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Aerofoil and Euler theories applied to Axial Pumps and Fans.

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Axial Flow Pumps and fans.

Notes 1) 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.

2) The impeller vanes may be adjustable

3) The machine may be fitted with pre-entry vanes to ensure that there is no pre-rotation and that the flow is purely Axial.

4) The bottom diagram is produced by considering a Radius R of the impeller and drawing it out in a flat plane.

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

(1)

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

$Also\;\;\;\;\;V = V_f = V_f1\;\;\;\;( Blade\;area\;neglected)$

(2)

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Euler Theory

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

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

(3)

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

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Applying Bernoulli's equation across the Vanes:-

$\frac{p}{w} + \frac{V^2}{2g}\;+\frac{V_{w1}v}{g} = \frac{p_1}{w} + \frac{v_1^2}{2g} (Neglecting\;losses\;in\;the\;Vanes)$

(4)

Therefore the Pressure rise across the Vanes is given by:-

$\frac{p_1 - p}{w} = \frac{V_{w1}v}{g} - \frac{V_1^2 - V^2}{2g}$

(5)

$But\;\;\;\;\;V_1^2 = V_{w1}^2 + V_{f1}^2 = V_{w1}^2 + V_{f}^2 = V_{w1}^2 + V^2$

(6)

$\therefore\;\;\;\;\;\frac{p_1 - p}{w} = \frac{V_{w1}v}{g} - \frac{V_{w1}^2}{2g}$

(7)

Aerofoil Theory Applied to Propeller Pumps.

The Combined Inlet and Outlet and Inlet Triangles.

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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.

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Force P exerted by the Fluid on the Vane element i the vector Sum of the liftL 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.

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

(8)

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

(9)

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

Resolving L and D in the direction of motion.

$P_t = L\;sin\,\theta _{ave.} + D\;Cos\,\theta _{ave.}\;\;\;\;lb./Vane$

(10)

$P_a = L\;Cos\,\theta _{ave.} - D\;Sin\,\theta _{ave.}\;\;\;\;lb./Vane$

(11)

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

For N Vanes the Total Tangential Force = $n\;P_t$

$= n\left(L\;Sin\,\theta _{ave.} + D\;Cos\,\theta _{ave.} \right)$

(12)

$\therefore\;\;\;\;\;Torque = nP_t\times R$

(13)

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

$The\; Efficiency \;\eta = \frac{W.H.P.\;out put}{H.P.\;in put}$

(14)

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

$\Delta p\times 2\pi \,R\,dr = nP_a$

(15)

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

Example 1

The figures below are for an Axial Flow Propeller fan pumping air. It is required to find the torque /ft. radius using both Aerofoil and Euler Theories.

Radius R = 3.2 ft. Speed N = 450 r.p.m.

$V = V_f = V_{f1} = 131\;ft./sec$

(16)

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

(17)

Theoretical Head = 114 ft. of air.

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

(18)

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

(19)

Chord = 1.1 ft.

It is required to find the Torque per ft radius

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

(20)

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

(21)

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

(22)

Velocity triangle.

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

(23)

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

(24)

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

(25)

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

(26)

$= 20.1 lb./ft radius$

(27)

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

(28)

For twelve Vanes the longitudinal Force = 12 Pt

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$\therefore\;\;\;\;\;Torque\;required = 12\;P_t\;R$

(29)

$= 12\times 13.85\times 3.2 = 532\'lb.ft./ft.\,Radius$

(30)

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NOTE

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

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

(31)

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

If $\delta\,p$ is the pressure rise

$Then\;\;\;\;\;\Delta p\times 2\pi \times R\,dR = 174.6 lb./ft.$

(32)

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

(33)

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

(34)

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

The weight of Flow through an annular element (W ) $= 2\pi \;R\,dR\times V_f\times w$

$= 2\times 3.2\times \pi \times 1\times 131\times 0.074 = 195\;lb./sec.$

(35)

Work done per lb per lb. $= \frac{V_{w1}v_1}{g} = \frac{H_M}{\eta }$

$= \frac{114}{0.9} = 126.7\;ft.lb./lb.$

(36)

Therefore the work done per second per ft. radius

$= W\times \frac{V_{w1}v_1}{g} = 195\times 126.7\;ft.\;lb.\;/sec.$

(37)

$= Torque\times Angular\;Velocity = \frac{T\times 2\pi N}{60}$

(38)

$\therefore\;\;\;\;\;T = \frac{195\times 126.7\times 60}{2\times \pi \times 450} = 524\;lb.ft.$

(39)

The Pressure rise across the Vanes:-

$\frac{p_1 - p}{w} = \frac{V_{w1}v_1}{g} - \frac{V_{w1}^2}{2g}$

(40)

$= 126.7 - \frac{27^2}{64.4} = 115.4 \;ft.\; of\; air$

(41)

Axial Flow Pumps and Fans

Axial Flow Pumps and fans.

Aerofoil Theory Applied to Propeller Pumps.

Example 1

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