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Fluid with viscosity as a function of Pressure
Posted 3 set 2015, 03:20 GMT-4 Fluid & Heat Version 4.4, Version 5.1 4 Replies
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my function of viscosity is
mu=((0.3839*(p)/spf.sr)*(1-exp(-m*spf.sr))+((0.241*0.0001*(p))/(spf.sr*0.001+0.3*((p)/2600)^0.5)))
where p=pressure, m=parameter and spf.sr=shear rate
Gives an error of non-integral power of negative number which is of Pressure in this case.
I have also tried with (p+eps) but it won't work.
So how to solve it?
mu=((0.3839*(p)/spf.sr)*(1-exp(-m*spf.sr))+((0.241*0.0001*(p))/(spf.sr*0.001+0.3*((p)/2600)^0.5)))
where p=pressure, m=parameter and spf.sr=shear rate
Gives an error of non-integral power of negative number which is of Pressure in this case.
I have also tried with (p+eps) but it won't work.
So how to solve it?
4 Replies Last Post 4 set 2015, 18:30 GMT-4
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Posted:
9 years ago
Hi
You have not told us which pressure "p" your formula is valid, is it for the pA absolute pressure or the Gauge pressure around "0", and if the latter which Gauge definition ?
I suspect that is the issue you are facing (but it might be something else), if you check the materials dependent on the pressure they are mostly defined w.r.t. the absolute pressure pA in the material section, to allow COMSOL to solve for the Gauge pressure "p" which is situated close to "0" to minimise numerical scaling error.
--
Good luck
Ivar
You have not told us which pressure "p" your formula is valid, is it for the pA absolute pressure or the Gauge pressure around "0", and if the latter which Gauge definition ?
I suspect that is the issue you are facing (but it might be something else), if you check the materials dependent on the pressure they are mostly defined w.r.t. the absolute pressure pA in the material section, to allow COMSOL to solve for the Gauge pressure "p" which is situated close to "0" to minimise numerical scaling error.
--
Good luck
Ivar
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Posted:
9 years ago
Ivar thanks for your suggestion
p used is gauge pressure
so now what can I do?
p used is gauge pressure
so now what can I do?
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Posted:
9 years ago
Hi
well then if it is really "p" that is becoming negative (something often normal if its "small") and your formula does not like a negative "p" then you should check your BC to see if there is not something in the starting point that might induce a negative pressure "p".
Or move your initial values to a clearly positive "p", or make your formula -p compatible ...
I do not see many other ways
--
Good luck
Ivar
well then if it is really "p" that is becoming negative (something often normal if its "small") and your formula does not like a negative "p" then you should check your BC to see if there is not something in the starting point that might induce a negative pressure "p".
Or move your initial values to a clearly positive "p", or make your formula -p compatible ...
I do not see many other ways
--
Good luck
Ivar
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Posted:
9 years ago
Hi Mahesh,
make sure that your initial value of p is not set to zero. By numerical fluctuations it might always become negative in this case. If this does not help, abs(p). If this does not help either, provide the model. NOte that also 0.3839*(p)/spf.sr is critical, as for the initial model spf.sr might be zero.
Best regards,
Sven
make sure that your initial value of p is not set to zero. By numerical fluctuations it might always become negative in this case. If this does not help, abs(p). If this does not help either, provide the model. NOte that also 0.3839*(p)/spf.sr is critical, as for the initial model spf.sr might be zero.
Best regards,
Sven