Axial Flow Pumps and Fans

Leonhard Euler (15 April 1707 - 18 September 1783) was a pioneering Swiss mathematician and physicist. He made important discoveries in fields as diverse as infinitesimal calculus and graph theory.

Torque, also called moment or moment of force, is the tendency of a force to rotate an object about an axis, fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist.

Aerofoil and Euler theories applied to Axial Pumps and Fans.

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Axial Flow Pumps and fans.

The figures below are for an Axial Flow Propeller fan pumping air.

Notes:

The fixed diffuser vanes are used to remove the whirl component of the discharge velocity of the impeller and to convert the energy to Pressure.
The impeller vanes may be adjustable
The machine may be fitted with pre-entry vanes to ensure that there is no pre-rotation and that the flow is purely Axial.
The bottom diagram is produced by considering a Radius $R$ of the impeller and drawing it out in a flat plane.

$v_1 = v = \frac{2\pi RN}{60}$

The flow through the machine, $q = \pi \left(r_0^2 - r_{boss}^2 \right)\times V_f$ . (The boss area can be neglected.)

Also, $V = V_f = V_f1$ (Blade area neglected.)

Euler Theory

Work done by the Vanes per lb. of water $= \displaystyle\frac{V_{w1}v_1 - V_wv}{g} = \displaystyle\frac{V_{w1}v}{g}\;ft.$ ( $V_w = 0$ )

Hydraulic or Manometric $\eta = H_m\div \displaystyle\frac{V_{w1}v}{g}$

Where $H_m$ = The Manometric head minus the Head developed by the Pump across the Flanges.

Applying Bernoulli'\b{s equation} across the Vanes: $\frac{p}{w} + \frac{V^2}{2g}\;+\frac{V_{w1}v}{g} = \frac{p_1}{w} + \frac{v_1^2}{2g}$ (Neglecting losses in the Vanes)

Therefore the Pressure rise across the Vanes is given by: $\frac{p_1 - p}{w} = \frac{V_{w1}v}{g} - \frac{V_1^2 - V^2}{2g}$

But, $V_1^2 = V_{w1}^2 + V_{f1}^2 = V_{w1}^2 + V_{f}^2 = V_{w1}^2 + V^2$ $\therefore\;\;\;\;\;\frac{p_1 - p}{w} = \frac{V_{w1}v}{g} - \frac{V_{w1}^2}{2g}$

Aerofoil Theory Applied to Propeller Pumps.

The Combined Inlet and Outlet and Inlet Triangles.

The Blades are considered to be aerofoils in cascade in a fluid stream of Velocity $V_r$ Average in a direction $\theta$ to the tangent.

Force $P$ exerted by the Fluid on the Vane element i the vector Sum of the lift L and the force exerted on the Vane by the Fluid. The components of P are $P_t$ in the Tangential Direction and $P_a$ in the Axial Direction.

Consider an elemental Vane thickness $dR$ at Radius $R$ .

From Aerofoil Theory Lift, $L = C_L\;\times \;\displaystyle\frac{1}{2}\rho V_{r\;ave.}^2\;\times C\;dr\;\;lb.$

And Drag, $D = C_D\;\times \frac{1}{2}\rho V_{r\;ave.}^2\;\times C\;{dR}\;\;lb.$

$C_L$ and $C_D$ are lift and drag coefficients depending upon the aerofoil section and the angle of incidence.

Resolving $L$ and $D$ in the direction of motion. $P_t = L\;sin\,\theta _{ave.} + D\;Cos\,\theta _{ave.}\;\;\;\;lb./Vane$ $P_a = L\;Cos\,\theta _{ave.} - D\;Sin\,\theta _{ave.}\;\;\;\;lb./Vane$

$D$ is commonly small compared with $L$ and the second terms are often neglected.

For $N$ Vanes the Total Tangential Force = $n\;P_t$ $= n\left(L \sin \theta _{ave.} + D \cos \theta _{ave.} \right)$ Therefore, Torque $= nP_t\times R$

The Horse-Power required to rotate the Vane elements = $\displaystyle\frac{2\pi NT}{33000}$

The Efficiency, $\eta$ = W.H.P. out put / H.P. in put

Where W = Weight of flow through the annular ring = $w\times 2\,\pi\;RdR \times V_f$

The Total Axial Force = $n Pa$ If $\Delta p$ is the pressure drop across the vanes, $\Delta p\times 2\pi \,R\,dr = nP_a$

This equation can be solved to find $\Delta p$ .

Example 1 [imperial]

Problem

For the Axial Flow Propeller it is required to find the torque /ft. radius using both Aerofoil and Euler Theories.

We know the radius $R$ is $3.2 ft$ and the speed $N$ is $450 r.p.m.$

Workings

$V = V_f = V_{f1} = 131\;ft./sec$ and $\rho = 0.074\;lb/ft^3$

Theoretical Head = 114 ft. of air.

$\eta _H = 0.9\;\;\;\;\;\;\;\;\;\;\;\;\;\;n = 12$ Vanes

$C_L = 0.44\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;C_d = 0$

Chord = 1.1 ft.

It is required to find the Torque per ft radius

$\eta _H = H_M\div \frac{V_{w1}V_1}{g} = 114\div \frac{V_{w1}v_1}{g}$

But, $v_1 = v = \displaystyle\frac{2\pi RN}{60} = 151\;ft./sec.$

$\therefore\;\;\;\;\;V_{w1} = \frac{114\times 32.2}{0.9\times 151} = 27\;ft./sec.$

Velocity triangle:

$V_{r\, ave.} = 131^2 + (151 - 13.5)^2 = 190\;ft./sec.$

$\theta _{ave.} = Tan^{-1}\frac{131}{151 - 13.5} = 43^0\;36'$

$\textup{Lift}= C_L\times \frac{1}{2}\rho V_{r\,ave.}\times C\times dR$

$= 0.44\times \frac{0.074}{2\times 32.2}\times 190^2\times 1.1\;lb./ft.radius$

$= 20.1 lb./ft. radius$

$P_t = 20.1\times Sin\theta _{ave.} = 13.85\;lb./ft. radius$

For twelve Vanes the longitudinal Force = 12 Pt

Therefore, Torque required $= 12P_tR = 12\times 13.85\times 3.2 = 532lb.ft./ft. radius$

Note:

Axial thrust / blade $P_a = L\,cos\,\theta _{ave.}$

$= 20.1\times Cos\,45^0\;36' = 14.55\;lb./ft.\;width$

Total axial Force for 12 Vanes = $14.55 X 12 = 174.6 lb./ft. width$

If $\delta\,p$ is the pressure rise, then $\Delta p\times 2\pi \times R\,dR = 174.6 lb./ft.$

$\therefore\;\;\;\;\;\Delta p = \frac{174.6}{2\pi \times 3.2} = 8.68\;lb./ft^2$

Or, $\;\;\;\;\Delta p = \displaystyle\frac{8.68}{0.074} = 117.3\;ft.\;of\;air$

\subsection{An alternative approach is:} The weight of Flow through an annular element, $W = 2\pi \;R\,dR\times V_f\times w$ $= 2\times 3.2\times \pi \times 1\times 131\times 0.074 = 195\;lb./sec.$

Work done per lb per lb. $= \frac{V_{w1} v_1}{g} = \frac{H_M}{\eta }= \frac{114}{0.9} = 126.7\;\;\;[ft.lb./lb.]$

Therefore the work done per second per ft. radius $= W\times \frac{V_{w1}v_1}{g} = 195\times 126.7\;\;\;[ft.\;lb.\;/sec.]$

= Torque * Angular Velocity = $\displaystyle\frac{T\times 2\pi N}{60}$ $\therefore\;\;\;\;\;T = \frac{195\times 126.7\times 60}{2\times \pi \times 450} = 524\;lb.ft.$

The Pressure rise across the Vanes: $\frac{p_1 - p}{w} = \frac{V_{w1}v_1}{g} - \frac{V_{w1}^2}{2g}= 126.7 - \frac{27^2}{64.4} = 115.4 \;ft.\; of\; air$

Solution

The Torque required is $532 lb.ft./ft.radius$

The above question can also be done using the Euler Theory.

The weight of Flow through an annular element, $W = 2\pi \;R\,dR\times V_f\times w$ $= 2\times 3.2\times \pi \times 1\times 131\times 0.074 = 195\;lb./sec.$

Work done per lb per lb. $= \frac{V_{w1} v_1}{g} = \frac{H_M}{\eta }\= \frac{114}{0.9} = 126.7\;\;\;[ft.lb./lb.]$