Fall 2015 HW3 Solutions PDF

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GEORGIA INSTITUTE OF TECHNOLOGYSCHOOL OF CIVIL AND ENVIRONMENTAL ENGINEERINGCEE 2300 – Environmental Engineering Principles S. G. Pavlostathis/Fall 2015PROBLEM SET #3 – SOLUTIONS4-1 (10 pts) A municipal landfill has available space of 16 ha at an average depth of 10 m. Seven hundred sixty-five (765)...

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GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL OF CIVIL AND ENVIRONMENTAL ENGINEERING CEE 2300 – Environmental Engineering Principles

S. G. Pavlostathis/Fall 2015

PROBLEM SET #3 – SOLUTIONS 4-1 (10 pts)

A municipal landfill has available space of 16.2 ha at an average depth of 10 m. Seven hundred sixty-five (765) m3 of solid waste is dumped at the site 5 days a week. This waste is compacted to twice its delivered density. Draw a massbalance diagram and estimate the expected life of the landfill in years.

Solution: a. Mass balance diagram Solid Waste 765 m3 · d-1

Accumulated Solid Waste

b. Total volume of landfill (16.2 ha)(104 m2 · ha-1)(10 m depth) = 1.620 x 106 m3 c. Volume of solid waste is ½ delivered volume after it is compacted to 2 times its delivered density (765 m3 d-1)(0.5) = 382.5 m3 d-1 d. Annual volume of solid waste placed in landfill (382.5 m3 d-1)(5 d · wk-1)(52 wk · y-1) = 9.945 x 104 m3 · y-1 e. Estimated expected life 1.620 10 6 m 3  16.29 or 16 years 9.945  10 4 m 3  y 1 NOTE: the actual life will be somewhat less due to the need to cover the waste with soil each day (typical practice).

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4-8 (10 pts)

In water and wastewater treatment processes a filtration device may be used to remove water from the sludge formed by a precipitation reaction. The initial concentration of sludge from a softening reaction (Chapter 10) is 2% (= 20,000 mg L-1) and the volume of sludge is 100 m3. After filtration the sludge solids concentration is 35%. Assume that the sludge does not change density during filtration and that liquid removed from the sludge contains no sludge. Using the mass-balance method, determine the volume of sludge after filtration.

Solution: a. Mass balance diagram

Filter

Vin = 100 m3 Cin = 2%

Vout = ? Cout = 35%

b. Mass balance equation Cinin = Coutout c. Solve for out

Vout 

Cin Vin C out

d. Substituting values Vout 

0.02 100m 3   5.71 m 3 0.35

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4-11 (10 pts) To remove the solution containing metal from a part after metal plating, the part is commonly rinsed with water. This rinse water is contaminated with metal and must be treated before discharge. The Shinny Metal Plating Co. uses the process flow diagram shown in Figure P-4-11. The plating solution contains 85 g L-1 of nickel. The parts drag out 0.05 L min-1 of plating solution into the rinse tank. The flow of rinse water into the rinse tank is 150 L min-1. Write the general massbalance equation for the rinse tank and estimate the concentration of nickel in the wastewater stream that must be treated. Assume that the rinse tank is completely mixed and that no reactions take place in the rinse tank.

Solution:

a. Mass balance diagram Qdragout = 0.05 L · min1

Qin = 0.05 L · min-1 Cin = 85 g · L-1

Qrinse = 150 L · min-1 Crinse = 0 Qrisne = 150 L · min-1 Cnickel = ?

b. Mass balance equation QinCin + QrinseCrinse – QdragoutCnickel – QrinseCnickel = 0 c. Because Crinse = 0 this reduces to QinCin = QdragoutCnickel + QrinseCnickel d. Solving for Cnickel C nickel 

Qin Cin Q dragout  Q rinse

e. Substituting values

0.05 L  min 85 g  L   28 mg  L  1

C nickel

(0.05  150) L  min

1

1

1

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4-14 (10 pts) If biodegradable organic matter, oxygen, and microorganisms are placed in a closed bottle, the microorganisms will use the oxygen in the process of oxidizing the organic matter. The bottle may be treated as a batch reactor, and the decay of oxygen may be treated as a first-order reaction. Write the general mass-balance equation for the bottle. Using a computer spreadsheet program you have written, calculate and then plot the concentration of oxygen each day for a period of 5 days starting with a concentration of 8 mg L-1. Use a rate constant of 0.35 d-1. Solution:

a. General mass balance equation for the bottle is Ct = Coe-kt b. With Co = 8.0 mg · L-1 and k = 0.35 d-1, the plotting points for oxygen remaining are: Day 0 1 2 3 4 5

Oxygen Remaining, mg/L 8.0 5.64 3.97 2.79 1.97 1.39

-1

Oxygen Remaining [mg - L ]

Problem 4-14 BOD Decay 9 8 7 6 5 4 3 2 1 0 0

1

2

3

4

5

6

Day

Figure S-4-14: BOD decay (BOD: Biochemical Oxygen Demand)

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4-16 (10 pts) A water tower containing 4000 m3 of water has been taken out of service to install a chlorine monitor. The concentration of chlorine in the water tower was 2.0 mg L-1 when the tower was taken out of service. If the chlorine decays by first-order kinetics with a rate constant k = 1.0 d-1, what is the chlorine concentration when the tank is put back in service 8 hours later? What mass of chlorine (in kg) must be added to the tank to raise the chlorine back to 2.0 mg L-1? Although it is not completely mixed, you may assume the tank is a completely mixed batch reactor.

Solution:

a. Because there is no influent or effluent, the concentration is described by first order decay. Ct  e  kt Co

b. Substituting values and solving for Ct (Note: 8 h = 0.33 d) Ct = 2.0 exp [-(1.0 d-1)(0.33 d)] Ct = 2.0(0.72) = 1.44 mg · L-1 c. Mass of chlorine to raise concentration back to 2.0 mg · L-1 Concentration change required 2.0 mg · L-1 – 1.44 mg · L-1 = 0.56 mg · L-1 Mass required in kg

0.56mg  L 4000m 1000L  m   2.25 1

3

10 6 mg  kg 1

3

or 2.3kg

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4-23 (10 pts) A 1900-m3 water tower has been cleaned with a chlorine solution. The vapors of chlorine in the tower exceed allowable concentrations for the work crew to enter and finish repairs. If the chlorine concentration is 15 mg m-3 and the allowable concentration is 0.0015 mg L-1, how long must the workers vent the tank with clean air flowing at 2.35 m3 s-1?

Solution:

a. Assume the water tower behaves as CMFR (= CSTR) Use equation 4-40; for a non-reactive, i.e., conservative substance, k = 0 Cout = Co exp[-(t/to)] Eq. 1 to 

1900 m 3  808.51 s 2.35 m3  s1

Convert concentration to similar units (0.0015 mg · L-1)(1,000 L · m-3) = 1.5 mg · m-3 Substitute all known values to Eq. 1, above and solve for t:   t  1.5 mg  m 3  15 mg  m 3 exp   808.51 s   t  0.10  exp   808.51 s 

Take the natural log of both sides  t    2.303   808 . 51 s  

t = 1,861.66 s or 31 min or 30 min

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B1 (10 pts)

Calculate the volume (m3) of an ideal plug flow reactor for 95% conversion of a pollutant. The volumetric flow rate is 6,500 m3/day and the pollutant is converted according to a first-order reaction with a rate constant, k = 9,000 days-1.

Solution:

Ct = (1 – 0.95) Co = 0.05 Co From Table 4-1 and a PFR: to = [ln (Co/Ct)]/k = [ln (Co/0.05 Co)]/k = [ln 20]/9000 d-1 = 3.33 x 10-4 days But, to = V/Q ==> V = to Q = (3.33 x 10-4 days)(6,500 m3/d) = 2.16 m3

B2 (10 pts)

For a continuous-flow, completely mixed reactor (i.e., CSTR) with a flow rate (Q) equal to 1,000 m3/day, an influent contaminant concentration (Co) equal to 200 mg/L, and a second-order reaction rate constant (k) equal to 0.2 L/mg · day, calculate the steady-state detention time (to, days) and the reactor volume (V, m3) necessary to achieve a contaminant removal efficiency equal to 90%.

Solution:

For an influent contaminant concentration of 200 mg/L and a reactor removal efficiency of 90%, Ct = (1 - 0.9) 200 = 20 mg/L From Table 4-1, second-order reaction and an ideal CSTR: to = [1/(k Ct)] [(Co/Ct) - 1] 1 200 to = ----------- ( ------ - 1) = 2.25 days 0.2(20) 20 But, to = V/Q ==> V = to Q = (2.25 d) 1000 m3/d = 2,250 m3

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B3 (20 pts)

Consider the air over a city to be a box 100 km wide, 100 km long and 1 km high. Clean air is blowing into the box along one of its sides with a speed of 4 m/s. A reactive air pollutant is emitted into the box at a rate of 10 kg/s and decays with a decay rate constant k = 0.20 h-1. a) Find the steady-state concentration of the air pollutant if the air is assumed to be completely mixed. b) If the wind speed suddenly drops to 1 m/s, and all other conditions remain the same, estimate the air pollutant concentration 2 hours later.

Solution: a)

NOTE: C∞ = Ct at steady-state (= Css) b)

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