2 replies to this topic

[Ben2009]

How do i calculate heat loss of non insulated vessel by only measuring the outside temperature of the vessel, i don't know the inside temperature of the wall.

any thought?

Thanks

[Art Montemayor]

Ben:

The inside pipe wall temperature can be computed from the calorie temperatures when both the inside and outside film coefficients are known for the pipe in question. You don’t state where you are located, so I can only tell you that you should obtain a copy of Don Q. Kern’s famous 1950 text book, “Process Heat Transfer” and look in page 97.

The equation derived by Kern is:

t_w = T_c – [h_o / (h_io + h_o)] (T_c – t_c)

Where,

t_w = Outside pipewall temperature, ^oF
T_c = Caloric temperature of hot fluid, ^oF
h_o = outside film coefficient, Btu/hr-ft²-^oF
h_io = inside film coefficient referred to outside surface, Btu/hr-ft²-^oF
t_c = Caloric temperature of cold fluid, ^oF (the temperature of the properties at which the film coefficients are computed.

I hope this helps you out with your calculations. Get a copy of Kern'a book and keep it with you - even after memorizing it.

[kkala]

How do i calculate heat loss of non insulated vessel by only measuring the outside temperature of the vessel, i don't know the inside temperature of the wall.any thought?

A practical approach (used in my graduating work for a rotary kiln) is to estimate heat losses to the environmental air, since you know the outside metal temperatures - probably measured by a contact thermometer. You can divide the external surface into several areas of constant temperature, then find heat loss for every area and sum the total.
Heat loss to base ground is comparatively small and can be neglected (conclusion based on guidance for heated large fuel tanks, comment from others would be welcomed).
Heat loss to environmental air is by radiation and convection.
Concerning natural convection G. G. Brown (Unit Operations, Wiley 1950, Chapter 29 - Heat transfer 2) gives following correlation for heat transfer coefficient (h, apparently in Btu/(hr*ft2* 0F)) to environment (vertical plates over 3 ft high).
h=0.3*(Ts-Tair)^0.25, where Ts=surface temperature, Tair=air temperature, both in 0F. A point to start, assuming e.g. 60 0F as ambient air temperature. Detecting local temperature drops of external temperature measured (irregularities of temperature distribution in general) makes a sign of local wind currents, multiplying the convection losses.
Heat loss by radiation calculation adopts surface emissivity (from tables, e.g 0.85), uses absolute temperatures, and often does not consider influence of neighboring objects. At least here, radiation calculations are simplified (as in convection), but do not neglect them as insignificant unless you have verified it (such error was made here).
Well, do not expect precise values, yet calculation of such heat loss is an exciting and probably useful experience.

Edited by kkala, 24 January 2010 - 09:42 AM.