Energy Balance for the FE Exam
The first law of thermodynamics is an energy balance: energy in minus energy out equals the change in stored energy. Applying this principle correctly to both closed and open systems is essential for thermodynamics problems on the FE exam.
Closed Systems (No Mass Crossing the Boundary)
For a closed system undergoing a process from state 1 to state 2:
Q - W = delta_U = m * (u_2 - u_1)
Where Q is the net heat added to the system, W is the net work done by the system, and u is the specific internal energy.
Sign Convention
- Q positive: Heat is added to the system
- Q negative: Heat is rejected by the system
- W positive: Work is done by the system on the surroundings
- W negative: Work is done on the system by the surroundings
Be consistent with your sign convention. The convention above is the most common, but some textbooks use the opposite convention for work.
Boundary Work for Closed Systems
Boundary work occurs when the system boundary moves (e.g., a piston expanding or compressing):
W = integral from V_1 to V_2 of P dV
For specific processes:
- Constant pressure: W = P * (V_2 - V_1)
- Constant volume: W = 0 (no volume change, no boundary work)
- Constant temperature (ideal gas): W = nRT * ln(V_2/V_1) = P_1*V_1 * ln(V_2/V_1)
- Polytropic (PV^n = constant): W = (P_2*V_2 - P_1*V_1) / (1 - n), for n not equal to 1
Open Systems (Steady-State, Steady-Flow)
For an open system at steady state (no accumulation of mass or energy), the energy balance per unit mass is:
q - w = (h_2 - h_1) + (V_2^2 - V_1^2)/2 + g*(z_2 - z_1)
Where q is heat transfer per unit mass, w is shaft work per unit mass, h is specific enthalpy, V is velocity, and z is elevation.
For rate form (using mass flow rate m_dot):
Q_dot - W_dot = m_dot * [(h_2 - h_1) + (V_2^2 - V_1^2)/2 + g*(z_2 - z_1)]
Simplifications for Common Devices
Many open system devices allow simplifications of the general energy equation:
Turbines
A turbine extracts work from a flowing fluid (steam, gas). Typically:
- Neglect kinetic and potential energy changes
- Adiabatic (Q = 0) for ideal turbines
- Energy equation simplifies to: w_turbine = h_1 - h_2 (work output)
Compressors and Pumps
Compressors (gases) and pumps (liquids) add energy to a fluid. Typically:
- Neglect kinetic and potential energy changes
- Adiabatic for ideal devices
- Energy equation: w_in = h_2 - h_1 (work input)
- For liquid pumps: w_pump = v_f * (P_2 - P_1), where v_f is the specific volume of the liquid
Heat Exchangers
Heat exchangers transfer heat between two fluids with no work interaction. For each fluid stream:
- W = 0 (no work)
- Neglect kinetic and potential energy changes
- For hot stream: Q_rejected = m_hot * (h_hot_in - h_hot_out)
- For cold stream: Q_absorbed = m_cold * (h_cold_out - h_cold_in)
- Energy balance: Q_rejected = Q_absorbed (no heat loss to surroundings)
Throttling Devices (Valves, Orifices)
Throttling is a process where fluid passes through a restriction with no work or heat transfer:
- W = 0, Q = 0
- Neglect kinetic and potential energy changes
- Result: h_1 = h_2 (isenthalpic process)
- Pressure drops, but enthalpy remains constant
Nozzles and Diffusers
These devices convert between pressure energy and kinetic energy:
- W = 0, Q = 0 (adiabatic)
- Potential energy change negligible
- Energy equation: h_1 + V_1^2/2 = h_2 + V_2^2/2
- Nozzle: pressure decreases, velocity increases
- Diffuser: velocity decreases, pressure increases
Mixing Chambers
Two or more streams combine in a mixing chamber:
- W = 0, Q = 0
- Mass balance: m_dot_1 + m_dot_2 = m_dot_3
- Energy balance: m_dot_1 * h_1 + m_dot_2 * h_2 = m_dot_3 * h_3
Problem-Solving Strategy
- Step 1: Identify the system as closed or open.
- Step 2: Draw a schematic and label all inlet and outlet states.
- Step 3: Write the appropriate energy balance equation.
- Step 4: Identify simplifications (adiabatic, no work, negligible KE/PE changes).
- Step 5: Look up properties at each state using ideal gas relations or property tables.
- Step 6: Solve for the unknown.
Isentropic Efficiency
Real devices are not ideal. Isentropic efficiency compares actual performance to ideal (isentropic) performance:
- Turbine efficiency: eta_t = (h_1 - h_2_actual) / (h_1 - h_2s), where h_2s is the isentropic exit enthalpy
- Compressor/pump efficiency: eta_c = (h_2s - h_1) / (h_2_actual - h_1)
Energy balance problems are the bread and butter of thermodynamics on the FE exam. Master the general equation and know the simplifications for each device type, and you will handle these problems confidently.