Watt..BTU-horsepower confusion 2

[DesignerMike]

I don't do these calculations often enough, and would like someone to verify my conversions please.

I need to heat a 150 gallon tank of oil from ambient to 300 deg F.
I come up with 250,000 BTU

I want to hit temp after about 1 hour.

I have a single shell, insulated, SS tank (bottom head not insulated).

I think the easiest (and most economical) would be to use a heat blanket (like something on page 474 of McMaster-Carr)

IF I had 100% efficiency (SS will not be great)...but for simplicity lets imagine.

I come up with requiring 1100 watts to heat the tank in one hour (If the heat blanket was submerged in the oil, which it won't). We have a small mixer to help with convection.

I would probably use (3) @ 1200 watt each and hope to hit temp in 45-55 minutes.

Am I on the right track?

Thanks!

 

[MintJulep]

Don't forget that you will also be heating up the material of the tank.

 

[mpritchett]

DesignerMike -

Your numbers seem a bit off.

(3.412 BTU/Watt-hr)(1 hr)(3 blankets)(1200 Watt/blanket) = 12,283 BTU.

This is well below the 250,000 BTU you say you need.

This also assumes all the blanket energy goes into the oil, which is unlikely.

Cheers!


MPritchett

 

[owg]

See if you can find a copy of The Electric Heater Handbook, Omega Engineering Inc. 1992, 1 800 872 432. They spell out the methodology with supporting curves for heat loss, etc.

HAZOP at

http://www.curryhydrocarbons.ca

 

[smah]

If you're planning to heat from the outside as you describe, to heat the oil will require:

BTU=(Vol X Density X Specific Heat X ?T)/t
where: Vol is volume in gallons
Density is in lbs/gallon
Specific Heat is in btu/lb/°F
?T is temperature rise in °F
t is required heat-up time in hours
Multiply this by at least 15% for a safety factor (depending on the shape of the tank). I think that you'll find that you're in the vicinity of 140,000 BTU's. Now, this is what the inside surface of the tank will have tranfer to the oil. Your blanket will have to heat the outside of the tank to something much greater than that. I think that you may have missed some decimal places in your conversion from BTU's to watts. You'll need 40KW+ to heat the oil and probably that much again to heat the tank.

Steve
Eichenauer, Inc.

 

[25362]

Apparently a typical oil would need about

150 gal [×] 8.4 lb/gal [×] 0.55 Btu/lb ^oF [×] 200 ^o = 138,600 Btu or about 140,000 Btu (as estimated by smah) just for the oil.

Thus your estimate of 250,000 Btu in one hour would take:

250,000 Btu/h [÷] 3.412 Btu/W-h = 73.3 kW

and the heat-up duration with the three 1.2 kW blankets would take 73.3[÷]3.6 ~ 20 h as estimated from mpritchett's posting.

 

[bewdley]

Put the other (right for me) way round :

0.57 m3 x 850 kg/m3 x 3.6 kJ/kg x 130 degK = 222,768 kJ

Add 5 % for radiation loss ---> 234,493 kJ/3600 = 65.13 kWh

So you need this kind of energy to heat to 150 degC; up to you to decide the time...

 

[25362]

To bewdley, what particular oil has a heat capacity of 3.6 kJ/(kg.K) ?
2.3 kJ/(kg.K) seems more adequate for a lubricant at an average temperature of, say, 370 K. Do you agree ?

 

[DesignerMike]

I thought my numbers looked crazy (thus the reason for posting). Something didn't make any sense...and after a good night sleep (and some helpful posts), I can see where I blew it.

Thanks for all your help!