3 In a desalination plant seawater is pumped to a high press

3. In a desalination plant, seawater is pumped to a high pressure (65 atm) and passed through a reverse osmosis (RO) unit that removes the salt from the water. A simplified diagram of this process is shown below Potable water RO unit water return 10 m The inlet to the RO unit is located 10 m above sea level and the water enters the unit at a pressure of 65 atm. At normal operating conditions, the RO unit processes 100 m3/min of seawater. Of this amount, 55% becomes brackish water and is returned to the sea while the remaining 45% is potable (drinkable water. For this set up answer the following: (a) What power (in kW) must be supplied to the pump if its efficiency is 93% and the frictional losses in the piping are 210 J/kg? Clearly state any assumptions that you make. (b) In order to recover some power, it is proposed to hook a turbine up to the brackish water return pipe. If the brackish water leaves the RO unit at 60 atm and is sent directly to the turbine, how much power can be recovered if the turbine is 72% efficient? You may assume that the outlet pressure from the turbine is 1.2 atm, and there are no frictional losses in the pipes connecting the RO unit to the turbine? Clearly state any assumptions that you make. (c) What is the net cost of operating this RO unit per m3 of potable water (for part b if the cost of electricity is $0.05/kWh?

Solution

(a) Pump outlet pressure = 65 atm and RO height is 10 m w.r.t. sea level.

So difference in pressure around the pump = (65 - 1) atm = 64 atm = 64x1.013x100000 Pa = 6483.2x1000 Pa

Therefore difference in pressure head = [(6483.2x1000) / (1029x9.81)] m = 642.25 m of sea water

Assumed that, density of sea water,D = 1029 kg/m3

difference between gravity head = 10 - 0 = 10 m [sea level is 0 m]

Therefore total head generated by pump, H = 642.25 + 10 = 652.25 m

Efficiency of the pump, E = 93% = 0.93

Flow rate of water through pump, Q = 100 m3/min = (100/60) m3/s = 1.67 m3/s

P = power required

As we know, E= DQgH / P

So, P = DQgH / E = (1029x1.67x9.81x652.25) / 0.93 W = 11823.12 x 10³ W = 11823.12 kW

Also there is friction loss in pipe = 210 J/kg

friction power loss = 210 J/kg x 1029 kg/m³ x 1.67 m³/s = 360.87 x 10³ W = 360.87 kW

Therefore total power supplied to the pump = (11823.12 + 360.87) kW = 12184 kW

(b) Here flow rate, Q = 55% of 1.67 m³/s = 0.9185 m³/s

Assumed, density of backish water, D = 1029 kg/m³

Difference of pressure head = [{(60 - 1.2) x 1.013 x10⁵} / (1029 x 9.81)] m = 590.07 m of backish water

Difference between gravity head = 10 - 0 m = 10 m

So total head around the turbine, H = (590.07 + 10) m = 600.07 m

Efficiency of the turbine, E = 72% = 0.72

Power generated by the turbine = P

As we know, E = P / DQgH

or, P = DQgHE = 1029 x 0.9185 x 9.81x 600.07 x 0.72 W = 3979.71 x 10³ W = 3979.71 kW

Therefore power can be recovered is 3979.71 kW.

(c) Potable water flow rate = 1.67 x 0.45 m³/s = 0.7515 m³/s

Net power required = pump power - turbine power = (12184 - 3979.71) kW = 8204.29 kW

Net power required per m³ of potable water = Net power required / potable water flow rate

= 8204.29 / 0.7515 kWs

= 10917.2189 /3600 kWh

= 3.0325 kWh

Therefore the net cost of operating the RO unit per m3 of potable water = $3.0325 x 0.05 = $0.1516.