Respuesta :
Answer:
26.9 Pa
Explanation:
We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:
[tex]A_1 v_1 = A_2 v_2[/tex] (1)
where
[tex]A_1[/tex] is the cross-sectional area of the 1st section of the pipe
[tex]A_2[/tex] is the cross-sectional area of the 2nd section of the pipe
[tex]v_1[/tex] is the velocity of the 1st section of the pipe
[tex]v_2[/tex] is the velocity of the 2nd section of the pipe
In this problem we have:
[tex]v_1=0.15 m/s[/tex] is the velocity of blood in the 1st section
The diameter of the 2nd section is 74% of that of the 1st section, so
[tex]d_2=0.74d_1[/tex]
The cross-sectional area is proportional to the square of the diameter, so:
[tex]A_2=(0.74)^2 A_1=0.548 A_1[/tex]
And solving eq.(1) for v2, we find the final velocity:
[tex]v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s[/tex]
Now we can use Bernoulli's equation to find the pressure drop:
[tex]p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2[/tex]
where
[tex]\rho=1025 kg/m^3[/tex] is the blood density
[tex]p_1,p_2[/tex] are the initial and final pressure
So the pressure drop is:
[tex]p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa[/tex]
Following are the equation of continuation expression:
[tex]\to A_1 V_1= A_2 V_2 \\\\[/tex]
[tex]A_1\ and \ V_1[/tex] are the initial area and velocity, respectively, and [tex]A_2\ and \ V_2[/tex] are the end area and velocity.
[tex]\to A_1 V_1= A_2 V_2 \\\\\to v_2= \frac{A_1}{A_2} \times v_1\\\\[/tex]
[tex]= \frac{\frac{\pi}{4} d_1^2}{ \frac{\pi}{4} d_2^2 } \times v_1\\\\= (\frac{d_1}{d_2})^2 \times v_1\\\\[/tex]
[tex]\to d_2= 0.74\ d_1 \\\\[/tex]
Therefore,
[tex]\to v_2 = (\frac{d_1}{d_2})^2 \times v_1\\\\[/tex]
[tex]= (\frac{d_1}{0.74 \ d_1})^2 \times 0.15 \ \frac{m}{s}\\\\= 0.2739 \ \frac{m}{s}\\\\[/tex]
Using the Bernoulli equation at the height of constant.
[tex]\to P_1+ \frac{\rho v_1^2}{2} + \rho gh_1 = P_2 + \frac{\rho v_2^2}{2} + \rho gh_2\\\\\to P_1-P_2 =\frac{\rho}{2} (v_2^2-v_1^2)\\\\\to \Delta P =\frac{\rho}{2} (v_2^2-v_1^2)\\[/tex]
Let the Human bool is the average density then:
[tex]\to \rho =1060 \ \frac{kg}{m^3}\\\\[/tex]
[tex]\to \Delta P =\frac{\rho}{2} (v_2^2-v_1^2)\\\\[/tex]
[tex]= \frac{(1060 \frac{kg}{m^3})}{2} ((0.2739 \frac{m}{s})^2 -(0.15 \frac{m}{s})^2) \\\\= \frac{(1060 \frac{kg}{m^3})}{2} (0.07502121- 0.0225) (\frac{m}{s})^2) \\= \frac{(1060 \frac{kg}{m^3})}{2} (0.05252121) (\frac{m}{s})^2) \\\\= 27.836 \ Pa\\\\[/tex]
Learn more about the blood viscosity:
brainly.com/question/8321796
