**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).

There was one comment on the cell phone technology. Keep these coming (that is why we are trying these, to get student and instructor feedback):** 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).

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,

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,

9) * No mud* (26 students).