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Energy transfer in the impeller, specific work and head head

Dalam dokumen Learn about Centrifugal Pumps (Halaman 106-109)

3 Pump hydraulics and physical concepts

3.2 Energy transfer in the impeller, specific work and head head

No swirl α1 = 90° (c1u = 0)

a) b)

c) d)

Pressure surface

Suction surface

Counter rotation: α1 > 90° (c1u < 0)

Pre-rotation α1 < 90° (c1u > 0) α1

w1u

β1' β1

w1'

i1' c1'

β1B

c1u

-c1u

c1m' c1m

w1' w1

β1B α1

β1' β1

i1'

β1 β1' β1B

w1' w1

α1

c1mc1m' i1'

u1

β1B

i1' β1' w1' u1

u1

α1'

Fig. 3.1. Velocity triangles at the impeller inlet

shockless entry, the incidence is negative and the stagnation point is located on the blade suction surface.

Outlet triangle: The velocity relationships at the impeller outlet are shown in Fig. 3.2a and Table 3.2. The meridional velocity downstream of the impeller is obtained from Eq. (T3.2.2). Blade blockage is still present immediately upstream of the impeller outlet and the velocity is correspondingly greater than downstream of the trailing edge: c2m' = c2m×τ2, Eq. (T3.2.3). Again, the blockage does not af- fect the circumferential component. The absolute velocity c2 and outflow angle α2

are relevant for the design of the diffusing elements, Eqs. (T3.2.10 and 3.2.13).

3.2 Energy transfer in the impeller, specific work and

3.2 Energy transfer in the impeller, specific work and head 73

• The flow QLa enters the impeller through area 1 with the angular momentum ρ×QLa×r1m×c1u.

• The flow leaves the impeller through area 2 with the angular momentum ρ×QLa×r2m×c2u.

• The control surface cuts through the shaft. Therefore, external forces in the form of M = Msch + ΣMRR must be applied to this interface.

• Viscous shear stresses cause friction moments MRR on front and rear shrouds that can be estimated according to Chap. 3.6.1. But these need not be consid- ered when calculating the moment Msch acting on the blades.

• Turbulent shear stresses occur in the areas 1 and 2 in circumferential direction, since the control surfaces are not positioned perpendicularly to the velocity vectors c1 and c2. These shear stresses generate a moment Mτ, which is ne- glected when calculating according to the streamline theory. From numerical calculations it can be inferred that Mτ amounts to typically 1% of the moment transmitted by the impeller when no recirculation is present1.

The static pressures in the areas 1 and 2 do not generate any forces in the circumferential direction and therefore do not enter the momentum balance.

Likewise, the radial velocity components do not contribute to the angular momen- tum acting on the blades; consequently, only the circumferential components c1u

and c2u need to be considered. When applying one-dimensional streamline theory, the non-uniform velocity distributions at the control surfaces are replaced by rep- resentative average velocities. They generate exactly the same angular momentum on the mean radii r1m, r2m as the integral over the real flow distributions. Subject to all these prerequisites, the conservation of momentum provides the angular mo-

u2

c2

α2 c2u

β2B β2

w2

δ

β1B a)

QLa

r2m

MRR

r1m

Control volume

M = Msch +Σ MRR

1

2 b)

r2m

r1m

c1u

c1

c1m

i1

w1

u1 α1

w2

Fig. 3.2. Balance of angular momentums acting on the impeller. a velocity vectors; b con- trol volume

1 Recirculation normally occurs at partload below q* = 0.5 to 0.7 (Chap. 5). It may, how- ever, occur even near the best efficiency point when extremely wide impellers are used (for example in dredge pumps).

mentum acting on the blades in the form of “Euler's turbine equation”:

) c r c (r Q

Msch =ρ La 2m 2u1m 1u (3.1)

The mean radii are defined in such a way that they divide inlet and outlet cross sections into two areas which both receive identical flow rates (where cm is as- sumed constant over the cross section):

) r (r r

and ) r (r

r1m= 12 1a2 +12i 2m= 21 2a2 + 22i (3.2) Msch is the torque which must be applied to the shaft in order to create the flow conditions shown in Fig. 3.2a. According to Newton's law (“action equals reac- tion”), Msch is also equal to the moment transmitted to the fluid. At an angular ve- locity ω of the shaft the corresponding driver power is (with u = ω×r):

) c u c (u Q M

Psch = schω=ρ La 2m 2u 1m 1u (3.3) The specific work produced by the blades is obtained from dividing Psch by the mass flow rate m passing through the impeller (m = QLa ȡ), Chap. 1.2.2:

1u m 1 2u La 2m

th sch

sch u c u c

Q Y P

Y =

=ρ

= (3.4)

As demonstrated by Eq. (3.1), (3.3) and (3.4) a pre-swirl (α1 < 90°) reduces blade moment, power consumption and head while a counter-swirl increases these val- ues. Since the specific work of the blades according to Eq. (3.4) does not include any losses (although it applies to flows with losses!), it is also called “theoretical specific work”. Accordingly, Hth = Ysch/g is the “theoretical head”.

The energy Ysch transmitted to the fluid comprises both the useful energy in the discharge nozzle Y according to Eq. (2.1) and the hydraulic losses (see Chap. 1.2.2 and Eq. (1.8)). The following therefore applies: Ysch = Y + g×ΣZh. These losses are described by the “hydraulic efficiency” ηh defined by Eq. (3.5):

h th

h sch H Z

H H

H Y

Y

Σ

= +

=

=

η (3.5)

The hydraulic efficiency includes all hydraulic losses between the suction and discharge nozzles, i.e. in the inlet, impeller, diffuser and discharge casing. Details concerning ηh can be found in Chap. 3.9, Eq. (3.28a) and Figs. 3.27 to 3.29.

From geometrical relationships of the velocity triangles u×cu = ½ (u2 + c2 - w2) is obtained. Inserting this expression into Eq. (3.4) produces Eq. (T3.3.2). Accord- ing to this equation, the specific work consists of three components: centrifugal component u22 - u12, deceleration of the relative velocity w12 - w22 and accelera- tion of the absolute velocity c22 - c12. The impeller outlet velocity c2 is decelerated in the collector. There, its kinetic energy is for the most part converted into static pressure. With reference to Eq. (1.15) the first two terms stand for the increase of the static pressure in the impeller plus the impeller losses, Eq. (T3.3.3).

The energy Ysch transmitted to the fluid by the impeller causes a total pressure increase Ytot,La at the impeller outlet. A fraction of Ysch is dissipated in the impeller

Dalam dokumen Learn about Centrifugal Pumps (Halaman 106-109)