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: Identify what each blade row does.
Station 1 is ahead of the rotor, station 2 is between the rotor and the stator, and station 3 is after the stator. The rotor (R) is the moving blade row that does work on the flow, and the stator (S) is a stationary blade row that only redirects and diffuses the flow.
Step 2: Track total pressure and total temperature across the rotor (station 1 to 2).
The rotor absorbs shaft work and feeds it into the air as it spins the flow up (imparts swirl). Real work is added to the fluid, so its total temperature must rise: \(T_{02}>T_{01}\), always true, with or without losses, since it follows directly from the energy actually put into the flow. Total pressure also rises, \(p_{02}>p_{01}\), because useful work raises stagnation pressure even though blade friction eats into part of the ideal (isentropic) rise.
Step 3: Track total pressure and total temperature across the stator (station 2 to 3).
The stator does no work at all; it just turns and slows the flow down to raise static pressure. Since no work and no heat is added, the stagnation enthalpy, and hence the total temperature, stays essentially unchanged: \(T_{02}\approx T_{03}\). Total pressure, however, always falls slightly across any real blade row because of skin friction and secondary flow losses, even though no work is involved: \(p_{02}>p_{03}\).
Step 4: Track the absolute flow speed \(C\) across the stage.
The rotor, by adding tangential swirl velocity to the flow, increases the absolute speed as seen by a stationary observer: \(C_2>C_1\). The stator's whole job is to diffuse this fast, swirling flow, converting some of that kinetic energy into static pressure rise, which slows the flow back down: \(C_3<C_2\).
Step 5: Rule out the other options.
Option (B) assumes the stator is lossless (\(p_{02}=p_{03}\)), but the question explicitly says to consider losses, so this cannot be right. Option (C) has the temperature rising across the stator (\(T_{02}<T_{03}\)), which would require the stator to do work on the flow, and it also has the speed falling across the rotor (\(C_1>C_2\)), the opposite of what work addition does. Option (D) has total pressure rising across the stator (\(p_{02}<p_{03}\)) with no work input, which violates the second law for a real, lossy blade row, and it wrongly keeps the speed constant through the whole stage.
