Help on integrating differential dynamic pressure (kinetic energy per unit volume) for 1D radial flow towards a line sink

Question

A quick introduction to my question and then the question asked at the end. For this problem the cross-sectional area normal to flow is the surface of a cylinder, $A=2\pi r L$, where $r =$ radial distance from the axis of the cylinder (line sink). The dynamic pressure (kinetic energy per unit volume) for a parcel of fluid is $$\tag{1} 0.5 \rho v^2$$where $\rho =$ volumetric-mass density and $v=$ velocity. The derivative of the dynamic pressure with respect to radial position is $$\tag{2} \frac{d}{dr}(0.5 \rho v^2)=\rho v \frac{dv}{dr}$$ I want to find the integral of the change in dynamic pressure with respect to $r$ (the change in dynamic pressure as the fluid moves towards or away from the line sink). In doing so I make the following steps, $$\tag{3} \int_{r_1}^{r_2} \rho v \frac{dv}{dr} dr$$ Since $v = w/(\rho A)$, where $w=$ mass flow rate, then, $$\tag{4} \int_{r_1}^{r_2} \rho \frac{w}{\rho A} \frac{dv}{dr} dr$$ $$\tag{5} \int_{r_1}^{r_2} \frac{w}{2\pi r L} \frac{dv}{dr} dr$$ $$\tag{6} \frac{w}{2\pi L} \frac{dv}{dr} \int_{r_1}^{r_2} \frac{1}{r} dr$$ $$\tag{7} \frac{w}{2\pi L} \frac{dv}{dr} \ln\left(\frac{r_2}{r_1}\right)$$

I have values for all variables in last equation above except for $dv/dr$. How do I determine the value for $dv/dr$?

Answer 1

At a radius $r$ the velocity is $v$ and the mass flowing across the surface per second is $2\pi rLv\rho$

At a radius $r+dr$ with velocity $v+dv$ it's $2\pi (r+dr)L(v+dv)\rho$

we can put these equal, assuming constant density, so $$ rv = (r+dr)(v+dv)$$

Expanding and simplifying and ignoring the term with $dv dr$ gives

$$\frac{dv}{dr} = -\frac{v}{r}$$

so

$$\int \frac{dv}{v} = -\int \frac{dr}{r}$$

$$ln(v) = -ln(r) +ln(k)$$

the constant of integration has been put as $ln(k)$ to help with the following steps.

$$ln(v) = ln(\frac{k}{r})$$

$$v=\frac{k}{r}$$

so the kinetic energy for the parcel of fluid is

$$K.E. = \frac{k'\rho}{r^2}$$

the constant $k'$ can be found from the kinetic energy or speed of the parcel of fluid at any radius.

Answer 2

Assuming w is constant, you need to substitute A in v formula and then take the derivative of v. dv/dr is not constant, it's function of r so you have to keep it within integral. Although, eventually you will get same kinetic energy term and integral would simply be difference of kinetic energy at different values of r.