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diff --git a/530/CH9/EX9.4.b/example_9_4b.sce b/530/CH9/EX9.4.b/example_9_4b.sce
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+++ b/530/CH9/EX9.4.b/example_9_4b.sce
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+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 9
+// Mass Transfer
+
+
+// Example 9.4(b)
+// Page 356
+printf("Example 9.4(b), Page 356 \n \n");
+
+L = 1 ; // [m]
+D = 0.005 ; // [m]
+Pa1 = 1 ; // [atm]
+Pa2 = 0 ; 
+R = 8314 ;
+T = 298 ; // [K]
+
+// Since the tank is large and the pressure and temperature at the two ends of the same tube are same, we are assuming Equimolal counter diffusion
+// From Table 9.1
+Dab = 2.80*10^-5 ; // [m^2/s]
+// Substituing in eqn 9.4.12
+Na = -[Dab/(R*T)*(Pa2-Pa1)*(1.014*10^5)/L]*(%pi*(D/2)^2);
+
+// Since equimolal counter diffusion is taking place
+Nb = - Na ;
+// therefore rate at which air enters the tank
+R_air = abs(Nb)*29 ; // [kg/s]
+
+printf("Rate at which air enters the tank = %e kg/s",R_air);
+\ No newline at end of file