Questions: Throttling and Isenthalpic Expansion Processes

Question 1

A gas flows steadily through an adiabatic throttle valve. No work is done and kinetic energy changes are negligible. Which thermodynamic constraint correctly characterizes this process?

• A. Temperature is constant, because no heat is transferred across the valve
• B. Entropy is constant, because the process is both adiabatic and involves no shaft work
• C. Enthalpy is constant, because the steady-flow energy equation reduces to h₁ = h₂ under these conditions
• D. Internal energy is constant, because the fluid is in steady state and no chemical reactions occur

Answer: (C)

The steady-flow energy equation for an open system includes flow work (PV) in addition to internal energy — together these equal specific enthalpy h = u + Pv. For an adiabatic device with no shaft work and negligible kinetic/potential energy changes, this reduces directly to h₁ = h₂: enthalpy is conserved. The process is NOT isentropic (option B) — entropy increases because the irreversible pressure drop through a restriction generates entropy, making throttling one of the classic examples of irreversible adiabatic processes. NOT isothermal (option A) — temperature may change for real gases.

Question 2

An engineer proposes liquefying an ideal gas by cooling it to −50°C and then throttling it through an expansion valve to low pressure. What does thermodynamics predict about the temperature change during throttling?

• A. The gas will cool further during throttling, facilitating liquefaction, because lower pressure always corresponds to lower temperature
• B. The ideal gas temperature will not change during throttling, because enthalpy of an ideal gas depends only on temperature
• C. The gas will warm during throttling because the pressure drop reduces PV work done on the gas
• D. The temperature change depends on whether the gas is above or below its normal boiling point

Answer: (B)

For an ideal gas, enthalpy depends only on temperature: h = h(T). Since throttling conserves enthalpy (h₁ = h₂), and h depends only on T, it follows that T₁ = T₂ — ideal gases do not change temperature when throttled. The Joule-Thomson coefficient μ_JT = (∂T/∂P)_h = 0 for an ideal gas. Liquefaction by throttling only works for real gases whose intermolecular forces cause temperature to drop when pressure drops (μ_JT > 0). An ideal gas, by definition, has no intermolecular interactions and therefore no temperature change — it cannot be liquefied by throttling.

Question 3

During a throttling process, entropy always increases even though enthalpy is conserved, because the irreversible pressure drop generates entropy.

• T. True
• F. False

Answer: True

Throttling is adiabatic (Q = 0) and isenthalpic (h₁ = h₂), but it is NOT isentropic. The flow through a constriction is highly irreversible — eddies, friction, and turbulence dissipate mechanical energy into thermal energy internally. By the second law, any irreversible adiabatic process must increase entropy (ΔS > 0). This entropy generation represents the thermodynamic 'cost' of the pressure drop and is what distinguishes a throttle from an isentropic turbine, which can also drop pressure but does so reversibly while producing shaft work.

Question 4

Most gases cool when throttled at room temperature, because reducing pressure typically causes temperature to decrease in an expanding gas.

• T. True
• F. False

Answer: False

Hydrogen and helium have negative Joule-Thomson coefficients (μ_JT < 0) at room temperature, meaning they actually warm when throttled. For these gases, intermolecular repulsions dominate at typical conditions, and expanding molecules that repel each other gain kinetic energy (not lose it), raising temperature. To use throttling for liquefaction, hydrogen must first be pre-cooled below its inversion temperature (~204 K) — where μ_JT changes sign from negative to positive — so that subsequent throttling produces cooling rather than heating. Helium has an even lower inversion temperature.

Question 5

Why does enthalpy remain constant across a throttle valve even though pressure drops dramatically and no heat is exchanged?

Model answer: The steady-flow energy equation accounts for flow work: a fluid element entering a control volume does work pushing against upstream pressure (contributing +Pu·v per unit mass in), and the element leaving does work against downstream pressure (contributing −Pd·v per unit mass out). For an adiabatic throttle with no shaft work, the energy balance becomes: h₁ = h₂ (since h = u + Pv combines internal energy and flow work). Even though P drops (reducing the Pv term), the internal energy adjusts to compensate — for real gases, increased molecular separation raises internal energy. The total h is conserved even though neither u nor Pv is individually constant.

Enthalpy, not internal energy, is the conserved quantity in steady-flow processes because flowing fluids carry flow work (Pv) as well as internal energy. This is the key distinction between open and closed system analyses. For a closed system (piston-cylinder), internal energy and work are the relevant variables. For an open system like a throttle, enthalpy is the natural conserved quantity. The Joule-Thomson effect — whether temperature rises or falls — then tells us how internal energy and Pv redistribute to keep h constant as P changes.