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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?

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  • $\begingroup$ 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. $\endgroup$
    – Solar Mike
    Jul 7 '20 at 8:29
  • $\begingroup$ I edited my question with both details. Thanks $\endgroup$
    – Skawang
    Jul 7 '20 at 9:18
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If you examine a derivation of the Euler work equation (http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node91.html), you will see that the change in angular momentum of the fluid is proportional to the change in enthalpy of the fluid. Remember the definition of enthalpy is the flow work and the internal energy. Therefore, the enthalpy contains the change in pressure via the flow work term. This is why pressure does not directly appear in the Euler work equation, but is physically accounted for. I recommend stepping through a derivation of the equation yourself to gain a better understanding.

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  • $\begingroup$ 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? $\endgroup$
    – Skawang
    Jul 7 '20 at 13:29
  • $\begingroup$ Look at the definition of enthalpy. Enthalpy is the sum of the flow work (which includes pressure and density) and the internal energy. Enthalpy includes the pressure in its definition. The enthalpy change and angular momentum are proportional because the power, work, and torque are directly related. This connects conservation of angular momentum to the energy equation, as shown in the link. Power is the work done per unit time. Work is in the energy equation. $\endgroup$
    – mechcad
    Jul 7 '20 at 15:08
  • $\begingroup$ 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? $\endgroup$
    – Skawang
    Jul 8 '20 at 11:59
  • $\begingroup$ I am an aero engineer, so I understand how it works. Consider flow in a control volume around a turbine rotor airfoil and passage. Flow leaves the CV with less energy than when it entered due to the work done on the airfoil (i.e. the pressure differences between the pressure and suction sides). You can integrate the static pressure distribution on the airfoil surface to obtain the tangential force applied on the airfoil. In an adiabatic flow, the work done on the airfoil by the fluid is reflected in the enthalpy difference between the inlet and exit of the CV. Does this help at all? $\endgroup$
    – mechcad
    Jul 8 '20 at 12:34
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    $\begingroup$ Yes, the expression given is the work done by the fluid on the blade. However, enthalpy is the summation of the internal energy (U) and the flow work (PV), not the pressure and the flow work. The Euler turbine equation is valid for a single blade row. So, in other words, if the turbine has more than 1 stage, then the equation would need to be applied for each stage. If it is a 4 stage machine, the equation would need to be applied 4 times. Is this more clear? $\endgroup$
    – mechcad
    Jul 8 '20 at 18:59

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