---
title: "Design of an ion-exchange water softener"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:water-softener-design"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/water-softener-design
site: BCT Engineering
---

# Design of an ion-exchange water softener

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2073 Shrawan, 8 marks
- Video: [Ion Exchange - Water Treatment and Analysis - Applied Chemistry 2](https://www.youtube.com/watch?v=x8OBVQVGXDM) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/water-softener-design
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Design a water softener for a flow of 20,000 l/hr, hardness = 450 mg/l as CaCO$_3$, allowable hardness after treatment = 50 mg/l as CaCO$_3$, ion exchange capacity of the resin = 20 kg/m$^3$ of the resin, regeneration period = 7.5 hours.

## Short answer

Remove 400 mg/L hardness; hourly load = 8 kg/h, per cycle 60 kg. Resin volume = 60 / 20 = 3.0 m³. Vessel: bed depth 1.5 m, area 2.0 m², diameter ~1.6 m.

## Answer

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.

```text
        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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How much hardness must be removed?
   A. 450 mg/l
   B. 50 mg/l
   C. 500 mg/l
   D. 450 - 50 = 400 mg/l as CaCO3
   Answer: D. Only the excess over the allowable 50 mg/l is removed. (Key point: Hardness to remove)

2. What is the hardness load per regeneration cycle?
   A. 8.0 kg/h x 7.5 h = 60 kg
   B. 8.0 kg
   C. 20 x 7.5 = 150 kg
   D. 400 kg
   Answer: A. Load per hour = 20 m3/h x 400 g/m3 = 8 kg/h; multiply by 7.5 h. (Key point: Load per cycle)

3. What is the resin volume?
   A. 8 / 20 = 0.4 m3
   B. 60 / 20 = 3.0 m3
   C. 60 x 20 = 1200 m3
   D. 7.5 m3
   Answer: B. Volume = load per cycle / exchange capacity. (Key point: Resin volume)

4. With a bed depth of 1.5 m, what is the vessel diameter?
   A. About 3.0 m
   B. About 0.8 m
   C. About 1.6 m, from A = 3.0 / 1.5 = 2.0 m2
   D. About 2.0 m
   Answer: C. D = sqrt(4A/pi) = sqrt(2.55) = 1.6 m. (Key point: Vessel diameter)

## Related in Water Supply Engineering

- [Hardness from ion analysis (Ca 65, Mg 35)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-c71) — Water Supply Engineering, chapter 4
- [Hardness of well water with strontium 25.26 mg/l](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-bk81) — Water Supply Engineering, chapter 4
- [Hardness from ion analysis with strontium](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-b81) — Water Supply Engineering, chapter 4
- [Effects of hardness; zeolite softener](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-zeolite-softener) — Water Supply Engineering, chapter 6
- [Ca and Mg concentration, alkalinity (300 mg/l hardness)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-ca-mg-equal-300) — Water Supply Engineering, chapter 4, asked 2×
- [Hardness with Ca, Mg, Sr equal, TH 150](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ahardness-as70) — Water Supply Engineering, chapter 4

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