0
$\begingroup$

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 $dv/dr$?

$\endgroup$
1
  • 1
    $\begingroup$ You differentiated and then integrated with respect to r? That just leaves you with the original kinetic energy term. $\endgroup$
    – Triatticus
    Dec 15, 2021 at 6:22

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ So?: $$\frac{d}{dr}(0.5 \rho v^2)=\rho v \frac{dv}{dr}=-\rho v \frac{v}{r}=-\rho \frac{v^2}{r}$$ $$\int_{r_1}^{r_2} -\rho \frac{v^2}{r}=\int_{r_1}^{r_2} \frac{-w^2}{\rho A^2 r} dr = \int_{r_1}^{r_2} \frac{-w^2}{\rho (2 \pi r L)^2 r} dr$$ $$\frac{-w^2}{4 \pi^2 \rho L^2} \int_{r_1}^{r_2} r^{-3} dr=\frac{w^2}{8\pi^2 \rho L^2}(r_2^{-2}-r_1^{-2})$$ $\endgroup$
    – Armadillo
    Dec 15, 2021 at 15:01
  • 1
    $\begingroup$ @ Armadillo seems ok if you meant that the density was constant, are you trying to find the change in KE of part of the fluid as it goes from one radius to another? $\endgroup$ Dec 15, 2021 at 17:32
  • $\begingroup$ Yes, for this question I want to assume constant density. And yes, I am trying to find the change in KE as the fluid accelerates due to change in area normal to flow. If OK to ask here, what would change in the derivation if we consider the fluid's compressibility ($\rho=f(p)$)? $\endgroup$
    – Armadillo
    Dec 15, 2021 at 19:14
  • 1
    $\begingroup$ @Armadillo, not sure about the changing density situation, the answer will be edited to get the kinetic energy in terms of $r$ $\endgroup$ Dec 15, 2021 at 19:33
  • 1
    $\begingroup$ @Armadillo that's all I've got time for, over to you now, all the best with it. $\endgroup$ Dec 15, 2021 at 22:34
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ So subbing $A$ in $v$ formula, I get (after integrating, see derivation in comment to @JohnHunter) $$\frac{w^2}{8\pi^2 \rho L^2}(r_2^{-2}-r_1^{-2})$$ Is this what you mean? $\endgroup$
    – Armadillo
    Dec 15, 2021 at 15:06
  • $\begingroup$ Yes, this seems correct $\endgroup$
    – hp007
    Mar 11 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.