The model is quite short relative to its diameter. I am wondering if you have tried an expansion vessel with a larger length. My intuition is that the prescribed inflow (125 ml/s) has a sufficient momentum to reach outlet even though outlet pressure is higher by 150 dyn/cm^2 which is only 0.1 mmHg. If you increase the length say ~10D, do you have the same result?
Thanks
Problem simulating simple geometries (negative pressures)
-
Weiguang Yang
- Posts: 110
- Joined: Mon Apr 07, 2008 2:17 pm
-
Florian Gartner
- Posts: 12
- Joined: Tue Dec 12, 2023 7:13 am
Re: Problem simulating simple geometries (negative pressures)
Hi Weiguang,
thank you very much for your reply! You are absolutely right! I have come to a similar conclusion. I think the negative pressure values calculated by Simvascular might indeed be correct since the average flow velocity at the outlet is smaller than the inlet velocity due to the tube expansion. According to Bernoulli's Equation, lower velocity means higher pressure; therefore, a negative pressure gradient in the flow direction is reasonable.
To make it more quantitative, we may assume the stationary flow profile is approximately parabolic everywhere in the tube. The maximum velocity in the center of the tube is then given by V_max = 2Q/(pi*R^2), where R is the tube radius and Q = 125 ml/s the flow. According to the Bernoulli Equation (neglecting viscosity at first), we get a pressure difference along the center streamline, p1 - p2 = rho/2 * (V2^2 - V1^2), where index 1 denotes the inlet and 2 the outlet. Using the above formula for the center velocity and R1 = 1cm and R2= 1.2cm, we obtain a negative pressure difference p1-p2 = -1738 dyn/cm^2.
Counteracting this pressure increment due to the velocity decrease, one must also consider the pressure drop due to viscous friction ("extended Bernoulli Equation"). To estimate the viscous pressure drop, we assume that, for an infinitesimal segment of the tube, the pressure drop is given by the Hagen-Poiseuille formula: dp/dl = 8*eta*Q/(pi*R(l)^4). Integrating l from 0 to 5cm with R(l) interpolating linearly from R1 to R2, we obtain a pressure difference p1 - p2 = 44.7 dyn/cm^2. So, the pressure gain due to the velocity difference by far outweighs the pressure loss due to friction, and adding both contributions, we would obtain a negative overall pressure difference. Of course, this calculation is not very accurate since the flow profile is not precisely parabolic, and one would have to average over all streamlines (not only the center streamline). However, it gives an idea about the relative strength of both contributions, showing that velocity changes have a much more substantial impact on pressure gradients than friction in the laminar regime.
So, it makes absolute sense to obtain negative pressure differences, as the fluid loses kinetic energy due to the velocity drop in the expanding tube. At first glance, it's just a bit unintuitive that the flow goes in the direction opposing the pressure gradient. I think this should resolve the issue. Thank you very much for your help!
Cheers, Flo
thank you very much for your reply! You are absolutely right! I have come to a similar conclusion. I think the negative pressure values calculated by Simvascular might indeed be correct since the average flow velocity at the outlet is smaller than the inlet velocity due to the tube expansion. According to Bernoulli's Equation, lower velocity means higher pressure; therefore, a negative pressure gradient in the flow direction is reasonable.
To make it more quantitative, we may assume the stationary flow profile is approximately parabolic everywhere in the tube. The maximum velocity in the center of the tube is then given by V_max = 2Q/(pi*R^2), where R is the tube radius and Q = 125 ml/s the flow. According to the Bernoulli Equation (neglecting viscosity at first), we get a pressure difference along the center streamline, p1 - p2 = rho/2 * (V2^2 - V1^2), where index 1 denotes the inlet and 2 the outlet. Using the above formula for the center velocity and R1 = 1cm and R2= 1.2cm, we obtain a negative pressure difference p1-p2 = -1738 dyn/cm^2.
Counteracting this pressure increment due to the velocity decrease, one must also consider the pressure drop due to viscous friction ("extended Bernoulli Equation"). To estimate the viscous pressure drop, we assume that, for an infinitesimal segment of the tube, the pressure drop is given by the Hagen-Poiseuille formula: dp/dl = 8*eta*Q/(pi*R(l)^4). Integrating l from 0 to 5cm with R(l) interpolating linearly from R1 to R2, we obtain a pressure difference p1 - p2 = 44.7 dyn/cm^2. So, the pressure gain due to the velocity difference by far outweighs the pressure loss due to friction, and adding both contributions, we would obtain a negative overall pressure difference. Of course, this calculation is not very accurate since the flow profile is not precisely parabolic, and one would have to average over all streamlines (not only the center streamline). However, it gives an idea about the relative strength of both contributions, showing that velocity changes have a much more substantial impact on pressure gradients than friction in the laminar regime.
So, it makes absolute sense to obtain negative pressure differences, as the fluid loses kinetic energy due to the velocity drop in the expanding tube. At first glance, it's just a bit unintuitive that the flow goes in the direction opposing the pressure gradient. I think this should resolve the issue. Thank you very much for your help!
Cheers, Flo
-
Weiguang Yang
- Posts: 110
- Joined: Mon Apr 07, 2008 2:17 pm
Re: Problem simulating simple geometries (negative pressures)
Yes, I also realized that it could be explained by the conservation of energy. For a short diverging pipe, the energy loss can be neglected. The flow slows down when it enters a region with a larger cross-sectional area. The kinetic energy is converted into a potential. So the dynamic pressure is reduced while the static pressure increases. The larger R2 in your model results in lower velocity and higher pressure.