open loop question

[nnjunger]

I have a process problem that I am trying to solve. I will try to provide all the background I can.

I have a tank, volume is approximately 10,000 gallons, surface area is 628 ft^2. This tank is initially filled with water at 60 degrees F. I have a constant supply of water to the tank at 60 degrees F and 9,658 gallons per hour. I also have the same volume of water leaving the tank to go to the process. The problem is we are now proposing to add a heat source of 2,000,000 btu/hr to the tank. The tank is in a room that is at a constant 60 degrees F for the purpsoses of this excercise. My question is, what temperature will the tank and the exit water be expected to stabalize at, and how do you work the calculations for this.

 

[friartuck]

Hi nnjunger

Sorry I have to convert to metric

I am allowing 4.54 litres per gallon (so you have 9658x4.54=43847litres/hr or 12.1798l/sec)

your heat input is 2000000btu/hr or 586.17kW (586.17kJ/sec)

Q=mc DT
m = mass flow kg/sec
c=SHC of water 4.2kJ/kgK
DT=temp rise in Kelvin or C
Q=kW or kJ/sec

The tank has an area of only 58m2 and the room is more or less at the same temp so the heat loss will be small. If the loss is important, simply insulate the tank and you can eliminate this part from the problem.

Rearrange Q=mc DT

Dt= Q/mc

586/4.2*4.2=11.45 C rise

So the water will increase by 11.45 C at steady state and ignoring tank losses. The 60F or 15.55C will heat to 27C which isn't hot and the loss from the tank QL-will be:
QL= UA DT Watts

U = thermal transmittance of tank wall
A = area of tank
DT = temp diff i.e. 11.45 C in this case since water will be 27 and the room is at 15.55

I dont know what the U value will be but I would guess around 7 or 8 for a simple steel tank. or only 0.3 or so for a well insulated vessel.

I also ignore heat loss thro the vent.

I would gueass a loss of only 58 x 7 x 11.45=4648Watts which pails into insignificance when you compare with the 586kW input

Hope this helps



Friar Tuck of Sherwood

 

[nnjunger]

Thanks for your help Friar Tuck,

Actually, the tank is fiberglass, and I am slightly worried about the heating affecting other portions of the process. However, I also think the heat input is a little higher than what will actually be observed, so I think the process will be okay.

 

[poetix99]

Friar Tuck appears to have made an accurate calculation of 11.5 C° temperature rise for a well-mixed tank.

There are MANY other factors that ought to be considered:
What is the temperature of the 2E6 BTU/hr heat source?
How, for that matter, are you sure that the heat input is 2E6 BTU/hr?
Where is the heat source with respect to the tank? Top? Bottom? Sides? Immersion heaters?
Where is the water input?
Where is the water exit?

The warmer water is less dense. The tank will immediately stratify; you could unwittingly heat a lesser amount of water to a much higher temperature.

Do you want to mix the entire volume? A tank is typically not as good a heat exchanger as a, well... a heat exchanger is. If you have changed your process, you might need to change your equipment (or add another component).

 

[friartuck]

Good point poetix, what i also wanted to add was what if the flow stopped but the heat did not. And rightly as you say, my calcs are for perfect mixing.

Putting 2MBTU in a tank would probably mean steam injection (guess)and steam injection wouldn't be the most accurate method of achieving 2MBTU. Also, a coil would be sited probably at the base of the tank and water below it would not be heated satisfactorily.(hence mixing would not be perfect.

PS if you need to calculate the temp of water with no flow, use the mc DT equation again.

Friar Tuck of Sherwood

 

[sailoday28]

Is the tank always flowing full?

 

[nnjunger]

Lots of good questions. The heat input is actually from a heat exchanger. The water for the heat exchanger will be pulled from the bottom, returned to the top. The amount of heat generated as given is worse case scenario. The pull to and from the tank should be relatively constant, however, there will be times when we are not pulling the water flow, however then the heat input should drop. The heat input was shown to be 150 degrees F. The high flow rate water enters on the bottom and leaves on the bottom. The tank can be assumed to be full, and there is alarming for the process to indicate when this is not the case.