Tank Heating Coil Design 1

[JDN]

Hi

I'm in the process of sizing up a tank coil for heating heavy viscous crude. I'm targeting a temp difference of about 36 F. The tank is relatively large (5000 bbl)and once full will remain full as it is set up such that fluid entering the tank overflows to a second tank via an internal riser. Heavy crude is constantly entering the tank at a significant rate. I've done the math on heat duties and have selected a conservative overall U. I'm using a heat medium system at 385 F. My question is this. How can I take credit for the bulk fluid temp in order to reduce my required coil area. The way is stands now, I would require alot coil in order to accomplish the task. My gut tells me that if the bulk fluid temp is already at 176F and my incoming fluid is 140F, there must be some way to apply credit for the bulk temp. Has anyone come accross this before?

 

[willard3]

U x Area x delta T = BTU/hr is a simple approach and relatively accurate.

You know the btu/hr and U will be constant. Delta T and area are the only variables and you know the delta t between the cold product and final temp.

This is not very complicated or am I missing something?

 

[JDN]

Hi

The problem seems rather simple if I dismiss the credit for the large volume of hot fluid and just assume the tank coil has to deliver all the heat. However, I want to optimize the coil area by taking credit for the tank heat. The inlet flow volume is approaching 22500 BBl/d at 140 F. The tank though should be at about 176 F. As fluid enters the tank and is mixed with the hot fluid there should be some temperaure gain even without the coil, I'm not too sure how to determine what the gain is.

 

[JimCasey]

The coil has to deliver all the heat.

If you look at it on an instantaneous basis, if you have a preheated tank and you add new oil to it, it will heat up the new oil, but it will cool the existing mass of the tank by the same number of BTUs.
On a continuous basis, you have to supply enough heat continuously to heat the inlet stream to the setpoint temperature.

Where I went to school, energy was conserved. Q in=Q out.

 

[sshep]

Not only is Jim absolutely correct, but for coil sizing you must use the bulk tank temperature for your heat transfer calculations (dT=385F-176F=209F, which seems very good). Slightly less heat transfer area (ignoring viscosity, thermal conductivity impact on U) might be required if you were only heating the incoming fluid (140F) since the LMTD will be better. As usual the entropy effects of mixing are working against you, so whatever coil size you calculate using the bulk tank temperature is what what you need.

If you want less coil length then you should consider a coil using some sort of extended area design.

best wishes,
sshep

 

[JDN]

HI

I agree with all of the above, what I failed to mention was that I am willing to live with a 5-10 C deviation from setpt due to imperfect mixing in the tank and variable flow of the fluid entering (could swing as low as 50% of max). I wanted to ensure that I'm not installing a coil that's suppling 100% of the duty 100% of the time.

Thanks for all your help, but I believe the problem is solved.

regards

 

[thermcool]

Accuracy within +/- 5 or 10 C can only be achieved with the help of some active control element. In your design the liquid flow temperature is changing randomly which can makes it difficult to control the temperature within accuracy you mention.

Even the accuracy of 2-3 C require capillary thermostats.

Try to put the stirrer inside the system for better mixing.

If you are atleast sure about the type of deviation of the incoming fluid (positive or high temp, negative or low temperature) than the fluid in the tank, you can accordingly account for better design of the heating coils.

If the application really require high precision and the tank size is really large try putting some active control.

 

[chicopee]

If you have adiabatic condition around the tank, then theoretically your surface area of the heating element will be based on the flow rate and temperature difference of that flow into the tank, temperature drop and flow rate of the heating media, scale build up inside and outside the heating element. Once you got that figured out then apply a 50% increase in surface area for unaccounted factors such as inefficient mixing within the tank, irregular flow rates, design flaws.

 

[mechprocess]

The process appears to be similar to a jacketed vessel being heated with continuous feed of known fluid temperature and with a non-isothermal heating medium (Assuming Dowtherm). These methods were developed by Donald Q. Kern (page 624). Although Kern only provides the initial differential energy balance equation with the final solution equation and can also be found in Perry’s. The differential equation for energy balance is

(M + w?)c dt/d? = wc(tf – t) + WC(T1 – T2)

where

M = mass of liquid in tank at time ?

W = mass flow rate of cold liquid entering tank
? = time in hours
t = temperature of liquid being heated (generally the cold fluid)
c = specific heat of cold fluid being heated
C = same for hot fluid doing the heating
T = temperature of hot fluid
tf = temperature of liquid entering tank
W = mass flow rate of hot liquid
T1 = inlet hot liquid temp
T2 = outlet hot liquid temp

The solution is based on

WC(T1 – T2) = UA?Tlm
Solve this equation for T2 and substitute it in the ODE and you should be able to separate the variables and integrate over the definite interval between t = t initial to t = final temp

If you have fluid leaving, then add that as energy leaving on the right side of the first equation above.

Fun stuff. Working on something like that right now.

 

[mechprocess]

Looks like all my Greek symbols turned into ? marks.

UA?lm should be UA DeltaT log-mean

otherwise ? refers to time in hours