I am simulating a single phase flow problem in which the numerical pressure drop obtain is less than theoretical one. So I think this is due to the decrease in viscosity of fluid.
Did your theoretical pressure drop discuss the various of viscosity with temperature?
No.
I have not considered the variation of viscosity with temperature in my theoretical pressure drop calculation.
With this consideration (I mean variation of viscosity with temp),Now things just seem to fall into place. Thanx dear Hulin Huang.
yes, the pressure loss will be less as the viscosity decrees.
however, you need to keep the velocity same. if the increase of temperature will cause increase in volume, the pressure drop will increase. in this case, you need to consider the combined effect of all variables affecting the pressure drop calculation.
Hello everybody,
in my opinion the Reynolds number, which has not been mentioned so far, is the critical quantity. Its influence on the non-dimensional pressure loss is extremely helpful for further discussions.
For some clarifications i would suggest the following: In a reference flow situation a viscosity mu_1 prevails. This viscosity now is decreased (possibly due to heating) to mu_2=x*mu_1 with 0100...1000, elbow, bend, orifice) the pressure drop is not affected by the viscosity change. This, however, is surprising at first glance only:
Delta p_2= K*rho*u_2^2=K*rho*u_1^2=Delta p_1
If K now is affected by the Reynolds number, the situation is more complex. For a developed flow K=C/Re is assumed, as described above.
Delta p_2= C/Re_2*rho*u_2^2 K*rho*u_2^2 = C/(Re 1/x) * rho*u_1^2 = x*Delta p_1
In this case, the pressure difference is decreased linearly with x.
In the more general case K=C_1/Re + C_2 special consideration is necessary.
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