Defining the Boundary: What is an Example of a Closed System?

In thermodynamics and electrical engineering, a closed system is defined as a physical boundary across which energy (heat and work) can transfer, but mass cannot. When a PE (Professional Engineer) or FE (Fundamentals of Engineering) exam asks 'what is an example of a closed system', the classic electrical answer is a sealed, rigid IP68 electronics enclosure containing heat-generating components. No air or moisture enters or leaves (mass is constant), but electrical work enters via wires, and heat leaves via conduction and convection through the enclosure walls.

This is distinct from an open system (like an actively liquid-cooled server rack where coolant flows in and out) and an isolated system (a perfectly insulated thermos where neither mass nor energy crosses the boundary). Understanding this boundary definition is the critical first step before writing any energy balance equations.

Exam Problem Walkthrough: Sealed Battery Enclosure Thermal Check

Problem Statement:
A sealed, rigid IP67 aluminum enclosure (acting as a thermodynamic closed system) houses a LiFePO4 battery pack and a BMS. During a continuous 1C discharge, the internal components generate 45 W of constant Joule heating. The enclosure has a total thermal capacitance (heat capacity) of Cth = 1500 J/K and a combined thermal resistance to the ambient environment of Rth = 1.2 K/W. The ambient temperature (T∞) is 20°C, and the initial internal temperature (T0) is 20°C.

Find:
1. The steady-state internal temperature (Tss).
2. The internal temperature at t = 3000 seconds.

Step 1: Identify the Governing Theorem

We apply the First Law of Thermodynamics for a Closed System (Conservation of Energy), coupled with the Lumped Capacitance Model for transient heat transfer. The First Law states that the net energy transfer into the system equals the change in internal energy: Ein - Eout = ΔEsys. We use the lumped capacitance method because the Biot number for standard thin-walled aluminum enclosures is well below 0.1, meaning internal temperature gradients are negligible compared to the surface-to-ambient resistance.

Step 2: Execute the Algebra (No Skipped Steps)

First, we set up the differential energy balance equation. Power generated (Pgen) enters the system, and convective heat loss leaves it:

Pgen - [(T(t) - T∞) / Rth] = Cth · (dT/dt)

Part A: Steady-State Temperature (t → ∞)

At steady state, the temperature is no longer changing, so dT/dt = 0. The equation simplifies to:

  1. Pgen - [(Tss - T∞) / Rth] = 0
  2. Pgen = (Tss - T∞) / Rth
  3. 45 W = (Tss - 20°C) / 1.2 K/W
  4. 45 · 1.2 = Tss - 20
  5. 54 = Tss - 20
  6. Tss = 74°C

Part B: Transient Temperature at t = 3000s

We solve the first-order linear ordinary differential equation. The standard solution for a step-input thermal response is:

T(t) = T∞ + Pgen · Rth · [1 - e(-t / τ)]

  1. Calculate the thermal time constant (τ):
    τ = Rth · Cth = 1.2 K/W · 1500 J/K = 1800 seconds.
  2. Substitute known values into the transient equation:
    T(3000) = 20 + (45 · 1.2) · [1 - e(-3000 / 1800)]
  3. Simplify the exponent:
    -3000 / 1800 = -1.6667
  4. Calculate the exponential term:
    e-1.6667 ≈ 0.18887
  5. Complete the bracket:
    1 - 0.18887 = 0.81113
  6. Calculate the final temperature rise:
    54 · 0.81113 = 43.8°C
  7. Add ambient temperature:
    T(3000) = 20 + 43.8 = 63.8°C

The Trap, Sanity Checks, and Independent Verification

The Trap: The most common mistake on this exam problem is confusing a closed system with an isolated system. If you treat it as isolated, you assume no heat leaves (Rth = ∞), leading to a linear temperature rise to infinity. Remember: closed means mass is trapped, but heat must escape to the environment.

Sanity Check: Units and Order of Magnitude

  • Units: Pgen (Watts) · Rth (K/W) = Kelvin (or °C). The units cancel perfectly. Time constant τ = (K/W) · (J/K) = Joules / (Joules/sec) = seconds. Units are verified.
  • Order of Magnitude: 74°C is hot to the touch but sits safely below the 85°C thermal shutdown threshold of standard commercial LiFePO4 BMS units. A 54°C rise over ambient for a 45W dissipation in a passive enclosure is physically realistic for bench-scale hardware.

How to Verify the Answer Independently

We can verify the t=3000s answer by performing a total energy balance check. Total electrical energy injected over 3000 seconds is:

Ein = 45 W · 3000 s = 135,000 Joules.

Energy stored in the thermal mass at 63.8°C is:

Estored = Cth · ΔT = 1500 J/K · (63.8 - 20)K = 65,700 Joules.

Energy dissipated to ambient is the integral of the heat loss rate over time. Using the exact integral of the convective loss equation yields 69,300 Joules. Summing stored and dissipated energy: 65,700 + 69,300 = 135,000 Joules. The energy balance closes perfectly, verifying the algebra.

Decision Path: Choosing Your Thermal Model

When analyzing a closed system enclosure, you must decide whether to use the simple Lumped Capacitance model (as we did above) or a complex 1D Spatial Conduction model (which requires partial differential equations). Use this decision tree to make a concrete pick:

Condition Physical Meaning Required Action
Biot Number (Bi) < 0.1 Internal conduction resistance is negligible compared to surface convection resistance. The enclosure heats uniformly. Pick: Lumped Capacitance Model (Use the RthCth ODE above).
Biot Number (Bi) ≥ 0.1 Significant temperature gradient exists across the enclosure wall or internal potting compound. Pick: 1D Spatial Conduction (Use Heisler charts or FEA software like ANSYS).
Enclosure is plastic (low k) Thermal conductivity (k) of ABS/Polycarbonate is very low (~0.2 W/m·K), driving Bi up. Pick: 1D Spatial Conduction. Lumped capacitance will fail and under-predict internal junction temps.

Concrete Default: For standard die-cast aluminum IP68 enclosures (like Hammond 1590 series) dissipating under 100W, the Biot number is universally < 0.01. Always default to the Lumped Capacitance Model for these specific parts unless you are modeling the internal silicon junction of the semiconductor itself.

FAQ: Clearing Up Terminology Overlaps

Is a closed-loop control system the same as a closed thermodynamic system?

No. In control theory (like a PID temperature controller or a Phase-Locked Loop), a 'closed-loop' refers to an information feedback path where the output is measured and compared to a setpoint. In thermodynamics, a 'closed system' refers strictly to a physical mass boundary. An IP68 enclosure is a closed thermodynamic system; the BMS inside it might use closed-loop control to manage charging. Do not mix these definitions on a fundamentals exam.

What if the enclosure has a pressure relief valve?

If the internal pressure rises and the valve vents gas to the atmosphere, mass has crossed the boundary. The system instantly transitions from a closed system to an open system (specifically, a transient open system). You must then abandon the simple closed-system First Law equation and use the open-system enthalpy formulation (accounting for the mass flow rate and specific enthalpy of the venting gas). For standard exam problems, assume the valve remains sealed unless a specific venting pressure and mass flow rate are provided.

For further reading on thermal capacitance values for specific enclosure materials, refer to the Engineering Toolbox specific heat database. For deeper mathematical proofs of the lumped capacitance derivation, review the MIT OpenCourseWare Thermal-Fluids Engineering I lecture notes.