Thermodynamic Open System- Example Problems and Solutions
What Is a Thermodynamic Open System?
An open system exchanges both mass and energy with its surroundings. Unlike closed systems (energy only) or isolated systems (neither), open systems have material crossing their boundaries constantly.
Real-world examples include:
- Internal combustion engines
- Heat exchangers
- Turbines and compressors
- Boilers
- Refrigeration evaporators
If you're working with any of these devices, you need open system analysis. Full stop.
The Governing Equations
For steady-state open systems (most engineering applications), two equations matter:
Mass Conservation (Continuity Equation)
Mass in - Mass out = Accumulation
At steady state: ṁ₁ = ṁ₂ (mass flow rate stays constant)
For multiple streams: Σṁᵢₙ = Σṁₒᵤₜ
Energy Conservation (First Law for Open Systems)
The general energy balance:
Q̇ - Ẇ = Σṁₒᵤₜ(h + ½V² + gz) - Σṁᵢₙ(h + ½V² + gz)
Where:
- Q̇ = heat transfer rate
- Ẇ = work transfer rate (shaft work)
- h = specific enthalpy
- V = velocity
- gz = potential energy term
In most engineering problems, you drop the kinetic and potential energy terms unless told otherwise. They only matter for aerospace or hydroelectric applications.
Example Problem 1: Steam Turbine
Problem: Steam enters a turbine at 3 MPa, 400°C with a mass flow rate of 2 kg/s. Steam exits at 0.1 MPa. The turbine produces 500 kW of work. Heat loss is 20 kW. Find the exit velocity if inlet velocity is negligible.
Solution:
Step 1: Write the energy equation (steady-state)
Q̇ - Ẇ = ṁ(h₂ - h₁ + ½(V₂² - V₁²)/1000)
Step 2: Plug in values
-20 - (-500) = 2(h₂ - h₁) + 2(½V₂²/1000)
480 = 2(h₂ - h₁) + V₂²/1000
Step 3: Find enthalpy values from steam tables
h₁ at 3 MPa, 400°C = 3230.9 kJ/kg
h₂ at 0.1 MPa (saturated mixture) = 2675.5 kJ/kg (assuming isentropic for estimation)
Step 4: Solve for V₂
480 = 2(2675.5 - 3230.9) + V₂²/1000
480 = -1110.8 + V₂²/1000
V₂² = 1590.8 × 1000
V₂ = 39.9 m/s
Example Problem 2: Heat Exchanger
Problem: Hot oil (cp = 2.1 kJ/kg·K) enters a heat exchanger at 150°C, 5 kg/s. Cold water (cp = 4.18 kJ/kg·K) enters at 20°C, 3 kg/s. Exit temperature of oil is 80°C. Find the exit temperature of water.
Solution:
For a heat exchanger with no work and negligible kinetic/potential changes:
Energy lost by oil = Energy gained by water
ṁₒᵢₗ·cpₒᵢₗ·(Tₒᵤₜ - Tᵢₙ) = ṁᵥₐₜₑᵣ·cpᵥₐₜₑᵣ·(Tₒᵤₜ - Tᵢₙ)
5 × 2.1 × (80 - 150) = 3 × 4.18 × (Tₜᵥₐₜₑᵣ - 20)
5 × 2.1 × (-70) = 3 × 4.18 × (Tₜᵥₐₜₑᵣ - 20)
-735 = 12.54(Tₜᵥₐₜₑᵣ - 20)
-58.6 = Tₜᵥₐₜₑᵣ - 20
Tₜᵥₐₜₑᵣ = 38.6°C
Example Problem 3: Compressor
Problem: Air enters a compressor at 100 kPa, 300 K with negligible velocity. Exit pressure is 600 kPa. The compressor requires 200 kJ/kg of work input. Heat transfer to surroundings is 20 kJ/kg. Assuming ideal gas behavior, find the exit temperature.
Solution:
For compressors, use specific energy form:
w_in - w_out = h₂ - h₁ + q
For air (ideal gas, cp = 1.005 kJ/kg·K):
200 - 0 = cp(T₂ - T₁) + (-20)
200 = 1.005(T₂ - 300) - 20
220 = 1.005(T₂ - 300)
219.9 = T₂ - 300
T₂ = 519.9 K ≈ 520 K
Open System vs Closed System: Key Differences
| Aspect | Open System | Closed System |
|---|---|---|
| Mass flow | Crosses boundary | Stays constant |
| Energy transfer | Heat + Work | Heat + Work |
| Work term | Flow work + Shaft work | Shaft work only |
| Typical devices | Turbines, pumps, nozzles | Pistons, rigid containers |
| Analysis method | Control volume | Control mass |
Common Mistakes to Avoid
- Forgetting flow work: In open systems, there's an extra work term (Pv) associated with mass entering and leaving. Closed system equations don't apply here.
- Ignoring units: Mass flow rate ṁ has units kg/s. Don't mix it up with plain mass (kg).
- Wrong enthalpy values: Always check your steam tables or property tables. Interpolation is often necessary.
- Sign errors: Heat INTO the system is positive. Work done BY the system is positive. Be consistent with your sign convention.
- Assuming steady-state when it's not: Transient analysis is harder. Make sure accumulation term is actually zero before simplifying.
Quick Reference: When to Use Which Equation
| Situation | Equation to Use |
|---|---|
| Steady-state turbine/compressor | Q̇ - Ẇ = ṁ(h₂ - h₁) |
| Heat exchanger (no work) | ṁ₁cp₁ΔT₁ = ṁ₂cp₂ΔT₂ |
| Nozzle (high velocity change) | h₁ + ½V₁² = h₂ + ½V₂² |
| Pump (negligible Δh) | Ẇ ≈ vΔP/η |
| Boiler/Evaporator | Q̇ = ṁ(h₂ - h₁) |
Getting Started: Step-by-Step Approach
When facing any open system problem:
- Identify the system boundaries. Draw it if needed. Know what's entering and leaving.
- Check if it's steady-state. Accumulation = 0? Then you can simplify.
- List known values. Mass flow rates, pressures, temperatures, work, heat.
- Apply mass balance first. Find any unknown mass flow rates.
- Apply energy balance. Start with the general form, then strip out terms that don't apply.
- Look up properties. Use steam tables, ideal gas tables, or given cp values.
- Solve algebraically first. Don't plug numbers in until you have the variable isolated.
- Check your answer. Does the sign make sense? Is the magnitude reasonable?
The Bottom Line
Open system thermodynamics isn't complicated. The equations are straightforward. The hard part is knowing which terms to keep and which to drop based on your specific problem.
Most engineering problems strip out kinetic and potential energy. Most assume steady-state. Once you internalize those defaults, solving open system problems becomes routine.