Rework the falling film problem for a position dependent viscosity with or mu = mu0(1-ax^2). (Hint: see solutions in Chapter 2 & 3 for falling film problem and also see Example 2.2.2) Here a is a constant and or mu0 is viscosity at the surface.

Transport phenomena. Below is example 2.2.2. Please help me. I will rate it. Thank you.

Rework the falling film problem for a position-dependent viscosity u = Moe-az/8, which arises when the film is nonisothermal, as in the condensation of a vapor on a wall. Here po is the vis- cosity at the surface of the film and a is a constant that describes how rapidly u decreases as x increases. Such a variation could arise in the flow of a condensate down a wall with a linear temperature gradient through the film. The development proceeds as before up to Eq. 2.2-13. Then substituting Newton's law with variable viscosity into Eq.2.2-13 gives Moe uxl8 do dx P8x cos B (2.2-26) This equation can be integrated, and using the boundary conditions in Eq.2.2-17 enables us to evaluate the integration constant. The velocity profile is then pg82 Vz cos B Mo [-(3-3)--(1-3) , (2.2-27) As a check we evaluate the velocity distribution for the constant-viscosity problem (that is, when a is zero). However, setting a = 0 gives - in the two expressions within parentheses. This difficulty can be overcome if we expand the two exponentials in Taylor series (see &C.2). as follows: po (v.)a=0 2882 cos B Mo ling 1 + a + a? + - 2! 3! Q? to +...) - ) 2) 1 + ax + 2x2 + 2!82 3!83 +... 8 TO une go 2882 cos B Mo lim [li 1 + Q+... 2 3 Q0 1 x?_1x? a +... (28 3 83 :) P882 cos B 1 2mo (2.2-28) which is in agreement with Eq. 2.2-18. From Eq. 2.2-27 it may be shown that the average velocity is COS B 2382 (0) [-(-2+2)-2) Mo (2.2-29 The reader may verify that this result simplifies to Eq. 2.2-20 when a goes to zero