A lake with volume 107 m 3 is fed by a polluted stream having a flowrate of 5 m3

Question

A lake with volume 107 m 3 is fed by a polluted stream having a flowrate of 5 m3 /s and pollutant concentration of 10 mg/L. Also directly discharging to the lake is a wastewater treatment plant effluent having a flowrate of 0.5 m3 /s and concentration of 500 mg/L of the same pollutant. A single stream leaves the lake. Assume the pollutant in the lake has a decay rate constant of 1.20 mg/Lday. Make the additional following assumptions: (i) steady state, (ii) no evaporation or other water losses or gains, and (iii) the pollutant is completely mixed in the lake. ) Write the two generalized mass balance equations for the control volume in your diagram from part (a). Clearly state all assumptions needed to simplify the equations consistent with the problem statement above. (c) What is the flowrate and pollutant concentration in the stream leaving the lake? (d) What is the residence time of the water in the lake? What is the half-life of the pollutant in the lake?

Explanation / Answer

2. Global Energy Balance: No Atmosphere

a) Energy Balance:

“Rate of Solar Energy Absorbed” = “Rate of Infrared Energy Emitted”

E(1A)S

  

(1-A) S RE2 = E 4 RE2

4(1.3)(1,360Watts/m2)

4238Watts/m2

b) Global Average Surface Temperature:

ET4

TE

  

1/4

238Watts/m2

5.67108Watts/(m2K4)

1/4

254.5K

Compared to 280 K, the actual average surface temperature. The calculated value is low because of the

greenhouse effect was omitted.

3. Global Energy Balance: with a Greenhouse Gas Atmosphere

At the earth’s surface, a radiation balance requires that

(irradiance in = irradiance out)

0.9E + y = x

while for the atmosphere layer, the radiation balance is

E + x = 0.9E + 2y + .2x

3. Solving these equations simultaneously for y and x results in

y.82

1.2E.82

1.2(238W/m2)162.6W/m2

x1.583E1.583(238W/m2)376.8W/m2

4. Surface temperature:

1/4

Tx

x376.8W/m2T4

376.8W/m2

5.67108W/(m2K4)

1/4

285.5K

Atmosphere temperature:

Ty

1/4

162.6W/m2

.8

y.8T4

(.8)(5.67108W/(m2K4))

1/4

244.7K

5. In order for the global average surface temperature of the earth to rise by 1 C above the value

calculated in part b (285.52K), the infrared absorbtivity would need to increase to 0.8166 from 0.80.

4. Global Carbon Dioxide Mass Balance

a) A mass balance for CO2 at the earth's surface is

Accumulation of CO2 in atmosphere = rate of CO2 release from surface -

rate of CO2 removal by surface

Accumulation of CO2 in atmosphere (metric tons C/yr) = (60+6+1.6) - (60+2+.5+1.8)

(3.3109metrictons/yr)44gCO2

=   

12gC

= 3.3 metric tons C/yr

103kg

metricton

1.211013kgCO

2/yr

6. +10% change in emissions from fuel combustion is +.6% for a total of 6.6%

Accumulation of CO2 in atmosphere (metric tons C/yr) = (60+6.6+1.6) - (60+2+.5+1.8)

(3.9109metrictons/yr)44gCO2

=   

12gC

= 3.9 metric tons C/yr

103kg

metricton

1.431013kgCO

2/yr

+1% change in emissions from release by microorganisms is +.6% for a total of 60.6%

Accumulation of CO2 in atmosphere (metric tons C/yr) = (60.6+6+1.6) - (60+2+.5+1.8)

(3.9109metrictons/yr)44gCO2

=   

12gC

= 3.9 metric tons C/yr

103kg

metricton

1.431013kgCO

2/yr

7. Rate of change for CO2 concentration in atmosphere (ppm/yr)

Change in number of moles CO2 from part a =

(1.211013kgCO

  

2/yr)103g

1moleCO2

kg

44gCO2

2.751014molesCO

2.751014molesCO

1.51020moles air

2/yr

2/yr

1ppm

10-6molesCO2

Change in mole fraction (ppm) of CO2 =   

This rate of change compares well with the observed rate of change of 0.5%/yr, which at the current

concentration of CO2 is (.005)(360 ppm) = 1.80 ppm/yr.

8. The rate of CO2 accumulation would decrease if the processes of CO2 fertilization and forest growth

were enhanced by a future global temperature rise. On the other hand, the rate of CO2 accumulation

would increase if the processes of CO2 release were accelerated, for example, by microbial

metabolism in soil.

moles air

1.83ppm/yr

For tris(2-chloroethyl) phosphate, NOAEL = 22 mg/kg-day and LOAEL = 44 mg/kg-day for increased

weights of the liver and kidneys in rats.

Using the NOAEL:

RfDNOAEL

FAFHFS22 mg/kg-day

  

10 x 10 x 10.022 mg/kg-day

Using the LOAEL:

RfDLOAEL

FAFHFSFL44 mg/kg-day

  

10 x 10 x 10 x 10.0044 mg/kg-day

The lesser of the two values would be chosen for the RfD.

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