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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+
+// Illustration 1.3
+// Page: 9
+
+printf('Illustration 1.3 - Page:9 \n\n');
+
+//*****Data*****//
+// Blood contains two parts a-blood cells b-plasma
+f_a = 45; // [percent of blood cells by volume]
+f_b = 55; // [percent of plasma by volume]
+r = 1200; // [Rate of blood which is pumped through artificial kidney, mL/minute]
+m_urine = 1540; // [mass of urine collected, g]
+x_u = 1.3; // [urea concentration, percent by weight]
+// Data for sample of blood plasma
+c_urea = 155.3; // [mg/dL]
+d = 1.0245; // [specfic gravity of plasma]
+//*****//
+
+printf('Illustration 1.3 (a) - Page:9 \n\n');
+// Solution (a)
+
+// Basis: 4 hours
+// Assuming that the rate of formation and decomposition of urea during the procedure is negligible and that no urea is removed by the patient’s kidneys
+// Therefore urea in “clean” blood = urea in “dirty” blood - urea in urine
+
+m_u = m_urine*(x_u/100); // [mass of urea in urine, g]
+// total volume of plasma that flows through the artificial kidney in 4 hours
+V_b = r*60*(f_b/100)*(1/100)*4; // [dL]
+// urea in dirty blood from given plasma concentration
+m_ud = c_urea*(1/1000)*V_b; // [g]
+// urea removal efficiency
+n = (m_u/m_ud)*100;
+printf("Urea removal efficiency is %f\n\n",n);
+
+printf('Illustration 1.3 (b) - Page:10 \n\n');
+// Solution (b)
+
+m_uc = m_ud-m_u; // [mass of urea on clean blood, g]
+m_p = d*100*V_b; // [Mass of plasma entering, g]
+m_rem = m_p-m_urine; // [Mass of plasma remaining, g]
+V_brem = m_rem/(d*100); // [Volume of plasma remaining, dL]
+c_y = (m_uc*1000)/V_brem; // [urea concentration in remaining plasma, mg/dL]
+printf("urea concentration in the plasma of the cleansed blood is %f mg/dL",c_y); \ No newline at end of file