3 Pump hydraulics and physical concepts
3.6 Calculation of secondary losses
of the coupling power – as with pumps having more than 80% overall efficiency – such an uncertainty is not too serious. However, in the case of small pumps and low specific speed it is significant. Loss analysis is necessary not only for new developments; it can also be very helpful in determining the cause of a perform- ance deficit observed in a plant or on the test stand. When performing this task, often the focus is not so much on the absolute value of the calculated loss; the goal is rather to determine trends caused by a deviation from design or to evaluate the effect of an intended or observed modification. In such cases, consistent con- sideration of the losses is more important than the choice between different calcu- lation methods.
3.6 Calculation of secondary losses
In the following, disk friction losses and leakage losses are considered as separate phenomena that can be measured in special test stands outside, or even inside, a pump. However, a close interaction is in effect between the main flow at the im- peller outlet, the flow in the impeller sidewall gaps, leakage through annular seals and the disk friction losses. The interaction is difficult to capture either experi- mentally or numerically (this topic is discussed extensively in Chap. 9.1).
3.6.1 Disk friction losses
When a circular disk or a cylinder rotates in a fluid, shear stresses corresponding to the local friction coefficient cf occur on its surface, Chap.1.5.1. On a disk rotat- ing in an extended stationary fluid (without the influence of a casing) the shear stress is τ = ½ρ×cf×u2 with u = ω×R, Eq. (1.31). The friction force on a surface element dA = 2π×r×dr is then dF = 2π×τ×r×dr and the torque exercised by friction becomes: dM = r×dF = π×ρ×cf×ω2×r4×dr. The friction power per side of the disk PRR = ω×M is obtained from the integral PRR = ω׳dM (between inner radius r1
and outer radius r2) as PRR = (π/5)×ρ×cf×ω3×r25×(1 - r15/r25). The friction coefficient cf depends on the Reynolds number and the surface roughness. It has roughly the same magnitude as found on a plate in parallel flow, Eq. (1.33).
In the case of a rotating cylinder (with the length L and radius r2) no integration is necessary since the radius is constant. In the same way as above the power PRZ = π×ρ×cf×ω3×r24×L is obtained.
In agreement with the similarity laws in Table 3.4, the disk friction power in- creases with the third power of the speed and the fifth power of the diameter (as- suming geometrical similarity L/r2 = constant and neglecting the influence of the Reynolds number).
If the body rotates in a casing (as is the case in a pump) the velocity distribu- tion between casing and rotating body depends on the distance between the impel- ler shroud and the casing wall as well as on the boundary layers which form on
the stationary and rotating surfaces. A core flow with approximately cu = ½ω×r is obtained (in other words, u = ω×r can no longer be assumed). In the case of turbu- lent flow the power absorbed by a disk in a casing therefore amounts to roughly half of the power of a free disk rotating in a stationary fluid.1
The disk friction losses of impeller shrouds or of cylindrical bodies are calcu- lated according to Eqs. (T3.6.2 and 3.6.4). The friction coefficients kRR and kRZ
required for this purpose are determined from correlations derived from experi- ments. All relevant equations are compiled in Table 3.6. They were derived from measurements on hydraulically smooth disks.2
The equations for turbulent flows were expanded by correction factors fR and fL
by means of which the effect of the roughness and the leakage flow through the impeller sidewall gaps can be estimated. The increase of the friction on rough disks compared with hydraulically smooth disks is determined as follows: the fric- tion coefficient cf is calculated for the rough surface and for the hydraulically smooth surface cf,o (ε = 0) from Eq. (1.33) with Re = u×r2/ν and ε/r2. This pro- duces the correction factor fR = cf/cf,o according to Eq. (T3.6.6).
Equation (T3.6.3) produces similar friction coefficients as Eqs. (T3.6.8 to T3.6.11), Fig. 3.10. However, Eq. (T3.6.3) has the advantage that the entire range from laminar to turbulent can be covered with a single equation without any un- steadiness at the transitions.
Numerous tests confirm that the axial casing clearance sax has only a very small effect on the disk friction; Eq. (T3.6.11) overestimates this effect. With sax tending to zero it yields too small and with sax tending to infinity too large friction coeffi- cients. In contrast, Eq. (T3.6.3) produces sensible values even with extreme axial clearances sax. With sax tending towards infinity Eq. (T3.6.3) supplies factors roughly corresponding to a free disk whose friction factors are given by kRR = 0.0365/Re0.2, [1.11]. Refer to Table 3.6 for further definition.
In [3.30] a method was developed to calculate disk friction losses by taking into account the fluid rotation in the impeller sidewall gap, Eqs. (T3.6.12 and T3.6.13). This equation covers the effects of the impeller sidewall gap geometry as well as the roughness of the casing and the impeller shrouds. It is useful par- ticularly when the casing and the impeller disk have different roughness values.
For a detailed discussion refer to chap. 9.1.
The various correlations from Table 3.6 are compared in Fig. 3.10. It can be seen that in particular Eqs. (T3.6.3) and (T3.6.12) are equivalent and suitably re- flect Eqs. (T3.6.8 to T3.6.11). The effect of the roughness is appropriately cov- ered by Eq. (T3.6.6).
1 This experimental finding follows from data in [1.11]; it cannot be deduced by inserting half the angular velocity in the equation given earlier.
2 Equations (T3.6.3) and (T3.6.5) are based on [3.6] and [3.7], Eqs. (T3.6.8) to (T3.6.11) on [3.8], Eq. (T3.6.12) on [3.30].
3.6 Calculation of secondary losses 87
1.E-04 1.E-03 1.E-02 1.E-01 1.E+00
1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 Re 1.E+07
kRR
Eq. (T3.6.11) Eq. (T3.6.8) Eq. (T3.6.9) Eq. (T3.6.10) Eq. (T3.6.3)
Eq. (T3.6.3) İ/R = 1.5E-4 [3.28] İ/R = 1.5E-4 Eq. (T3.6.3) İ/R = 4E-3 [3.28] İ/R = 4E-3 [3.9] İ/R= 1E-3
Eq. (T3.6.3) İ/R = 1E-3 Eq. (T3.6.12)
Fig. 3.10. Disk friction coefficients for smooth and rough surfaces calculated for sax/R = 0.08 (without leakage through the impeller sidewall gap); İ = roughness
Literature overviews concerning disk friction losses can be found in [3.29, 3.30 and 3.10]; the basic theory is treated in [1.11]. When comparing with other sources, the different definitions of the torque coefficients must be considered (frequently the coefficients are defined in a way that they comprise the torque or the friction power for both sides of the disk).
The disk friction losses in a pump depend on the following parameters:
1. Reynolds number: As with the flow over a plate or through a pipe, the friction coefficient drops with increasing Reynolds number. When pumping water, the flow in the impeller sidewall gap is generally turbulent. In the case of oils or other fluids with high viscosity the flow becomes laminar and the disk friction increases greatly (Chap. 13.1).
2. Roughness of the rotating disk: The roughness of the stationary or the rotat- ing surface increases the friction power, provided the roughness peaks protrude from the boundary layer thickness. In this context, grooves in circumferential direction such as generated by machining the shrouds on a lathe are less harm- ful than an unstructured roughness as typically found on a cast surface, [3.17].
According to the measurements in [3.17] the disk friction of a machined disk with εmax = 120 μm was practically identical in magnitude to that of a polished disk with ε≈ 0. Machined disks can therefore be calculated in general with ε≈ 0. For cast or sandblasted disks ε = εmax/ceq can be assumed. The equiva-
lence factor is ceq = 2.6 and εmax is the maximum roughness height, Chap. 3.10.
Disk friction losses are minimized if hydraulically smooth surface conditions are achieved. To this end, the permissible roughness that must not be exceeded can be evaluated according to Table D1.1 where w = cu≈ ½×ω×r must be em- ployed. The roughness problem is also discussed in Chap. 3.10.
3. Roughness of the casing wall: When rotor and casing have the same rough- ness the rotation of the fluid in the impeller sidewall gap is independent of the roughness, Eq. (T3.6.13). However, if εimpeller≠εcasing, the roughness has an ef- fect on the rotation and, consequently, the disk friction. With εimpeller > εcasing the rotation increases, while it is reduced if εimpeller < εcasing. At the same time the disk friction changes as per Eqs. (T3.6.12) and (T3.6.13). For a more detailed calculation refer to Chap. 9.1.
4. Axial sidewall gap sax: With a very small distance between impeller shroud