Homework set 3. Problems 4.1,16,27,40,47,56,62,70,83,85
4.1 (a) For the velocity field V = 4 t x i - 2 t^2 y j +
4 x z k the flow is unsteady because the components depend on time.
(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) a is (-4(1+4t^2) , 4t(1-t^3) , 0)
(d) eg. k is orthogonal.
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 ] (ui + vj) = rho (2xy^2 i + 3y^3 j)
. Thus, d,x p = - rho 2xy^2 .
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 = Uj
with a source m at the origin is psi = -Ux + m tan^-1 (y/x) . The
flow forms a Rankine half body aligned vertically.
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
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.