Engineering Water Flow from Pressurized Tank through Heat Exchanger

[AxisCat]

View attachment 330554

Here is what I'm coming up with for the EOM ( assumes uniform properties distributed across any particular section - that approximately the case for turbulent flow )

$$ \rho \left( l \frac{A}{A_p} + z \right)\ddot z + P_t(z) + \rho g \left( H + z\right) + \frac{1}{2} \rho \dot z ^2 = P_{atm} + \frac{1}{2} \rho \left( \frac{A}{A_p} \right)^2 \dot z^2 + \Delta P_L(Q) $$

Then from continuity you have:

$$ Q = -A \dot z \implies \dot Q = -A \ddot z $$

So you can sub that in and get a first order nonlinear equation mostly in terms of $Q$ and its derivative. There is still a $z$ buried in the pressure function to contend with, but with regards to a numerical solution its not an issue. Just note that:

$$ z = z_o + \int \dot z dt = z_o -\frac{1}{A}\int Q dt $$

The pressure function in the tank $P_t(z)$ is a thermodynamics problem, I'd go with adiabatic expansion of an ideal gas, but I would defer to @Chestermiller for the theory on that.

Also, I left the loss function $\Delta P_L(Q)$ general for now. You should be able to get away with either what you(or I) polynomial curve fit, or the power law fit Chester suggests. If you want to include the loss characteristics of other components, it's not too late for that.

$l$ is the length of all the plumbing from the tank to the exit.

Anyhow, you put $z$ where you want it initially, specify the initial condition in the gas, set the volumetric flow to zero and start the clock using standard numerical integration techniques.

The derivation was long winded, so if I've bungled it, and or anyone wants to examine it...let me know.

Wow! That looks like what I was trying to arrive at. I will need to dissect this and try and understand all the bits and pieces. Especially the calculus which is a big weakness of mine. I have a couple of busy days coming up but I hope I can reach back out to everyone after I get some time in working on it.

Nice work my friend!

Axis

 

[Chestermiller]

Sorry, the manufacturer doesn't provide that information. Just the pressure drop as a function of volumetric flow

And the exit pipe length and diameter?

 

[AxisCat]

And the exit pipe length and diameter?

Since I am just trying to understand flow through the heat exchanger as the bladder tank empties I am ignoring losses due to the pipe. But to get velocity you have to have an area so let's use a 1" diameter pipe 12" long.

 

[Chestermiller]

For steady flow in a smooth tube, the overall force balance on the fluid is $$\frac{\pi D^2}{4}\Delta P=\pi DL\tau_w$$or$$\Delta P=\frac{4L}{D}\tau_w$$where $\tau_w$ is equal to the shear stress at the wall. Dimensional analysis tells us that the shear stress at the wall can be expressed as $$\tau_w=\frac{\rho v^2}{2}f$$where $\rho$ is the fluid density, v is the cross section average fluid velocity, and f is the "Fanning friction factor," a function of the so-called Reynolds number Re for the flow: $$\frac{\rho v D}{\mu}=\frac{4\dot{m}}{\pi D\mu}$$with $\mu$ equal to the fluid viscosity and $\dot{m}$ equal to the mass flow rate. For laminar flow (Re < 2200), the friction factor f is related to the Reynolds number by $$f=\frac{16}{Re}$$which is equivalent to the well-known Hagen-Poiseuille relationship. For turbulent flow in the Reynolds number range 2800 < Re < 100000, the Blasius equation describes the relationship between the friction factor f and the Reynolds number: $$f=\frac{0.0791}{Re^{0.25}}$$

In my next post, I'll carry out a sample calculation for a volumetric flow rate of 12 GPM to predict the pressure drop, assuming at tube of 1" diameter and a water temperature of (nominally) 70 F.

 

[Chestermiller]

Sample Calculation

Q = 12 gpm, D = 1", $\mu=0.01 Poise = 0.01\frac{gm}{cm.s}$

Volume Flow Rate = $Q=12\ gpm=0.0267\frac{f^3}{s}=46.2\frac{in^3}{s}=757\ cc/s$

Mass flow rate = $\dot{m}=\rho Q=757\frac{gm}{s}$

Cross sectional area = A = $\pi\frac{D^4}{4}= 5.07\ cm^2$

Fluid velocity v = Q/A= 149.3 cm/s = 58.8 in/s = 4.9 ft/s

Reynolds number Re = $\frac{4\dot{m}}{\pi D\mu}=\frac{4(757)}{\pi (2.54)(0.01)}=37950$

friction factor f = $\frac{0.0791}{Re^{0.25}}=0.0057$

wall shear stress $\tau=\frac{\rho v^2}{2g_c}f=\frac{62.4(4.9)^2}{2(32.2)}0.0057=0.133\ psf$

Pressure gradient = $\frac{\Delta P}{L}=\frac{4}{D}\tau_w=6.38\ psf/ft=0.044\ psi/ft$

This seems much smaller than the observed pressure gradient.

 

[Chestermiller]

Continuation of previous response.

Going to a smaller diameter would increase the pressure gradient and shorten the required length of the tube. I'm going to try a 0.5" diameter and see how that plays out. For 0.5" diameter, cross sectional area is 4x smaller, velocity v is 4x larger: v = 4.9 x 4 = 19.6 ft/s, and Reynolds number is 2x larger: Re = 75900.

friction factor $f=\frac{0.0791}{75900^{0.25}}=0.00477$

wall shear stress $\tau=\frac{(62.4)(19.6)^2}{2(32.2)}(0.00477)=1.776\ psf$

pressure gradient = $\frac{\Delta P}{L}=\frac{4}{D}\tau=170\ psf/ft=1.18\ psi/ft$

So, to obtain the observed 1.5 psi with a 0.5" ID tube would require a tube length of 1.27 ft.

 

[AxisCat]

Continuation of previous response.

Going to a smaller diameter would increase the pressure gradient and shorten the required length of the tube. I'm going to try a 0.5" diameter and see how that plays out. For 0.5" diameter, cross sectional area is 4x smaller, velocity v is 4x larger: v = 4.9 x 4 = 19.6 ft/s, and Reynolds number is 2x larger: Re = 75900.

friction factor $f=\frac{0.0791}{75900^{0.25}}=0.00477$

wall shear stress $\tau=\frac{(62.4)(19.6)^2}{2(32.2)}(0.00477)=1.776\ psf$

pressure gradient = $\frac{\Delta P}{L}=\frac{4}{D}\tau=170\ psf/ft=1.18\ psi/ft$

So, to obtain the observed 1.5 psi with a 0.5" ID tube would require a tube length of 1.27 ft.

Thank you Chester. I am experiencing a bit of information overload at the moment. I may have misunderstood your question about the diameter and length. In regards to the heat exchanger that is information I won't be able to obtain. It is actually coiled up in a donut shape with spray foam insulation. It is probably obvious where my math skills land on the scale of things. My algebra back in the day was pretty decent but not my calculus. It is going to take some time to digest everything in this post. And I hope to get back to everyone after I wrap my head arounds some things.