Fluid Mechanics and Thermodynamics of Turbomachinery, 5e

352 Fluid Mechanics, Thermodynamics of Turbomachinery
c = c ( -aF).
x2 x1
1
From this we see that at the tips of the blades c x2 = c x1, because F is zero at that radius.
N.B. It was explained earlier that the limit of application of the theory occurs when
a Æ 0.5, (i.e. cx2 = cx1(1 - 2a), and, as the earlier calculations have shown, a is usually
greatest towards the blade tip. However, with the application of the tip correction factor
F, the limit state becomes aF = 0.5. As F progressively reduces to zero as the blade tip
is approached, the operational result gives, in effect, some additional leeway in the convergence
of the iterative procedure discussed earlier.
Performance calculations with tip correction included
In accordance with the previous approximation (to reduce the amount of work
needed), e is ascribed the value zero, simplifying the above equations for determining
a and a¢ to
( )= ( )
2
a 1 - a l cosf Fsin
f
a¢ ( 1 + a¢ )= l ( Fcosf)
(10.47a)
(10.48a)
When using the BEM method an extra step is required in Table 10.1 between steps 1
and 2 in order to calculate F, and it is necessary to calculate a new value of CL for each
iteration which, consequently, changes the value of the blade loading coefficient l as
the calculation progresses.
Example 10.9. This example repeats the calculations of Example 10.7 using the
same blade specification (i.e. the pitch angle b = b(r)) but now it includes the Prandtl
correction factor. The results of the iterations to determine a, a¢, j and CL and used as
data for the summations are shown in Table 10.5. The details of the calculation for one
mid-ordinate radius (r/R = 0.95) are shown first to clarify the process.
Solution. At r/R = 0.95, F = 0.522, using eqns. (10.44b) and (10.43). Thus, with Z
= 3, l = 1.0, then
F l = 63. 32 CL
In the BEM method we start with a = a¢ =0 so, initially, tanj = (R/r)/J = (1/0.95)/5 =
0.2105. Thus, j = 11.89deg and CL = (j - b)/10 = (11.89 - 1.59)/10 = 1.03. Hence,
F/l = 60.5. With eqns. (10.47a) and (10.48a) we compute a = 0.2759 and a¢ =0.0172.
The next cycle of iteration gives j = 8.522, CL = 0.693, F/l = 89.9, a = 0.3338 and
a¢ =0.0114. Continuing the series of iterations we finally obtain
a= 0. 351, a¢ = 0. 010, j = 7. 705, CL=
0. 6115.
For the elements of force,
1 DX= ZlRc { ( -a)
} C D(
r R)
2
2
r 1 sinj cosj
2
Where, in Table 10.5, Var. 1 = {(1 - a)/sinj} 2 cosjCLD(r/R)
SVar. 1 = 6. 3416.
x1L As in Example 10.6, 1 – 2 ZlRc 2 x1 = 1518.8, then