Energy and Environment

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Conventional and alternative energy source concepts and the environmental impacts of energy production, including greenhouse gas emissions, carbon footprint, and thermal and water demands.

2 concepts

Exam areas

A. Energy sources concepts

Conventional and Alternative Energy Sources

Compare conventional and renewable energy sources by their environmental trade-offs, capacity factor and energy density, and the conversion efficiencies that cap how much usable energy each delivers.

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B. Environmental impact of energy sources and production

Greenhouse Gases and Carbon Footprint

Normalize greenhouse gases to CO2-equivalent via global warming potential, account for emissions with emission factors and life-cycle energy, and quantify the thermal/water burden of power production.

Energy and environment questions on the FE Environmental exam rarely hinge on a single handbook equation — the Reference Handbook is deliberately thin here — so points are won or lost on whether you can reason cleanly through unit-bearing accounting: how much carbon dioxide a kilowatt-hour really carries, how to put methane and carbon dioxide on the same scale, and where the two-thirds of a power plant's fuel energy that never becomes electricity actually goes. This concept builds that fluency. The few governing relations that the handbook does supply — combustion stoichiometry and the Carnot efficiency bound in the Thermodynamics chapter, plus the energy-unit conversion table — anchor everything else, and we lean on them explicitly.

Greenhouse gases and radiative forcing

A greenhouse gas absorbs and re-emits outgoing longwave (infrared) radiation, warming the lower atmosphere. The principal anthropogenic agents are carbon dioxide (

\text{CO}_2

, from combustion of fossil carbon), methane (

\text{CH}_4

, from landfills, livestock, and gas leaks), nitrous oxide (

\text{N}_2\text{O}

, from fertilizer and combustion), and the engineered fluorinated gases (HFCs,

\text{SF}_6

). They differ enormously in two ways: how strongly each molecule traps heat, and how long it survives in the atmosphere. Carbon dioxide is the reference because it dominates the total mass emitted, even though molecule-for-molecule it is one of the weakest.

Global warming potential and CO2-equivalent

To add unlike gases you convert each to the mass of carbon dioxide that would cause the same integrated warming over a chosen horizon (conventionally 100 years). That ratio is the global warming potential,

GWP

, a dimensionless number with

GWP_{\text{CO}_2}\equiv 1

. Representative 100-year values: methane

\approx 28

, nitrous oxide

\approx 265

\text{SF}_6\approx 23{,}500

. Multiply each gas's mass by its

GWP

and sum to get the carbon-dioxide equivalent,

\text{CO}_2e

. This single step is the most-tested skill in the topic: a small mass of a high-

GWP

gas can dominate a footprint.

\text{CO}_2e = \sum_i m_i\,GWP_i

Emission factors and carbon footprint

A carbon footprint is the total

\text{CO}_2e

attributable to an activity over a period. You almost never measure emissions directly; instead you multiply an activity level (kWh consumed, liters of fuel burned, km driven) by a published emission factor,

EF

, expressed as mass of

\text{CO}_2e

per unit of activity. The grid electricity factor is the workhorse — a coal-heavy grid runs near

0.9\ \text{kg CO}_2e/\text{kWh}

, a gas-and-renewables grid near

0.4

, a nuclear-and-hydro grid below

0.05

. Footprinting is then bookkeeping: tally every activity, apply its factor, and sum.

E_{\text{CO}_2e} = \sum_j A_j\,EF_j

Combustion stoichiometry sets the carbon floor

The emission factor of a fuel is not arbitrary — it is fixed by chemistry. Every carbon atom in the fuel leaves complete combustion as one molecule of

\text{CO}_2

, so the carbon mass scales up by the molar-mass ratio

44/12 = 3.67

. Burning methane,

\text{CH}_4 + 2\,\text{O}_2 \to \text{CO}_2 + 2\,\text{H}_2\text{O}

, releases

44/16 = 2.75\ \text{kg CO}_2

per kg of methane. This is why fuels are ranked by carbon intensity: per unit of useful energy, natural gas emits roughly half the

\text{CO}_2

of coal because more of its energy comes from oxidizing hydrogen to water rather than carbon to

\text{CO}_2

m_{\text{CO}_2} = m_{\text{C}}\times\frac{44}{12} = 3.67\,m_{\text{C}}

Life-cycle and embodied energy

A fair comparison counts emissions across the whole life cycle — extraction, manufacture, transport, operation, and disposal — not just the smokestack. The energy consumed to build and deliver a technology before it ever produces a kilowatt-hour is its embodied energy, and the matching metric is the energy payback time: how long the device must run to generate the energy that was spent making it. Operational emissions dominate the life cycle of fossil plants, but for wind and solar — which burn no fuel — the manufacturing (embodied) stage is where almost all the life-cycle

\text{CO}_2e

lives. Life-cycle carbon intensity, in

\text{g CO}_2e/\text{kWh}

, is the number that lets you compare sources honestly: roughly

820

for coal,

490

for gas, and

10

–

50

for nuclear, wind, and solar.

Thermal and water impacts of power production

A thermal power plant is a heat engine, so the second law caps its efficiency at the Carnot bound and guarantees waste heat. With electrical efficiency

\eta = W/Q_H

, the fuel heat input is

Q_H = P/\eta

and the rejected heat is

Q_{rej} = P(1/\eta - 1)

— for a

40\%

plant,

1.5\times

the electrical output is dumped as low-grade heat. Most of that heat leaves through the condenser into cooling water — though a smaller share, roughly

10

–

15\%

of the fuel energy in a fossil plant, leaves hot with the flue gas rather than through the condenser. Once-through cooling draws a large flow from a river or lake and returns it warmer; the temperature rise is set by an energy balance,

Q_{rej} = \dot{m}\,c_p\,\Delta T

. The resulting thermal pollution lowers dissolved-oxygen saturation and stresses aquatic life, which is why discharge

\Delta T

is regulated and why evaporative cooling towers (trading water consumption for a smaller thermal plume) are common.

Q_{rej} = P\left(\frac{1}{\eta}-1\right) = \dot{m}\,c_p\,\Delta T

Exam strategy

First decide whether a gas or a footprint is asked. For mixed gases, build a table — mass,

GWP

, product — and sum to

\text{CO}_2e

; the high-

GWP

trace gas usually dominates, so never drop it because its mass is small. For a footprint, identify each activity and its emission factor and keep units explicit (

\text{kWh}\times\text{kg/kWh}

\text{L}\times\text{kg/L}

). For combustion, route carbon mass through

44/12

and methane mass through

44/16

. For thermal impact, get

\eta

first, then

Q_{rej}=P(1/\eta-1)

, then close the cooling-water balance with

Q_{rej}=\dot{m}c_p\Delta T

using

c_p = 4186\ \text{J/kg}\cdot\text{K}

for water. When asked for the theoretical efficiency ceiling, use Carnot with absolute temperatures.

Key equations

Carbon-dioxide equivalent

\text{CO}_2e = \sum_i m_i\,GWP_i

Common-scale total of a gas mixture; $m_i$ = mass of gas $i$ (any consistent unit), $GWP_i$ = its global warming potential (dimensionless, 100-yr: $\text{CO}_2\,1$ , $\text{CH}_4\,28$ , $\text{N}_2\text{O}\,265$ ).

Carbon footprint by emission factor

E_{\text{CO}_2e} = \sum_j A_j\,EF_j

Total emissions from activities; $A_j$ = activity (kWh, L, km), $EF_j$ = emission factor (mass $\text{CO}_2e$ per unit activity).

Carbon-to-CO2 mass ratio

m_{\text{CO}_2} = \frac{44}{12}\,m_{\text{C}} = 3.67\,m_{\text{C}}

Every fuel carbon atom becomes one $\text{CO}_2$ ; molar masses $44$ and $12\ \text{g/mol}$ . For methane, $m_{\text{CO}_2}=2.75\,m_{\text{CH}_4}$ ( $44/16$ ).

Methane combustion

\text{CH}_4 + 2\,\text{O}_2 \to \text{CO}_2 + 2\,\text{H}_2\text{O}

Stoichiometric complete combustion (FE Handbook, Thermodynamics — Combustion Processes); basis for the natural-gas carbon factor.

Heat-engine (thermal) efficiency

\eta = \frac{W}{Q_H} = \frac{Q_H - Q_L}{Q_H}

Fraction of fuel heat $Q_H$ converted to work/electricity $W$ ; $Q_L$ = rejected heat. Handbook, Thermodynamics — Basic Cycles.

Carnot efficiency limit

\eta_c = 1 - \frac{T_L}{T_H}

Upper bound on any heat engine between absolute temperatures $T_H$ and $T_L$ (K or R). Real plants reach a fraction of this.

Rejected (waste) heat

Q_{rej} = P\left(\frac{1}{\eta}-1\right)

Low-grade heat dumped per unit electrical output $P$ ; fuel input is $Q_H = P/\eta$ .

Cooling-water energy balance

Q_{rej} = \dot{m}\,c_p\,\Delta T = \rho\,Q\,c_p\,\Delta T

Temperature rise $\Delta T$ of cooling flow; water $\rho=1000\ \text{kg/m}^3$ , $c_p=4186\ \text{J/kg}\cdot\text{K}$ , $Q$ = volumetric flow ( $\text{m}^3/\text{s}$ ).

Heat rate

HR = \frac{3413}{\eta}\ \left[\frac{\text{Btu}}{\text{kWh}}\right]

Fuel heat per kWh delivered; inverse measure of efficiency ( $1\ \text{kWh}=3413\ \text{Btu}$ , per the handbook energy-unit table). Lower heat rate = higher efficiency.

Life-cycle carbon intensity

CI = \frac{E_{\text{CO}_2e,\,\text{life}}}{\text{lifetime energy generated}}

Cradle-to-grave $\text{CO}_2e$ per kWh ( $\text{g CO}_2e/\text{kWh}$ ); the metric for comparing sources fairly, capturing embodied as well as operational emissions.

Worked examples

Putting a mixed emission stream on one scale

Problem. An industrial site reports annual releases of $1000\ \text{t}$ of $\text{CO}_2$ , $50\ \text{t}$ of $\text{CH}_4$ , and $2\ \text{t}$ of $\text{N}_2\text{O}$ . Using 100-year global warming potentials of $1$ , $28$ , and $265$ , find the total $\text{CO}_2$ -equivalent and identify the dominant contributor by warming.

Solution. Apply

\text{CO}_2e=\sum m_i GWP_i

term by term:

\text{CO}_2:\ 1000\times1 = 1000\ \text{t}

\text{CH}_4:\ 50\times28 = 1400\ \text{t CO}_2e

\text{N}_2\text{O}:\ 2\times265 = 530\ \text{t CO}_2e

. Sum:

\text{CO}_2e = 1000+1400+530 = 2930\ \text{t CO}_2e

. Methane contributes

1400/2930 = 47.8\%

of the warming from just

50\ \text{t}

of mass. Sanity check: methane is only

\sim 4.8\%

of the emitted mass yet nearly half the impact — exactly the high-

GWP

trap the metric exists to catch. Final:

2930\ \text{t CO}_2e/\text{yr}

, methane-dominated.

\text{CO}_2e = (1000)(1)+(50)(28)+(2)(265) = 2930\ \text{t}

Household carbon footprint

Problem. A home consumes $10{,}000\ \text{kWh/yr}$ of grid electricity (emission factor $0.45\ \text{kg CO}_2e/\text{kWh}$ ) and $40\ \text{GJ/yr}$ of natural gas for heating (emission factor $50.3\ \text{kg CO}_2e/\text{GJ}$ ). Find the annual carbon footprint.

Solution. Electricity:

E_{elec} = 10{,}000\times0.45 = 4500\ \text{kg CO}_2e

. Natural gas:

E_{gas} = 40\times50.3 = 2012\ \text{kg CO}_2e

. Total:

E = 4500+2012 = 6512\ \text{kg} = 6.51\ \text{t CO}_2e/\text{yr}

. Sanity check:

\sim 6.5\ \text{t/yr}

is squarely in the range for a single household; electricity is the larger share, as expected on a moderately carbon-intensive grid. Final:

6.51\ \text{t CO}_2e/\text{yr}

(3 s.f.).

E = (10{,}000)(0.45) + (40)(50.3) = 6512\ \text{kg} = 6.51\ \text{t CO}_2e

Waste heat and cooling-water flow

Problem. A $600\ \text{MW}_e$ steam plant has an electrical efficiency of $40\%$ and uses once-through river cooling. The condenser rejects essentially all the non-electric heat to the river. For a permitted temperature rise of $\Delta T = 8\,^\circ\text{C}$ , find the rejected heat and the required cooling-water flow ( $\rho=1000\ \text{kg/m}^3$ , $c_p=4186\ \text{J/kg}\cdot\text{K}$ ).

Solution. Rejected heat:

Q_{rej} = P(1/\eta - 1) = 600(1/0.40 - 1) = 600(1.5) = 900\ \text{MW}

. Cooling-water flow from the energy balance

Q_{rej} = \rho\,Q\,c_p\,\Delta T

Q = \dfrac{Q_{rej}}{\rho c_p \Delta T} = \dfrac{900\times10^6}{(1000)(4186)(8)} = \dfrac{900\times10^6}{3.349\times10^7} = 26.9\ \text{m}^3/\text{s}

. Sanity check: units

\dfrac{\text{W}}{(\text{kg/m}^3)(\text{J/kg}\cdot\text{K})(\text{K})} = \dfrac{\text{J/s}}{\text{J/m}^3} = \text{m}^3/\text{s}

. A flow of

\sim 27\ \text{m}^3/\text{s}

is a sizeable river — realistic for once-through cooling and the reason such plants sit on large water bodies. Final:

Q_{rej}=900\ \text{MW}

Q=26.9\ \text{m}^3/\text{s}

Q = \frac{P(1/\eta-1)}{\rho\,c_p\,\Delta T} = \frac{900\times10^6}{(1000)(4186)(8)} = 26.9\ \text{m}^3/\text{s}

Common pitfalls

•Dropping a high- $GWP$ trace gas because its mass is small. A few tonnes of $\text{N}_2\text{O}$ or $\text{CH}_4$ can rival thousands of tonnes of $\text{CO}_2$ — always multiply mass by $GWP$ before judging significance.
•Confusing $GWP$ horizons. Methane's $GWP$ is $\approx 28$ over 100 years but $\approx 84$ over 20 years; use the horizon the problem states and do not mix horizons within one sum.
•Forgetting the $44/12$ carbon-to- $\text{CO}_2$ ratio. Reporting emitted $\text{CO}_2$ as the carbon mass understates it by a factor of $3.67$ ; for methane the fuel-to- $\text{CO}_2$ factor is $44/16 = 2.75$ .
•Treating efficiency as the rejected fraction. Waste heat is $Q_{rej}=P(1/\eta-1)$ , not $P(1-\eta)$ — the latter would be a fraction of fuel input $Q_H$ , not of electrical output $P$ . Keep the basis straight.
•Using $^\circ\text{C}$ in the Carnot formula. $\eta_c = 1 - T_L/T_H$ requires absolute temperature (K or R); plugging in Celsius gives nonsense.
•Counting only smokestack (operational) emissions. For wind, solar, and nuclear, nearly all life-cycle $\text{CO}_2e$ is embodied in manufacturing — an operational-only comparison wrongly makes them look like zero and overstates the gap versus what a fair cradle-to-grave intensity shows.
•Mismatching units in a footprint. Electricity factors are per kWh, fuel factors per L or GJ; convert the activity to the factor's basis before multiplying, and watch kg versus tonnes in the total.

References

NCEES FE Reference Handbook — Environmental Engineering
NCEES FE Reference Handbook — Thermodynamics: Combustion Processes, Basic Cycles (Carnot)
IPCC Fifth Assessment Report (AR5), Working Group I — 100-year GWP values for CO2, CH4, N2O
US EPA, Emission Factors for Greenhouse Gas Inventories — grid and fuel emission factors