Appendix
27
14. Appendix
14.1 Example: Evaluation
(sugar solution of 40 %)
Density of ball 2:
2.2 (g/cm
3
)
Density of the solution:
1.18 (g/cm
3
)
Ball constant K :
0.09 (mPa
s
cm
3
/g
s)
Falling time:
61 s
Measuring temperature: 20.0
C
The absolute viscosity is ...
20
C
= 0,09
(2,2 -- 1,18)
61 = 5,60 (mPa
s)
In most cases the densities of the test liquids are known. The
evaluation may be simplified by introducing a factor which
includes the densities. In our example of the sugar solution
the exact factor is ...
(
1
--
2
)
K = 0,1098 (mPa
s / s)
14.2 Example: Calibration of ball 1
K = ball constant to be found
E
= 4.63 [mPa
s] viscosity of the standard liquid
1
= 2.217 [g/cm
3
] density of the ball
2
= 0.81 [g/cm
3
] density of the standard liquid
t = 417.7 [s]
average of the falling time
For the calculation of the ball constant K the following for-
mula applies:
K
=
η
E
(
Ã
1
−
Ã
2
)
⋅
t
=
4.63
(2.217
−
0.81)
⋅
417.7
=
0.00788
The determination of the constants of the other balls should
be done in the same way.
