Accounting for pressure energy in Euler turbine/pump equation

For all the analysis to find work done by a compressor or work done on a turbine, the book I'm reading (Fundamentals of Turbomachinery by Venkanna B.K) uses the Euler turbine and pump equation, $$W=\dot{m}(V_{w1}U_1\pm V_{w2}U_2)$$ where $$V_w$$ is the whirl velocity of fluid at inlet and exit, and $$U$$ is the mean rotational speed of the rotor blades and inlet and exit. It is based on the conservation of angular momentum of the fluid by drawing velocity triangles.

While this might give the value of work done due to momentum of the fluid, what about the work done by the pressure energy in the fluid or work done to increase the pressure energy of the fluid? Especially in cases like Francis turbine and axial compressors where change in pressure energy plays a big role, how can we consider only the momentum of the fluid in our analysis? I'm guessing work needs to be done to increase the pressure energy as well/work is done by pressure energy in turbines like Francis turbine.

Maybe because of complicated aerofoil shapes of the blades its hard to do an analytical approach but shouldn't we at least account for a factor of change in pressure energy?

• It would be helpful if you stated the book, but as we may not have your book you should show the formula you are looking at. Jul 7 '20 at 8:29
• I edited my question with both details. Thanks Jul 7 '20 at 9:18

• In the derivation you linked, they substitute the power $P=\omega \dot{m} (v_cr_c-v_br_b)$ in $P=\dot{w_s}=\dot{m}(h_{T_c}-h_{T_b})$. So they are assuming the total power is $P$ which is obtained from conservation of momentum and equate that to change in enthalpy. So it is only proportional because they assumed it was. My question is how is the total power obtained from considering only change in momentum in cases like Francis reaction turbines where a significant portion of energy comes from converting pressure energy to work? Jul 7 '20 at 13:29
• I understand enthalpy includes flow work, but I'm not talking about flow work. As the blades act like airfoils, there will be force acting on the blade due to pressure differences across each side of the blade. This causes work to be done on the blade as a result of pressure difference $between$ two flows. Flow work is the work done to push the fluid into the system so how can it include the work done by the lift forces acting on the airfoil? Jul 8 '20 at 11:59