Interactive Lecture
Closed, Open, and Isolated Systems
by Dr Arnaldo Delli Carri
Thermodynamic systems are classified by what they can exchange with their surroundings — mass, energy, or neither. Choosing the right system boundary is often the first and most important step in solving a thermodynamics problem.
A thermodynamic system is a quantity of matter or a region in space chosen for study. Everything external to the system is called the surroundings, and the real or imaginary surface that separates them is the boundary.
Mass Transfer: YES
Energy Transfer: YES
Both mass and energy can cross the system boundary.
Examples: Turbines, compressors, pumps, heat exchangers, boilers
Mass Transfer: NO
Energy Transfer: YES
Fixed mass; energy can cross the boundary as heat or work.
Examples: Piston-cylinder, pressure cooker, sealed can being heated
Mass Transfer: NO
Energy Transfer: NO
Neither mass nor energy can cross the boundary.
Examples: Insulated rigid tank, thermos bottle (ideal), the universe
Conservation of Mass (General):
\[\frac{dm_{system}}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}\]
For closed systems: \(\dot{m}_{in} = \dot{m}_{out} = 0\), so \(m_{system} = \text{constant}\)
For open systems (steady flow): \(\sum \dot{m}_{in} = \sum \dot{m}_{out}\)
First Law of Thermodynamics:
Closed Systems:
\[\Delta E_{system} = Q - W\]
Where: \(Q\) = heat transfer, \(W\) = work transfer, \(\Delta E\) = change in total energy
Open Systems (Steady Flow):
\[\dot{Q} - \dot{W} = \sum \dot{m}_{out}\left(h + \frac{V^2}{2} + gz\right)_{out} - \sum \dot{m}_{in}\left(h + \frac{V^2}{2} + gz\right)_{in}\]
Where: \(h\) = specific enthalpy, \(V\) = velocity, \(z\) = elevation
Second Law (Entropy):
Isolated System: \(\Delta S_{isolated} \geq 0\) (entropy never decreases)
Closed System: \(\Delta S_{system} = \int \frac{\delta Q}{T} + S_{gen}\), where \(S_{gen} \geq 0\)
Observe how mass and energy flow in different thermodynamic systems. Use the controls to switch between system types and observe their behaviour.
| System Type | Conservation Law | Equation |
|---|---|---|
| Open System | Mass Conservation | \(\frac{dm}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}\) |
| Energy Conservation | \(\dot{Q} - \dot{W} = \sum \dot{m}_{out}h_{out} - \sum \dot{m}_{in}h_{in} + \Delta KE + \Delta PE\) | |
| Closed System | Mass Conservation | \(m = \text{constant}\) |
| Energy Conservation (1st Law) | \(\Delta E = Q - W\) or \(\Delta U = Q - W\) (neglecting KE, PE) | |
| Isolated System | Mass + Energy | \(m = \text{constant}, \quad E = \text{constant}\) |
| Entropy (2nd Law) | \(\Delta S_{isolated} \geq 0\) | |
| General | Total Energy | \(E = U + KE + PE = U + \frac{1}{2}mV^2 + mgz\) |
| Steady Flow | Continuity | \(\sum \dot{m}_{in} = \sum \dot{m}_{out}\) and \(\frac{dm_{CV}}{dt} = 0\) |
Work and Heat Sign Conventions:
A piston-cylinder device contains 0.5 kg of air initially at 100 kPa and 27°C. The air is compressed in a reversible adiabatic process until the pressure reaches 500 kPa. If \(c_v = 0.718\) kJ/(kg·K) and \(k = 1.4\), calculate the final temperature and the work done on the air.
Show Solution
Solution:
Given:
Step 1: For isentropic process of ideal gas:
\[\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = \left(\frac{500}{100}\right)^{\frac{1.4-1}{1.4}} = (5)^{0.286} = 1.5838\]
Step 2: Calculate final temperature
\[T_2 = T_1 \times 1.5838 = 300 \times 1.5838 = 475.1 \text{ K} = 202.1°C\]
Step 3: Calculate work (closed system, adiabatic)
For adiabatic process: \(Q = 0\), so from first law: \(W = -\Delta U\)
\[W = -m c_v (T_2 - T_1) = -0.5 \times 0.718 \times (475.1 - 300)\]
\[W = -0.5 \times 0.718 \times 175.1 = -62.9 \text{ kJ}\]
Answer: Final temperature \(T_2 = 475.1\) K (202.1°C); work \(W = -62.9\) kJ (negative confirms work is done on the air).
Steam enters a turbine at 3 MPa, 400°C with a velocity of 50 m/s and exits at 50 kPa, 100°C with a velocity of 200 m/s. The mass flow rate is 5 kg/s. Heat loss from the turbine is 20 kW. Calculate the power output. Use: \(h_1 = 3231\) kJ/kg (inlet), \(h_2 = 2676\) kJ/kg (exit).
Show Solution
Solution:
Given:
Step 1: Apply steady flow energy equation
\[\dot{Q} - \dot{W} = \dot{m}\left[\left(h_2 - h_1\right) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)\right]\]
Neglecting potential energy change (\(\Delta PE \approx 0\))
Step 2: Calculate kinetic energy change
\[\Delta KE = \frac{V_2^2 - V_1^2}{2} = \frac{200^2 - 50^2}{2} = \frac{40000 - 2500}{2} = 18750 \text{ J/kg} = 18.75 \text{ kJ/kg}\]
Step 3: Calculate enthalpy change
\[\Delta h = h_2 - h_1 = 2676 - 3231 = -555 \text{ kJ/kg}\]
Step 4: Calculate power output
\[\dot{W} = \dot{Q} - \dot{m}(\Delta h + \Delta KE)\]
\[\dot{W} = -20 - 5(-555 + 18.75) = -20 - 5(-536.25)\]
\[\dot{W} = -20 + 2681.25 = 2661.25 \text{ kW}\]
Answer: Power output = 2661 kW or 2.66 MW
An insulated rigid tank is divided into two compartments. One contains 2 kg of water at 80°C, and the other contains 3 kg of water at 20°C. The partition is removed and the waters mix. Assuming constant specific heat \(c_p = 4.18\) kJ/(kg·K), find the final equilibrium temperature.
Show Solution
Solution:
Given:
Step 1: Apply energy conservation
For isolated system: \(\Delta E_{total} = 0\)
Energy lost by hot water = Energy gained by cold water
\[m_h c_p (T_h - T_f) = m_c c_p (T_f - T_c)\]
Step 2: Solve for final temperature
\[m_h (T_h - T_f) = m_c (T_f - T_c)\]
\[2(80 - T_f) = 3(T_f - 20)\]
\[160 - 2T_f = 3T_f - 60\]
\[160 + 60 = 3T_f + 2T_f\]
\[220 = 5T_f\]
\[T_f = 44°C\]
Step 3: Verify energy balance
Energy lost by hot: \(Q_h = 2 \times 4.18 \times (80-44) = 300.96\) kJ
Energy gained by cold: \(Q_c = 3 \times 4.18 \times (44-20) = 300.96\) kJ ✓
Answer: Final temperature = 44°C
Classify each of the following as an open, closed, or isolated system:
(a) A cup of hot coffee cooling on a table
(b) A steam turbine in a power plant
(c) A sealed, insulated thermos bottle
(d) An electric water heater with flowing water
Show Solution
Solution:
Answer (a): Closed system. The coffee's mass stays constant (no flow in/out), but heat is lost to the surroundings, so energy crosses the boundary.
Answer (b): Open system. Steam continuously flows in and out, so mass crosses the boundary, and energy is transferred as heat loss and shaft work.
Answer (c): Isolated system (approximately). Sealing prevents mass transfer and insulation minimises heat transfer; a real thermos has a small heat leak, so it's nearly but not perfectly isolated.
Answer (d): Open system. Water flows in cold and out hot, so mass crosses the boundary, and electrical energy enters as work/heat.
Always check for both mass and energy transfer across the boundary — one alone doesn't fix the classification.