I'm trying to expand my theoretical knowledge on turbomachinery, but I cannot decipher the equation derivation steps in the book I'm using. I attached the derivation steps bellow as an image.

I understand the eq (51), mass through the inlet ring is the same as the mass through the outlet ring. The part that boggles my brain is how the eqs (52) and then (53) were derived.

Perhaps someone could help me with some pointers to understand this completely.

enter image description here


1 Answer 1


It looks like Eq. (52) is a general statement of torque and angular momentum transfer across the control volume boundary. Presumably, that was discussed previously. It is not derived from Eq. (51) which is a conservation of mass, the flow in equals the flow out.

In $\rho_1 \vec c_1 d\vec S_1$ (and subscript 2 case) the vector multiplication is understood to use the component of $\vec c_1$ perpendicular to vector area $d \vec S_1$ to obtain volume flow through the surface.

Eq. (53) removes the integration symbols of Eq. (52), perhaps we are looking at a small part of the rings A and D. In Eq. (52), $F_B dA$ is a force perpendicular to the radius r, so $rF_B dA$ is an increment of torque (also called moment of force). Eq. (53) shows the small increment of torque $dM$ equal to $rF_B dA$. Looking at Eq. (51), $\rho_1 \vec c_1 d\vec S_1$ is an increment of mass flow so call it $d \dot m$, and the same amount at exit (subscript 2). The right side of Eq. (52) now becomes the right side of Eq. (53).

Excuse me if I explain too much, $c_u$ is the component of $c$ in the plane perpendicular to the axis x and perpendicular to the radius r. Some books will use a $\theta$, but u is common because it is in the wheel speed direction and U is the wheel speed, that is, the speed of the blade at a given radius.

  • $\begingroup$ thank you so much!!! $\endgroup$ Sep 21, 2023 at 5:55

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