General comments
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).
Responses to 'Muddiest Part of the Lecture Cards'
(11 respondents, 65 students in class)
1) Is r-cross-u always just u-theta ( in the first PRS question )? (1 student) No. u-theta is the axial component of the vector. Could you draw a picture or do some math to show what u-theata is? It was a little unclear in the 2D picture on the slide. Thanks? (1 student) First, this is a right-handed cylindrical coordinate system, so if you align your hand with the direction of r and roll it towards the direction of theta (out of the page) your thumb points in the direction of x. Now consider the cross product of two vectors one only has a component in the radial direction, r, the second vector u, can have components in the radial, tangential (i.e. it is coming out of the page) and axial directions. If you take the cross product between the two you get a new vector with components in the theta direction (-r times usubx) and the axial direction (r times usubtheta). You do this the same way you do any cross product. The way to show this with your hand is to assume that you only have a velocity component in the theta direction -- then r crossed with usubtheta points in the positive axial direction. You can do the same thing for the theta component of the vector.
2) You mentioned that the flow moves axially, then rotates, which I guess is due to the mobving and stationary blades. Is the rotation cause by the stationary blades? (1 student) Both the moving and stationary blades can add or remove a swirling component of velocity. Since the both change the angular momentum, both have forces. However, only the moving blade rows do work (changing the enthalpy of the flow).
3) Why is v-theta rotation times diameter? (1 student) You are referring to the second PRS question. Here the problem specified that the velocity leaves the rotor at a speed equal to the speed of the rotor disk edge. This speed is omega times radius.
4) Can you please put the transparencies/notes on-line? Why are we always concerned about momentum? (1 student) The lecture materials are posted. Relative to momentum -- in fluids and propulsion there are a few key concepts (conservation of mass, momentum, and energy, Second Law of Thermodynamics, etc.), conservation of momentum is one of them.
5)Why does the m-dot term only contain the x-component of velocity? (1 student) The mass flow is the flux across the surface = rho (u-dot-n) dA. The x-component of velocity comes from the dot product with a surface normal that is aligned with the x-direction.
6) For those who know stuff about engines, can you give an example of what the unsteady stuff is? (1 student) I would point you to the first few pages of Chapter 5 of Kerrebrock's book "Aircraft Engines and Gas Turbines", it is on reserve in the library. For non-ideal enginies, does bleeding air aaft from the compressor and adding it in the HPT blades affect performance since mdot is no longer constant? (1 student) Very much so. This flow is part of what is termed the "secondary flow system". These flows can be as high as 30-35% of the main core flow and thus are very significant in terms of their impact on cycle performance.
7) Does the fact that the working fluid is a liquid enable you to impart more power and thus more pressure rise than iif it was a gas? 16.05 would suggest that it would, but you didn't use cp in the PRS answers. How does that work out? (1 student) In the PRS answers I stopped with change in enthalpy -- and it was specific enthalpy. The power added is related to the change in angular momentum flux across a moving blade row times the mass flow. The mass flow is a function of the density (different for different fluids, gases/liquids). And the relationship between changes in specific enthalpy and changes in temperature is a function of the type of working fluid.
8) No mud (3 students).