1. Consider the behavior of the chemical hexamethyldeath (HMD) in a lake. The la

Question

1. Consider the behavior of the chemical hexamethyldeath (HMD) in a lake. The lake has a constant volume of 108 L, and it has one outflow and two inputs:

• a factory waste: flow = 7 L/s and HMD = 100 mg/L

• a river: flow = 50 L/s and HMD = 0 HMD decays with a rate k = 0.2/d. Assume well mixed conditions in the lake. Calculate the HMD concentration (mg/L) in the outlet of the lake.

2. Revisit the lake in Problem 1 above. Suppose the HMD concentration in the waste flow suddenly decreases to 0 mg/L.

a) Calculate the HMD concentration (mg/L) in the lake outlet 2 days later.

b) After the waste concentration decreases, is the change in HMD mainly due to reaction (decay) or dilution? Briefly explain.

Explanation / Answer

1. Constant volume of lake = 108 L

Inputs to the lake -

flow = 7 L/s and HMD = 100 mg/L , &

flow = 50 L/s and HMD = 0

Assuming lake is already full, to maintain constant volume,

outlet flow rate = total input flow rate

= 50 + 7 = 57L/s

Amount of HMD in the lake = 100* 7 + 0*50

= 700 mg

HMD concentration (mg/L) in the lake = 700 mg/s /108 L

= 6.48 mg/L

Let the concentration of HMD in outflow be M2.

M1*V1 = M2*V2

6.48*108 = M2*57

M2 = 12.27 mg/L

So, HMD concentration (mg/L) in the outlet of the lake = 12.27 mg/L

2. decay constant of HMD = 0.2/day

Time = 2 days

Using decay law

N = No * e-k*t

N = 6.48 * e- 0.2 * 2

N = 6.48 * 0.67

N = 4.34 mg/L

HMD concentration (mg/L) in the lake outlet 2 days later = 4.34 * 108 / 57

= 8.22 mg/L

So, HMD concentration (mg/L) in the lake outlet 2 days later = 8.22 mg/L

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.