Lecture P9: Energy Exchange with Moving Blades

You guys were great and I am going to miss teaching you. It is the very best part of my job.

Why do we have so little of the interesting subjects like propulsion, and so much of the boring subjects like signals and systems and dynamics? (1 student) They are all important (the politically correct answer), but I agree with you--I think propulsion and fluid mechanics are more interesting (which is why I do research in these areas)! However, others feel equally strong about their disciplines. If you like this stuff, take 16.100 and 16.50 next year.

I finished covering the material in Chapter IX of the notes. This included discussing how blade-level design parameters impact the power input/output of a turbomachine and the use of velocity triangles. There are a few key points:

1) Stators convert energy from one form to another. Even though the flow exerts a torque on the blades (see homework P3) since the blades are not moving (omega=0) the rate of doing work is zero thus the poweradded or removed is zero,

2) Moving blade rows do add or remove power from the flow. To evaluate how much it is necessary to determine how much the tangential velocity (in the stationary frame) changes across the moving blade row,

3) To evaluate the change in tangential velocity, you need to use velocity triangles.

4) There are a few rules we will apply for the velocity triangles.

• Rule #1: The axial velocity is approximately constant across the blade rows.
• Rule #2: The flow always leave the blade at the approximately the trailing edge angle of the blade--in the frame of reference of the blade,
• Rule #3: It is important to check to make sure the inlet flow is within roughly plus or minus 20 degrees of the inlet flow angle or else the flow is likely to separate.
• Rule #4: Always consider the flow in the frame relative to the blade---convert between the stationary and moving blades by adding or subtracting the blade velocity (omega times r) to the flow velocity.

We did two PRS questions (PRS #1, PRS#2).

Miscellaneous:

I couldn't concentrate because I was pre-occupied with 16.070. You should let the 16.070 instructors know how evil they are. (1 student)

Responses to 'Muddiest Part of the Lecture Cards'

(29 respondents, 52 students in class)

1) I definitely got a little lost with the coordinate frames and which arrow were which in the triangles. Could you go over it real quick? (1 student) It was clear but I need more practice. Hopefully the homework will help. (1 student) Ahh! Can we have more practice drawing these diagrams in recitation? (1 student) The homework will help, we can also go over this in recitation, I will also be available during the Monday office hours. And, although I won't have time to do it until this weekend, I am going to give you a step-by-step "instant replay" of how I did the examples in class. This should help clarify things. Also PLEASE FEEL FREE TO MEET WITH ME IF YOU ARE CONFUSED (send me an email to schedule an appointment).

2) How can you use this to get engine performance? (1 student) Good question. Given blade angles and speeds you have the tools to estimate the stagnation temperature change across the stage (and using isentropic relations the stagnation pressure change). If you do this for blade row after blade row after blade row (e.g. the whole compressor or the whole turbine) you can then find out the stagnationi temperature change across the whole compressor for instance. This is the tau sub c that appears in the thrust equation and the efficiency equations in Chapter 8--these tell you engine performance!

3) When doing velocity triangles across rows of rotors and stators, is the omega-r in the same direction each time as you move to the next relative row?. (2 students) Yes it is, but to get from the stationary frame to the moving frame you need to subtract the vector omega-r from the stationary frame velocity vector, and to get from the moving frame back to the stationary from you need to add the vector omega-r to the moving frame velocity vector (so the sign keeps switching).

4) What is the difference between using cambered blades to change the angle of the gas off the blade and just changing the angle of attack of a flat plate to achieve the same angle? (1 student). There is a limit to how much you can turn the flow with a flat plate (without the flow separating). Some of the turning done in turbomachinery blades can be quite high (like 90 degrees). Thus for many applications is is best to use airfoils with very strong camber. There are however, many applications (e.g. blades that run with supersonic relative flow velocities) for which the blades look very much like flat plates.

5) If we have a velocity triangle like this (nice picture drawn), this (pointing to tangential component of stationary frame velocity) becomes our tangential velocity right? So what was the point of finding the velocity in the frame of the blade, if we don't use it? (1 student) You don't need it to find the tangential component of velocity in the stationary frame at inlet to the moving blade row (one of the quantitites in the Euler Turbine Equation). However, for the outlet of the moving blade, you need to find it in the frame of the blade (leaving at the trailing edge angle) so you can then add omega-r to it and determine the stationary frame velocity (from which you get the tangential component--the other term in the Euler Turbine Equation).

6) What happens if the blade rows have different in and out areas? Do the axial velocity components still stay the same? (1 student) No they don't. You will learn how to deal with this if you take 16.50.

7) Can one learn the velocity of spinning blade rows by the lift on the blades? (1 student). Only if the blade row is free-wheeling (no other power being added or subtracted). However, in most turbomachines other things are usually attached (like a compressor on one end and a turbine on the other). However, the torque on the blade row that we used to form the Euler turbine equation is identically equivalent to the radius times the force Ry you found in homework problem P3 (check it out it is true!) and Ry is closely related to the "lift" on the blade (it is the tangential component of all the forces--lift and drag).

8) I don't quite understand exactly how the triangle shows the u sub theta in and the u sub theta out as those 1/2 legs of the triangles? (1 student) u sub theta is the tangential component of the stationary frame velocity vector.

9) No mud (20 students).