Mass flow rate of a fluid with non-uniform density

Relevant wiki: Continuity Equation (Fluids)

Mass flow rate is $\dot{m} = \rho Av.$ Since the density $\rho$ varies, an integral of the form $\dot{m} = \int\rho(r) v dA$ needs to be evaluated.

$\dot{m} = \int\rho(r) v dA = \int_0^1 \Big( \frac{1000}{\pi(1+2r)} \Big) v (2\pi r dr) = 2000v \int_0^1 \frac{r}{1+2r} dr$

In order to evaluate $\int \frac{r}{1+2r} dr$ set $u = 2r+1.$

$\int \frac{r}{1+2r} dr = \int \frac{u-1}{2u} \big( \frac12 du \big) = \frac14 \int (1 - \frac{1}{u} ) du = \frac14 (u - \ln u) = \frac14 [2r+1 - \ln(2r+1)]$

Evaluate the definite integral.

$\frac14 \int_0^1 \frac{r}{1+2r} dr = \frac14 [ (3-\ln 3) - (1-\ln 1) ] = \frac14 (2 - \ln 3)$

Using this along with the give $v = \frac{1}{2-\ln3}$ gives

$\dot{m} = 2000 \big( \frac{1}{2-\ln3} \big) [\frac14 (2 - \ln 3)] = 500 \frac{\text{kg}}{\text{s}}.$

First principles help no end on problems like this. Imagine taking an infinitesimally thin ring, centred around the central axis of the pipe. Give the inner radius of this ring as $r$ and the outer radius $r+ \delta r$ . Then the area of the ring is:

$\delta A = \pi(r+ \delta r)^{2} - \pi r^{2} = \pi(2r \delta r + (\delta r)^{2}) \approx 2\pi r \delta r$

In one second, the distance travelled along the axis of the pipe by this ring is $v$ metres. So:

$\delta V = v \delta A = 2\pi rv \delta r$

Then the mass of this ring's flow is:

$\delta m = \rho(r)\delta V = 2\pi rv \rho(r) \delta r$

Substituting the known function $\rho(r)$ in:

$\delta m = 2\pi rv \delta r\frac{1000}{\pi (1+2r)} = \frac{2000rv\delta r}{1+2r}$

We need to sum all these small ring flows within the range $0\leq r \leq 1$ :

$\frac{dm}{dt} = \int_{0}^{1}\frac{2000rv}{1+2r}dr = 1000v\int_{0}^{1}\frac{2r}{1+2r}dr = 1000v\int_{0}^{1}\frac{(1+2r)-1}{1+2r}dr = 1000v\int_{0}^{1} 1 - \frac{1}{1+2r}dr = 1000v\left.(r-\frac{1}{2}\ln(1+2r))\right|_0^1$

$= 1000v(1-\frac{1}{2}\ln3) = 500v(2-\ln3) = \frac{500(2-\ln3)}{2-ln3} = \boxed{500 \text{kgs}^{-1}}$