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During peak systole, the heart delivers to the aorta a blood flow that has a velocity of 100cm/sec at a pressure of 120mmHg. The aortic root has a mean diameter of 25mm. Determine the force (Rz)acting on the aortic arch if the conditions at the outlet are a pressure of 110mmHg and a diameter of 21mm. The density of blood is 1050 kg/m3. Assume that aorta is rigid non-deformable and blood is incompressible and steady state. Ignore the weight of blood vessel and the weight of blood inside the blood vessel (i.e. body force is zero).

Respuesta :

Solution :

Given :

Velocity, [tex]$V_1 =100$[/tex] cm/sec

Pressure,  [tex]$P_1 = 120 $[/tex] mm Hg

Then, [tex]$P_1 = \rho_1 g h$[/tex]

[tex]$P_1 = 0.120 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 16.0092 kPa

[tex]$P_2 = 110 $[/tex] mm Hg

[tex]$P_2 = \rho_2 g h$[/tex]

    [tex]$= 0.110 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 14.675 kPa

Then blood is incompressible,

[tex]$A_1v_1=A_2v_2$[/tex]

[tex]$\frac{\pi}{4}(25)^2\times 100=\frac{\pi}{4}(21)^2\times v_2$[/tex]

[tex]$v_2=141.72 \ cm/s$[/tex]

Then the linear momentum conservation fluid :

(Blood ) in y - direction

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$m_1=m_2=P_1A_1v_1$[/tex]

              [tex]$=1.50 \times \frac{\pi}{4}\times (0.025)^2 \times 1.00$[/tex]

              = 0.515 kg/ sec

Then the linear conservation of momentum of blood in y direction.

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$16.0092 \times 1000 \times \frac{\pi}{4} \times (0.025)^2+14.675 \times 1000\times \frac{\pi}{4}\times (0.021)^2$[/tex]

[tex]$-F_y=0.515(-1.4172-1)$[/tex]

7.858+5.0828- Fy = 0.515(-2.4172)

Fy = 14.1856 N

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