• Assumptions: Constant density and viscosity, steady flow (u and v independent of time).
• Approach: Substitute the known (u, v, w) into Eqs. (4.38) and solve for the pressure gradients. If a unique pressure function p(x, y, z) can then be found, the given solution is exact.
• Solution step 1: Substitute (u, v, w) into Eqs. (4.38) in sequence:
\rho (0) – \frac{\partial p}{\partial x} + \mu (2a – 2a + 0)=\rho \left(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}\right)=2a^2 \rho(x^3+xy^2)
\rho (0) – \frac{\partial p}{\partial y} + \mu (0 + 0 + 0)=\rho \left(u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}\right)=2a^2 \rho(x^2 y+y^3)
\rho (-g) – \frac{\partial p}{\partial z} + \mu (0 + 0 + 0)=\rho \left(u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}\right)=0
Rearrange and solve for the three pressure gradients:
\frac{\partial p}{\partial x}=-2a^2 \rho(x^3+xy^2) \frac{\partial p}{\partial y}=-2a^2 \rho(x^2 y+y^3) \frac{\partial p}{\partial z}=-\rho g (1)
• Comment 1: The vertical pressure gradient is hydrostatic. (Could you have predicted this by noting in Eqs. (4.38) that w = 0?) However, the pressure is velocity-dependent in the xy plane.
• Solution step 2: To determine if the x and y gradients of pressure in Eq. (1) are compatible, evaluate the mixed derivative,(\partial^2 p/ \partial x \partial y); that is, cross-differentiate these two equations:
\frac{\partial}{\partial y}\left(\frac{\partial p}{\partial x}\right)=\frac{\partial}{\partial y}[-2a^2 \rho(x^3+xy^2)]=-4a^2 \rho xy
\frac{\partial}{\partial x}\left(\frac{\partial p}{\partial y}\right)=\frac{\partial}{\partial x}[-2a^2 \rho(x^2 y+y^3)]=-4a^2 \rho xy
• Comment 2: Since these are equal, the given velocity distribution is indeed an exact solution of the Navier-Stokes equations.
• Solution step 3: To find the pressure, integrate Eqs. (1), collect, and compare. Start with \partial p / \partial x. The procedure requires care! Integrate partially with respect to x, holding y and z constant:
p=\int{\frac{\partial p}{\partial x}dx}|_{y,z}=\int{-2a^2 \rho(x^3+xy^2)dx}|_{y,z}=-2a^2 \rho\left(\frac{x^4}{4}+\frac{x^2 y^2}{2}\right)+f_1(y,z) (2)
Note that the “constant” of integration f_1 is a function of the variables that were not integrated. Now differentiate Eq. (2) with respect to y and compare with \partial p / \partial y from Eq. (1):
\frac{\partial p}{\partial y}|_{(2)}=-2a^2 \rho x^2y+\frac{\partial f_1}{\partial y}=\frac{\partial p}{\partial y}|_{(1)}=-2a^2 \rho(x^2y+y^3)
Compare: \frac{\partial f_1}{\partial y}=-2a^2 \rho y^3 or f_1=\int{\frac{\partial f_1}{\partial y}dy}|_{z}=-2a^2 \rho \frac{y^4}{4}+f_2(z)
Collect terms: So far p=-2a^2 \rho\left(\frac{x^4}{4}+\frac{x^2 y^2}{2}+\frac{y^4}{4}\right)+f_2(z) (3)
This time the “constant” of integration f_2 is a function of z only (the variable not integrated). Now differentiate Eq. (3) with respect to z and compare with \partial p / \partial z from Eq. (1):
\frac{\partial p}{\partial z}|_{(3)}=\frac{\partial f_2}{\partial z}=\frac{\partial p}{\partial z}|_{(1)}=-\rho g or f_2=-\rho gz+C (4)
where C is a constant. This completes our three integrations. Combine Eqs. (3) and (4) to obtain the full expression for the pressure distribution in this flow:
p(x, y, z) = -ρgz – \frac{1}{2} a^2 \rho (x^4 + y^4 + 2x^2y^2) + C Ans. (5)
This is the desired solution. Do you recognize it? Not unless you go back to the beginning and square the velocity components:
u^2 + v^2 + w^2 = V^2 = a^2(x^4 + y^4 + 2x^2y^2) (6)
Comparing with Eq. (5), we can rewrite the pressure distribution as
p + \frac{1}{2} \rho V^2 + pgz = C (7)
• Comment: This is Bernoulli’s equation (3.54). That is no accident, because the velocity distribution given in this problem is one of a family of flows that are solutions to the Navier-Stokes equations and that satisfy Bernoulli’s incompressible equation everywhere in the flow field. They are called irrotational flows, for which curl V = ∇ × V ≡ 0. This subject is discussed again in Sec. 4.9.
\frac{p_1}{\rho}+\frac{1}{2}V^2_1+gz_1=\frac{p_2}{\rho}+\frac{1}{2}V^2_2+gz_2 = const (3.54)