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PE Mechanical Thermal Fluids study materials
PE Mechanical Thermal Fluids study resources

PE Mechanical Thermal Fluids: Thermal and Fluid Systems: Integrated Mass, Energy, and Heat-Transfer Calculations

PE Mechanical Thermal Fluids · Work through your exam, one topic at a time.

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Thermal and Fluid Systems: Integrated Mass, Energy, and Heat-Transfer Calculations

What this domain tests

Thermal and Fluid Systems is the integrated fluid-mechanics, thermodynamics, and heat-transfer domain. Items usually ask you to pick the right conservation statement - mass, momentum, or energy - and carry it through a pipe, pump, heat exchanger, or power cycle with consistent units. Expect property lookups (density, viscosity, cp) that must be evaluated at the correct bulk or film temperature. Many items are multi-step: a flow rate from part one feeds a later heat-transfer calculation.

Core relations

  • Continuity: ṁ = ρAV; incompressible flow in a constant-area pipe gives constant velocity.
  • Extended Bernoulli: p₁/γ + z₁ + V₁²/2g = p₂/γ + z₂ + V₂²/2g + h_L, with h_L = [f(L/D) + ΣK]V²/2g.
  • Reynolds number: Re = ρVD/μ = VD/ν. Laminar below ~2300; fully turbulent above ~4000.
  • Friction factor: f = 64/Re in laminar flow; in turbulent flow use Colebrook or the explicit Swamee–Jain approximation.
  • Pump operating point: intersection of the pump curve with the system curve H = static lift + [f(L/D) + ΣK]V²/2g.
  • Affinity laws at fixed impeller diameter: Q ∝ N, H ∝ N², P ∝ N³.
  • Heat exchangers: q = U·A·F·ΔT_lm; use ε-NTU when outlet temperatures are unknown.
  • Cycles: η_Carnot = 1 − T_L/T_H, with both temperatures absolute.

Worked example

Water (ν = 1×10⁻⁶ m²/s) flows at 0.010 m³/s through a 0.10 m ID pipe, 50 m long, f = 0.020, fittings neglected. Area = π(0.05)² = 0.00785 m², so V = 0.010/0.00785 = 1.27 m/s. Re = 1.27(0.10)/1×10⁻⁶ = 1.3×10⁵, turbulent, so the stated f is valid. Then h_L = 0.020 × (50/0.10) × (1.27²/19.62) = 0.020 × 500 × 0.0826 = 0.83 m of head, or Δp = ρg·h_L ≈ 8.1 kPa. Note that head loss scales with V², so doubling flow roughly quadruples friction loss.

Mark-losing traps

  • Mixing absolute and gauge pressure in NPSH or Bernoulli terms.
  • Using diameter where radius belongs in area, or 2r vs D in the Reynolds number.
  • Dropping minor losses when ΣK is large, or applying affinity laws to a trimmed impeller.
  • Taking water properties at 20 °C when the problem states 80 °C.
  • Using ΔT_lm without the F correction for multipass exchangers.