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johnwandering

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Q=Q
So as the capillaries have a larger total radius, the blood velocity slows down in the capillaries.

But according to Bernoulli's equation

k = proprtional to P+v^2

wouldn't pressure increase as velocity decreases??
But the capillaries have the lowest pressure of the blood vessels......


Also, what does k mean in Bernoulli's equation?
 

milski

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Bernoulli is applicable only if the flow stays the same. With the capillaries the flow is split in separate flows - you cannot apply Bernoulli's equation directly to that.

k in Bernoulli's equation is just a constant - it gives how one of the velocity of the flow and the pressure on the walls will change when you change the other. k will change for different situations, in other words it does depend on Q, ρ, etc.
 

EnginrTheFuture

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Think of the blood as electric current, it is a viscous fluid:

V = IR
is analogous to
P = Q * Resistance

Hypothetical:

So as your going through a circuit. You go through a slightly resistive wire the whole time... and at certain points you also hit resistor 1 2 then 3. There is a drop in voltage equal to V=IR at each resistor right? When you go through a wire, before an after these resistors, there is also a small loss of voltage to the wire itself. The whole LENGTH of the thin wire you are slowly losing voltage to the wire itself whether you hit the "resistors" or not. This loss is dependent on the resistivity of the wire which depends on its cross sectional area and a few other factors multiplied by the length of the this stuff your travel through. The "resistors" you pass can be viewed as specially coiled wires of specific length and radius... packaged into a "black box" that drops a specific (usually relatively high compared to circuit's wire) "resistance". So really the resistors we spoke of before are no different than the wire we are traversing along the way, but they are packaged into a cool black box.

So how does this relate to blood?

As you travel through the body's "wires", the blood loses its "voltage" (pressure) every bit of length you travel just like the charges passing through a circuit lose voltage. The voltage after a specific resistor is the original voltage minus that of the voltage loss over the resistor right? You can never get more voltage in the middle of a circuit unless you have a battery (an electron "pump") which is equivalent to a heart. The voltage of a circuit is analogous to the pressure for a circ. system.

The pressure when blood exits the heart is the "battery voltage". The voltage you lose over the wire is the pressure loss over the length of the vein/artery or even a system like all the capillaries (a resistor of sorts if you collect them and consider the system as a block box). The voltage along the journey of a circuit always goes from high to low as you go... so does the pressure in the circulatory system, THAT IS WHY THE PRESSURE DROPS. You start at the heart with 140mmhg (10Volts) let's say, you go through 10 meters of artery (the wire) then pass through something like the capillary network (a resistor which is really just a collection of wires)... plot the voltage in your head vs distance traveled. This a conceptual tool that can make blood flow seem more intuitive. Just remember that the resistance of the circulatory system obeys rules similar but not mathematically identical to the resistivity of wire.


Hope the analogy helps
 
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johnwandering

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Yes conceptually, pressure and E are obviously related.

But im saying that the lower pressure is not something that can be determined through the mcat fuild formulas.
 

EnginrTheFuture

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Yes conceptually, pressure and E are obviously related.

But im saying that the lower pressure is not something that can be determined through the mcat fuild formulas.

This exact pressure at a distance L would be impossible to solve for given MCAT formulas and conditions. You would need calculus and most likely mathematica if you want to sleep at night. BUT the idea that the pressure will drop over the capillaries can be proven easily through MCAT formulas. The amount, not likely solvable, but the direction of pressure change... totally can be reasoned out like so:

deltaP = - Q x (8 nu L) / (pi r^4)

The change in pressure of a viscous fluid is always negative as you progress through a pipe (in this case the capillaries or veins) due to this formula. Why? Because your L is getting bigger and bigger as you go further through the system. Whether or not your r increases or decreases on the journey only changes how QUICKLY your pressure DROPS as your L gets bigger but the pressure will assuredly drop because L gets larger and the rest of that formula (whether r goes to 0 or infinite) will always be positive.

delta p = - (constant) L

imagine it looks like this.... where the constant isn't exactly constant but it is always positive despite what r is.

The integral of the right side of this equation with respect to small changes in L, with r as a function of L can explain this more clearly but screw it, that's beyond the scope. What matters is that pressure decreases as you travel further away from the heart and towards the venous system and the materialistic repercussions of this can be seen via MCAT formulas.
 
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