General comments

If the flow spins too fast does it get dizzy? (1 student)

I covered the material in Chapter IX of the notes. This included deriving the relationship between shaft power added to a flow and change in flux of angular momentum. There are a few key points: 1) Without heat addition the only way to change the total enthalpy of a fluid is through an unsteady process (like a moving blade row), 2) In an analogous fashion to the integral form of the linear momentum equation for a steady flow through a control volume of fixed mass (sum of forces is equal to net flux of linear momentum), for a steady flow through a control volume of fixed mass, the sum of the torques is equal to the net flux of angular momentum, 3) Shaft power is equal to angular velocity times torque and is also equal to the change in total enthalpy (from the steady flow energy equation for adiabatic flow). Setting these two relationship equal to one another results in the Euler Turbine Equation. We did two PRS questions (PRS #1, PRS#2).

I like the phones. (1 student)

Responses to 'Muddiest Part of the Lecture Cards'

(34 respondents, 52 students in class)

1) What are the b and c components? (1 student) These are the notation I will use when we get to axial flow compressors. For now just think of them as "inlet" nad "exit".

2) Maybe I'm missing something, but I didn't see the integrals & equation in the notes ....where are they in there? (1 student) You are referring to the integral form of the angluar momentum equation. A link to this appears in two places (near the linear momentum theorem and also where we discuss the Euler Turbine Equation). Maybe it is just some terminology I missed, but how did M^2 suddenly turn into a bunch of u^2's? Isn't M Mach #? (same student) I think you are referring to this equation which just expresses the magnitude of the velocity vector in terms of its components. I don't think I used Mach number in class.

3) I don't see why a compressor and a turbine don't do the same thing. Is it because the compressor has lots of kinetic energy but that entering the turbine doesn't? Being on the same shaft, I'd expect them to do the same thing. (1 student) We will talk about this in detail in the last lecture. They do the same thing but in reverse order. Roughly you can think about it as follows. A compressor uses a spinning blade row to add energy to the flow (e.g. swirling kinetic energy) and then a stationary blade row to convert the swirling kinetic energy into internal energy (without adding anything--it just converts the energy from one form to another). The compressor continues to do this stage after stage, rotating blade row, stationary blade row, rotating blade row, stationary blade row. For a turbine (designed to extract energy), flow with high internal energy is passed through a stationary set of swirl vanes (these do not add or remove energy, but they convert the energy from internal energy to swirling kinetic energy). The swirling flow then drives a set of spinning blades which extract energy (removing the swirling kinetic energy), leaving the flow with little or no swirl. The flow then passes to the next stage (a stationary blade followed by a moving blade). This is all discussed in greater detail in the notes.

4) On the first PRS I did not follow r x u = ruq. (1 student). The vector r x u has three components. We are interested in the axial component. To form the axial component, the relevant flow component must be perpendicular to both r and x --- hence uq.

5) How does the flow leave a centrifugal pump? (1 student) The flow leaves a centrifugal pump into a volute (a large chamber at the exit of the spinning blade rows). In some designs there are vanes that help convert the swirling kinetic energy to internal energy. Hill and Peterson, Mechanics and Thermodynamics of Propulsion, 2nd edition (available in the library) has a much more extensive discussion (and pictures -- if I get a chance I will scan a few in).

6) Could we find a better way to mark variables for T=temperature, T=thrust and T=torque? (1 student) Well, we could probably start using other alphabets, but this is a common problem--too many variables for the 26 letters we have (even when we use lower and upper case and Greek). As you become more familiar with the various equations, you will be able to recognize which is which (e.g. by comparing units).

7) You can just multiply omega (the angular velocity) by the radius to get the exit velocity of the turbopump? (on the PRS) I was uncertain how to find vc. (1 student). Yes, since the problem says that the tangential velocity at the exit is approximately equal to the rim speed (r times omega).

8) Is the blade shaped like this (nice picture drawn of swept turbopump blades) so that the velocity is relatively constant from r=0 to r=rexit (less swirl)?. (1 student) The various design trades associated with sweeping the turbopump blades (forwards and backwards) are a more advanced topic, but one that you can tackle once we talk about velocity triangles in the next lecture. Due to limited time, we will focus on axial compressors and turbines in the next lecture, but if you are interested in finding out more on turbopumps, there is a good discussion in Hill and Peterson, Mechanics and Thermodynamics of Propulsion, 2nd edition (available in the library). Why not have counter rotating blades instead of stationary ones? (same student). Good question. Advanced "Unducted Fan Engines" that NASA has studied have counter-rotating blades instead of stationary ones. And Professors Epstein and Kerrebrock have proposed similar configurations for advanced military engines for a current research effort in the Gas Turbine Laboratory. Some of the difficulty comes in having every other blade on both ends of the spool (compressor and turbine) counter rotating--this presents an aerodynamic design challenge--but not an insurmountable one.

9) No mud (26 students).