2.3 Engineering Sciences

Scope and the Conservation Theme

Engineering Sciences is 4–6 questions on the FE Civil exam and gathers the cross-disciplinary physics that other areas assume: work, energy, and power, basic electricity, and basic thermodynamics/heat. Civil engineers meet these in pumps (power), site power and instrumentation (circuits), and concrete curing or pavement temperature (heat). The single most powerful idea is conservation of energy — energy is neither created nor destroyed, only converted among mechanical, electrical, and thermal forms.

Because the exam is open to the NCEES FE Reference Handbook, success means selecting the correct relation and keeping a consistent unit system; SI (joules, watts, kelvin) is cleanest, but be ready for US Customary units (ft·lb, Btu).

Work, Energy & Power

Work is force acting through a distance in the direction of motion: W = F·d (more generally W = ∫F·dx, and W = F·d·cosθ when force and motion are not aligned). Units: joule (J = N·m) in SI, ft·lb in USCS. Energy is the capacity to do work and shares those units:

Power is the time rate of work, P = W/t, measured in watts (W = J/s); 1 horsepower = 746 W. For a steady force moving at velocity v, P = F·v.

The work-energy principle ties these together: the net work done on a body equals its change in kinetic energy, W_net = ΔKE = ½mv₂² − ½mv₁². For conservative systems, mechanical energy is conserved: KE₁ + PE₁ = KE₂ + PE₂, so a mass dropped from height h arrives with v = √(2gh). These relations let you solve impact, braking-distance, and falling-object problems without integrating forces. The recurring trap is mixing weight and mass: in SI, mass is in kilograms and weight W = mg in newtons; in US Customary, a 'pound' can be force (lbf) or mass (lbm), and you must insert g_c = 32.2 lbm·ft/(lbf·s²) to keep units consistent.

Basic Electricity — Ohm's Law and Resistance

Direct-current circuits obey Ohm's law: V = I·R, where V is voltage (volts), I is current (amperes), and R is resistance (ohms, Ω). Electrical power has three equivalent forms: P = V·I = I²R = V²/R (watts). Combine resistors with these rules:

• Series: R_eq = R₁ + R₂ + R₃ + … (same current through each; voltages add). Series resistance is always larger than any one resistor.
• Parallel: 1/R_eq = 1/R₁ + 1/R₂ + … (same voltage across each; currents add). Parallel resistance is always smaller than the smallest resistor.

For just two parallel resistors a handy form is R_eq = (R₁·R₂)/(R₁ + R₂). The classic trap is adding parallel resistors directly instead of adding reciprocals.

Two conservation laws govern circuits and are listed in the Handbook. Kirchhoff's current law (KCL): the sum of currents into a node equals the sum out (charge conservation). Kirchhoff's voltage law (KVL): the sum of voltage rises and drops around any closed loop is zero (energy conservation). In a series circuit the same current flows everywhere and the source voltage divides across the resistors (voltage divider: V_R = V_source·R/R_total); in a parallel circuit the same voltage appears across every branch and the current divides. Energy consumed over time is E = P·t, billed in kilowatt-hours (1 kWh = 3.6 MJ).

Basic Thermodynamics & Heat

The first law of thermodynamics is conservation of energy: ΔU = Q − W (internal energy change equals heat added minus work done by the system). On the FE Civil exam the heat concept most likely to appear is sensible heat: Q = m·c·ΔT, where m is mass (kg), c is specific heat (J/kg·°C; water ≈ 4186 J/kg·°C), and ΔT is the temperature change. 8·T(°C) + 32. Heat transfers by conduction (Fourier's law, q = −kA·dT/dx), convection, and radiation. A ΔT is identical in °C and K (a 10 °C rise equals a 10 K rise), but absolute temperatures are not — a frequent mix-up.

Worked Example — Pump Power

A pump lifts water to height h = 20 m at a mass flow rate of ṁ = 50 kg/s. Find the ideal power required.

Step 1 — energy per second (power) added equals the rate of potential-energy gain: P = ṁ·g·h.

Step 2 — substitute: P = (50 kg/s)(9.81 m/s²)(20 m) = 9,810 W ≈ 9.81 kW.

If the pump is 80% efficient, the input power is P_in = P/η = 9.81/0.80 ≈ 12.3 kW.

Cross-check (electricity): if that 12.3 kW pump runs on a 480-V circuit, the current is I = P/V = 12,300/480 ≈ 25.6 A — a direct use of P = VI. Keeping units consistent (watts, not kilowatts, when dividing by volts) prevents a factor-of-1000 error.

Worked Example — Steady Conduction

For one-dimensional steady conduction through a flat wall, Fourier's law integrates to Q = k·A·ΔT/L, where k is thermal conductivity (W/m·K), A is the cross-sectional area, ΔT is the temperature difference across the wall, and L is the wall thickness. The combination L/(k·A) is the thermal resistance R_th, so Q = ΔT/R_th — a direct analogy to Ohm's law (Q ↔ I, ΔT ↔ V, R_th ↔ R).

A concrete wall has k = 1.7 W/m·K, area A = 10 m², thickness L = 0.20 m, with inside surface at 25 °C and outside at 5 °C (ΔT = 20 °C = 20 K).

Step 1 — apply Fourier's law: Q = k·A·ΔT/L = (1.7)(10)(20)/0.20.

Step 2 — evaluate: Q = (1.7 × 10 × 20)/0.20 = 340/0.20 = 1,700 W of heat flow through the wall.

Because a ΔT of 20 °C equals 20 K, the result is identical whether you work in Celsius or kelvin — only temperature differences enter the equation.

Worked Example — Series & Parallel Resistance

Resistors of R₁ = 4 Ω and R₂ = 12 Ω are placed in parallel, and that combination is then put in series with R₃ = 5 Ω across a 24-V source. Find the total resistance and the source current.

The parallel result (3 Ω) is correctly smaller than the smaller branch (4 Ω), confirming the reciprocal rule. The full 3 A flows through R₃, but it divides between the parallel branches in inverse proportion to their resistances: the 4-Ω branch carries three times the current of the 12-Ω branch.