This is bascially a theory problem. Consider the axisymmetric flow through two c

Question

This is bascially a theory problem.

Consider the axisymmetric flow through two conical diffusers as shown in Figure 1. In diffuser (a) the flow exits radically and in diffuser (b) the flow exits horizontally. For both diffusers, assume that the magnitude of the velocity vector at all points on the exit plane is the same and that the radius R of the exit plane is the same. In which case do you expect the volume flow rate to be greater? Give reasons. Obtain an expression for volume flow rate in diffuser (a) and show that it is a function of the radius R of the exit plane and the distance P of the exit plane from the imaginary point where the walls of the diffuser meet (see figure below, P and R determine the angle of the cone). Plug in sample values of P and R in the expression you obtained in (ii) and comparing with the volume flow rate in diffuser (b) check that it is consistent with your answer to (i).

Explanation / Answer

i) Volume flow rate will be greater for case (b) because flowrate = area * normal velocity. Even though velocity magnitude and exit plane area is same for both cases, the component of velcoity normal to the exit plane is more for case b.

ii) Let us say semi-cone angle of diffuser is such that Tan = R/P

Normal component of velocity (perpendicular to exit area) = VCos = V*P/(R^2 + P^2)

Area = R^2

Volume flow rate = (R^2)*V*P/(R^2 + P^2)

So, flowrate is a function of R,P and V.

iii) Let P = 3 m and R = 1 m

Tan = R/P = 1/3

So, = 18.4 deg and Cos = 0.95

Area = R^2 = 3.14 m^2

For diffuser (a), flowrate = 3.14*VCos = 3.14*V*0.95 = 2.978*V

For diffuser (b), flowrate = 3.14*V = 3.14*V

Hence, diffuser b has more flow.

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