Beware of the Cream Filling

The velocity of blood in an artery is inversely proportional to the cross-sectional area of the artery.

In other words, $A_1 v_1=A_2 v_2$ .

If plaque closes off $80\%$ of the area of the artery, the area is reduced to $\frac{1}{5}$ of the original area.

So, $A_2=\frac{A_1}{5}$ .

And our new velocity, $v_2$ is equal to $5v_1=5\times 0.14$ $\text{m/s}$ $=0.7$ $\text{m/s}$ .

Now we are going to use something called Bernoulli's equation which tells us that the sum of the pressure, the gravitational potential energy per volume and the kinetic energy per volume of a fluid always remains the same [assuming we have no viscosity].

We can write the equation like this:

$p_1+\rho g h_1 + \frac{1}{2} \rho {v_1}^2=p_2+\rho g h_2 + \frac{1}{2} \rho {v_2}^2$

where $p$ is the pressure, $\rho$ is the density, $h$ is the height and $v$ is the velocity of the fluid.

This equation can be rewritten like this:

$p_1-p_2=\rho g (h_2-h_1) +\frac{1}{2}\rho ({v_2}^2- {v_1}^2)$ .

Notice that the artery is horizontal implying $h_2=h_1$ .

So our equation can be further simplified into:

$p_1-p_2= \frac{1}{2}\rho ({v_2}^2- {v_1}^2)$ .

We know all the values of the terms in the right hand side of this equation.

So, plugging them in gives us the pressure drop, $p_1-p_2= \boxed{246.96}$ $\text{Pascals}$ .

new cross-sectional area = 0.14/5 = 0.028m/s^2 , using equation of continuity, new velocity = 0.7m/s, Now, use bernoulli's equation neglecting height difference(since the artery is horizontal) to get the answer.

According to continuity equation:

$A_i v_i = A_f v_f$

And using Bernoulli principle (neglecting elevation difference):

$\frac{v_i^2}{2} + \frac{P_i}{\rho} =\frac{v_f^2}{2}+ \frac{P_f}{\rho}$

Since $\Delta P = P_f-P_i$ we have that:

$\frac{ \Delta P}{\rho} = \frac{v_f^2-v_i^2}{2}$

$\Delta P = \frac{\rho}{2} (v_f^2-v_i^2)$

$\Delta P = \frac{\rho}{2} (v_i^2\frac{A_i^2}{A_f^2} -v_i^2)$

$\Delta P = \frac{\rho}{2}v_i^2(\frac{A_i^2}{A_f^2}-1 )$

Substituting $\rho = 1050\:kg/m^3$ , $A_f = 0.2 A_i$ and $v_i = 0.14\:m/s$ :

$\Delta P = \frac{1050 }{2}0.14^2 24$

$\Delta P = 246.96\: Pa$

The flow rate must be constant, so we have $A_1v_1=A_2v_2$ , or $A_1(0.14)=0.2A_1v_2$ , getting that $v_2=0.7$ .

Next, we use Bernoulli's Equation, $p_1+\rho g y_1+\frac{1}{2} \rho v_1^2=p_2+\rho g y_2+\frac{1}{2} \rho v_2^2$ . Since $y_1=y_2$ , we have that $p_1+\frac{1}{2} \rho v_1^2=p_2+\frac{1}{2} \rho v_2^2$ , or that $|p_2-p_1|=\frac{1}{2} \rho (|v_2^2-v_1^2|)$ . Plugging in our values, we get that the answer is $\frac{1}{2}(1150)(|0.7^2-0.14^2| )= \fbox{246.96}$

First we wish to relate the speeds of blood in the two areas of the vessels. We use the continuity equation:

$A_{1}v_{1} = A_{2}v_{2}$

$A_{1} (0.14 m/s) = (1 - 0.8)A_{1}v_{2}$

$v_{2} = 0.70 m/s$

Now we use Bernoulli's Principle to find the pressure difference. We note that the artery is horizontal so the $\rho gh$ cancels giving us:

$P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} + \frac{1}{2} \rho v_{2}^{2}$

$| P_{1} - P_{2} | = \frac{1}{2} \rho (v_{1}^{2} - v_{2}^{2})$

$= \frac{1}{2} * (1000 kg/m^{3}) * ((0.7 m/s)^{2} - (0.14m/s)^{2})$

$= 246.96 Pa$

From equation of continuity, $a_1 v_1 = a_2 v_2$ , [ $a$ is area of CS] calculate $v_2$ .Use Bernoulli equation:

$\frac{v_1^2}{2} + \frac{P_1}{\rho} = \frac{v_2^2}{2} + \frac{P_2}{\rho}$

Evaluate $P_1 - P_2 = \rho(\frac{v_2^2 - v_1^2}{2})$ . Substitute to get answer.

Let a normal artery has an area, A and Let the dirty artery has an area A' . We know that A x v (the blood velocity) is equal to A' x v'(the blood velocity in the other side of artery). A x v = A' x v' ==> where A' is only 20 % left. This can make us to know that 5 x v = v'. Then now we use the Bernoulli equations : 0.5 x ρ x v² + P(v) = 0.5 x ρ x v'². ==> substitute 5 x v = v' to this sequence and we get : 0.5 x ρ x (v'² - v²) = P(v). Substituting the number and we get P(v) = 0.5 x 1050 x (25 x v² - v²) ==> P(v) = 0.5 x 1050 x (24 x 0.14²) = 246.96 Pa.

Let the area at the healthy portion of the artery be A then, the area of the weak portion is A/5. By equation of continuity,

                                                A1v1=A2v2

                                          i.e A*0.41=A/5 *v2

                                               v2=0.7 m/s.

By Bernoulli's theorem,

                                     P1 + d*g*v1^2 = P2 + 1/2*d*v2^2                                                                   

                                               (d=density=1050 kgm^-3)


                       Pressure drop = P1-P2 = 1/2*d*( v2^2-v1^2)

Plug in the values and get the answer!!!!!!!!!!

Using Bernoulli's equation:

$P_{healthy} - P_{unhealthy} = \frac {1}{2} \rho (v_{unhealthy}^2 - v_{healthy}^2)$

But $v_{unhealthy}$ is 5 times $v_{healthy}$ because the cross section at the unhealthy part is 5 times smaller.

Thus $\Delta P = 12 \rho v_{healthy}^2$

Because the flow of the blood is constant, the velocity of blood flow where the area of the artery decrease five times is $v_2=5v_1=0.7m/s$ .

According to Bernoulli's equation, the magnitude of the pressure drop is: $\Delta P=\frac{1}{2} \rho (v^2_2-v^2_1)=246.96 Pa$ .