Clarification following lime-soda softening

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Chemical feed rate from dosage — Clarification following lime-soda softening

A plant treating 7.2 MGD requires a 3.0 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 7.2 MGD$
$Dose = 3.0 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 7.2(3.0)(8.34)$
Evaluate — 180.1 lb/day
At 65% available strength — 277.1 lb/day of product

Answer:

≈ 180.1 lb/day

Why the other options are there

21.6 lb/day (8.34 omitted)
2.59 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 2

Chemical feed rate from dosage — Clarification following lime-soda softening (2)

A plant treating 2.0 MGD requires a 11.5 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 2.0 MGD$
$Dose = 11.5 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 2.0(11.5)(8.34)$
Evaluate — 191.8 lb/day
At 65% available strength — 295.1 lb/day of product

Answer:

≈ 191.8 lb/day

Why the other options are there

23.0 lb/day (8.34 omitted)
2.76 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 3

Chemical feed rate from dosage — Clarification following lime-soda softening (3)

A plant treating 0.9 MGD requires a 4.5 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 0.9 MGD$
$Dose = 4.5 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 0.9(4.5)(8.34)$
Evaluate — 33.8 lb/day
At 65% available strength — 52.0 lb/day of product

Answer:

≈ 34 lb/day

Why the other options are there

4.1 lb/day (8.34 omitted)
0.49 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 4

Chemical feed rate from dosage — Clarification following lime-soda softening (4)

A plant treating 5.4 MGD requires a 3.5 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 5.4 MGD$
$Dose = 3.5 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 5.4(3.5)(8.34)$
Evaluate — 157.6 lb/day
At 65% available strength — 242.5 lb/day of product

Answer:

≈ 157.6 lb/day

Why the other options are there

18.9 lb/day (8.34 omitted)
2.27 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 5

Chemical feed rate from dosage — Clarification following lime-soda softening (5)

A plant treating 2.3 MGD requires a 4.5 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 2.3 MGD$
$Dose = 4.5 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 2.3(4.5)(8.34)$
Evaluate — 86.3 lb/day
At 65% available strength — 132.8 lb/day of product

Answer:

≈ 86 lb/day

Why the other options are there

10.4 lb/day (8.34 omitted)
1.24 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 6

Chemical feed rate from dosage — Clarification following lime-soda softening (6)

A plant treating 6.8 MGD requires a 4.0 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 6.8 MGD$
$Dose = 4.0 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 6.8(4.0)(8.34)$
Evaluate — 226.8 lb/day
At 65% available strength — 349.0 lb/day of product

Answer:

≈ 226.8 lb/day

Why the other options are there

27.2 lb/day (8.34 omitted)
3.26 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 7

Chemical feed rate from dosage — Clarification following lime-soda softening (7)

A plant treating 7.5 MGD requires a 2.5 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 7.5 MGD$
$Dose = 2.5 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 7.5(2.5)(8.34)$
Evaluate — 156.4 lb/day
At 65% available strength — 240.6 lb/day of product

Answer:

≈ 156.4 lb/day

Why the other options are there

18.8 lb/day (8.34 omitted)
2.25 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 8

Chemical feed rate from dosage — Clarification following lime-soda softening (8)

A plant treating 6.7 MGD requires a 7.0 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 6.7 MGD$
$Dose = 7.0 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 6.7(7.0)(8.34)$
Evaluate — 391.1 lb/day
At 65% available strength — 601.8 lb/day of product

Answer:

≈ 391.1 lb/day

Why the other options are there

46.9 lb/day (8.34 omitted)
5.62 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 9

Chemical feed rate from dosage — Clarification following lime-soda softening (9)

A plant treating 1.1 MGD requires a 2.0 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 1.1 MGD$
$Dose = 2.0 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 1.1(2.0)(8.34)$
Evaluate — 18.3 lb/day
At 65% available strength — 28.2 lb/day of product

Answer:

≈ 18 lb/day

Why the other options are there

2.2 lb/day (8.34 omitted)
0.26 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening

Example 10

Chemical feed rate from dosage — Clarification following lime-soda softening (10)

A plant treating 2.5 MGD requires a 6.0 mg/L dose. What is the daily chemical demand in pounds?

Given

$Q = 2.5 MGD$
$Dose = 6.0 mg/L$
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$lb/day = Q(MGD) \times dose(mg/L) \times 8.34$
Substituting
$lb/day = 2.5(6.0)(8.34)$
Evaluate — 125.1 lb/day
At 65% available strength — 192.5 lb/day of product

Answer:

≈ 125.1 lb/day

Why the other options are there

15.0 lb/day (8.34 omitted)
1.80 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Clarification following lime-soda softening