Chapter 3 Unidirectional transport - Chemical Engineering ...

3.3. EFFECT OF PRESSURE ON MOMENTUM TRANSPORT 35We can make use of one of the boundary conditions, that the stress is equalat the interface, to relate C (I) and C (II) .Atx = 0τ xz(I) = τ xz(II)C (I) = C (II) (3.140)Using Newton’s law of viscosity to relate the stress to the strain rate, we get( )(I) dv(I) z p0 − p Lµ = − x + C (I)dxL( )(II) dv(II) zp0 − p Lµ = − x + C (II) (3.141)dxLThis can be integrted to give( )v z (I) p0 − p L= −v (II)z = −2µ (I) L( )p0 − p L2µ (II) Lx 2 + C(I) x+ C (I)µ (I) 2x 2 + C(I) x+ C (II)µ (I) 2(3.142)There are three constants in the above equations, which are determined usingthe three available boundary conditions.Atx = 0Atx = −bAtx = bv z(I)v (I)= v (II)zz = 0v z (II) = 0 (3.143)These boundary conditions provide three relationships between the constantsC (I) , C (I)2 and C (II)2 .C (I)These can be solved to obtain2 = C (II)2( )p0 − p L0 = −2µ (I) L( )p0 − p L0 = −2µ (II) LC 1 = (p 0 − p L )b2LC (I)2 = (p 0 − p L )b 22µ (I) Lb 2 − C(I) b+ C (I)µ (I) 2b 2 + C(I) bµ (I) + C (I)2 (3.144)( µ (I) − µ (II) )µ (I) + µ (II)(µ (I) )2µ (I) + µ (II)(3.145)