A city of 19,698 grows at 2.3% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 410 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

   $$P = 19698(1 + 0.023)^26 = 35,579 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 19698(1 + 0.023 \times 26) = 31,477 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 35,579(410)/1000 = 14,587 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.9 \times 14,587 = 42,303 m^{3}/d$$

Why the other options are there

• 11,779 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 153,726 grows at 3.3% per year. Project the population in 25 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 304 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

   $$P = 153726(1 + 0.033)^25 = 346,143 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 153726(1 + 0.033 \times 25) = 280,550 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 346,143(304)/1000 = 105,227 m^{3}/d$$

7. Peak day

   $$Q_peak = 3.0 \times 105,227 = 315,682 m^{3}/d$$

Why the other options are there

• 126,824 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 45,832 grows at 3.3% per year. Project the population in 30 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

   $$P = 45832(1 + 0.033)^30 = 121,389 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 45832(1 + 0.033 \times 30) = 91,206 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 121,389(490)/1000 = 59,480 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.9 \times 59,480 = 172,493 m^{3}/d$$

Why the other options are there

• 45,374 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 148,713 grows at 2.2% per year. Project the population in 16 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 456 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

   $$P = 148713(1 + 0.022)^16 = 210,651 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 148713(1 + 0.022 \times 16) = 201,060 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 210,651(456)/1000 = 96,057 m^{3}/d$$

7. Peak day

   $$Q_peak = 1.9 \times 96,057 = 182,508 m^{3}/d$$

Why the other options are there

• 52,347 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 109,668 grows at 2.3% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 445 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

   $$P = 109668(1 + 0.023)^26 = 198,083 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 109668(1 + 0.023 \times 26) = 175,249 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 198,083(445)/1000 = 88,147 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.7 \times 88,147 = 237,996 m^{3}/d$$

Why the other options are there

• 65,581 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 175,663 grows at 3.2% per year. Project the population in 17 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 311 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

   $$P = 175663(1 + 0.032)^17 = 300,079 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 175663(1 + 0.032 \times 17) = 271,224 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 300,079(311)/1000 = 93,324 m^{3}/d$$

7. Peak day

   $$Q_peak = 1.8 \times 93,324 = 167,984 m^{3}/d$$

Why the other options are there

• 95,561 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 38,651 grows at 1.5% per year. Project the population in 25 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 255 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

   $$P = 38651(1 + 0.015)^25 = 56,080 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 38651(1 + 0.015 \times 25) = 53,145 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 56,080(255)/1000 = 14,301 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.6 \times 14,301 = 37,181 m^{3}/d$$

Why the other options are there

• 14,494 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 158,110 grows at 1.4% per year. Project the population in 10 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 385 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

   $$P = 158110(1 + 0.014)^10 = 181,693 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 158110(1 + 0.014 \times 10) = 180,245 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 181,693(385)/1000 = 69,952 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.9 \times 69,952 = 202,861 m^{3}/d$$

Why the other options are there

• 22,135 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 140,673 grows at 2.4% per year. Project the population in 24 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 319 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

   $$P = 140673(1 + 0.024)^24 = 248,548 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 140673(1 + 0.024 \times 24) = 221,701 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 248,548(319)/1000 = 79,287 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.4 \times 79,287 = 190,288 m^{3}/d$$

Why the other options are there

• 81,028 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth

A city of 157,204 grows at 2.1% per year. Project the population in 11 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 303 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

   $$P = 157204(1 + 0.021)^11 = 197,582 people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P = 157204(1 + 0.021 \times 11) = 193,518 people$$

5. Formula

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

6. Substituting

   $$Q_avg = 197,582(303)/1000 = 59,867 m^{3}/d$$

7. Peak day

   $$Q_peak = 2.4 \times 59,867 = 143,681 m^{3}/d$$

Why the other options are there

• 36,314 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Decreasing-Rate-of-Increase Growth