Question

A stream of flowing water at 20°C initially has an ultimate BOD in the mixing zone of 10 mg/L. The saturated oxygen concentration is 8.9 mg/L, and the initial dissolved concentration rate is 8.5 mg/L. The reaeration rate is 2.00/d, the deoxygenation rate constant is 0.1/d, and the velocity of the stream is 0.11 km/min. Estimate the dissolved oxygen in the flowing stream after 160 km.

Answer 1

To estimate the dissolved oxygen concentration in the flowing stream after 160 km, we need to consider factors such as reaeration, deoxygenation, and temperature. The dissolved oxygen concentration is approximately 2772.349 mg/L.

Step-by-step explanation:

To estimate the dissolved oxygen concentration in the flowing stream after 160 km, we need to consider factors such as reaeration, deoxygenation, and temperature. First, let's calculate the change in dissolved oxygen concentration due to reaeration:

Reaeration rate = 2.00/d

Velocity of stream = 0.11 km/min

Distance traveled = 160 km

Time traveled = (160 km)/(0.11 km/min) = 1454.55 min

Dissolved oxygen added due to reaeration = Reaeration rate * Time traveled = 2.00/d * 1454.55 min = 2909.1 mg/L

Next, let's calculate the change in dissolved oxygen concentration due to deoxygenation:

Deoxygenation rate constant = 0.1/d

Dissolved oxygen removed due to deoxygenation = Deoxygenation rate constant * Time traveled = 0.1/d * 1454.55 min = 145.455 mg/L

Finally, let's calculate the change in dissolved oxygen concentration due to temperature:

Change in temperature = 20°C - 15°C = 5°C

Temperature correction factor = 1.024

Initial dissolved oxygen concentration = 8.5 mg/L

Dissolved oxygen concentration at 15°C = Initial dissolved oxygen concentration * Temperature correction factor = 8.5 mg/L * 1.024 = 8.704 mg/L

Total change in dissolved oxygen concentration = Dissolved oxygen added due to reaeration - Dissolved oxygen removed due to deoxygenation + Dissolved oxygen concentration at 15°C = 2909.1 mg/L - 145.455 mg/L + 8.704 mg/L = 2772.349 mg/L

Therefore, the dissolved oxygen concentration in the flowing stream after 160 km is approximately 2772.349 mg/L.

Answer 2

The estimated dissolved oxygen in the flowing stream after 160 km is 8.27 mg/L.

Step-by-step explanation:

We can estimate the dissolved oxygen using the Streeter-Phelps model, which considers the initial oxygen concentration, BOD, reaeration rate, deoxygenation rate, and stream velocity.

Steps:

Calculate the critical dissolved oxygen deficit (Dcrit):

Dcrit = (Cs - Cu) / (1 + k1/k2)

where:

Cs = Saturated oxygen concentration (8.9 mg/L)

Cu = Ultimate BOD concentration (10 mg/L)

k1 = Reaeration rate (2.00/d)

k2 = Deoxygenation rate constant (0.1/d)

Dcrit = (8.9 mg/L - 10 mg/L) / (1 + 2.00/d / 0.1/d) ≈ -0.61 mg/L (negative due to convention)

Calculate the deoxygenation rate coefficient (K):

K = k2 * u / d

where:

u = Stream velocity (0.11 km/min)

d = Stream depth (assumed constant)

Assuming a stream depth of 5 meters:

K = 0.1/d * 0.11 km/min * 60 min/h = 0.066 h^-1

Calculate the time of travel (t) for 160 km:

t = L / u

where:

L = Distance traveled (160 km)

t = 160 km / 0.11 km/min * 60 min/h ≈ 1454.55 h

Use the Streeter-Phelps model equation:

C = Cs - (Dcrit + Cu * exp(-Kt)) * exp(-k1 * t)

where:

C = Dissolved oxygen concentration at time t

C = 8.9 mg/L - (-0.61 mg/L + 10 mg/L * exp(-0.066 * 1454.55)) * exp(-2.00/d * 1454.55)

C ≈ 8.27 mg/L

Therefore, the dissolved oxygen concentration in the flowing stream after 160 km is estimated to be 8.27 mg/L.