 Let's try another example. This time through a pipe. Now we're considering incompressible flow flowing through a pipe. That pipe has a radius of r, that's uppercase r, not to be confused with the lowercase r, which is one of my axes. And I want to know how the velocity profile is developed if we have a pressure gradient again. So I will use the same set of assumptions that I applied to the previous problem. I have incompressible flow. I'm assuming that it's developed in a laminar direction, which means that there's no velocity in any direction other than what I'm calling the z-axis now. I'm neglecting gravity. I have axial symmetry, which means that del of anything with respect to theta is zero. So regardless of which angle of the pipe you look at, it's all the same. And I'm assuming a constant viscosity. With that set of assumptions in place, we can look at the conservation of mass again. But this time, our conservation of mass includes a bunch of scary thetas. Worse yet, our conservation of momentum includes even more scary thetas. Those are a result of the polar coordinate system that we're using here. And even though it's intimidating looking, it's not actually that bad to work with because just like in the Cartesian coordinate system, most of these terms are going to cancel. So looking back at our pipe, we'll deploy our conservation of mass and we'll recognize that for steady state, nothing can change with respect to time, including the density. Also the density is constant because of incompressibility. I have no velocity in the r-direction. I have no velocity in the theta direction. I only have velocity in the z-direction. So the only turn that's left is del vz, that is the rate of change of the z component of velocity with respect to z is zero. So just like in the flat plate setup, I'm assuming that the velocity only changes in one direction. It only changes as a function of radius. So I'm saying that my velocity profile only changes as you get further away from the center of the pipe. It doesn't change with respect to z. It doesn't change with respect to theta. Now I can look at my conservation of momentum. For that, I will consider the r-direction first. In the r-momentum equation, everything disappears except for the pressure gradient driving any flow. So I have no r component of velocity. Therefore, I have no r component of velocity changing with respect to anything else. And that leaves me with del p del r is zero. Therefore, pressure is only a function of z. Next, I can consider my z-momentum. In the z-momentum, I'm saying that the velocity in the z-direction only changes as a function of r. As a result, everything disappears, including the gravity term because we're neglecting gravity, except for the pressure gradient in the z-direction and everything associated with the velocity in the z-direction changing with respect to r. That means that zero is equal to negative del p del z plus the viscosity times the quantity, 1 over r times del r del vz del r with respect to r. Rearranging this equation in terms of del p del z, I get del p del z is equal to viscosity times 1 over r times the quantity rate of change of r del vz del r with respect to r. I can begin to work my way to a velocity profile by unwrapping the terms on the left-hand side of my equation here. First of all, I can divide by viscosity, and then I can multiply by radius, and then I can integrate, and then I can divide by radius, and then I can integrate a second time, at which point I have 1 over 4 mu times del p del z times r squared plus c1 times the natural log of r plus c2. My two boundary conditions are that the velocity at the wall is zero because of no slip, and as a result of axial symmetry, I'm saying that the velocity doesn't change with respect to r in the middle. So the velocity profile with respect to r is exactly vertical. For the first boundary condition, or rather for the first constant that we solve as a result of the boundary condition, it's easiest if I look back at the first derivative form with respect to r, at which point I can solve for c1, which is zero. For the other boundary condition, I can apply that at the end form, wherein I see that c2 is equal to negative the radius of our pipe squared divided by 4 times the viscosity times del p del z. Just like before, it's going to be convenient for me to express this in terms of a maximum velocity, as that is something that we either care about or have most of the time, so I'm going to be determining that at an r position of zero, wherein I see that my maximum velocity in the z direction is 1 over 4 times mu times del p del z times the quantity zero minus r squared, which means that it's a negative quantity again, negative r squared over 4 mu times del p del z for the same reasons as in the previous example that the velocity is going to be in the opposite direction of a positive pressure gradient. Plugging this back in, I have the velocity in the z direction as a function of r is the maximum velocity times 1 minus little r squared over big r squared, where the maximum velocity is negative radius of the pipe squared divided by 4 times the viscosity times the pressure gradient driving the flow.