Thermal-Fluid Design and Energy Cases

Define the boundary before the equation

Thermal-fluid FE items are often lost before the first number is entered. Decide what you are analyzing: a closed piston-cylinder, a steady-flow control volume, a pipe run, a pump, a nozzle, a turbine, a heat exchanger, or a wall. That choice decides whether mass crosses the surface, whether shaft work appears, whether flow work is embedded in enthalpy, and whether kinetic or potential energy is negligible.

For a closed system use the first law Q - W = dU with boundary work W = integral P dV. For a steady control volume use the steady-flow energy equation, q - w = (h2 - h1) + (V2^2 - V1^2)/2 + g(z2 - z1), where enthalpy h already contains flow work. For incompressible pipe flow the mechanical energy (extended Bernoulli) equation is fastest.

Worked problem: pump shaft power

A pump delivers water (rho = 1000 kg/m^3) at Q = 0.030 m^3/s while adding H = 18 m of head. Pump efficiency is eta = 0.75. Find the required shaft input power.

Hydraulic (fluid) power: W_fluid = rhogQ*H = 1000 * 9.81 * 0.030 * 18 = 5297 W ≈ 5.30 kW. Shaft input power: W_shaft = W_fluid / eta = 5297 / 0.75 = 7.06 kW. The classic FE distractor multiplies by 0.75 instead of dividing — that would give 3.97 kW, which is physically backwards because the shaft must supply more than the fluid receives.

Worked problem: steady-flow turbine

Superheated steam enters an adiabatic turbine at h1 = 3214 kJ/kg and leaves at h2 = 2470 kJ/kg with mass flow m_dot = 4.0 kg/s; velocity and elevation changes are negligible. The steady-flow energy equation with q = 0 reduces to w = h1 - h2 = 3214 - 2470 = 744 kJ/kg. Power output W_dot = m_dot * w = 4.0 * 744 = 2976 kW ≈ 2.98 MW. The trap here is reaching for internal energy u; for a flow device the proper property is enthalpy h, which already includes flow work P*v.

Property and basis discipline

The FE Reference Handbook supplies steam, refrigerant, and ideal-gas tables. Do not assume ideal-gas behavior for every vapor. Identify the state region first: if P and T are given for water, compare to saturation; if quality x is given, the state is a saturated mixture and any property y is y = y_f + x*(y_g - y_f). Property tables use absolute pressure, so convert gauge readings (P_abs = P_gauge + P_atm) before lookup. Absolute temperature in kelvin or rankine is mandatory in ideal-gas (PV = mRT) and radiation (T^4) relations; a Celsius value substituted into a T^4 term is a guaranteed miss.

For heat transfer, pick the mode before the formula: conduction q = kAdT/L, convection q = hAdT, radiation q = epsilonsigmaA*(T_hot^4 - T_cold^4). Finish with a physical-direction sanity check: heat flows hot to cold, pumps add head, turbines remove energy from the flow, and friction reduces available mechanical energy.

Worked problem: composite-wall conduction

A plane wall has two layers in series: layer A is k_A = 0.8 W/m·K, L_A = 0.10 m, and layer B is k_B = 0.04 W/m·K, L_B = 0.05 m, with area A = 5 m^2. Inner surface is 200 °C, outer surface is 30 °C. Conduction resistances add in series: R = L/(kA). R_A = 0.10/(0.85) = 0.025 K/W; R_B = 0.05/(0.04*5) = 0.25 K/W; R_total = 0.275 K/W. Heat rate q = dT/R_total = (200 - 30)/0.275 = 618 W. Notice layer B, the insulator, carries about 91% of the resistance even though it is thinner — low conductivity dominates. The trap is averaging the two k values or adding conductances instead of resistances.

Worked problem: heat exchanger temperature difference

A counterflow heat exchanger cools oil from 110 °C to 60 °C using water entering at 25 °C and leaving at 45 °C. For counterflow, the end temperature differences are dT1 = 110 - 45 = 65 °C and dT2 = 60 - 25 = 35 °C. 5 °C**. Then q = UALMTD with the overall coefficient U. The classic error is using a plain arithmetic average (50 °C) or pairing the wrong terminal temperatures, which happens when parallel-flow and counterflow logic are mixed. Always match inlet-to-outlet ends correctly for the stated flow arrangement before computing LMTD.

As a cross-check, an energy balance on either stream also yields q: for the oil, q = m_dot_oil * cp_oil * (110 - 60), and that same q must equal the water-side gain m_dot_water * cp_water * (45 - 25). If the two balances disagree, a mass-flow or specific-heat value was misread. Pairing LMTD on the heat-transfer side with conservation of energy on the stream side catches a wrong lookup before it propagates into the final answer — the kind of self-verification the FE rewards under time pressure.