Drag and Viscosity

[Monkeymaniac]

I was reading Examkracker Physics book, and I got stuck at "non-ideal fluids" section. The book talks about fluid flowing through a pipe and states, "Notice that if the radius of a pipe is reduced by a factor of 2, the fluid volume is reduced by a factor of 4, but the surface area is only cut in half. Thus, the more narrow the pipe, the greater the effect of drag."

Here, since the drag occurs because of the fluid colliding against the surface of the pipe, and since the amount (volume) of fluid flowing across the pipe at a given point has decreased more than the pipe's surface area at a given point has, shouldn't there be less drag? Am I missing something here? Thank you.

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[Airplay355]

To be honest I'm not sure how to answer your question because I'm still in my first semester of physics

But based on what you said I would think it would have to do with a volume:surface area ratio. I think, based on what you said, it seems the volume decreased more then surface area so you'd have more surface area/volume and thus more collisions=more drag.

This is because of the equation for the area of a circle, we will ignore the fact that it's a pipe because the area for a pipe just means you multiply by the height or length of the pipe, tube, blood vessel, whatever. The equation is A= pi x r^2. So if your pipe has a radius of 10 units then you have an area of 314 units. If you decrease the radisu of the pipe to 5 then your new area is only 78.5 or 4 times less. So cut the radius of the pipe in half and you have 4 times less the area (or volume).

If you look at the equation for circumference which will give us surface area it is C = pi x 2r or C = pi x d. So if we take our same circles the first one with r=10 has a C=62.8 but when we cut the radius to 5 we have C=31.4.

Again I'm not exactly positive this is the answer you need but it seems to me that since there is now more surface area/unit volume you will have more drag since there will be more area for collisions. SA to volume rations would be 62.8:314 for the circle with r=10 and 31.4:78.5.

I hope this helps

 

[TRN1983]

Area of a circle = pi*r^2 so if you have r = r/2 then the are of the circle will be pi(r/2)^2 or pi*(1/4)*r^2; i.e. the area is reduced by a factor of 4. Vol = A1 x L, so if A1 is reduced by 4 then Vol is reduced by 4.

Drag is a "resistance", R.

R=(resistivity)*L/A. Notice, area is included in there. If the area decreases by 4 (volume decreases by 4) then that means that the resistance (drag) increases by a factor of 4.