Question

Calculate the flux and the rate of removal of urea at steady-state in g/h from blood in a cellophane membrane dialyzer at 37oC. Membrane thickness is 0.025 mm thick and has an area of 2m2. The mass transfer coefficient on the blood side is estimated as kc1=1.25x10-5 m/s and on the aqueous side is 3.33x10-5m/s. The permeance of the membrane is 8.73x10-6 m/s. The concentration of urea in blood is 0.02 g urea/100mL and in the dialyzing fluid urea concentration is assumed to be 0.

Answer 1

Answer:

$J_{AS} = 3.22 gram/raw- m^2$

Rate of removal of urea = 6.44 gram/hr

Explanation:

Given Data:

we know flux formula is given as

$J_{AS} = \frac{C_1 - C_2}{\frac{1}{KC_1} +\frac{1}{\rho_m} +\frac{1}{KC_2}}$

$C_1 =0.02 g urea/100 ml$

      =200 gram/m^3,

$J_{AS} = \frac{200 - C_2}{\frac{1}{1.25\times 10^{-5}} +\frac{1}{8.73\times 10^{-6}} +\frac{1}{3.35\times 10^{-5}}}$

${AS} =\frac{200}{10^5(0.80+0.11+0.33)}$

$J_{AS} = 8.91\times 10^{-4} g/s -m^2$

$J_{AS} = 3.22 gram/raw- m^2$

Rate of removal of urea

$= Area \times Flux$

$= 2 \times 3.22$

Rate of removal of urea = 6.44 gram/hr