Question

ore than 80 ml, and that the total pressure drop across the fibers should be no more than 105 dyne/cm2 at a total flow rate of 50 ml/s. If the blood viscosity is 3.5 cP and the density of blood is 1.05 g/cm3 , how many fibers should be used, and of what diameter should they be, so as to meet the design conditions

Answer 1

Complete Question

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Answer:

The number of fibers is $N \approx 540$

Explanation:

From the question we are told that

  The  length of the fibers is $L = 30 \ cm = 300 \ mm$

  The hold up  volume is  $V = 80 \ mL =0.08 L = 0.08 *10^{6}\ mm^3$

  The total pressure that must not be exceeded   $P_t = 10^5 \ dync/cm^2$

  The total flow rate is  $\r m = 50 ml/s = 0.05 \ l/ s = 0.05 *10^{6} \ mm^3 /s$

   The blood viscosity is  $\eta = 3.5\ cP = 0.0035\ kg \cdot s^{-1} \cdot m^{-1} = 0.0035 *10 = 0.035 \ kg \cdot s^{-1 } \cdot cm^{-1}$

   The density of the blood is  $\rho = 1.05 \ g/cm^3$

     

Generally the volume of blood a single fiber can contain at a time is mathematically represented as

      $V = \frac{\pi}{4} * d^2 * L$

Here diameter of the fiber

So

     $d = \sqrt{\frac{4 V}{\pi * L}}$

=>  $d = \sqrt{\frac{4 * 0.08 *10^{6} }{3.142 * 300}}$      

=>  $d = 18.42 \ mm$

Converting to cm

=>  $d = 18.42 * 10 = 1.842 \ cm$

Generally the pressure in the fiber is mathematically represented as

    $P = \frac{32 * \eta * \r v * L }{ d^2 }$

Generally the velocity of the blood flow which is mathematically represented as

    $\r v = \frac{ 4 *0.05 *10^{6} }{3.142 * 18.42 }$

=>  $\r v = 187.62 \ mm/s$

Converting to cm

=>  $\r v = \frac{187.62 }{10} = 18.762 \ cm/s$

So

     $P = \frac{32 * 0.035 * 18.76 * 30 }{ 1.842^2 }$

=> $P = 185.77 \ dync / cm^2$

Generally the number of  fibers required is mathematically represented as

     $N = \frac{P_t}{P}$

=>  $N = \frac{10^5}{185.77}$

=>  $N = 538.3$

=>  $N \approx 540$