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At this stage, it's not possible to say which of the two options incurs the greatest cost. However, it's reasonable to assume in both cases that energy consumption is a significant contributor.

The main contributors to water and wastewater process operating expenditure (or OPEX) generally are: energy and chemicals consumption, component replacement (including membranes, as well as hardware such as valves and pumps), waste disposal and labour.

Back to Section 1 Introduction

Flow has units of m3/s; pressure has units of kg/(ms2).

So, flow x pressure takes units of kg m/ (s.m.s2), or kg m2/s3. These are the same units as for power.

Back to Section 2 The fundamental and derived SI units

The units of power are kg m2/s3. Mass flow has units of kg/s. So power per mass flow has units of m2/s2.

The units of energy are kg m2/s2.  So, energy per unit mass has the same units as power per unit mass flow (i.e. m2/s2).

If this is multiplied by density, which has units of kg/m3, then it becomes kg/m3 x m2/s2 = kg/(m s2). These are the same units as those of pressure.

Back to Section 2 The fundamental and derived SI units

Power = flow x pressure.

In SI units: 1 bar is 10Pa or 100 kPa, so 0.5 bar pressure is 50 kPa; 1000 m3/h is 0.278 m3/s.

So, the power is 50 x 0.278 = 13.9 kW.

Feed flow rate = 2 x product flow rate =  2 x 0.278 = 0.556 m3/s.

Pressure = 54 bar = 5400 kPa.

So, the power required to generate the product is 0.556 x 5400 kPa = 3002 kW (or 3 Mega Watts, MW).

This power generates 2000 m3/h product flow.

Specific energy consumption (SEC, kWh/m3) = Power (kW) / Flow (m3/h).

So, SEC = 3000 kW / 2000 m3/h = 1.5 kWh/m3.

Flux J = Flow/Area, where the flow is in litres per hour and the area is provided by 200 modules at 30 m2.

So, J = 1000 x 150 / (200 x 30) = 25 LMH.

In SI units of m3/(m3.s) the flux would be 25 / (200 x 30 x 3600) = 1.16 x 10-6 m/s.

Note that flux has units of velocity (and in some old text books is referred to as 'filtration velocity').

Back to Section 4 Membrane process parameters

Permeability (LMH/bar) = Flux/TMP, for flux in LMH and TMP in bar.

So TMP = Flux/permeability = 25/80 = 0.313 bar.

Back to Section 4 Membrane process parameters

Converting flow to m3/h: Q = 43 L/s x 3600 (s/h) / 1000 (L/m3) = 155 m3/h.

So, recovery = permeate flow/feed flow = 150/155 = 96.9%

(a) Specific aeration demand with reference to membrane area SADm = Aeration rate / membrane area, where membrane area = 200 modules x 30 m2/module = 6000 m2.

So, SADm = 2400 Nm3/h / 6000 m2 = 0.40 Nm3/h per m2.

(b) Specific aeration demand with reference to permeate volume Q'a,m = SADm / flux, for flux in units of m3/(m2h).

A flux of 25 L/(m2h) is 0.025 m3/(m2h), so Q'a,m= 0.40 / 0.025 = 16 Nm3/m3 permeate.

Membrane aeration, biological aeration, sludge transfer, permeation and agitation.

Back to Section 5 Components of a membrane bioreactor

SEC for membrane air scour Ea,m = E'a,m Q'am

E'a,m = 0.011/ԑ kWh/Nm3 air, where ԑ is the pumping efficiency, assumed to be 60%. According to Question 10(b), Q'a,m= 16.

So, Ea,m =  (0.011 / 0.6) x 16 = 0.293 kWh/m3 permeate.

Back to Section 5 Components of a membrane bioreactor

SEC for biological aeration Ea,b = E'a,b Q'a,b.

Q'a,b = 0.023 O2 Nmair per m3 permeate; E'a,b = 0.013/ԑ kWh per Nm3 air; O2 =  0.67 x COD; ԑ is the pump efficiency as before, assumed to be 60%

So, for COD = 800 mg/L, Ea,b =  0.67 x 800 x 0.013 x 0.023 / 0.6 = 0.271 kWh per m3 permeate.

Back to Section 5 Components of a membrane bioreactor

(a) Sludge pumping and stirring energy Es = 0.032 / ԑ, where ԑ = 60% as before.

So, Es = 0.032 / 0.6 = 0.053 kWh/m3 permeate.

(b) Permeate pumping energy Ep = (P + Ploss,filt)/(36 ԑ), where P is the transmembrane pressure, calculated in Question 8, and Ploss,filt the pressure losses, assumed to be 0.15.

So, Ep = (0.15 + 0.313)/(36 x 0.6) = 0.021 kWh/m3 permeate.

Back to Section 5 Components of a membrane bioreactor

The total energy must be divided by the conversion, calculated from Question 9 to be 96.9%.

So, the total SEC of the MBR, EMBR = (Ea,m + Ea,b + Es + Ep + Epi) / R = (0.293 + 0.271 + 0.053 + 0.021 + 0.10 )/96.9% = 0.76 kWh/m3 permeate.

The aeration components (Ea,m + Ea,b) / R make up around 76% of the total SEC.

Back to Section 5 Components of a membrane bioreactor

The spreadsheet should be used to determine the impact on the SEC of incrementally increasing the CF value. Accordingly, a minimum in the SEC can be determined at a CF of ~1.8. Note that recovery R = 1 - 1/CF = 44%.

Back to Section 6 Components of the reverse osmosis process

Concentration = 1000 mg/L = 1 g/L = 0.1 g per 100 mL, which is ~ 0.1 g per 100g, and so 0.1 wt%.

(a) Using the spreadsheet in the same way as in Question 16, the optimum CF is ~3.5 when SEC = 0.74. So, the recovery is much higher than for RO - around 71% cf. 44%.

(b) Given the probable cost of the energy recovery turbine, investing in it is unlikely to be justified to recover a fraction of a kWh/m3. You can check the loss of recovered energy from removing the energy recovery turbine by entering a value of 0% in the turbine efficiency value and noting the change in SEC.

Back to Section 6 Components of the reverse osmosis process

Assuming the SEC for the UF/MF membrane pre-treatment to be the same as the permeate pumping component of the MBR (from Question 14(b)), then Ep = 0.021 kWh/m3 for the MBR filtration.

According to the RO calculation from Question 16, only 44% of the UF/MF filtrate is converted to product. So, Ep = 0.021/0.44 = 0.048 kWh/m3 overall.

Back to Section 6 Components of the reverse osmosis process

The SEC for the RO step, according to Answer 16, is 4.17 kWh/m3. The UF/MF pre-treatment, according to Answer 18, has a SEC of 0.048 kWh/m3.

So, the total SEC for RO is 4.17 + 0.048 = 4.22 kWh/m3.

Back to Section 6 Components of the reverse osmosis process

For the MBR, as with the UF/MF pre-treatment step in Answer 17a, around 71% of the MBR permeate can be assumed to form the RO permeate product. So, the SEC for the MBR step = 0.76 kWh/m3 (according to Answer 15) / 0.71 = 1.07. For the RO downstream of the MBR the SEC is 0.74, also from Answer 17a.

So, the total SEC for the MBR and the RO post-treatment = 1.07 + 0.74 = 1.81 kWh/m3.

Back to Section 6 Components of the reverse osmosis process