## Answers to questions

**Answer to Question 1:**

**Answer to Question 1:**

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

**Answer to Question 2:**

Flow has units of m^{3}/s; pressure has units of kg/(ms^{2}).

So, flow x pressure takes units of kg m^{3 }/ (s.m.s^{2}), or kg m^{2}/s^{3}. These are the same units as for power.

Back to Section 2 The fundamental and derived SI units

**Answer to Question 3:**

The units of power are kg m^{2}/s^{3}. Mass flow has units of kg/s. So power per mass flow has units of m^{2}/s^{2}.

The units of energy are kg m^{2}/s^{2}.
So, energy per unit mass has the same units as power per unit mass flow (i.e. m^{2}/s^{2}).

If this is multiplied by density, which
has units of kg/m^{3}, then it becomes kg/m^{3} x m^{2}/s^{2} = kg/(m s^{2}). These are the same units as those of pressure.

**Answer to Question 4:**

Power = flow x pressure.

In SI units: 1 bar is 10^{5 }Pa or 100 kPa, so 0.5 bar pressure is 50 kPa; 1000 m^{3}/h is 0.278 m^{3}/s.

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

Back to Section 3 Specific energy consumption

**Answer to Question 5:**

Feed flow rate = 2 x product flow rate = 2 x 0.278 = 0.556 m^{3}/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).

Back to Section 3 Specific energy consumption**Answer to Question 6:**

This power generates 2000 m^{3}/h product flow.

Specific energy consumption (SEC, kWh/m^{3}) = Power (kW) / Flow (m^{3}/h).

So, SEC = 3000 kW / 2000 m^{3}/h = 1.5 kWh/m^{3}.

**Answer to Question 7:**

Flux J = Flow/Area, where the flow is in litres per hour and the area is provided by 200 modules at 30 m^{2}.

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

In SI units of m^{3}/(m^{3}.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').*

**Answer to Question 8:**

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

**Answer to Question 9:**

Converting flow to m^{3}/h: Q = 43 L/s x 3600 (s/h) / 1000 (L/m^{3})^{ }= 155 m^{3}/h.

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

**Answer to Question 10:**

(a) Specific aeration demand with reference to membrane area SAD

_{m}= Aeration rate / membrane area, where membrane area = 200 modules x 30 m

^{2}/module = 6000 m

^{2}.

^{}

So, SAD

_{m}= 2400 Nm

^{3}/h / 6000 m

^{2}= 0.40 Nm

^{3}/h per m

^{2}.

(b) Specific aeration demand with reference to permeate volume Q'

_{a,m}= SAD

_{m}/ flux, for flux in units of m

^{3}/(m

^{2}h).

A flux of 25 L/(m

^{2}h) is 0.025 m

^{3}/(m

^{2}h), so Q'

_{a,m}= 0.40 / 0.025 = 16 Nm

^{3}/m

^{3}permeate.

Back to Section 4 Membrane process parameters

**Answer to Question 11:**

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

Back to Section 5 Components of a membrane bioreactor

**Answer to Question 12:**

SEC for membrane air scour E

_{a,m}= E'

_{a,m}Q'

_{am}

E'

_{a,m}= 0.011/ԑ kWh/Nm

^{3}air, where ԑ is the pumping efficiency, assumed to be 60%. According to Question 10(b), Q'

_{a,m}= 16.

So, E

_{a,m}= (0.011 / 0.6) x 16 = 0.293 kWh/m

^{3}permeate.

Back to Section 5 Components of a membrane bioreactor

**Answer to Question 13:**

SEC for biological aeration E_{a,b} = E'_{a,b} Q'_{a,b}.

Q'_{a,b} = 0.023 O_{2} Nm^{3 }air per m^{3} permeate; E'_{a,b} = 0.013/ԑ kWh per Nm^{3} air; O_{2} = 0.67 x COD; ԑ is the pump efficiency as before, assumed to be 60%

So, for COD = 800 mg/L, E_{a,b} = 0.67 x 800 x 0.013 x 0.023 / 0.6 = 0.271 kWh per m^{3} permeate.

**Answer to Question 14:**

(a) Sludge pumping and stirring energy E_{s} = 0.032 / ԑ, where ԑ = 60% as before.

So, E_{s} = 0.032 / 0.6 = 0.053 kWh/m^{3} permeate.

(b) Permeate pumping energy E_{p} = (P + P_{loss,filt})/(36 ԑ), where P is the transmembrane pressure, calculated in Question 8, and P_{loss,filt }the pressure losses, assumed to be 0.15.

So, E_{p} = (0.15 + 0.313)/(36 x 0.6) = 0.021 kWh/m^{3} permeate.

**Answer to Question 15:**

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

So, the total SEC of the MBR, E_{MBR} = (E_{a,m} + E_{a,b} + E_{s} + E_{p} + E_{pi}) / R = (0.293 + 0.271 + 0.053 + 0.021 + 0.10 )/96.9% = 0.76 kWh/m^{3} permeate.

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

**Answer to Question 16:**

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

**Answer to Question 17:**

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/m^{3}. 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.

**Answer to Question 18:**

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 E

_{p}= 0.021 kWh/m

^{3}for the MBR filtration.

According to the RO calculation from Question 16, only 44% of the UF/MF filtrate is converted to product. So, E

_{p}= 0.021/0.44 = 0.048 kWh/m

^{3}overall.

Back to Section 6 Components of the reverse osmosis process

**Answer to Question 19:**

The SEC for the RO step, according to Answer 16, is 4.17 kWh/m

^{3}. The UF/MF pre-treatment, according to Answer 18, has a SEC of 0.048 kWh/m

^{3}.

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

^{3}.

Back to Section 6 Components of the reverse osmosis process

**Answer to Question 20:**

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/m

^{3 }(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/m

^{3}.

Back to Section 6 Components of the reverse osmosis process

**Answer to Question 21:**

Wastewater recycling incurs less than half the energy of seawater desalination. So, based on energy efficiency alone, wastewater reuse would always be the preferred choice of water supply.

Back to Section 6 Components of the reverse osmosis process