heat transfer coefficient

[JP123]

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.

 

[EdStainless]

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

 

[JP123]

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?

 

[TD2K]

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