Engineering Water Flow from Pressurized Tank through Heat Exchanger

[AxisCat]

Homework Statement

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Relevant Equations

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Hi All-

New member here. As the title states I am trying to solve the flow volume of water from a pressurized storage tank through a heat exchanger and then discharging out the end of a pipe. I have been working on this for several days now without any luck, seems like now I am just throwing equations at it and hope something sticks. So I could use a nudge in the right direction.

The storage tank is a bladder tank in that air pressure is one side of the bladder and water on the other.

From the bottom of this tank the water flows through a heat exchanger. All I have on the heat exchanger to work with is (3) data points relating change in pressure with flow rate. I used curve fitting software to come up with a quadratic equation to describe this.

The water exits the heat exchanger and just empties into the open.

I have been using Bernoulli’s Equation to try and solve this. All my units should be a match. Volume in cubic feet, Area in square feet, pressure in pounds per square foot, density in pounds, time in seconds. Attached is a spreadsheet of my latest attempt. I apologize for how messy it is.

Thanks for you help!
Axis

 

[berkeman]

Welcome to PF. I'm not seeing an attachment...

It's best not to try to attach files that can contain macros (Excel, Word, etc.). Instead, try to attach text files or PDF / JPEG copies of the work.

Thanks.

 

[AxisCat]

Welcome to PF. I'm not seeing an attachment...

It's best not to try to attach files that can contain macros (Excel, Word, etc.). Instead, try to attach text files or PDF / JPEG copies of the work.

Thanks.

Gotcha. Didn't think about that. The only way I could do some of the math easily is using excel.. But here is a PNG file of my work

 

[erobz]

Gotcha. Didn't think about that. The only way I could do some of the math easily is using excel.. But here is a PNG file of my work

Put it right in the body of the response using the "insert image tool" (it looks like a little picture). It's too small to see (for me at least).

 

[AxisCat]

Try this one

 

[erobz]

Try this one View attachment 330525

hmmm, only a marginal improvement. Can you break it up into a few images? 1 for the diagram, 1 for the math, etc... they can all be inserted independently in the same way.

 

[AxisCat]

Still having trouble getting the resolution right, Third time??

 

[erobz]

Still having trouble getting the resolution right, Third time??

Sorry, no better. You are going to have to break it into a few separate images.

 

[AxisCat]

The website is compressing my image files to a point they are too hard to read... let me try something else

 

[AxisCat]

 

[erobz]

Incompressible flow, uniform properties across the pipe,is the pipe the same diameter at 3 and 4?

 

[AxisCat]

Incompressible flow, uniform properties is the pipe the same diameter at 3 and 4?

Correct on the statement and the diameters are the same

 

[AxisCat]

I am not a student... not really in the math field. My algebra and physics was about 30 years ago.

 

[erobz]

Correct on the statement and the diameters are the same

The pressure inside the jet, just outside the pipe almost immediately goes to atmospheric pressure. So $v_4$ is the jet velocity, and you can apply conservation of mass (continuity) to find $v_3$. What do you get for $v_3$ if the cross sectional areas of 4 and 3 are the same?

 

[Chestermiller]

You should not be using the Bernoulli equation, since the pressure drop in the heat exchanger is mainly frictional pressure drop. Please show us the exact pressure drop vs flow rate data points you have.

In your calculations, I hope you recognize that pressure is in lbs force units and density is in lbs. mass units, and that you are using $g_c=32\ \frac{lb_m\ ft}{lb_f\ sec^2}$

Are you just trying to add the pressure drop in the final discharge pipe and the gravitational heat in the tank to the pressure drop in the heat exchanger? What is the layout of the heat exchanger: single pass vs double pass, shell side vs tube side, tube diameter, number of tubes, total flow rate, etc.?

 

[AxisCat]

The pressure inside the jet, just outside the pipe almost immediately goes to atmospheric pressure. So $v_4$ is the jet velocity, and you can apply conservation of mass to find $v_3$.

That makes sense. So looking at how I set up my equation does it look right? The quadratic I am using is from curve fitting the pressure drop as a function of flow. I am not sure I can just set this equal to the change in pressure between point 4 and 1. And I don't know if it was correct to substitute the velocity by a volume/sec?

 

[AxisCat]

You should not be using the Bernoulli equation, since the pressure drop in the heat exchanger is mainly frictional pressure drop. Please show us the exact pressure drop vs flow rate data points you have.

In your calculations, I hope you recognize that pressure is in lbs force units and density is in lbs. mass units, and that you are using $g_c=32\ \frac{lb_m\ ft}{lb_f\ sec^2}$

Are you just trying to add the pressure drop in the final discharge pipe and the gravitational heat in the tank to the pressure drop in the heat exchanger? What is the layout of the heat exchanger: single pass vs double pass, shell side vs tube side, tube diameter, number of tubes, total flow rate, etc.?

This is all the data i have on the heat exchanger, it is a single pass tube inside of a tube. Refrigerant on one side, water on the other. It is a ground source heat pump.

Intuitively I know the pressure drop in the heat exchanger is due to friction I guess I was hoping I could convert it to just pressure drop as a function of flow and then use the Bernoulli equations...

 

[erobz]

That makes sense. So looking at how I set up my equation does it look right? The quadratic I am using is from curve fitting the pressure drop as a function of flow. I am not sure I can just set this equal to the change in pressure between point 4 and 1. And I don't know if it was correct to substitute the velocity by a volume/sec?

It doesn't look great. You should basically have a quadratic curve passing through the origin. i.e $\Delta P(Q) = \beta Q^2$ meaning if there is no flow there is no loss between points 2 and 3. You should be trying to force that form.

 

[AxisCat]

It doesn't look great. You should basically have a quadratic curve passing through the origin. i.e $\Delta P(Q) = \beta Q^2$ meaning if there is no flow there is no loss between points 2 and 3. You should be trying to force that form.

Right, it was close to passing through the origin, i think rounding got it off some.

 

[erobz]

Right, it was close to passing through the origin, i think rounding got it off some.View attachment 330536

format the trendline for your original data, and set the origin to 0 for good measure.

 

[AxisCat]

format the series, and set the origin to 0 for good measure.

Right... if i can loose the b and c coefficients it would be even simpler and just be (k)Q^2. but the K factor seemed to change with flow

 

[AxisCat]

Right... if i can loose the b and c coefficients it would be even simpler and just be (k)Q^2. but the K factor seemed to change with flow

Exactly what you posted ΔP(Q)=KQ^2

 

[AxisCat]

My goal is to determine a flow rate through this system based on a pressure inside the tank. As time moves forward the pressure in the tank will decrease along with the flow until the pressure inside the tank is equal to atmospheric pressure. I am trying to model that over time

 

[erobz]

There is what I get, it doesn't perfectly fit theory, because that linear term isn't negligible.

 

[AxisCat]

View attachment 330537

Nice! Never thought about adding (0,0) then fit the curve

 

[erobz]

Nice! Never thought about adding (0,0) then fit the curve

Its a trivial data point, you don't have to add it. You can just set the intercept to zero under formatting the trendline options.

 

[erobz]

My goal is to determine a flow rate through this system based on a pressure inside the tank. As time moves forward the pressure in the tank will decrease along with the flow until the pressure inside the tank is equal to atmospheric pressure. I am trying to model that over time

Unfortunately, not such a trivial model to tackle... Also, all of the components have losses, not just the heat exchanger.

 

[Chestermiller]

What is the tube ID and what is the length and diameter of the exit pipe? What is the temperature of the fluid?

 

[AxisCat]

Unfortunately, not such a trivial model to tackle... Also, all of the components have losses, not just the heat exchanger.

It sure has me stumped. Full disclosure. I am an HVAC contractor and have been installing geothermal heat pump systems since the early 1990's. I am probably counting piping buried in hundreds of miles by now. All closed loop. Closed loop is easy, just a circulator and component pressure loses. I have a customer who wants an open well and I said sure! I can do that. Bare in mind I have never done open loop and after hearing the well contractor start talking about 5 hp jet pumps and 100 gallon storage tanks I thought I better educate myself on the dynamics involved.

 

[AxisCat]

If it was just geo unit and the well it becomes easy again... he wants to use the well for potable water too. The heat pump doesn't care about the pressure as long as it is above zero. And the 100 gallon storage tank really only has a usable draw down of about 35 gallons. My unit will drain that in about 3 minutes then the pump cycles on. All this above is really just me trying to understand the theory.

 

[erobz]

It sure has me stumped. Full disclosure. I am an HVAC contractor and have been installing geothermal heat pump systems since the early 1990's. I am probably counting piping buried in hundreds of miles by now. All closed loop. Closed loop is easy, just a circulator and component pressure loses. I have a customer who wants an open well and I said sure! I can do that. Bare in mind I have never done open loop and after hearing the well contractor start talking about 5 hp jet pumps and 100 gallon storage tanks I thought I better educate myself on the dynamics involved.

Full disclosure, you probably should have done that before you accepted the work.

 

[AxisCat]

Full disclosure, you probably should have done that before you accepted the work.

Hey I couldn't agree more but there has to be a first time for everything, right? If other people can do it there is no reason I can't as well. This is actually a small project compared to some larger ones I do. Make sense?

 

[erobz]

If it was just geo unit and the well it becomes easy again... he wants to use the well for potable water too. The heat pump doesn't care about the pressure as long as it is above zero. And the 100 gallon storage tank really only has a usable draw down of about 35 gallons. My unit will drain that in about 3 minutes then the pump cycles on. All this above is really just me trying to understand the theory.

So you are trying to determine how long the storge tank will take to deplete to some minimum capacity when someone opens the tap?

 

[AxisCat]

So you are trying to determine how long the storge tank will take to deplete to some minimum capacity when someone opens the tap?

Not really... the storage tank just keeps the water pressure between 60 and 40 psi. And with the 35 gallon draw down a tap can be open for quite a while before our pressure drops below 40 and the pump cycles back on. My unit will drain the tank in 3 minutes (roughly) before the pump cycles back on. A 5 hp pump draws more power than my entire unit. On any HVAC unit you want to maximize run times for effiecency and comfort. I don't want to short cycle the pump and need to be sure the pumping costs will not be more than my heating and cooling cost to operate.

 

[AxisCat]

Also, I am not completely on my own with this. I plan to reach out to some of my experts for guidance on the design. I was just trying to understand the underlying math before I did so.

 

[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