A turbine operates adiabatically with a single inlet and single outlet. Which correctly applies the SFEE to find shaft work per unit mass flow?
Aẇ = h_out − h_in, because the fluid gains enthalpy as it does work
Bẇ = h_in − h_out, because the fluid loses enthalpy and that energy becomes shaft work
Cq̇ = h_out − h_in, because energy is conserved through heat transfer
Dẇ = u_in − u_out, because internal energy is the relevant energy variable for work calculations
For an adiabatic turbine, Q̇ = 0, so the SFEE simplifies to Ẇ = ṁ(h_in − h_out). The fluid's enthalpy drops as it expands, and that energy is extracted as shaft work. Option D is the classic error: internal energy u is the energy variable for closed systems. In open systems, mass crossing the boundary also carries flow work Pv, so enthalpy h = u + Pv is the correct energy variable — and replacing h with u would miss this contribution entirely.
Question 2 Multiple Choice
Why does enthalpy (h = u + Pv) rather than internal energy (u) appear in the steady-flow energy equation?
AInternal energy cannot be measured directly, while enthalpy is tabulated in steam tables
BEnthalpy includes the flow work (Pv) done to push fluid parcels across the control volume boundary
CEnthalpy is a more general form of energy that reduces to internal energy when velocities are low
DThe SFEE uses enthalpy by convention; either variable gives the same answer if applied consistently
When a fluid parcel crosses a control volume boundary, the fluid behind it must do work to push it across against the local pressure — this is flow work, equal to Pv per unit mass. The total energy transported per unit mass is therefore u (stored energy) + Pv (work of entry) = h. Internal energy alone would undercount the energy transfer. Enthalpy is not a convention or a measurement convenience — it arises necessarily from the physics of mass crossing a boundary.
Question 3 True / False
A throttle valve (like a pressure-reducing valve) operates with no shaft work, no heat transfer, and negligible kinetic and potential energy changes. Therefore, the fluid's enthalpy is the same at inlet and outlet.
TTrue
FFalse
Answer: True
This follows directly from the SFEE with all terms set to zero except the enthalpy terms: 0 − 0 = ṁ(h_out − h_in), so h_in = h_out. Throttling is isenthalpic. Note that pressure and temperature can both change dramatically across the throttle — but enthalpy is conserved. For real gases with Joule-Thomson cooling, this isenthalpic process produces a temperature drop that is exploited in refrigeration systems.
Question 4 True / False
For a nozzle, the kinetic energy terms in the SFEE can be neglected because nozzles are small and flow velocities are modest.
TTrue
FFalse
Answer: False
This is exactly backwards. A nozzle exists precisely to convert enthalpy into kinetic energy — the velocity increase is the entire purpose. The SFEE for a nozzle (adiabatic, no shaft work) is V²_out/2 − V²_in/2 = h_in − h_out. The kinetic energy term is not small; it is the output. In contrast, for devices like boilers and heat exchangers, kinetic and potential terms are genuinely negligible compared to enthalpy changes, but nozzles are the canonical exception.
Question 5 Short Answer
Explain in your own words why enthalpy (h = u + Pv) rather than internal energy appears in the steady-flow energy equation. What does the Pv term represent physically?
Think about your answer, then reveal below.
Model answer: When a parcel of fluid enters a control volume, the fluid behind it must exert pressure to push it across the boundary. This pushing requires mechanical work equal to P × v per unit mass (force per unit area times volume per unit mass = work per unit mass). This 'flow work' Pv is delivered to the control volume along with the parcel's stored internal energy u. The total energy arriving per unit mass is therefore u + Pv = h. Internal energy alone misses this mechanical contribution. For closed systems (fixed boundaries, no mass crossing), no flow work occurs and dU = Q − W is correct. Open systems require the enthalpy formulation precisely because mass — and the work of moving it — crosses the boundary continuously.