Theory of Bulk Flow
Chapter 5 Chapter 5 Numerical Results
5.1 The Seal of Marquette and Childs
Using the parameter values recommended by Marquette and Childs, ns
=
nr=
0.079 and an exit loss of 1, the rotordynamic force for the seal in the tangential direction is predicted very well by the current model as shown by the solid line in figure 5.l.The normal force, however, exhibits a large offset from the experimental data. The calculated nondimensionalized pressure drop is 2.35, slightly below the experimental measurement of 2.5.
Now examine the effects of the adjustable parameters. As seen in figure 5.1 in- creasing ns increases the magnitude of normal forces by a small amount for all whirl ratios. It also changes the slope of the tangential forces. The pressure drop increases linearly in the small range of ns that is used. Figure 5.3 shows that increasing nr and ns together had effects similar to increasing ns alone. The pressure drop varies linearly with the coefficients; the increase is about twice as large as those from in- creasing ns alone. The increase in the magnitude of the normal forces is also larger, but still deviates from the experimental results by a significant offset. The change in the coefficients also changes the slope of calculated tangential forces to the whirl ratio more than that caused by changing ns alone. The coefficient values of ns = nr = 0.079 still seem to give the best results on the tangential forces.
Varying nr alone, however, shows a different behavior as seen in figure 5.2. The effect on the pressure drop is almost exactly the same as that for ns , but the forces are affected differently. The normal forces exhibit almost no effects from the varying of nr •
The slope of the tangential forces, the direct damping coefficient, shows a very small effect. The main effect is the intercept of the tangential forces. So far, by varying only the coefficients, the discrepancy of the normal forces from the experimental data remain unexplained.
Introducing an entrance loss helps to reduce some of the discrepancy in the normal forces as shown in figure 5.6. With a reasonable value of 0.1, the discrepancy in the normal force is reduced by half. The change in entrance loss has only a slight effect on the tangential forces and the pressure drop. Figure 5.6 also shows that increasing the entrance loss coefficient to an unreasonable value like 0.3 produces unreasonable results, so only an entrance loss value of less than 0.1 seems to be warranted.
Reducing the exit loss shows the most promise in reducing the discrepancy in the normal forces (figure 5.4). A lowering of the exit loss coefficient increases the magnitude of the normal forces significantly. Indeed the forces seem very sensitive to small changes in the exit condition. The magnitude of change in the direct damping coefficient is also not small; it is similar to that of varying the two shear stress coefficients together.
5.5, it has no effect on the normal forces and the slope of the tangential forces, but it does shift the tangential force curve up or down by a uniform offset. The effects seem to be similar to that of nr , though there is not an obvious answer as to why. In addition to the parameters mentioned, the exponents on the turbulent shear stress could also be varied. Their effects, however, are very similar to that from the coefficients, and changing them would not contribute to the understanding of the problem.
In conclusion, by adjusting the various parameters, we can fit the numerical results to the experimental data. First the exit loss coefficient is lowered to eliminate the offset discrepancy in the normal forces. Then ns adjusted to fine tune the normal forces to the experimental data, and nr is tuned to achieve the correct slope on the tangential forces. Finally, the inlet swirl velocity is adjusted to match the tangential forces. The values which best fit the data are ns
=
0.05, nr=
0.20, an exit loss coefficient of 0.4, and zero inlet swirl velocity. Notice that no inlet loss is used because the introduction of an inlet loss causes the added mass to become too small to be recovered through varying other parameters. The forces for the best parameters are plotted in figure 5.7. The pressure drop, however, is 3.14, significantly higher than experimental value of 2.51.seal
2.SI----,---;.---.---r---,---~
2.6 ... ..... . .
2.4 a..
<l
2.2
2
1.S~---~---L---~---L---~---~
0.05 0.06 0.07 0.09 0.1 0.11
0 -20 -40
- '- . - --
c: -60 u..
b"
-SO b" b"
b"
b"
b"
-100 b" b"
b" b" b" b" b" b"
-120
-O.S -0.6 -0.4 -0.2 0 0.2 0.4 0.6 O.S
win 200
- - ~:: ~:~~
150
- -
... ... ._.
_.- ns = 0.11... b" expr
100 u..-
50
~ l5.
0 L:. b"
L:.
b"
-50 -O.S -0.6 -0.4 -0.2 0 0.2 0.4 0.6 O.S
win
Figure 5.1: Calculated rotordynamic forces on the seal versus turbulent shear stress coefficient ns , with no inlet losses and an exit loss coefficient of 1. Pressure drops are 1.99, 2.35, and 2.72, resp.
a..
<l
2.6
2.4 .............. -........... ; .... .
2.2 .. ...... .
. ........ : ................ : .......... .
1.S 0.05 0.06 0.07
-20 ....
~
-40
-60
c:: /':,.
U.
/':,.
-SO /':,.
/':,.
/':,.
-100 /':,.
-120
-O.S -0.6 -0.4 -0.2
200
150
100
u.-
50
0
-50 -O.S -0.6 -0.4 -0.2
/':,.
O.OS n r
/':,.
0 win
o
win
0.09
-'- .
- --
'-'-'-
/':,. /':,. /':,. /':,.
0.2 0.4
- -
/':,.
0.2 0.4
0.1
/':,.
/':,.
0.11
/':,. /':,.
0.6 O.S
/':,.
nr = 0.05 nr = 0.079 nr = 0.11 expr
/':,.
0.6 O.S
Figure 5.2: Calculated rotordynamic forces on the seal versus the turbulent shear stress coefficient nr, with no inlet losses and an exit loss coefficient of 1. Pressure drops are 1.99, 2.35, and 2.72, resp.
LL
seal
3.2 r - - - , - - - , , - - - , - - - , - - - - . - - - , 3 ................. .
2.8 2.6 a.. <l 2.4
c:
2.2 2 1.8
1.6·:-::----~---L---L---L---'---.J
0.05 0.06 0.07 0.08 0.09 0.1 0.11
n = n
r s
-20,-~~II---,,_---,---_,----._----~---~----~
-40
-
-
- .-60
-'-
.6.
-80 .6. .6.
.6.
.6. .6.
-100 .6.
.6. .6.
.6. .6. .6. .6. .6. .6.
-120L---~---~---L---L---~---L---L---~
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8
win
200.---~-.---.---~---~---~---~---~----~
150
100
50
o
n = n = 0.05 n'=n'=0.07 n~ = n: = 0.11
.6. expr
_50L-______ L -_ _ _ _ _ _ L -_ _ _ _ _ _ L -_ _ _ _ _ _ L -_ _ _ _ _ _ L -_ _ _ _ _ _ L -_ _ _ _ _ _ L -____ ~
-0.8 -0.6 -0.4 -0.2 o
win
0.2 0.4 0.6 0.8
Figure 5.3: Calculated rotordynamic forces on the seal, with no inlet losses and an exit loss coefficient of 1. Pressure drops are 1.65, 2.35, and 3.10, resp.
-30 -40 -50 -60
LL c: -70 -80 -90 -100 -110
-120L---~----~----~----~
-1 -0.5 o
rolf!
0.5
140 120 100 80
LL- 60 40 20 0 -20 -40 -1
\ 1:::,:
\
'. ~.
. \
I:::,
. \
".1:::,\
".~.
'. \
"1::,.' . \
'1::,.\
-0.5 o
rolf!
. \
'. \
0.6 . -. 1.0
1::,. expr
0.5
Figure 5.4: Calculated rotor dynamic forces on the seal versus the exit loss coefficient, with no inlet losses and shear stress coefficients of 0.79. Pressure drops are 2.17,2.29, and 2.35, resp.
-20
-30 "-
"-
-40
,
,
-50 ~ ~
...
=>-
-60
c: -70 1::,.
LL
1::,.
-80 1::,.
1::,.
-90 I:::,
1::,.
-100 1::,.
1::,.
-110 1:::,1::,.
-120
-1 -0.5 0
rolf!
160 140 120 100 80
LL- 60 40 20 0
1::,. -20
1::,.1:::,1::,.1::,.
0.5 -40 -1 -0.5
. \ b.. '. \
4\
o
rolf!
- A=0.3
A=O.4
. - . A=0.5
1::,. expr
0.5
Figure 5.5: Calculated rotordynamic forces on the seal versus the inlet swirl velocity, with no inlet losses and an exit loss coefficient of 1. Pressure drops are 2.28, 2.31, and 2.35, resp.
-20r---.---.---.----~
-30 -40 -50 -60
u. c -70 -80 -90 -100 -110
-120L---~---L----~----~
-1 -0.5
o
00/0.
0.5
250 r---,---.----;::=::!====:::;l - 0 . 200
150
50
o
,
\
\
\
is. . \
is. ... \ is. . \
is. ". \ is. ". \
is. ' .. ', is. .... ,
Ii··.\
K .
Ii
t::.
0.1 - 0.3
t::. expr
t::.
t::.
t::.
-50L---~----~----~----~
-1 -0.5 o
00/0.
0.5
Figure 5.6: Calculated rotor dynamic forces on the seal versus the inlet loss coefficient, with an exit loss coefficient of 1. Pressure drops are 2.35, 2.41, and 2.53, resp.
-50 160
t::.
140 -60
120
-70 100
-80 80
c u.- 60
u. t::.
-90 40 t::.
t::.
-100 t::. 20 t::.
t::. t::.
t::. 0
-110 t::. t::.
t::.t::. t::.t::.t::. -20 t::.
-120 -40 t::.
-1 -0.5 0 0.5 -1 -0.5 0 0.5
00/0. 00/0.
Figure 5.7: Calculated rotordynamic forces on the seal with nr