Watts needed to heat a water circulating system.

[JWS7]

Watts needed to heat a water circulating system.

I am trying to heat a closed recirculating fluid system from room temp to 194F, in less than an hour.

Currently with 2800 Watts into the 5 gallon tank of deionized water, the system peeks at 180F after three hours with the water flowing.
I’m sure the problem is the heat loss into the room. We keep adding insulation, but aren’t getting closer to the solution.

How can we calculate the watts need to do the job?
The system material is 316SS, I can estimate the mass of the piping system, surface areas etc.
Tank vented, Pressure after pump 12 psi; flow 6 gpm;

Any help would really be appreciated.

Thanks, JWS7

 

[IRstuff]

"recirculating" into where, exactly? If it isn't doing anything, then it wouldn't need to be "recirculating." Since it is, then the intent must be to transfer its latent heat into something else. What is that something else?

TTFN

FAQ731-376

 

[zekeman]

"..........Tank vented,"

Why is it vented? Maybe that's part of the problem.

Like IRstuff, what is the purpose of recirculating?

Please explain in sufficient detail what you are doing so we don't have to guess and give you wrong answers.

 

[JWS7]

Thanks for the response.
The water circulates to heat a complex rotary valve which is being tested. The valve is of 316SS, and weighs about 75 lbs. We are going to be testing the valve through numerous cycles which simulate the actual operation during production in the pharmaceuticals industry. All the plumbing are the “Sanitary type” (clamp w/ gaskets) fitting. JWS7

 

[IRstuff]

OK, so how did you account for the heat transfer to the valve? Given that your water volume occupies less than 1 cubic foot, and therefore has less than 1 m^2 of available heat loss area, it can't be losing much more than about 500W, even with no insulation.

You need to do the math. 6 gpm at only 3°F temperature drop would remove over 2500 W, which is nearly your entire heating capacity.

TTFN

FAQ731-376

 

[JWS7]

A centrifugal pump draws H2O from the heated tank, and pushes the water to a series of plumbing fittings and up into a couple ports of the motorized rotary valve. The water exits the valve through open ports at the bottom, falls into a drain system back to the pump. Actually we added some needle valves on the exit ports to keep the spec pressure at the valve input.
The tank is covered with Plexiglas and gasketed. I said vented because I meant it has no pressure.
Heat is lost from all parts of the system plumbing as well as the test valve. We also have a makeup water tank to account for evaporation. About ½ gallon per 10 hrs.
I am about to add a heater tape with control and wrap it around the test valve, adding 720 watts. I’m just looking for a formula to estimate the energy needed in watts to see if we are close or not. Thanks, JWS7

 

[ione]

Any chance of performing a measure of Tout (downstream the tank) and Tin (upstream the tank)?
Have you carried out any test without water flow, that is just heating the volume of water stored in your tank?

 

[JWS7]

Yes, the tank heats to 200F with the water not flowing. When we start to flow, the tank temp drops to 160F, and after a while it goes up and settles at 180F. Meanwhile, we have two other thermometers in the system, one close to the pump output, and another closer to the test valve input. These usually register 10F lower as you get farther from the tank.

 

[willard3]

Get an Engineer familiar with fluids and thermodynamics. This is a lot more complicated than you think it is.

 

[IRstuff]

There is no "formula," because you haven't even begun to provide sufficient information to do anything beyond sheer speculation.

What goes through the rotary valve?
How big is it?
You claim that your system is "closed" yet you're able to lose water from evaporation?
Where in the chain are these temp sensors and what do they see?

You need to get an engineer conversant with this type of problem. You're losing 10°F along the way, which is a substantial amount of heat loss.

TTFN

FAQ731-376

 

[ione]

Ops, I've forgotten to ask how long is the heating process up to 200 F (with no water flowing).
Anyway if you’ve detected a 10 F temperature drop along your line your heat source is definitely underrated (check previous IRstuff’s comment).

You have to heat up your pipeline, fittings, valves (specific heat for stainless steel is approx. 0.17 W/lb/ F) and compensate heat losses of the whole system. IRstuff has again given you a good figure for the heat loss throughout the tank surface, then considering only natural convection you can assume the heat transfer coefficient for your pipe is on the order of 6 W/(m^2 °C). Evaluation of heat loss though the valve is another matter.

 

[zekeman]

" The water exits the valve through open ports at the bottom, falls into a drain system back to the pump"



Looks like a wide open system to me.

Is that correst?

More like a cooling tower.


Why do you need an open system for this?

Well, if that is the best you can do structurally and you are willing to add more heat , then there is no need to hire an engineer for this.

I calculated that given your steady state losses of 2800W at 180 F and making the reasonable assumption that losses are proportional to (T-Tamb) and a system time constant of 1 hour, you would need 4500W to raise the water to 194 in 1 hour, using a 70 F starting temperature.

 

[JWS7]

Thanks Zekeman. The exit port falls into the return bucket because the test valve is motorized and rotating. The guard around this zone where I lose some water as the condensation drips out. The tanks heats to 200F in less than an hour.
I appreciate your 4500W estimate. This info what I was looking for. My current plan is to add 720 W heating tape to the test valve OD, ; and use a 1500W space heater on the metal bench top aimed up toward all this upper plumbing. That totals about 5020W. This should be enough to get our testing going.
Next I’m facing a filtering issue; because the point of the test is to measure particulate matter, and I cannot keep the 10 micron filters clean enough to keep the flow up to 6 gpm. The water system needs to be cleaner before we can measure accurately the PM. I wonder if there are experts somewhere on this forum for this subject. Thanks, JWS7

 

[ione]

JWS7,

If it takes approx 1 hour to reach 200 F (assuming 68 F as starting temperature), heat loss in your tank are on the order of 1200 W. It seems a bit high to me, anyway I suppose you are sure your heat input is 2800 W.

Frankly I doubt only another 4500 W could compensate the heat loss of the line (according to the info you’ve passed on to us, that is 10 F temperature drop detected with 6 usgpm flow). I would go for 12 kW overall heat input.

 

[JWS7]

Thanks ione. The tank is actually a heated water bath (rated to 200F) with 600W heater and rated at 2048 output BTU. We added two 1100W immersion heaters. It has insulation below in its cabinet, but not great insulation. We had a pipe welded to the tank bottom for the pump suction line. We changed the SS lid for Plexiglas so we could close-fit the fittings.
Another issue is the SS tank is showing some rust spots, and this is putting contamination into our system that we cannot tolerate. We covered the numerous spots with JB Weld, but we may need to have the tanks coated with PFA or something.
Your 12KW estimate is scary. My assistant says the static tank heats to 180F in 45 min, but it take another hour to reach 200F. Maybe we need to improve the tank insulation.

 

[zekeman]

"that is 10 F temperature drop detected with 6 usgpm flow). I would go for 12 kW overall heat input."

Ionne

Let's see, the 10 degree drop @6 gpm cannot square with 180 degree equilibrium and 2800Watts.


Jws,

Is 10 gpm with a 10 degree drop an equilibrium condition, and exactly where are the measurements made.

 

[JWS7]

Hi Zekeman. Yes, the 6 gpm and the 10F less at subsequent points seemed steady for three hours or so. The tank was 180F. The next measurement point is about 5 tube- ft away after a few large SS valves and tees. The last measurement point is after some more valves, tees, and hose another 5 feet away; just as the water enters the test-valve manifold system. The test valve itself is probably lower temp; I will need to measure it with a heat gun for the next test run.

 

[mjpetrag]

Easy experiment would be to run a smaller heater in your tank (maybe try one of those 1100W heaters you got), wait for it to get to steady state with the pump running at 6 gpm and record the temperature and time. You have one data point, steady state temp on the x-axis and wattage on the y-axis reaching steady state in a certain amount of time. You already have a second data point for 180 deg steady state with 2800 W in a certain amount of time. Extrapolate the Wattage to 194 F and multiply by the desired time you want to heat it in. I think that might get you in the ball park with an oversizing factor to be sure without doing the tedious heat transfer calcs.


Are you controlling the temperature somehow to the heater? You will overshoot if you want to heat it in one hour and keep having to turn the heater on and off manually if not.

What do you guys think?


-Mike

 

[zekeman]

" Zekeman. Yes, the 6 gpm and the 10F less at subsequent points seemed steady for three hours or so. The tank was 180F. The next measurement point is about 5 tube- ft away after a few large SS valves and tees. The last measurement point is after some more valves, tees, and hose another 5 feet away; just as the water enters the test-valve manifold system. The test valve itself is probably lower temp; I will need to measure it with a heat gun for the next test run."

This is very confusing.You use terms like tube-feet and now you say several valves, etc. As I stated earlier, the equilibrium 180 @ 2800 Watts is at an severe odds with the delta T of 10 degrees and 6gpm which doesn't make sense to me and resulted in the wildly different estimates of mine and Ione.... unless you are talking 2 different experiments. It should not be so difficult to draw what you are trying to say, since your explanations are not clear.

Accordingly, I would respectfully ask you to show us a sketch of what this thing is doing, including all of your heat inputs and flow rates,pump location, tank runs of piping and the valve(s) being tested. I think you have simple problem that is not being presented properly.

Very confused

Zeke

 

[IRstuff]

There's a gigantic disconnect:

4.1868(J/gm*K)*.9718(gm/cm^3)*10(°F)*6(gal/min) = 8.56kW

By this calculation, your tank should never get much above room temperature.

TTFN

FAQ731-376