# Flow through Pipe annulus with pressure controlled opening

I have a pipe with radius $$R_1$$ with a smaller pipe within (radius $$R_2$$). I want to find an expression for the flow through the pipe which flows in the annulus area. However, the known values are the dimensions of the pipe, including the $$R_1$$ and $$R_2$$, the length of the pipe (L), and the pressure at each side of the pipe ($$p_1$$ and $$p_2$$).

The fluid is fresh water at 10 degrees.

I tried to use Poisseulles law under the assumption of laminar flow: $$Q = \frac{(p_1-p_2) \cdot \pi \cdot (R_1^4-R_2^4)}{8 \cdot \mu \cdot L}$$

However, this yields an unrealistically high velocity, and flow rate, which I guess means that it is not laminar flow.

So to Calculate the flow in turbulent, I use Darcy-Weisbach : $$Q = \Delta p = f \cdot \frac{L}{D} \cdot\frac{\rho}{2} \cdot v_{avg}^2$$

Where I find the friction factor in a moody diagram from a Reynolds number where the velocity is neglected.

$$Re = \frac{\rho \cdot v_{avg} \cdot D_h}{\mu} = \frac{\rho \cdot D_h}{\mu} \approx 18000 \rightarrow f \approx 0.03$$

I am not sure this is a good way to estimate the Reynolds number. Does anyone have a better way?

Furthermore, the opening at the end of the pipe is controlled by the pressure how can I include this?

• The velocity is not neglected in the Reynolds number. Jan 18, 2023 at 6:21
• No thats just to have a reference
– Nil
Jan 18, 2023 at 15:46

For the pressure drop in pipe with annular section, you need to consider hydraulic diameter $$D_h$$, which is 4 times the ratio of section area to wetted perimeter:

$$D_h = 4\cdot \frac{\pi\cdot \left(R_1^2-R_2^2\right)}{2\cdot \pi\cdot \left(R_1+R_2\right)} = 2\cdot \left(R_1-R_2\right)$$

If $$p_2$$ is the pressure behind the spring valve, you need to define the pressure just before it $$p_{valve}$$. When all the pressures are equal, the valve is probably closed with force $$F_{closed}$$. Now in order to achieve some flow, you need to open the valve, so $$p_{valve}-p_2$$ has to generate a force $$F_p$$ higher than $$F_{closed}$$. For this threshold, you can use static pressure analysis, i.e. $$p_1 = p_{valve}$$:

$$F_p = \pi\cdot R_1^2\cdot \left(p_{valve}-p_2\right) > F_{closed}$$

Pressure drop at the valve will surely depend on how much the valve is opened (distance $$d$$), which can be controlled by linear spring with stiffness $$k$$:

$$d = \frac{F}{k} = \frac{F_p-F_{closed}}{k} = \frac{\pi\cdot R_1^2\cdot \left(p_{valve}-p_2\right) - F_{closed}}{k} = \frac{\pi\cdot R_1^2\cdot \Delta p_{valve} - F_{closed}}{k}$$

The local pressure drop at the valve can be expressed using minor losses formula and it depends on fluid density $$\rho$$, fluid velocity $$v_{valve}$$ and a minor loss coefficient $$\zeta$$:

$$\Delta p_{valve} = \zeta\cdot \frac{\rho}{2}\cdot v_{valve}^2$$

If you express fluid velocity at the valve using volumetric flow $$v_{valve} = \frac{Q}{d\cdot 2\cdot \pi\cdot R_1}$$ and combine the 2 equations above, you can derive relationship between volumetric flow and opening distance:

$$Q = \sqrt{\frac{8\pi}{\rho\cdot \zeta}\cdot \left(k\cdot d^3+F_{closed}\cdot d^2\right)}$$

As you can see, there is no easy direct solution for this problem, so you will have to use some iterative method. I would recommend using bisection method, where the volumetric flow has to be between zero and $$Q_{no\;valve}$$, a value you would obtain if there was no valve at all from the pressure difference $$p_2-p_1$$.

The biggest unknown in this is the minor loss coefficient $$\zeta$$. Also, be aware about the fact that dynamic pressure of the flow can also contribute to the opening of the valve.

I believe this may be solved by using Bernoulli's Equation Simply expressed Q1=Q2 Inflow must equal outflow regardless of size and shape of the pipe.

The annulus sectional area is all that matters. $$\frac{{^𝑃}{1}} {\rho g}+\frac{{^{V^2}}{1}}{2g}=\frac{{^p}{2}} {\rho g}+\frac{(\frac{{^A} {1}}{{^A} {2}}×{^V} {1})^2} {2g}$$ I hope this this will help answer your question.

Ref:Douglas, J. F., et al. (2005). Fluid Mechanics, Pearson Education Limited.

• How will you define flow from this? the velocities v1 and v2 must be identical right?
– Nil
Jan 18, 2023 at 15:45
• Velocities do not have to be identical. Jan 23, 2023 at 10:13