A city of 157,231 grows at 1.6% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 207 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 157231(1 + 0.016)^26 = 237,561 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 157231(1 + 0.016 \times 26) = 222,639 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 237,561(207)/1000 = 49,175 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.7 \times 49,175 = 132,773 m^{3}/d$$

Why the other options are there

• 65,408 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 118,315 grows at 1.0% per year. Project the population in 13 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 490 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 118315(1 + 0.010)^13 = 134,654 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 118315(1 + 0.010 \times 13) = 133,696 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 134,654(490)/1000 = 65,980 m^{3}/d$$

7. Peak day

   $$Q_peak = 1.9 \times 65,980 = 125,362 m^{3}/d$$

Why the other options are there

• 15,381 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 90,512 grows at 3.0% per year. Project the population in 17 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 495 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 90512(1 + 0.030)^17 = 149,603 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 90512(1 + 0.030 \times 17) = 136,673 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 149,603(495)/1000 = 74,053 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.7 \times 74,053 = 199,944 m^{3}/d$$

Why the other options are there

• 46,161 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 42,437 grows at 3.3% per year. Project the population in 13 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 406 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 42437(1 + 0.033)^13 = 64,722 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 42437(1 + 0.033 \times 13) = 60,642 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 64,722(406)/1000 = 26,277 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.6 \times 26,277 = 68,320 m^{3}/d$$

Why the other options are there

• 18,205 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 97,528 grows at 1.0% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 396 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 97528(1 + 0.010)^14 = 112,106 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 97528(1 + 0.010 \times 14) = 111,182 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 112,106(396)/1000 = 44,394 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.6 \times 44,394 = 115,424 m^{3}/d$$

Why the other options are there

• 13,654 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 116,224 grows at 3.0% per year. Project the population in 17 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 215 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 116224(1 + 0.030)^17 = 192,101 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 116224(1 + 0.030 \times 17) = 175,498 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 192,101(215)/1000 = 41,302 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.9 \times 41,302 = 119,775 m^{3}/d$$

Why the other options are there

• 59,274 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 71,844 grows at 1.9% per year. Project the population in 23 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 358 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 71844(1 + 0.019)^23 = 110,764 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 71844(1 + 0.019 \times 23) = 103,240 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 110,764(358)/1000 = 39,653 m^{3}/d$$

7. Peak day

   $$Q_peak = 1.9 \times 39,653 = 75,341 m^{3}/d$$

Why the other options are there

• 31,396 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 139,969 grows at 3.3% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 277 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 139969(1 + 0.033)^26 = 325,567 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 139969(1 + 0.033 \times 26) = 260,062 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 325,567(277)/1000 = 90,182 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.2 \times 90,182 = 198,400 m^{3}/d$$

Why the other options are there

• 120,093 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 121,647 grows at 2.0% per year. Project the population in 22 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 388 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 121647(1 + 0.020)^22 = 188,064 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 121647(1 + 0.020 \times 22) = 175,172 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 188,064(388)/1000 = 72,969 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.9 \times 72,969 = 211,609 m^{3}/d$$

Why the other options are there

• 53,525 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

A city of 154,296 grows at 2.0% per year. Project the population in 13 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 244 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P = P_{0}(1 + i)ⁿ$$

2. Substituting

   $$P = 154296(1 + 0.020)^13 = 199,598 people$$

3. Formula (arithmetic)

   $$P = P_{0}(1 + i n)$$

4. Substituting

   $$P = 154296(1 + 0.020 \times 13) = 194,413 people$$

5. Formula

   $$Q_avg = P \times per capita use$$

6. Substituting

   $$Q_avg = 199,598(244)/1000 = 48,702 m^{3}/d$$

7. Peak day

   $$Q_peak = 1.9 \times 48,702 = 92,534 m^{3}/d$$

Why the other options are there

• 40,117 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth