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Advanced Environmental Engineering

Major Theme

1. Mass and Energy Transfer

2. Environmental Chemistry

3. Water Pollution

4. Treatment of Water and Wastes

※ Include introduction of Membrane Separation Process

Chapter 1. Mass and Energy Transfer

1. Introduction

1) American unit verse SI units

2) Law of conservation of mass

- Important tool for quantitatively tracking pollutants

- Pollutant have to go somewhere

(Transport pollutants from one medium to another)

3) Law of conservation of energy

- Global climate change

- Thermal pollution

- Dispersion of air pollution

2. Units of Measurement

Table 1. Common Prefixes

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Quantity Prefix Symbol

1018 exa E

1015 peta P

1012 tera T

109 giga G

106 mega M

103 kilo k

102 hecto h

10 deca da

10-1 deci d

10-2 centi c

10-3 milli m

10-6 micro μ

10-9 nano n

10-12 pico p

1) Liquid : weight of Substance per unit volume of Mixture

(weight of Solute per unit volume of Solution)

1 mg/L = 1 g/m3 = 1 ppm (W/V or W/W by weight) (1)

1 µg/L = 1 mg/m3 = 1 ppb (W/V or W/W by weight) (2)

where ppm : part per million and ppb : part per billion.

But, in special case like high concentration of high specific gravity of

pollutant,

mg/L = ppmw (W/W by weight) × Specific Gravity (3)

2) Gas : Express pollutant concentrations in Volumetric terms

volumes of air

volume of gaseous pollutant ppmv by volume VV (4)

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×

× ×

×

or more simply

× ℃ and (5)

For other temperature and pressure,

× ×

×

(6)

【Example 1】Converting ppm to ㎍/m3

The federal air quality standard for carbon monoxide(based on an 8-hr

measurement) is 9.0 ppm. Express this standard as a percentage by volume as

well as in mg/m3 at 1 atm and 25 ℃.

<Solution>

There are 9.0 volumes of CO within a million volume of this air.

×

×

To find the concentration in mg/m3,

- Molecular weight of CO : 28

- ×

×

at 25 ℃(298 K) by Eq.(1.6)

※

× ℃ and

3. Material Balance

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Figure 1. A material balance diagram.

(Input) = (Output) + (Decay) + (Accumulation) (8)

1) Steady State Conservation Systems

At steady state or equilibrium, nothing is changing with time.

Simplest system : Input rate = Output rate (9)

(if there are no decay like chemical reaction)

Figure 2. A steady state conservative system. Pollutants enter and leave the

region at the same time rate.

【Example 2】Two Polluted Stream

A stream flowing at 10.0 m3/s has a tributary feeding into it with a flowing 5.0

m3/s. The stream’s concentration of chlorides upstream of the junction is 10.0

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mg/L and the tributary chloride concentration is 40.0 mg/L. Treating chlorides as

a conservative substance, and assuming complete mixing of the two streams, find

downstream chloride concentration.

<Solution>

Figure 3 Flow rate and chloride concentration for example stream and tributary.

From Eq.(1.10),

× × ·

2) Steady State Systems with Non-conservative Pollutants

Input rate = Out rate + Decay rate (11)

(Steady State → No Accumulation)

If the rate of loss of the substance is proportional to the amount of the

substance, it means that the dissipation model of the substance 1st order reaction,

dC/dt = - KC (13)

where K = reaction rate coefficient, (time)-1

C= pollutant concentration

which yields

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ln ln ln

Solving for concentration gives us

(13)

where C0 = initial concentration of pollutant at time=0.

Decay rate = - d(CV)/dt = - V dC/dt = KCV (14)

(if C is uniform throughout the fixed volume V like CSTR)

Then from Eq.(1.11) and Eq.(1.14),

Input rate = Output rate + KCV (15)

【Example 3】A Polluted Lake

Assuming

1. The pollution is completely mixed in the lake

2. No evaporation or other water losses or gain

Find the steady state concentration.

Figure 1.4 A lake with a non-conservative pollutant.

<Solution>

Input rate = QsCs + QwCw

= (5.0 m3/s×10.0 mg/L+0.5 m3/s×100.0 mg/L)×103 L/m3

= 1.0 × 105 mg/s

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Output rate = QmCm = (Qs + Qw) × C, since Cm = C.

= (5.0 + 0.5)m3/s × C mg/L × 103 L/m3

= 5.5 × 103 mg/s

Decay rate = KCV = × × × × ×

= 23.1 × 103 C mg/s

So from Eq.(1.15),

1.0 × 105 = 5.5 × 103 C + 23.1 × 103 C = 28.6 × 103 C

× ×

【Example 4】A Smoky Bar

Figure 5 Tobacco smoke in a bar

① 50 smokers in a bar and each smoking two cigarettes per hour.

② An individual cigarette emits about 1.4 mg of formaldehyde(HCHO).

③ Formaldehyde converts to CO2 with a reaction rate coefficient K = 0.40/hr

Estimate the steady state concentration of HCHO in the air and how does the

results compare with the threshold for eye irritation of about 0.05 ppm?

Assuming complete mixing, 25℃, 1 atm.

<Solution>

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Input rate = 50 smokers × 2 cigas/hr × 1.4 mg/ciga = 140.0 mg/hr

Output rate = 1000 m3/hr × C (mg/m3) = 1000 C mg/hr

Decay rate = KCV = (0.4/hr)×(C mg/m3)×(500 m3) = 200 C mg/hr

Input rate = Output + Decay rate

140 = 1000 C + 200 C = 1200 C

C = 0.117 mg/m3

Convert mg/m3 to ppm by using Eq.(1.7), MW of HCHO is 30,

×

×

This is more than enough to cause eye irritation.

3) Step Function Response

Figure 1.6 A box model for a transient analysis

Accumulation rate = Input rate – Output rate – Decay rate

(16)

where V = box volume(m3)

C = conc. in the box and in the exiting waste stream(g/m3)

S = total rate at which pollutants enter the box(g/hr)

Q = the total flow rate into and out of the box(m3/hr)

K = reaction rate coefficient(hr-1)

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At steady state, dC/dt = 0, which yields

∞

(17)

At unsteady state, from Eq.(1.16)

(18)

Let

(19)

Then

(20)

So, Eq.(1.18) becomes

(21)

(22)

where y0 = y at t = 0 and if C0 = C at t = 0, then from Eq.(1.19)

(1.23)

Eq.(1.19) & Eq.(1.23) → Eq.(1.22) yields

or

(1.24)

By inserting the steady state solution, C(∞) from Eq.(1.17)

∞ ∞ (1.25)

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Figure 7. Step function response for a complete-mix box model.

【Example 4】The Smoky Bar Revisited

∙ Volume (V) = 500 m3 ∙ Fresh air Q = 1,000 m3/hr

∙ K = 0.4 hr-1∙ S = 140 mg/hr ∙ C0 = 0 at 5 pm

<Question> C after 1 hr?

<Solution>

∙ Steady state solution

C(∞) =S

=140 mg/hr

= 0.117mg

Q + KV 1000 m3/hr+0.4/hr×500 m3 m3

∙ C(t) from Eq (1.25) with C0 = 0

∞ ∞

So, C(1 hr) = 0.106 mg/m3 after 1 hr(at 6 pm)

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[Example 5] Cooling tower improvement

<Given Data>

1) Q5 = 0 (only air flow)

2) C4 = 0 (only pure water is evaporating)

3) Q1, Q6, C1, W5

<Question> Q4 and C2 = ?

<Solution>

▶ Overall Mass Balance : Q4 = Q1 – Q6 ①▶ Component Balance : Accumulation = Input – Output

VdC/dt = (Q1C1 + W5) – (Q4C4 + Q6C6) = 0 at steady state

and C4 = 0

∴ Q6C6 = Q1C1 + W5 → C2 = C6 = (Q1C1 + W5) / Q6 ②

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<Given Data>

1) Q5 = 0 (only air flow)

2) C4 = 0 (only pure water is evaporating)

3) Q1, Q6, C1, W5

3) Recovery rate of filter = R = Qb/Qa → Qb = RQa ③4) Rejection rate of filter = η = (Ca – Cb)/Ca = 1 – Cb/Ca

→ Cb = (1 – η)Ca = (1 – η)C2 ④<Question> Q4 and C2 = ?

<Solution>

▶ Mass Balance :

Q1 = Q4 + Q6 → Q6 = Q1 – Q4 ⑤ Q6 = Qa – Qb ⑥ Eq③→Eq⑥ : Q6 = (1 – R)Qa → Qa = Q6 / (1 – R) ⑦▶ Component Balance : Accumulation = Input – Output

V∙dC/dt = (Q1C1 + W5) – (Q4C4 + Q6C6) = 0 at steady state

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and C4 = 0

Q6C6 = Q1C1 + W5 ⑧ Q6C6 = QaCa – QbCb = QaC2 – QbCb ⑨ Eq④→Eq⑨ : Q6C6 = QaC2 – (1 - η)QbC2 ⑩ Eq⑧→Eq⑩ : Q1C1 + W5 = QaC2 – (1 – η)QbC2 ⑪ Eq③→Eq⑪ : Q1C1 + W5 = QaC2 – (1 – η)RQaC2

= QaC2[1 – (1 – η)R] ⑫ Eq⑦→Eq⑫ : Q1C1 + W5 = Q6C2[1 – (1 – η)R] / (1 – R)

Let ζ = R / (1 – R) → Q1C1 + W5 = Q6C2(1 + ζη) ⑬ Rearrange Eq⑬ → C2 = (Q1C1 + W5) / Q6(1 + ζη) ⑭

If Q6 = Q6 and Q1 = Q1, C2 is much smaller than C2, because

C2 after improvement=

C2=

1≪ 1.0

C2 before improvement C2 1 + ζηwhere R = 0.90 ∼ 0.95(ζ = 9 ∼ 19) and η = 0.9 normally.

It means that the cooling water quality could be enhanced by adding the filter,

if the make-up water quality and flow rate were not changed.

<Question> Q1/Q1? when C2 = C2 after adding the filter.

<Solution>

From Eq②, C2(Q1 – Q4) = Q1C1 + W5

Q1(C2 – C1) = W5 + Q4C2

Q1 = W5 + Q4C2 ⑮C2 – C1

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From Eq⑭,C2 = W5 + Q1C1

(1 + ζη)(Q1–Q4)

(1 + ζη)(Q1–Q4)C2 = W5 + Q1C1

Q1[(1 + ζη)C2–C1] = W5 + (1 + ζη)Q4C2

Q1 = W5 + (1 + ζη)Q4C2 ⑯(1 + ζη)C2–C1

Divide Eq⑯ by ⑮, Q1=

(C2–C1)[W5 + (1 + ζη)Q4C2]

Q1 [(1 +ζη)C2–C1](W5 + Q4C2)

Because ζ = 9 ∼ 19 and C1 is normally very small, Q1≪Q1. It means most of

make-up water or waste water can be saved by installing the filter at cooling

tower. But it was not true if the filter could not removed soluble pollutants like

hardness, since they would be accumulated at the inside of cooling water.

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[Extra Example 6] RO system with feed-and-bleed mode operation

<Given Data> Q1, C1, Q3, R, η1) Recovery rate of RO : R = Q3/Q2 → Q2 = Q3 / R ①2) Rejection rate of RO : η = (C2 – C3) / C2 = 1 – C3 / C2

→ C3 = (1 – η)C2 ②<Question> C3 = ?

<Solution>

▶ Mass Balance :

Q6 = Q1 – Q3 ③Q2 = Q1 + Q5 = Q3 / R → Q5 = Q3 / R – Q1 ④Q4 = Q2 – Q3 = Q3 / R – Q3 = (1 / R – 1)Q3 ⑤Q6 = Q4 – Q5 ⑥▶ Component Balance (※ C4 = C5 = C6)

Q2C2 = Q3C3 + Q4C4 = Q3C3 + Q4C6

→ C2 = (Q3C3 + Q4C6) / Q2 ⑦Q2C2 = Q1C1 + Q5C5 = Q1C1 + Q5C6 ⑧Q6C6 = Q1C1 – Q3C3 → C6 = (Q1C1 – Q3C3) / Q6 ⑨From Eq ⑧ and ⑨,

Q2C2 = Q1C1 + Q5C6 = Q1C1 + Q5(Q1C1 – Q3C3) / Q6 ⑩From Eq ② and ⑩,

Q2C3 = (1 – η)Q2C2 = (1 – η)[Q1C1 + Q5(Q1C1 – Q3C3) / Q6]

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= (1 – η)(Q1C1 + C1Q1Q5 / Q6 – C3Q3Q5 / Q6)

= (1 – η)Q1C1 + (1 – η)C1Q1Q5 / Q6 – (1 – η)C3Q3Q5 / Q6

Multiply Q6 on both side and rearrange, then

Q2Q6C3 = (1 – η)Q1Q6C1 + (1 – η)Q1Q5C1 – (1 – η)Q3Q5C3

[Q2Q6 + (1 – η)Q3Q5]C3 = (1 – η)(Q5 + Q6)Q1C1 = (1 – η)Q1Q4C1

C3 = (1 – η)Q1Q4 × C1Q2Q6 + (1 – η)Q3Q5

= (1 – η)Q1(1 / R – 1)Q3 × C1(Q3/R)(Q1 – Q3) + (1 – η)Q3(Q3/R – Q1)

= (1 – η)Q1(1 – R)Q3 × C1Q3(Q1 – Q3) + (1 – η)Q3(Q3 – RQ1)

= (1 – η)(1 – R)Q1Q3 × C1Q3[(Q1 – Q3) + (1 – η)(Q3 – RQ1)]

= (1 – η)(1 – R)Q1 × C1(Q1 – Q3) + (1 – η)(Q3 – RQ1)

C3 = (1 – η)(1 – R)Q1 × C1 ∥Ans ⑪[1 – (1 – η)R]Q1 – ηQ3

From Eq ②, C2 = C3 / (1 – η) ⑫From Eq ⑪ and ⑫,

C2 = (1 – R)Q1 × C1 ∥Ans ⑬[1 – (1 – η)R]Q1 – ηQ3

From Eq ⑨ and ⑬,

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C6 = [Q1C1 – (1 – R)Q1Q3 ] / Q6[1 – (1 – η)R]Q1 – ηQ3

= [Q1C1 – (1 – R)Q1Q3 ] / (Q1 – Q3)[1 – (1 – η)R]Q1 – ηQ3

= Q1C1 – (1 – R)Q1Q3 ∥AnsQ1 – Q3 (Q1 – Q3){[1 – (1 – η)R]Q1 – ηQ3}

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4 Energy Fundamentals

1st Law of Thermodynamics(Energy Conservation) ⇒

Write Energy Balance EquationsAnalysis of Energy Flow

2nd Law of Thermodynamics(Some inefficiency in Work) ⇒ Heat is generated by doing Work

※ Interest is how the heat generated affected on the environment.

∙ Definition of Energy : Capacity for doing Work

∙ Definition of Work : Force × Displacement of Object

∙ Definition of Power : Rate of doing Work (Energy/Time)

= 1 J/s = 1 W = 3.412 Btu/hr

1) The 1st Law of Thermodynamics

▶ Open System

both energy and matter can flow across the boundary.

▶ Closed System

energy is allowed to flow across the boundary, but matter is not.

▶ Accumulation = Macroscopic forms + Microscopic forms

▶ Macroscopic forms : kinetic and potential energies

▶ Microscopic forms : atomic and molecular structure of the system

- kinetic energies of molecules.

- energies associated with the forces acting between molecules, between

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atoms within molecules, and within atoms.

∴ E(Total) = U(Internal) + KE(Kinetic) + PE(Potential) (47)

▶ Specific Heat(c)

The amount of energy needed to raise the temperature of a unit mass of a

substance by 1 degree.

▶ British Thermal Unit(Btu)

the energy required to raise 1 lb of water by 1 ℉.

▶ Kilocalorie(kcal)

the energy required to raise 1 kg of water by 1 ℃

※ Conversion Factor : 1 kcal/kg℃ = 1 BTU/lb℉ = 4.184 kJ/kg℃

▶ Specific heat at constant volume(cv) & Specific heat at constant pressure(cp)

- For incompressible substances(liquids and solids) : cv = cp

- For gases : cv < cp

▶ Enthalpy(H) = U(Internal Energy) + P(Pressure)∙V(Volume) (48)

※ Unit of H or U : energy, kJ or Btu

For solids or liquids, cv = cp and ΔU = ΔH,

Change in stored energy = m c ΔT (51)

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【Example 10】A Water Heater

∙Energy source : Water heater 5 kW

∙Boundary volume : 40 gallon

<Question> Time to raise the temperature 50 ℉ → 140 ℉ ?

<Assumption>

1) No energy loss during converting electrical energy to heat.

2) No energy losses from the tank.

3) No water withdrawn from tank.

<Solution>

∙ Rate of energy input = Power = 5 kW × Δt(hr) = 5∙Δt (kWhr)

∙ Energy output = 0

∙ Change in stored energy

=40gal×8.34lb/gal×1Btu/lb°F×(140-50)°F=30,060BTU=30×103BTU

※ specific gravity of water = 8.34 lb/gal

∙ By conservation, Input energy rate = Internal energy change rate,

5∙Δt (kWhr) × 3,412 ((BTU/hr)/kW) = 30 × 103 BTU

∴ Δt = 1.76 hr∥Answer

■ Phase Change

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Energy released or absorbed in phase change = m L (52)

where m = mass, L = latent heat of fusion or vaporization.

※ Melting energy of Ice at 0 ℃ = 333 kJ/kg

※ Vaporization energy of water at 100 ℃ = 2,256 kJ/kg

※ 15℃ is a useful number that can be used to estimate the amount of energy

required to cause surface water on the Earth to evaporate. The value of 15℃ has been picked as the starting temperature since that is approximately the

current average surface temperature of the globe.

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Figure 13 The latent heats of fusion and vaporization for water.

【Example 11】Power for the Hydrologic Cycle

∙ Global rainfall = about 1 m of water per year

∙ Earth’s surface = 5.10 × 1014 m2

<Question>

1) Energy required to cause that much water to evaporate each year.

2) Compare this to world energy consumption at 2007(4.7×1017 kJ).

3) Compare it to the average rate at which sunlight strikes the surface of the

Earth(168 W/m2).

<Solution>

In Table 4, Heat of vaporization at 15 ℃ = 2,465 kJ/kg

The total energy required to vaporize all of that water is

1) Increase in stored energy

= 1 m/yr × 5.10 × 1014 m2 × 103 kg/m3 × 2,465 kJ/kg

= 1.25 × 1021 kJ/yr ∥ Answer

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2) 4.7×1017 kJ ÷ 1.25 × 1021 kJ ≃ 2,700 Times∥ Answer

3) From Answer 1), the rate of sunlight striking on the Earth is

1.25×1021 kJ 1000 J yr day hr 1 W = 78 W/m2

yr kJ 365 day 24 hr 3600 sec 1 J/s 5.10×1014 m2

This can be compared with 168 W/m2.

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■ Open System

Rate of change of stored energy = m × c × ΔT

where m=Q×d : the mass flow rate across the system boundary.

【Example 12】Thermal Pollution of a River

∙ 1/3 of coal energy is converted to electric energy(1,000 MWt).

⇒ Total coal energy = 3,000 MWe

∙ Waste gas energy through stack = 15% of 2/3 of coal energy.

⇒ Waste energy through gas = 300 MWt

∙ Waste cooling water to river = 85% of 2/3 of coal energy.

⇒ Waste energy through water = 1,700 MWt

∙ River water flow rater, Q = 100 m3/s at 20 ℃.

<Question>

1) ΔT of cooling water = 10℃, Q of cooling water?

2) Temperature of river water after receiving the heated cooling water?

<Solution>

1) Rate of change in internal energy = m × c × ΔT

1,700 MWt = m(kg/s) × 4,184(J/kg℃) × 10(℃) × 1(MWt/106 J/s)

m = 40.6 × 103 kg/s = 40.6 m3/s ∥Answer

2) From Rate of change in internal energy = m × c × ΔT

So, the temperature of the river water will be 24. 1 ℃or

× ×

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Figure 14 Cooling water energy balance for the 33.3 percent efficient, 1000

MWe power plant in Example 12.

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2. The 2nd Law of Thermodynamics

The second law of thermodynamics says that there will always

be some waste heat; that is, it is impossible to devise a machine that can

convert heat to work with 100 percent efficiency

max

FIGURE 15 Definition of terms for a Carnot engine.

where T = absolute temperature,

K = ℃ + 273.15 (56)

R = ℉ + 459.67 (57)

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FIGURE 16. A fuel-fired, steam-electric power plant.

※ Efficiency by rule of thump about thermal power plant

▶ Chemical energy → Thermal energy(ex, combustion) : ∼90%

▶ Well designed turbine(Mechanical → Electric) : ∼90%

▶ Nuclear power plant : 33%

▶ Thermal(Fuel Fired Steam) power plant : max 66%

※ ηmax of thermal power plant by reasonable estimation

Th = 600 ℃, Tc = 20 ℃max

- Steam turbine(Rankin cycle) power plant : 30∼40%

- Combined power plant : max 60%, normal 50%

※ Combined : Gas Turbine(Brayton cycle) + Steam Turbine

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Electricity↑

Electricity Steam Turbine(Rankin Cycle)

↑ ↑

CombustionGas →

Gas Turbine(Brayton Cycle) →

SteamGenerator →

Off gasto stack

【Example 13】Mass & energy balance for a Coal-fired power plant

∙Energy of typical coal = 24 kJ/g

∙Ave. content of carbon in coal = 62%

∙SO2 emission standard = 260 g of SO2 / 106 kJ of heat input

(130 g of S / 106 kJ of heat input)

∙Particulate emission standard = 13 g / 106 kJ of heat input

<Assumption>

∙S content in coal : 2%

∙Amount of ash generated : 10% of feed coal

∙Fly ash : 70%, Bottom ash : 30%

∙Conversion efficiency(η) = 33%

<Question>

1) Emission of SO2, particulate, carbon / kWe-hr?

(assume all of the carbon in the coal is released to the atmosphere).

2) Removal efficiency of SO2 and Particulate emission control system to meet

limitations?

<Solution>

1) Pollutant emission limitation / kWe∙hr of electricity generation

∙Heatinput = 3 kWt∙hr 1 kJ/s 3600 s = 10,800 kJ input

1 kWe∙hr kWt hr kWe∙hr

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∙Sulfur

emission =130 g S 10,800 kJ input

= 1.4g S

106 kJ input kWe∙hr kWe∙hr

∙SO2

emission =260 g SO2 10,800 kJ input

= 2.8g SO2

106 kJ input kWe∙hr kWe∙hr

∙Particulateemission =

13 g Part. 10,800 kJ input= 0.14

g Part.106 kJ input kWe∙hr kWe∙hr

∙Coalinput =

10,800 kJ input 1 g coal= 450

g coalkWe∙hr 24 kJ kWe∙hr

∙Carbon

emission =0.62 g carbon 450 g coal

= 280g carbon

g coal kWe∙hr kWe∙hr

2) Removal efficiency to meet limitation

∙S % of coal = 2%

∙S removalefficiency = 1 -

1.4 g S by limitation= 0.85 = 85%

9.0 g S by raw emission

∙Fly ash

generated = 10% × 70% × 450 g coal

= 31.5g fly ash

kWe∙hr kWe∙hr

∙Part. removal

efficiency = 1 -0.14 g Part. by limitation

= 0.885 = 99.5%31.5 g S by raw emission

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FIGURE 17. Energy and mass balance for a coal-fired power plant generating 1

kWe∙hr of electricity (see Example 13).

【Example 14】 Ocean Thermal Energy Conversion(OTEC) System Efficiency

∙Th = 30 ℃

∙Tc = 5 ℃

<Question> ηmax?

<Solution> ηmax = 1 - 5 + 273 = 0.08 = 8%30 + 273

※ Real efficiency in real OTEC plant : 2∼3%

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3. Conductive and Convective Heat Transfer

▶ Heat will be transferred from the hotter object to the colder one.

▶ Conductive heat transfer is usually associated with solids, as one molecule

vibrates the next in the lattice.

▶ Convective heat transfer occurs when a fluid at one temperature comes in

contact with a substance at another temperature.

▶ Radiation can take place even in the absence of any physical medium

between the objects.

FIGURE 18. Heat transfer through a simple wall.

(58)

where q = heat transfer rate through the wall (W) or (Btu/hr)

A = wall area (m2) or (ft2)

Ti = air temperature on one side of the wall (℃) or (℉)

To = ambient air temperature (℃) or (℉)

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R = overall thermal resistance (m2-℃/W) or (hr-ft2-℉/BTU)

※ The overall thermal resistance R is called the R-value.

【Example 15】Reducing Pollution by Adding Ceiling Insulation

∙Heating season = 8 month per year = 8 × 30 × 24 = 5,760 hr

∙To = 40 ℉

∙Ti = 70 ℉

∙Cost for insulation = 1,000 $ from 11 to 40 hr-ft2-℉/BTU

∙Heating cost = 8 ¢/kWe-hr

<Question>

1) Annual money saving? and pay back period?

2) Annual reduction in particulate, and carbon emissions?

Basis : 1 million homes, coal plants in Example 13

<Solution>

1) Annual saving and pay back period

∙ Heat loss rate with the existing insulation

q = A(Ti - To) = 1500 ft2 × (70-40)℉ = 4090 BTUR 11 (hr-ft2-℉/BTU) hr

∙ Heat loss rate after adding insulation

q = A(Ti - To) = 1500 ft2 × (70-40)℉ = 1125 BTUR 40 (hr-ft2-℉/BTU) hr

∙ Annual energy saving

Energysaved

=(4090–1125)BTU kWe-hr 5,760 hr

= 5005kWe-hr

hr 3412 BTU yr yr

∙ Annual saving

AnnualSaving

=5005 kWe-hr 8 ¢

= 40,040¢≃ 400

$yr kWe-hr yr yr

∙ Pay back period : 1000/400 ≃ 2.5 year

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∙ Saving energy for 1 million same home = 5×109 kWe-hr/yr

2) Apply to 1 million home and reduce pollutants at power plant

∙Carbon

reduction =280 g C 5×109 kWe-hr MT

= 1400×103 MTkWe∙hr yr 106 g yr

∙SO2

reduction =2.8 g SO2 5×109 kWe-hr MT

= 14×103 MTkWe∙hr yr 106 g yr

∙Particulatereduction =

0.14 g Part. 5×109 kWe-hr MT= 0.7×103 MT

kWe∙hr yr 106 g yr