DavidPE
Active member
I don't understand the solution to problem 129 of the 2011 NCEES Practice Exam, and I'm hoping someone can 'splain it to me. The problem:
A 4.0-MGD wastewater treatment plant discharges a secondary-treated effluent into a receiving stream. The wastewater has the following characteristics:
BOD5: 20 mg/L
DO: 2.0 mg/L
Temp: 24C
The receiving stream upstream from the point of wastewater discharge has the following characteristics:
Flow Rate: 18 cfs
BOD5: 4.0 mg/L
DO: 6.0 mg/L
Temp: 27C
Cross-sectional area: 36 sf (uniform)
The reaeration rate is estimated to be 0.4/day (base e at 20C) and the temperature correction coefficient is 1.024. The deoxygenation rate constant is estimated to be 0.23/day (base e at 20C) and the temperature correction coefficient is 1.047. Reaeration and deoxygenation are the only major factors affecting the dissolved oxygen concentration in the stream after mixing with wastewater effluent. Assume the time of travel in the stream to reach the maximum dissolved oxygen deficit is 3 days. The dissoved oxygen deficit (mg/L) at this location downstream from the point of discharge is most nearly:
a) 3.2
b) 5.2
c) 7.2
d) 8.4
According to the solution:
1. Calculate the temperature of the mixed flows. Mass balance. T = 26.2C.
2. Correct the rate constants for temperature. kt = k20 * theta^(T-20). kd = 0.305; kr = 0.46.
3. Calculate the BOD5 of the mixed flow. Mass balance. BOD5 = 8.09 mg/L.
4. Calculate the BODu of the mixed flow. BOD5 = BODu * (1-e^(-k*t)). They say that BODu = 11.76 mg/L.
5. Calculate the dissolved oxygen deficit at 3 days (Dc). Dc = [(kd * BODu)/kr] * e^-(k*t). They say that Dc = 3.12 mg/L.
I am with them on everything except the BODu calculation. When they calculated the BODu, they used the uncorrected kd value of 0.23/day. Why is that? Why not use the corrected kd value of 0.305/day? I understand everything else.
Thanks very much for the help, I really appreciate it.
Dave
A 4.0-MGD wastewater treatment plant discharges a secondary-treated effluent into a receiving stream. The wastewater has the following characteristics:
BOD5: 20 mg/L
DO: 2.0 mg/L
Temp: 24C
The receiving stream upstream from the point of wastewater discharge has the following characteristics:
Flow Rate: 18 cfs
BOD5: 4.0 mg/L
DO: 6.0 mg/L
Temp: 27C
Cross-sectional area: 36 sf (uniform)
The reaeration rate is estimated to be 0.4/day (base e at 20C) and the temperature correction coefficient is 1.024. The deoxygenation rate constant is estimated to be 0.23/day (base e at 20C) and the temperature correction coefficient is 1.047. Reaeration and deoxygenation are the only major factors affecting the dissolved oxygen concentration in the stream after mixing with wastewater effluent. Assume the time of travel in the stream to reach the maximum dissolved oxygen deficit is 3 days. The dissoved oxygen deficit (mg/L) at this location downstream from the point of discharge is most nearly:
a) 3.2
b) 5.2
c) 7.2
d) 8.4
According to the solution:
1. Calculate the temperature of the mixed flows. Mass balance. T = 26.2C.
2. Correct the rate constants for temperature. kt = k20 * theta^(T-20). kd = 0.305; kr = 0.46.
3. Calculate the BOD5 of the mixed flow. Mass balance. BOD5 = 8.09 mg/L.
4. Calculate the BODu of the mixed flow. BOD5 = BODu * (1-e^(-k*t)). They say that BODu = 11.76 mg/L.
5. Calculate the dissolved oxygen deficit at 3 days (Dc). Dc = [(kd * BODu)/kr] * e^-(k*t). They say that Dc = 3.12 mg/L.
I am with them on everything except the BODu calculation. When they calculated the BODu, they used the uncorrected kd value of 0.23/day. Why is that? Why not use the corrected kd value of 0.305/day? I understand everything else.
Thanks very much for the help, I really appreciate it.
Dave