Can you help with the calculation of grease produced by a small delicatessen serving lunch only and prepared foods? We are preparing to go before the sewer district board to request a variance from the 265 gallon, 170 lb external grease trap requirement they have given us. Need info by Monday, June 5, 2000. Thank you in advance for any assistance you may be able to provide.
To determine the cubic contents of the fixture you need to multiply L W D Then determine the actual drainage load (normally filled to 75% of capacity with waste water) due to plates and pots would equal 25%
Determine the flow rate and the drainage period (in general figure 1 minute drain time some codes allow 2 minutes)
I would need the ACTUAL size and number of sinks AND floor drains emptying to this grease interceptor to properly size this grease trap. At 170 pound requirement your looking at a 4″ inlet and outlet piping with a flow rate up to 100 GPM. By getting the actual sizes and not allowing for any future hook ups you can ask for a variance BUT you must have the real fixture unit demands and the actual number of cubic feet connected to this grease trap.
Once you get Lenth Width Depth and divide it by 2.31 (2.31 inches in a gallon) then you can know your fixture units because one cubic foot would equal this FU (7.48 gallons) Then the rest is just plugging in these numbers into sizing chart provided by the manufacturer of ther grease trap. Basic plumbing math.
Remember most model codes do demand a means of washing the floors in this kind of establishment and that is where the floor drain sizing comes into play and this of course increases the sizing of the grease trap.