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heat transfer coefficient

heat transfer coefficient

heat transfer coefficient

(OP)
Hi,

I need to know the heat transfert coefficient to calculate heat losses for a tank.

here some informations about the tank:
-vertical cylinder
-diameter: 86.5 ft
-height: 34.12 ft
-wind speed: 4.027 m/s
-temperature of ambiant air : -30 degree C
-temperature of the interior surface: 5 degree C
-the tank is empty
-thickness: 0.25 in
-shell material: steel 44W

Do i have to use equations for a cylinder in cross flow or a plane wall?

thx.

RE: heat transfer coefficient

At that large of a diameter I can't see that it would make much difference.  Use the easy one.

= = = = = = = = = = = = = = = = = = = =
Corrosion never sleeps, but it can be managed.
http://www.trenttube.com/Trent/tech_form.htm

RE: heat transfer coefficient

(OP)
If i use plane wall equations, heat losses are 309.1 kW, and with cylinder in cross flow equations, heat losses are 241.7 kW.

I dont know whats going on with these results. There's a big difference.

Is there someone who try this problem to compare my results?

RE: heat transfer coefficient

Kern recommends the following for external heat transfer coefficients (natural convection).  If you look in heat tracing catalogues, they will typically have factors to adjust these for wind velocities as a multiple over still air heat loss.

horizontal pipes hc = 0.5*(dt/do)^0.25
long vertical pipes hc = 0.4*(dt/do)^0.25
vertical plates less than 2 ft high hc = 0.28*(dt/z)^0.25
vertical plates more than 2 ft high hc = 0.3*dt^0.25
horizontal plates facing up, 0.38*dt^0.25
horizontal plates facing down, hc = 0.2*dt^0.25

do pipe diameter inches
z is the height in feet
dt is the temperature difference, F
hc BTU/hrft2F

For a tank, I would assume it's a vertical plate.

You may also want to look at "predict storage tank heat transfer coefficients precisely", Chemical engineering magazine, March 11, 1982

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