Trying to better understand Thermal Fluid BTU Loss to Calculate GPM Requirements 1

[Rbang]

Good Day All !

Although I have perused the site on many occasions and shared in the knowledge posted, this is the first time for me to request assistance.
Should I not be in the proper section , please inform me so that it may be posted accordingly. Thank You.

I have an application where I am attempting to confirm the GPM requirements of thermal fluid to convey a given amount of BTU's into a liquid.
In addition, to confirm the Surface Area and Pipe Length required to achieve the goal.

Conditions:

Liquid: Water
Weight: 8.33 lb/Gal
Volume: 10,000 Gal
Container: 8' Dia. x 28' Vertical Carbon Steel
Insulation: 3" Fiberglass

Start Temp: 70F
Ending Temp: 130F
Temp Rise : 60F

(1 BTU x 1 lb = 1 F rise in temp.)
BTU's Required: 10,000 x 8.33 x 60 = 4,998,000 BTU

Tank Heat Loss:
(K-Factor/Insulation Thickness x Surface Area x (Liquid Temp - Air Temp))
.29 / 3 x 804 x (130-60) = 5,440 BTU/Hr

Time to Heat: 4 hrs
4,998,000 BTU / 4 hrs = 1,249,500 + 5,440 = 1,254,940 BTU/Hr

Thermal Fluid Temp In: 450F
Thermal Fluid Temp Out: 425F

I have Calculated "Q" as follows:
A x Delta x U = Q
A=.622 , Delta = 337.5, U = 20
.622 x 337.5 x 20 =4,198.5 BTU per LF of 2" Sch. 40 CS Pipe

To calculate the LF of 2" Pipe required:
BTU-Hr / Q = LF
1,254,940 / 4,198.5 = 186 LF

The Thermal Fluid Properties are listed as:

Specific Heat = 0.56 Btu/Lb/F
Thermal Conductivity = .07 Btu-Ft/Ft2/Hr/F
Wt. Per Gal = 7.0

This is where I have a bit of confusion. Calculating the GPM's required.

My instincts tell me that:

For each gallon of Thermal Fluid that passes through the length of coil with a 25F Temperature drop I would lose 411.6 BTU.

Wt. per Gallon x Specific Heat x Temp Drop x Thermal Conductivity x 60 min.
7 x .56 x 25 * .07 * 60= 411.6 BTU Loss Per Gallon.

To Transfer 1,254,940 BTU's per hr.
1,254,940 / 411.6 = 3,049 GPH
3,049 GPH / 60 = 51 GPM (Could this be Correct?)

Would someone be willing to assist with a confirmation / correction and direction.

I sincerely Thank You for Sharing your Knowledge.
Rbang

 

[SNORGY]

This looks like a classic case of Newton's Law Of Cooling (Heating) for a batch operation. I think if you have access to Perry's Chemical Engineers Handbook, the equations you need are in Chapter 11.

 

[chicopee]

"Wt. per Gallon x Specific Heat x Temp Drop x Thermal Conductivity x 60 min.
7 x .56 x 25 * .07 * 60= 411.6 BTU Loss Per Gallon."
That equation should be instead:
Qloss= (7 lb/gal)* (1 gal)* (.56 btu/lb-dF) *(25 dF)= 98 btu ; the .07 value is not needed since you already know that you have a 25 dF temperature drop in the one foot line.

 

[MintJulep]

Do you know that the heating fluid temperature drop is 25 degrees, or are you just hoping that it will be?

Heat transfer from fluid through pipe into water will not be a constant value, since the temperature of the water in the tank is not a constant.

Likewise, heat loss from tank to surrounding air will not be constant. Is the air temperature 60F, and it's just a coincidence that this is numerically the same as your stated desired water temperature increase?

 

[Rbang]

I thank All for the participation with my inquiry.

That equation should be instead:
Qloss= (7 lb/gal)* (1 gal)* (.56 btu/lb-dF) *(25 dF)= 98 btu

Thank you Chicopee. This was also one of the equations I toiled with.

MintJulip.

Do you know that the heating fluid temperature drop is 25 degrees, or are you just hoping that it will be?

25F Temp. Drop into the calculations was a goal value in the worst case scenario.

Inserting a Heat transfer from fluid through pipe into water will not be a constant value, since the temperature of the water in the tank is not a constant.

The purpose was to determine the minimum GPM's required to maintain a temperature loss within the 25F range, however you have triggered some thoughts that I should back into the values from another direction to calculate the temp drop through the coil.

Heat Transfer Rate =
.07 x dT x Wt. Per Gal x SF
.07 x 337.5 x 7 x (186 x .622) = 19,132 BTU/Hr/Gal
19,132/60 = 32.2 BTU/Min/Gal

Coil Volume =
Pi x R^2 x L x 7.48
3.1416 x .00694 x 186 x 7.48 = 30.33 Gal.

30 GPM = 32.2 / .56 = 57.5 F Temp Loss
60 GPM = 32.2 / .56 = 57.5 / 2 = 28.75 F Temp Loss.

Am I following your Lead Correctly? Please Confirm / Correct. I like the results of the calculations.

Is the air temperature 60F, and it's just a coincidence that this is numerically the same as your stated desired water temperature increase?

I comprehend your inquiry because it does seem oddly coincidental. The location is in South Florida, so realistically, the Average Mean Temperature is well in excess of 60F. But, with 3" of insulation and the small dT (Liquid Temp - Air Temp), the BTU's lost are miniscule in the scheme of things.

Thank you Again for your Assistance,
Rbang

 

[MintJulep]

Am I following your Lead Correctly? Please Confirm / Correct. I like the results of the calculations.

Not quite yet.

It's a time dependent problem, but all of your equations are for steady state.

If you just want to bound the problem then you can go with that approach at the coldest tank condition and the hottest thank condition, then put in an appropriate control system.

But if your system design is constant flow and constant fluid temperature in then your fluid temperature out will be what it needs to be to satisfy the energy balance, but you will have no control over it. In this case you need a time dependent set of equations to solve, or you could probably get "close enough" with steady-state equations and arbitrarily small calculation intervals.

 

[IRstuff]

I'm thinking he's trying to calculate the required flow to achieve that drop.

5 MBTUh / 98 BTU/gal = 213 gpm Given the larger delta T than what I've seen on a previous project, it's a factor of 3 optimistic in flow rate, i.e., 400 W @ 2gpm @ 5 ºCdT. So, I think something more like 700 gpm would be more apropos. Part of the problem is that this is not a static problem, so the htc's are not necessarily appropriately applied. This is where the thermal conductivity plays into it, but that becomes a much more complicated problem that requires so finite distributed analysis.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[abeltio]

Process Heat Transfer by Donald Q. Kern has very practical equations for batch processes.

Using steady state equations, as mentioned already, will not provide the right solution.

It is also important to know if the fluid will be agitated or not, the heat transfer coefficient and the equations are quite different.

I am a few thousand miles away from my PHT copy... more later.

saludos.
a.

 

[Rbang]

I had started on a lengthy response ..but.. I'm going to save it until it is compiled comprehensively.

I will be back after I have implemented the suggestions. (to my best ability)

Thank you again for your responses !