4.1 (a) For the velocity field V = 4 t x

(b) The flow is three dimensional because all three velocity components are generally nonzero.

(c) Use the definition of the substantive derivative to evaluate the acceleration vector. It is ( 4x + 16 t^2 x , -4 t y + 4 t^4 y , 16 t x z + 16 x^2 z ). At ( -1,+1,0)

(d) eg.

4.16 Substitute the given velocity field in the equation expressing the incompressibility assumption; that is, d,i v_i = 0. This gives 0 = 3x^2 + d,y v + (-y) . Integrating this gives v = y^2 - 3 x^2 y + function (x,z) .

4.27 Substitute the given 2D velocity field in the Navier Stokes equation (constant density, inviscid). Neglecting gravity gives -grad p = rho [u d,x + vd,y ] (u

4.40 For the parabolic velocity profile of laminar flow between parallel plates, use the energy equation to find the temperature profile. Assume incompressible flow. The energy equation form we want is for internal energy, where du = c_p dT . Thus,

rho D,t u = k d,x_m^2 T -p d,i v_i + rho epsilon = rho D,t u = k d,x_m^2 T + rho epsilon , where epsilon is 2 nu e_ij e_ij . Write out the terms of the rate of strain tensor e_ij = [(0, d,y u , 0) , (d,y u , 0,0) , (0,0,0)]. Find epsilon = nu (d,y u )^2 . The left side of the equation is zero, leaving d^2,dy^2 T = - ( mu / k ) (d,y u)^2 . Substitute u(y) and integrate. Use T = T_w at y=0 and y=h, and d,y T =0 at y=h/2. Thus T = T_w + [8 mu u_max ^2 / k][y/3h - y^2/h^2 + 4y^3 / 3h^3 - 2y^4 / 3h^4 ].

4.47 Given u=2y and v=3x, first check continuity; that is d,x u + d,y v = 0 + 0 . OK. Then find psi by integrating u = d,y psi and v = - d,x psi to give psi = y^2 - 3x^2 / 2 . This represents flow in unequal corners. Plot psi values, show the flow direction.

4.56 Given phi = Kxy , the streamlines form a net similar to that of 4.47, except for equal corners.

4.62 Given u = Vy/h , v = 0 , check continuity. It is satisfied. Therefore psi exists. From definitions of the velocity components we find psi = V y^2 / 2h . Check irrotationality. Omega_z = d,x v - d,y u = - V/h which is not zero, so the flow is not irrotational. Therefore phi cannot exist.

4.70 The stream function psi for a vertical uniform stream V = U

4.83 Use the incompressible Navier Stokes equations. rho D,t u is zero. Thus dp/dx = mu d^2 u / dy^2 . Integrate this twice and substitute the b. c. that u = U at y=0, and u=0 at y=h. The result is the given equation. Note the solution is the sum of a linear and a parabolic profile, the two linear solutions.

4.85 Substitute the assumptions into the incompressible Navier Stokes equations. This gives rho d,t u on the left and mu d,y^2 u on the right, subject to u(0,t) = U_o sin ( omega t ) and u( infinity , t) = 0.