Apr 9, 2010
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Pre-Medical
Either my brain is fried, I'm grossly misunderstanding these concepts....or, there is an error.

1) Chapter 6, Passage III, page 78
If a more viscous fluid is used with the same mass density, what would be observed?
Well, Poiseuille's eq. says that flow rate is proportional to delta P and inversely proportional to viscosity. If a more viscous solution was used, one would predict that the overall flow rate would decrease throughout the system, given that the pressures have not changed. So at point D, velocity should decrease. However the solutions state that the answer is C, and that the fluid heights in the columns are unchanged. How can this be? If the fluid velocity is slower at the outlet, it should be slower everywhere, due to this increased viscosity. No pressures have increased, so the pressure differential, the driving force of flow, is the same as before. Therefore, to me, the fluid heights should be higher in the individual pipes at the same time t after opening the valve. The pressure difference is the same, but there is a greater resistance to flow.

I had another question but already figured out my mistake, so this fluid flow one is the only one bothering me at the moment.
 
Dec 23, 2009
352
0
0
Australia or NYC or CT
Status
Pre-Medical
Either my brain is fried, I'm grossly misunderstanding these concepts....or, there is an error.

1) Chapter 6, Passage III, page 78
If a more viscous fluid is used with the same mass density, what would be observed?
Well, Poiseuille's eq. says that flow rate is proportional to delta P and inversely proportional to viscosity. If a more viscous solution was used, one would predict that the overall flow rate would decrease throughout the system, given that the pressures have not changed. So at point D, velocity should decrease. However the solutions state that the answer is C, and that the fluid heights in the columns are unchanged. How can this be? If the fluid velocity is slower at the outlet, it should be slower everywhere, due to this increased viscosity. No pressures have increased, so the pressure differential, the driving force of flow, is the same as before. Therefore, to me, the fluid heights should be higher in the individual pipes at the same time t after opening the valve. The pressure difference is the same, but there is a greater resistance to flow.

I had another question but already figured out my mistake, so this fluid flow one is the only one bothering me at the moment.
I was going to answer, but I'm not sure about the problem's set up. Is this a statics or a dynamics problem? If it is statics, i.e. columns are vertical and connected or something like that, they would remain unchanged because the pressure depends on the mass density*volume*gravity of the tube, and if the mass density is unchanged, then it would have the same height in a statics problem. if this is a dynamics problem i cant quite visualize what you mean
 
Apr 9, 2010
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Status
Pre-Medical
well, the mass densities are said to be the same. And yes, three vertical pipes are attached to a horizontal pipe which is receiving fluid from a large reservoir column all open to the atmosphere. The figure shows that at a short time after a valve is opened in the horizontal pipe (leading to a fluid outlet/waste), the fluid in the column closest the outlet has had a significant decrease in its height, the middle column has had a mild decrease in height, and the column closest to the reservoir column has had virtually no decrease in height. I understand this is because fluid pressure decreases throughout a pipe, thus the fluid pressure at the top of each column is the same (atmospheric), but the fluid pressure at the junction of each vertical column with the horizontal column decreases the closer you get to the outlet (P2 closest to outlet < P2 closest to reservoir). This is why heights are different at a short time after opening the valve.

So why, if a more viscous fluid was used instead of the original fluid, with the same mass density, would the fluid heights be the same? Flow has decreased (as velocity has decreased at the output but cross-sectional area in the pipes is the same) and the pressure in the vertical columns is the same (due to the same mass densities /same pressure), but viscosity has increased. The flow of fluid in all the pipes should be decreased. The only variable changing is viscosity in Poisuilles eq. I feel like the same trend would exist, but if we were examining the same point in time shortly after the valve is opened, the fluids should all be at a relatively higher point in their corresponding column, because they are flowing at a slower rate.

attached image is shortly after valve was opened, when closed all fluid heights were equal.
 

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Dec 23, 2009
352
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0
Australia or NYC or CT
Status
Pre-Medical
well, the mass densities are said to be the same. And yes, three vertical pipes are attached to a horizontal pipe which is receiving fluid from a large reservoir column all open to the atmosphere. The figure shows that at a short time after a valve is opened in the horizontal pipe (leading to a fluid outlet/waste), the fluid in the column closest the outlet has had a significant decrease in its height, the middle column has had a mild decrease in height, and the column closest to the reservoir column has had virtually no decrease in height. I understand this is because fluid pressure decreases throughout a pipe, thus the fluid pressure at the top of each column is the same (atmospheric), but the fluid pressure at the junction of each vertical column with the horizontal column decreases the closer you get to the outlet (P2 closest to outlet < P2 closest to reservoir). This is why heights are different at a short time after opening the valve.

So why, if a more viscous fluid was used instead of the original fluid, with the same mass density, would the fluid heights be the same? Flow has decreased (as velocity has decreased at the output but cross-sectional area in the pipes is the same) and the pressure in the vertical columns is the same (due to the same mass densities /same pressure), but viscosity has increased. The flow of fluid in all the pipes should be decreased. The only variable changing is viscosity in Poisuilles eq. I feel like the same trend would exist, but if we were examining the same point in time shortly after the valve is opened, the fluids should all be at a relatively higher point in their corresponding column, because they are flowing at a slower rate.

attached image is shortly after valve was opened, when closed all fluid heights were equal.
So, I understand why you might think that viscosity might change the height of the columns, but one must recognize that changing the viscosity does not change the pressures in the tube. Thus, since those heights are totally dependent on pressure, the heights can not change, since indeed higher viscosity causes lower flow rate, but flow rate depends on pressure and viscosity, while those column heights will depend only the pressure, which is independent of the viscosity.

I think you can understand this by using the same reasoning you used to understand the initial change in heights.
 
Apr 9, 2010
4
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Status
Pre-Medical
I get that the pressure in a static column is rgV (r = density). But what you are saying is that, in essence, flow rate has not changed at all in the entire system.

The amount of fluid that has descended the columns = the amount of fluid that has left the system. However, the fluid velocity is acknowledged as having decreased. Therefore, flow rate has decreased and less fluid has left the system over the same period of time. If you say that fluid heights are equal at the same point in time, then i feel like you are violating the previous statement, as well Poiseuille's eq:
Q (flow rate) = &#916;P &#960; r^4 / 8 &#951; l.

Pressure is the impetus for the fluid flow, but factors such as pipe radius, viscosity, pipe length affect its overall rate. The higher viscosity fluid experiences a slower flow rate, and therefore a slower change in height (volume) over time.

Qwater = (rVg)/L * (pi)r^4/(8*visc1)
Qviscous = (rVg)/L * (pi)r^4/(8*visc2)
visc 1 < visc 2 Qwater > Qviscous

How does this not apply to the vertical pipes as well?
 
Dec 23, 2009
352
0
0
Australia or NYC or CT
Status
Pre-Medical
I get that the pressure in a static column is rgV (r = density). But what you are saying is that, in essence, flow rate has not changed at all in the entire system.

The amount of fluid that has descended the columns = the amount of fluid that has left the system. However, the fluid velocity is acknowledged as having decreased. Therefore, flow rate has decreased and less fluid has left the system over the same period of time. If you say that fluid heights are equal at the same point in time, then i feel like you are violating the previous statement, as well Poiseuille's eq:
Q (flow rate) = &#916;P &#960; r^4 / 8 &#951; l.

Pressure is the impetus for the fluid flow, but factors such as pipe radius, viscosity, pipe length affect its overall rate. The higher viscosity fluid experiences a slower flow rate, and therefore a slower change in height (volume) over time.

Qwater = (rVg)/L * (pi)r^4/(8*visc1)
Qviscous = (rVg)/L * (pi)r^4/(8*visc2)
visc 1 < visc 2 Qwater > Qviscous

How does this not apply to the vertical pipes as well?

hmm, to be be honest, I'm not sure how to answer this question then because I hadn't thought about that.

does anyone else know?