Question

High pressure steam is expanded adiabatically and reversibly through a well-insulated turbine which produces some shaft work. If the enthalpy change and entropy change across the turbine are represented by $\Delta H$ and $\Delta S$ respectively for this process, then:

• A. $\Delta H = 0 \text{ and } \Delta S = 0$
• B. $\Delta H \neq 0 \text{ and } \Delta S = 0$
• C. $\Delta H \neq 0 \text{ and } \Delta S \neq 0$
• D. $\Delta H = 0 \text{ and } \Delta S \neq 0$

Hint:

An ideal turbine or compressor operates under an isentropic process, meaning $\Delta S = 0$. However, because work is exchanged with the surroundings, the enthalpy change $\Delta H$ is never zero.

Answer 1

The Correct Option is B

Concept: Let us analyze the given descriptors for the turbine expansion process:
• "Well-insulated" and "Adiabatic": This means there is no heat exchange between the steam and the surroundings ($Q = 0$).
• "Reversible": A process that is both adiabatic and reversible is, by definition, isentropic. This means the total entropy of the fluid remains constant throughout the expansion: \[ \Delta S = 0 \] Now, let us examine the enthalpy behavior using the open-system steady-state control volume energy balance (First Law of Thermodynamics) for a turbine, neglecting kinetic and potential energy changes: \[ \Delta H = Q - W_{\text{shaft}} \]

Step 1: Analyze the enthalpy change ($\Delta H$) using the energy balance.
Since the process is adiabatic, we substitute $Q = 0$: \[ \Delta H = -W_{\text{shaft}} \] The problem states that the turbine "produces some shaft work," meaning $W_{\text{shaft}} > 0$. Therefore: \[ \Delta H = -W_{\text{shaft}} \neq 0 \] As the steam expands to drive the turbine blades, its enthalpy decreases to produce work, so $\Delta H$ is non-zero (specifically, it is negative).

Step 2: Combine the results for enthalpy and entropy.
From our analysis, we have $\Delta H \neq 0$ and $\Delta S = 0$. This matches Option B.