Page 256 - Membranes for Industrial Wastewater Recovery and Re-Use

System design aids 22 5

v~ = tD X Qp/tir = 108 X 600/2160 = 30 m3 dayp1
The total volume demand is further increased by the use of permeate product
for backflushing for 20 seconds at three times the flux every 12 minutes which,
ignoring the cleaning downtime, equates to a volume demand of

Vbackf = Qp X (3 X 0.33/12.33) X (1 - &/ti,)
= 600 x 0.0811 x (1-50/2160) = 47.5 m3 day-'

where the (1 - t,/ti,) factor accounts for the cleaning period over which no
backflushing takes place.
So, the total volume demand, or feed flow, is:
Qf = 600 + 30 + 47.5 = 678 m3 day-'

which implies a conversion of 88%.
For a flux Jof 2 5 1 m-2 h-' , the membrane area required is:

A = 678 x 1000/(25 x 24) = 1130 m2
Thus the total number of membrane modules required is:
N = A/Am = 1130/46 = 24.6

which implies 25 membrane modules are required. The capital cost, for
membrane costs of €40 per m2, is then given by:

CAPEX, €k = €80 + 25 x 40 x 46/150 = €386 K

Operational costs are obtained from the pumping and aeration energy demands.
According to Equation (2.23), and based on a 40% pumping efficiency, the liquid
pumping energy demand El, in kWh per m3 permeate relates directly to the
average TMP (in Pa) over the whole cycle:
Ep (kWh m-3) = P,,,/(0.4 x 1000 x 60 x 60) = Pa,,, ,,p,/2.16 x 10'

where Pa,, = (P,,, + Pmi,)/2 = (lo5 + FJR,)/~ = 52 800 Pa.
The pumping demand energy (operating cycle) is thus given by:

EP = 52 800/2.16 x lo6 = 0.024 kWh m-'

At 8p/KWh, this relates to a cost of f0.002 per m3.
Aeration energy demand is given by:

EA = 0.02QJJA = 0.02 x 50/(25 x x 46) = 0.87 kWh m-'

which relates to a cost of 50.070 per m3.