heat transfer from pipe coil in tank

[khardy]

Ladies and Gentlemen,

I don’t do heat transfer very often so I thought I post my problem and make sure I’m handling this correctly.

I need to heat asphalt in a tank using pipe coils circulating hot oil. The tank will be agitated and the one-day temperature change would be 307-F to 310-F or 3-F per 24-hr period. I’ve got a good handle on the heat load on the tank: 600,000-btu/hr. For determining the length of pipe necessary I proceeded as follows:

Product temp, 1 307-F Assumed. Client gave me a 3-F per 24 temp change but didn’t give me an initial temp. This assumption gives the largest heat loss through the tank and insulation.
Product temp, 2 310-F given
Hot oil tem 1 425-F given
Hot oil temp 2 325-F Initial Guess
LMTD 50-F
Overall heat transfer 15-btu/hr/sf/F Assumed. Oil to oil with agitation
Area required 800-sf calculated with A=Q/(U*LMTD)
OD of pipe 2.375” Client wants to use 2” pipe
Length of pipe 1287-ft

Now I calculate the heat loss from the hot oil.

Flow rate 30-gpm initial guess
Mass flow rate 15,017-lb/hr calculated with specific gravity of 0.90
Specific heat of hot oil 0.41-btu/lb/F Assumed. This will be verified.
Heat loss 615,705 -btu/hr Calculated using hot oil delta T.

Putting this into a spreadsheet I keep changing hot oil temp 2 and the flow rate until the heat losses equal and the flow rate isn’t too high for the pipe coil.

Am I on the right track?

 

[Latexman]

Yes, you are on the right track. Doing it iteratively in a spreadsheet is fine. If you can lay your hands on Kern's Process Heat Transfer or Perry's Chemical Engineers' Handbook, they have a closed form solution of the differential equations. This would be an excellent way to verify your solution.

I have never heated asphalt before. How does your scenario compare to melting the asphalt? Which would be worst case?

Good luck,
Latexman

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[zekeman]

You can do it by writing one differential equation for a single pass assuming the mass temperature is almost constant. Get the slug of heatdelivered ,Q during that pass=rhoAVc(Tf-Ti)
Next write a difference equation for each increment of transit time
W(Tn-Tn-)=Q(Tn-1,Ti)-hA(Tn-1-Tout)
I'll be more explcit tomorrow,when I'll have more time.

 

[speco]

Khardy,

Tank heaters of this type often used finned tubes or finned pipe. If you have that option, you could do the same job with muchy fewer linear feet of pipe.

Regards,

Speco

 

[zekeman]

The problem is easier than I thought. (The usual problem is to heat from a low temperature within a prescribed time)
No need for iteration in this case.
First I solved the differential equation and got
(1) T-T0=(Ti-T0)exp(-kx)
T0= asphalt temp
T = temperature in pipe at position x
Ti entering hot oil
k defined =2*Pi*r*U/w'c, 1/ft
r=pipe radius
U overall conductance, 15 BTU/hr-ft^2 F
w' mass rate of oil flow, lb/hr
The heat transferred is
integral of 2*Pi*r*(T-T0)*dx
Substituting in eq (1)and integrating from x=0 to x=L
(2) q=2*Pi*r*kU(Ti-T0)[1-exp^-kL]/k
which after substitution becomes
q=w'c*(Ti-T0)[1-exp^-kL]
from this equation , if you assume a w', then you can directly get L, the length of the pipe.Setting q= 600000, the approximate load , I got
for W'=15017 (OP assumed )
L=1382 ft (looks suspiciously close to your "first guess")
After you get an acceptable w',L combination, you almost have a solution, since that combination only sustains the temperature . You must add additional heat to cause 3deg/hr increase or
WCadT/dt=3WCa
where
W is weight of fill
Ca= specific heat asphalt,
so q now becomes
600000+3WCa. Go back to eq (2) for new combination of w',L.

 

[zekeman]

Correction

The heat transferred is
integral of 2*Pi*r*(T-T0)*dx
Substituting in eq (1)and integrating from x=0 to x=L
(2) q=2*Pi*r*kU(Ti-T0)[1-exp^-kL]/k

Should read
"The heat transferred is
integral of 2*Pi*r*U*(T-T0)*dx
Substituting in eq (1)and integrating from x=0 to x=L
(2) q=2*Pi*r*U(Ti-T0)[1-exp^-kL]/k"