Water Flow from Pressurized Tank through Heat Exchanger

[AxisCat]

What got me down this path was the equation for an open tank draining through a hole of some diameter by gravity alone. This was easy math. So I thought I could expand this to a closed tank under pressure from a bladder by incorporating Boyle's law. The equation looked correct. I then used a spreadsheet to break it down into seconds.

 

[AxisCat]

I probably should just let this rabbit get away and start working on the real world engineering calculations.

 

[erobz]

Ok, so you want to find the volumetric flowrate vs time. If you did the experiment in a reasonably controlled manner with reasonably precise measurement equipment, then my guess is the linear factor in the trendline fit for the heat exchanger is probably due to the fact that the loss coefficient for a pipe of fixed properties depends on the Reynolds number, which depends on the flow velocity...i.e. it generally is not a constant as flow changes, perhaps your experiment has captured that reasonably well over the measure range of flows.

If you want to look at the transient behavior its basically a second order ordinary differential equation. Is it something you are interested in, I'll produce it under some simplified assumptions?

 

[erobz]

I probably should just let this rabbit get away and start working on the real world engineering calculations.

Its not "simple math" ( that's a relative term, there are people here that would find it a bore for example @Chestermiller ...I can assure you ). But its more than Algebra... its one of the last math coursework encountered in a bachelors of engineering (at least it was in my own). That being said, given the non-linear behavior of the gas in the tank as it expands, I could at best find a numerical result...not a general one.

 

[Chestermiller]

$$psi=.017816(GPM)^{1.7847}$$I can't read the column labels on your post #22.

 

[AxisCat]

Ok, so you want to find the volumetric flowrate vs time. If you did the experiment in a reasonably controlled manner with reasonably precise measurement equipment, then my guess is the linear factor in the trendline fit for the heat exchanger is probably due to the fact that the loss coefficient for a pipe of fixed properties depends on the Reynolds number, which depends on the flow velocity...i.e. it generally is not a constant as flow changes, perhaps your experiment has captured that reasonably well over the measure range of flows.

If you want to look at the transient behavior its basically a second order ordinary differential equation. Is it something you are interested in, I'll produce it under some simplified assumptions?

Sure, I had calculus, again 30 years ago and never really did understand it. I could work the mechanics but never had the opportunity to apply it to something useful for me. So basically my calculus sucks. I did however look at my quadratic and recognize it could be be solved with with calculus. I took the derivative and saw it is linear but couldn't see how to incorporate that with my other work. Even my spreadsheet seems to be using calculus because of time. That is on my bucket list, to at least understand some calculus.

 

[AxisCat]

$$psi=.017816(GPM)^{1.7847}$$I can't read the column labels on your post #22.

I was having problems getting decent resolution... do you mean post #10?

 

[AxisCat]

$$psi=.017816(GPM)^{1.7847}$$I can't read the column labels on your post #22.

Where did this equation come from?

 

[AxisCat]

Also, I do know how difficult fluid dynamics can be. I don't need to be dead on with the results, my unit really wont care if it gets 1.5 gpm/ton or 2. It will be happy with either one.

 

[AxisCat]

Gentlemen. I have to get some time in bidding a plan spec project or I am going to get in trouble. I look forward to more responses. I will check back later this evening. Thanks again for helping me get a better understanding of this.

 

[Chestermiller]

Where did this equation come from?

I fit your data to it.

 

[Chestermiller]

I was having problems getting decent resolution... do you mean post #10?

Sorry, #24

 

[Chestermiller]

$$psi=.017816(GPM)^{1.7847}$$I can't read the column labels on your post #22.

For turbulent flow in a smooth heat exchange tube at Reynolds numbers between 2800 and 100,000, the exponent in this equation, according to the Blasius equation, would be 1.75 (compared to 1.78). Please let me know the length and inner diameter of the heat exchanger tube, and an average value of the temperature of the water in the tube, and I will present a sample calculation of the pressure drop for 10 gpm so that we can compare with the theoretical prediction. This same approach can be used to predict the pressure drop in the discharge tube.

 

[erobz]

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 )

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

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.

 

[AxisCat]

For turbulent flow in a smooth heat exchange tube at Reynolds numbers between 2800 and 100,000, the exponent in this equation, according to the Blasius equation, would be 1.75 (compared to 1.78). Please let me know the length and inner diameter of the heat exchanger tube, and an average value of the temperature of the water in the tube, and I will present a sample calculation of the pressure drop for 10 gpm so that we can compare with the theoretical prediction. This same approach can be used to predict the pressure drop in the discharge tube.

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

 

[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.