Question

For a stage of a \(50%\) reaction ideal axial-flow compressor (symmetrical blading), select the correct statement.

• A. The \emph{stagnation} enthalpy rise across the rotor is \(50%\) of the rise across the stage.
• B. The \emph{static} enthalpy rise across the rotor is \(50%\) of the rise across the stage.
• C. Axial velocity at rotor exit is \(50%\) of that at rotor entry.
• D. The static pressure rise across the rotor is \(50%\) of the rise across the stator.

Hint:

Remember two separate "budgets": (i) Stagnation enthalpy rises only in the rotor (work input), (ii) Static enthalpy/pressure rises are split according to reaction \(R\). At \(R=0.5\) (symmetric triangles), rotor and stator each contribute half.

Answer 1

The Correct Option is B

Step 1: Reaction definition (compressor).
Stage reaction \(R\) is the fraction of the stage static enthalpy rise that occurs in the rotor: \[ R \;\equiv\; \frac{\Delta h_{\text{static, rotor}}}{\Delta h_{\text{static, stage}}} \text{with} \Delta h_{\text{static, stage}}= \Delta h_{\text{static, rotor}}+\Delta h_{\text{static, stator}} . \] A \(50%\)-reaction stage has \(R=\tfrac{1}{2}\).

Step 2: Consequence of \(R=0.5\).
\[ \Delta h_{\text{static, rotor}} = \Delta h_{\text{static, stator}} = \tfrac{1}{2}\,\Delta h_{\text{static, stage}} . \] Thus the rotor and stator each contribute half of the static enthalpy (and pressure) rise of the stage.

Step 3: Why options (A), (C), (D) are wrong.
(A) In an ideal compressor, the stagnation enthalpy increase occurs only in the rotor (work input), so \(\Delta h_{0,\text{rotor}}=\Delta h_{0,\text{stage}}\) (i.e., \(100%\), not \(50%\)). The stator ideally changes static pressure without changing stagnation enthalpy.
[2mm] (C) \(R=0.5\) stems from symmetric velocity triangles; a common design is nearly constant axial velocity (\(V_x\) in = \(V_x\) out), not "half". Reaction says nothing about halving \(V_x\).
[2mm] (D) With \(R=0.5\), \(\Delta p_{\text{static, rotor}}=\Delta p_{\text{static, stator}}\). Saying the rotor's rise is half of the stator's is incorrect wording; they are equal halves of the stage rise. (Helpful triangle view)
For symmetric blading with constant \(V_x\): the whirl components satisfy \(V_{\theta 2}-V_{\theta 1} = \text{const}\) for the stage (Euler). The rotor raises static enthalpy by diffusion of relative flow; the stator raises static enthalpy by diffusion of absolute flow. Symmetry \(\Rightarrow\) each diffuses half the stage static rise \(\Rightarrow R=0.5\).

Final Answer:
\[ \boxed{\Delta h_{\text{static, rotor}}=\tfrac{1}{2}\,\Delta h_{\text{static, stage}}} \]