Vol. % 914 30 35 10 A vendor has quoted that one of its standard plate packs would be capable of reducing the total oil and grease content of the effluent water to less than 200 mg/. The vendor's standard plate pack has the following geometric specification: H 3.25ft, W 3.25ft, L -5.75ft, h-0.69 in., 6-45° Calculate the total oil and grease content in effluent water from the plate pack to check the vendor's quoted performance.

Answer 1

We have to solve this using Stoke's law:

$V_r=\frac{g(\rho_w-\rho_o)d^2}{18\mu}$ --------------------(1)

where, $V_z=$ particle terminal velocity

$\rho_w=1.06*10^3 kg/m^3$ =density of feed water

$\rho_o$ = density of oil phase (unknown)

= diameter of droplet

$\mu=0.65 cp=6.5*10^{-4} kg/m-s$ = viscosity of feed water

Now we need to calculate the total volume of the needed plate pack $(V_P)$ , can be given as

$V_P=L*W*H*n*f$

where, L= length of the plate pack= 5.75 ft= 1.75 m

W= width of the plate pack=3.25 ft= 1 m

H= height of the plate pack= 3.25 ft 1 m

n= number of plate in the pack (not given)

f= factor for deducting the plate thickness of from the total pack volume (not given)

Now to solve this problem, we need to assume some of the standard value for this calculation

such as density of oil,   $\rho_o=0.845*10^3 kg/m^3$

and factor for deducting the plate thickness, f=0.8,

n can be calculated as= $\frac{H}{h}=\frac{3.25*12}{0.69}$ =100

Hence, $V_p=140m^3$

Now, residence time of oil-water in the plate pack= $\tau=\frac{V_p}{v_w}=$

where, $v_w=$ flow rate of feed water=25,000 bbl/day=25000*159/1000 m³=0.046 m/s

Hence, $\tau=3043 s$

Hence, minimum rise velocity of separated particle= $V_r=\frac{h}{\tau}=\frac{0.0175}{3043}=5.75*10^{-6}m/s$

Hence the diameter of the particle separable from the stream can be calculated from the equation 1 is

$d=\sqrt{V_r*18*\mu/g(\rho_w\rho_o)}$

=5.6 *10^-6 m= 5.6 micron

As, all the particle are above this size as per the given distribution, it can separate all the oil water droplet. However, the distribution data is ambiguous and very small fraction of particle may be less than the size range.

Hence, the claim of the maker is right.