Design of an ion-exchange water softener

A synthetic ion-exchange (cation) resin in sodium form removes hardness: $2R\text{-}Na + Ca^{2+} \rightarrow R_2Ca + 2Na^+$. The resin is regenerated with brine after each run.

Given data

• Flow $Q = 20{,}000$ l/h $= 20$ m³/h
• Raw hardness = 450 mg/l; allowable treated hardness = 50 mg/l (as CaCO$_3$)
• Exchange capacity of resin = 20 kg/m³; regeneration period (run time between regenerations) = 7.5 h

Step 1: Hardness to be removed

$$450 - 50 = 400\ \text{mg/l} = 400\ \text{g/m}^3$$

Step 2: Hardness load per hour

$$20\ \text{m}^3/\text{h} \times 400\ \text{g/m}^3 = 8000\ \text{g/h} = 8.0\ \text{kg/h}$$

Step 3: Hardness load per cycle

$$8.0\ \text{kg/h} \times 7.5\ \text{h} = 60.0\ \text{kg}$$

Step 4: Volume of resin

$$V = \frac{60.0}{20} = 3.00\ \text{m}^3$$

Step 5: Size of the softener vessel

Assume a bed depth $H = 1.5$ m (usual range 1 to 2 m).

$$A = \frac{V}{H} = \frac{3.00}{1.5} = 2.00\ \text{m}^2, \qquad D=\sqrt{\frac{4A}{\pi}} = 1.60\ \text{m}$$

Provide $D = 1.6$ m (area 2.01 m²). Service flow rate $= 20/2.00 = 10.0$ m/h (acceptable, since 5 to 20 m/h is normal). Add a free board of about 50 % of the bed depth for backwash expansion, giving a vessel height of about 2.5 m including the gravel support.

        raw water in
             |
   +---------v---------+
   |   free board      |  0.75 m
   |...................|
   |:::::: resin ::::::|  1.5 m      D = 1.6 m
   |:::::::::::::::::::|
   |ooooo gravel oooooo|  0.3 m
   +---------|---------+
             v
      soft water out       brine in (regeneration)

Answer: Resin volume = 3.0 m³, bed depth 1.5 m, area 2.00 m², diameter about 1.6 m, in one vessel regenerated every 7.5 h (a standby unit is provided so that supply continues during regeneration).

(If a bypass is used, only the fraction $\dfrac{450-50}{450}=0.889$ of the flow (17.8 m³/h) is passed through the softener and reduced to zero hardness. The hardness removed is still $17.8\times450 = 8000$ g/h, so the resin volume stays 3.0 m³; only the flow through the unit is smaller.)