An oil/water separator has been designed for the treatment of water from a car w
ID: 1714229 • Letter: A
Question
An oil/water separator has been designed for the treatment of water from a car
wash. The separator has dimensions of 6 ft x 2.5 ft x 2.5 ft (LxWxH). The flow rate
to the separator ranges from a high of 30 gpm to a low of 10 gpm. A new detergent
has just been purchased which keeps the size of the oil droplets to between 90 and
180 m. The separator must remove at least 50% of the incoming oil. If the system
must meet the removal requirements under all conditions, what is the minimum
value of the specific gravity of the oil?
Explanation / Answer
For the given question
Area of tank = 6*2.5 = 15 sqft
Maximum overflow rate = 30 gpm = 113.55 lpm = 113.55* 0.035 = 3.97 cfm
Minimum overflow rate = 10 gpm = 37.85 lpm = 1.32 cfm
Thus maximum overflow velocity = 3.97/15 = 0.265 ft/ min = 1.35 x 10^-3 m/s
Minimum overflow velocity = 1.32/15 = 0.088 ft/ min = 4.47 x 10^-4 m/S
Maximum particle size = 180 micro metre
Minimum particle size = 90 micro metre
Assuming 20 temperature and dynamic viscosity of water = 1.0016 mPa.s
For at least 50 percent removal at all times
Settling velocity of 90 micro metre particles = 0.5 * minimum overflow velocity = 2.235 x 10^ -4 m/s
Thus we get Yp- Yl = 0.497 KN/ m3
As we know that water is liquid thus Yl = 10 KN/ m3
Also oil is particle to be removed thus Yp = 0.497 + 10 = 10.497 KN/ m3 which is minimum specific weight and 1.0497 is minimum specific gravity of oil at 20