Questions: Control Volume Analysis and Steady-Flow Devices

Question 1

Steam enters a well-insulated turbine at high enthalpy h₁ and exits at lower enthalpy h₂. Kinetic and potential energy changes are negligible. What is the physical interpretation of the enthalpy drop (h₁ − h₂) per unit mass?

• A. It equals the heat lost through the turbine's insulation to the surroundings
• B. It equals the shaft work output per unit mass of steam flowing through the turbine
• C. It represents the increase in the steam's internal energy while it is inside the turbine
• D. It equals the kinetic energy the steam gains as it expands through the blades

Answer: (B)

For a steady-flow turbine with Q̇ ≈ 0 and negligible kinetic/potential changes, the energy equation simplifies to Ẇ_shaft = ṁ(h₁ − h₂), so the specific shaft work out equals h₁ − h₂. The turbine converts the enthalpy difference entirely into mechanical shaft output. Option A is wrong because the turbine is insulated (Q̇ = 0 by assumption). Option C is wrong because under steady-flow, there is no accumulation of internal energy inside the device — whatever enters must leave. Option D mistakes the nozzle case for the turbine case.

Question 2

Why does the steady-flow energy equation use specific enthalpy h = u + Pv rather than specific internal energy u to represent the energy carried by each stream?

• A. Enthalpy is always larger than internal energy, making engineering calculations more conservative
• B. Enthalpy accounts for both the fluid's internal energy and the flow work (Pv) required to push each parcel of fluid through the inlet or outlet against the local pressure
• C. Internal energy changes are negligible in most engineering devices, so enthalpy serves as a convenient approximation
• D. Enthalpy is easier to measure directly with sensors than internal energy

Answer: (B)

Every kilogram of fluid entering a control volume must push the fluid ahead of it to make room — that work is exactly Pv per unit mass (pressure times specific volume). Using h = u + Pv bundles this 'flow work' term into the enthalpy so you never have to track it separately. This is why steam tables list h so prominently: turbines, boilers, condensers, and compressors are all open systems where the enthalpy differences directly represent the net energy exchange. Using u instead would give the wrong answer unless you separately added Pv at each boundary.

Question 3

Under the steady-flow assumption, the thermodynamic properties (temperature, pressure, enthalpy) at any fixed point inside a control volume change continuously over time as fluid flows through.

• T. True
• F. False

Answer: False

The steady-flow assumption means precisely the opposite: properties at any fixed spatial location inside the control volume do NOT change with time. Fluid is flowing and properties may vary from point to point spatially, but any given point maintains a constant state. This is what allows the energy balance to be written as a simple rate equation without accumulation terms. If properties were changing with time at fixed points, the system would be unsteady and the accumulation term d(E_cv)/dt would be nonzero, significantly complicating the analysis.

Question 4

In a well-insulated converging nozzle, a drop in fluid enthalpy (due to falling pressure and temperature) must be accompanied by an increase in fluid velocity, because the steady-flow energy equation requires total energy to be conserved across the device.

• T. True
• F. False

Answer: True

For a nozzle with Q̇ = 0, Ẇ_shaft = 0, and negligible elevation change, the energy equation reduces to h₁ + ½V₁² = h₂ + ½V₂². If h falls (h₂ < h₁), then ½V₂² must rise — the enthalpy decrease is exactly converted to kinetic energy increase. This is the design purpose of a nozzle: trade pressure/temperature for velocity. A diffuser does the reverse. The principle follows directly from energy conservation applied to a steady-flow open system.

Question 5

Explain why engineers use enthalpy rather than internal energy in the steady-flow energy equation, and what physical quantity the Pv term represents.

Model answer: Pv represents 'flow work' — the work per unit mass that fluid at the inlet does pushing itself into the control volume against the existing pressure, and the work per unit mass the control volume does pushing fluid out at the outlet. Every kilogram crossing a boundary must displace its own volume against the local pressure. Using enthalpy h = u + Pv automatically accounts for this boundary-pushing work, so the energy balance only needs to track enthalpy differences between streams rather than separately computing internal energy plus two Pv terms at each port.

This is why open-system (control volume) analysis looks different from closed-system analysis even though both apply the first law. In a closed system, no mass crosses the boundary, so internal energy and heat/work are sufficient. In an open system, mass carries energy in two forms: its internal energy u and the flow work Pv needed to push it through the boundary. Enthalpy packages these together. The insight that 'flow work is already in h' is what makes steam tables so useful — you look up h at inlet and outlet conditions, subtract, and you have the work or heat per unit mass directly.