The figure below shows a compressor stage with station numbers 1, 2, and 3 as indicated. If \(p_{0i}\), \(T_{0i}\), and \(C_i\) refer to the average values of total pressure, total temperature, and absolute flow speed, respectively, at the \(i^{th}\) station, select the CORRECT option considering losses.
Show Hint
The rotor adds real work (so p and T both rise), the stator adds no work (so T stays fixed) but still loses total pressure to friction while it diffuses and slows the flow.
Step 1: Use the Euler turbine equation to see why the rotor raises pressure and temperature.
For a rotor with blade speed $U$, the specific work exchanged with the flow is $w = U(C_{w2}-C_{w1})$, where $C_w$ is the tangential (swirl) component of the absolute velocity. In a compressor rotor this work is positive (fed into the flow), which raises the stagnation enthalpy in line with the steady flow energy equation, $w = c_p(T_{02}-T_{01})$. A positive $w$ forces $T_{02}>T_{01}$ regardless of any losses, since this temperature rise is tied directly to the real work transferred. Because work is added and the process is only partially irreversible, $p_{02}>p_{01}$ too, just less than the ideal isentropic value.
Step 2: Use the Gibbs relation to see why the stator's total pressure must drop.
For steady adiabatic flow with no work, the Gibbs equation for the stagnation state reads
\[
T\,ds = dh_0 - \frac{dp_0}{\rho}
\]
Across the stator $dh_0=0$ (no work, no heat), so this reduces to
\[
dp_0 = -\rho T\,ds
\]
Since the flow always generates entropy through friction and secondary losses ($ds>0$ in a real machine), $dp_0$ must be negative. That gives $p_{03}<p_{02}$ directly from the second law, with no need to track individual loss mechanisms. Meanwhile $dh_0=0$ means the stagnation temperature is unchanged, $T_{02}\approx T_{03}$.
Step 3: Use the velocity triangle picture for the speed changes.
The rotor's whole purpose is to add tangential velocity, so the absolute velocity leaving the rotor is larger than entering it: $C_2>C_1$. The stator is a diffusing passage for this now-swirling flow: as the flow is straightened and slowed in the widening passage, the absolute speed drops again, $C_3<C_2$, while static pressure rises.
Step 4: Assemble the full picture.
\[
p_{01}<p_{02}>p_{03}, \qquad T_{01}<T_{02}\approx T_{03}, \qquad C_1<C_2>C_3
\]
This matches option (A) exactly, reached here from the Euler work equation and the Gibbs entropy relation rather than from qualitative reasoning about the blade rows alone.
\[
\boxed{\text{Option (A)}}
\]
