Question

In a single–spool aviation turbojet engine, which of the following is the correct relationship between the total work output \(W_t\) of a 2-stage axial turbine and the total work required \(W_c\) by a 6-stage axial compressor, neglecting losses?

• A. \(W_t = 2\,W_c\)
• B. \(W_t = 6\,W_c\)
• C. \(W_t = W_c\)
• D. \(W_t = 3\,W_c\)

Hint:

Single spool + negligible losses \(\Rightarrow\) turbine specific work equals compressor specific work. Stage count only sets how finely that work is shared, not the total.

Answer 1

The Correct Option is C

Step 1: One–shaft (single-spool) power balance.
Turbine and compressor sit on the same shaft. In steady state with no losses, \[ P_t = P_c \Rightarrow \dot m\,W_t = \dot m\,W_c \Rightarrow W_t = W_c, \] where \(W\) is specific shaft work (per unit mass flow \(\dot m\)). The number of stages only divides the same total work among stages.

Step 2: Euler turbomachinery view (why stage count is irrelevant).
Compressor: per stage \( \Delta h_{0,\text{stg}} = U\,(V_{\theta 2}-V_{\theta 1}) \) (positive for compressor).
Turbine: per stage \( \Delta h_{0,\text{stg}} = U\,(V_{\theta 1}-V_{\theta 2}) \) (delivers work).
Total compressor work \(W_c = \sum_{i=1}^{6}\Delta h_{0,\text{stg},i}\).
Total turbine work \(W_t = \sum_{j=1}^{2}\Delta h_{0,\text{stg},j}\).
On a single spool in lossless idealization, the sums must be equal so that the shaft neither accelerates nor decelerates: \[ \sum_{j=1}^{2}\Delta h_{0,\text{stg},j} \;=\; \sum_{i=1}^{6}\Delta h_{0,\text{stg},i}. \]

Step 3: What multipliers (2, 3, 6) would imply.
Any \(W_t \neq W_c\) would yield net shaft power \(\dot m (W_t - W_c)\neq 0\), violating steady single-spool operation (unless unmodelled losses/drives are present, which the problem excludes).

Final Answer:
\[ \boxed{W_t = W_c} \]