2010 Exam 3

53:050

Natural Environmental Systems

EXAM 3: Closed Book - Closed Notes

May 6, 2010

NAME

1. (1 pt for each blank = 15 pts)

(a) NBOD represents the amount of oxygen to oxidize in a wastewater sample.

(b) Pathogenic organisms are a water quality problem because they can potentially cause .

(c) Chemicals such as chlorinated solvents (PCE, TCE, etc.) that have the potential to sink to the bottom of aquifers,

contaminating the groundwater as they sink, are known as .

(d) Chemicals such as gasoline that have the potential to float on top to of aquifers, contaminating the groundwater

as they dissolve, are known as .

(e) + + = photochemical s mog.

(f) and are air pollutants that cause acid rain.

(g) True or False. An atmosphere with a prevailing lapse rate of -1.5o C/100 m is said to be stable.

(h) The average MSW generation rate in the US is about lb/capita-d.

(i) The average hazardous wastes generation rate in the US is about lb/capita-d.

(j) True or False. Darcy’s law can be used to describe leachate movement through a landfill liner.

(k) Currently, the most common method of disposal of MSW is .

(l) Currently, the most common method of disposal of newly generated hazardous waste is .

2. (5 pts)

In class, we discussed several aspects of greenhouse gasses and global climate change. We introduced the term global warming potential (GWP).

(a) We listed two of the major factors used in calculating GW P. What are they?

(b) We also listed the greenhouse gasses based on their GWP. List the 3 gasses with the highest GWP.

3. (5 pts)

Briefly explain in the space below why ozone (O3) is a human and environmental health problem when present in the troposphere but yet protective of human health when present in the stratosphere.

During manufacture of their products, GyNormous Corp. produces a wastewater containing 800 mg/L BOD5 (= 968 mg/L BOD L– both measured at 20o C). They build a wastewater treatment plant that removes 90% of this BOD prior to discharge into the Noscent River. Water is withdrawn 25 miles (40.2 km) downstream from GyNormous by a brewery to produce “Doc’s Perfect Suds” beer. The ri ver prior to treated wastewater discharge has a flow of 50 MGD (2.19 m3/s), a measured BOD L of 3 mg/L, a DO of 7.1 mg/L, and a temperature of 24o C. The treated wastewater effluent has a flow of 1 MGD (0.0438 m3/s), a DO of 1 mg/L, and a temperature of 24o C. Assume k = k d = 0.35 d-1 at 20o C. Assume k r = 0.25 d-1 at 20o C. Finally, assume a river velocity of 0.1 m/s (= 5.37 mile/d) after treated wastewater discharge.

(a) What is the dissolved oxygen (DO) of the river when the brewery takes it from the river?6.18 mg/L

(b) What is the BOD5(measured at 20o C) of the river when the brewery takes it from the river?0.53 mg/L

(c) Has the minimum (critical) DO been reached at the brewery? Show all calculations to justify your

answer – a simple yes or no answer without supporting calculations will receive no credit.Yes

You are to make some calculations about the movement of groundwater in an aquifer. The aquifer is a sandy loam with an estimated hydraulic conductivity of 3 x 10-3 cm/s (= 2.59 m/d), a porosity of 40%, a bulk density of 1.6 kg/L, and an organic carbon fraction of 0.001. A tracer study was conducted to assess groundwater velocity. This tracer study showed that a slug of a conservative tracer took 34 days to travel the 10-m distance between 2 wells.

(a) Calculate the groundwater velocity.0.294 m/d

(b) Calculate the Darcy velocity.0.118 md

(c) What is the hydraulic gradient?0.0454 m/m

(d) If the groundwater is contaminated with the herbicide alachlor (log K ow = 2.92), estimate the time in days it

would take the alachlor to travel the 10 m.105 d

You are having an argument with your neighbor DolliverDinglenarb about the impact of inversions on air quality. You tell Dolliver that it is better to live closer to the source than further away – that you are exposed to lower levels of pollution 1 km (≈ 0.62 mile) from the source than 100 km (≈ 62 miles) from the source,

especially on overcast days. He, of course, thinks you are nuts. You set out to prove that you are right by doing some model calculations for ground-level pollutant concentrations directly downwind from the source. You use the following data from a power plant:

?emission rate = 500 g/s ? average wind speed = 3 m/s

?effective stack height = 150 m ? inversion height = 425 m

Are you correct? Show all calculations necessary to justify your answer. A yes or no answer without supporting calculations will receive no credit.Yes, χ1 is less than χ100 (2.95 x 10-7 g/m3 vs. 3.75 x 10-5 g/m3)

53:050 – Exam 3

POTENITALLY USEFUL FORMULAE AND OTHER INFORMATION

1 m 3 = 1,000 L 1 kg = 1,000 g = 106 mg = 109μg 1 d = 86,400 s

T (o C) C s (mg/L) T (o C) C s (mg/L) 4 13.48 22 8.83 5 12.80 23 8.58 10 11.33 24 8.53 15 10.15 25 8.38 16 9.95 26 8.22 17 9.74 27 8.07 18 9.54 28 7.92 19 9.35 29 7.77 20 9.17 30 7.63 21 8.99

35

7.07

dt

dL = -kL BOD t = BOD L (1 – e -kt )

dt

dC = k r (C s – C) = k r D L t = t k o d e L -

D t = ()

t k -o t k -t k -d r o d r r d e D e - e k - k L k +

t c =

u

x c = ???? ?????? ??????

??

o d d r o d r d r L k )k - (k D - 1k k ln k - k 1

D c =

)e

(L k k c

d t k o r

d -

k d = k + ?

?

?

??h u η

k r =

5

.15

.0h

)

u (90.3

k d,T = k d,20(1.135)T-20or = k d,20(1.056)T-20k r,T = k r,20(1.024)T-20

v = K l

h ??

R = 1 + ???

?

??ηρb K oc f oc K oc = 0.63K ow

χ (x,y,0,H) = ????

? ?????? ?????

? ??-????

? ??????? ?????? ??-????

??π2

z 2y z

y s H 21exp s y 21exp u s s E

χ= π

2uL

s E y s z = 0.47(L – H) air

m mg 3

- =

)

T (08206.0)MW (ppm

H = h + ?H

?H = ???

?

?

????? ?????? ??+d T T - T (P)10 x 2.68 1.5u d v s a

s 2-s

s z = cx + f

s y = ax