transient heat transfer help

[shoot_j]

I am trying to understand the basics and have made it this far but I am sort of stumped.
Can someone help guide me through the formula? :confused: Please bare with me as I am new at this.

Outside properties: air density @ 70F(21.11_C) = .0745 lb/ft3 (.0012g/cm3)
Inside properties: 1lb of water liquid density @ 38F = 62.43 lb/ft3 (1g/cm3)
Insulative/container properties: thermal conductivity 15.5 mW(m·K), .107Btu/hr-ft2-oF, .375ft2(355.48 cm2) (surface area based on 3in (7.62 cm) cube) 6mm thick
Assuming the water is 38F(3.33_C) how long will it take for the outside air to raise the temperature to 39F in these conditions?

[autt]

Suppose water inside temperature is uniform all the time, the average container surface heat exchange coefficient is: inside Alpha_i=3 w/(sqm*K), outside Alpha_e=3 w/(sqm*K), then you can use this formula:
Total heat transfer coefficient
K=1/(1/Alpha_i+Delt/Lamda+1/Alpha_e));
Water temperature is:
T=Ta+(T0-Ta)*exp(-K*F*t/(m*cp));
All quantities is in metric units.
So you need to give the insulation thick Delt to complete the calculation.

[Brian_UK]

Hi Shoot, could you please use a slightly larger html font when you're posting.

We don't all have such young eyes as you ;)

Cheers....

[shoot_j]

knew I had to be missing something I edited the first post to reflect the thickness and adjusted the thermal conductivity value
sorry about the font size wasn't aware my text style wouldn't transfer, I pasted from an offline Word document I was working on. Think I got it fixed though?

[shoot_j]

I am still having difficulty with this equation, can someone explain the Delta/Lambda portion of the previously mentioned "heat transfer coefficient" formula?

[autt]

This formula is typical for two fluids heat transfer calculating.There are 3 main thermal resistances in this state: both fluids convection resistance, are the reciprocal of convection heat transfer coefficient, which are known, lamda is the isolated board(insalution or container) conductivity, so delta/lamda is the thermal resistance of board between fluids, heat flow through these 3 part one by one, so total resistance is sum of 3. For more boards as container and multilayer insalutions, just sum them.

[autt]

This formula is typical for two fluids heat exchange calculating.There are 3 main thermal resistances in this state:both fluids convection resistance, are the reciprocal of convection heat transfer coefficient, which are known; lamda is the isolated board(insalution or container) conductivity, so delta/lamda is the thermal resistance of board between fluids.Heat flow through these 3 part one by one, so total resistance is sum of 3. For more boards as container and multilayer insalutions, just sum them.

[nh3simman]

Suppose water inside temperature is uniform all the time, the average container surface heat exchange coefficient is: inside Alpha_i=3 w/(sqm*K), outside Alpha_e=3 w/(sqm*K), then you can use this formula:
Total heat transfer coefficient
K=1/(1/Alpha_i+Delt/Lamda+1/Alpha_e));
Water temperature is:
T=Ta+(T0-Ta)*exp(-K*F*t/(m*cp));
All quantities is in metric units.
So you need to give the insulation thick Delt to complete the calculation.


Careful with this.

The method quoted by autt is the lumped heat capacity method and is based on the assumption that the whole product is at the same temperature. This may be OK for small quantities of fluids where convection will ensure mixing.

But as you can imagine, the core temperature of a carcass of beef will respond much more slowly than the surface temperature when placed in a cold room.