Heat Transfer in Poultry Processing

I would first look at the process itself and get further understanding. Is the carcase movement and water movement in the same direction? (Or does the cold water come in at the end where carcases are removed?)

Contraflow means that the most heat is removed from the bath from the end where carcases are introduced, while same direction flow means that the water already warmed by the hot carcase then travels with it through the process.

You might like to do some additional experiments to further understand the process.

What is the typical temperature of carcases at each 2 minute point? You would then be able to project tank length/residence time yourself.

What is the total heat amount involved? It seems currently that you only have 50% of the necessary cooling capability. You will also need to have a lower temperature for your cooled water. (As previously pointed out you cannot get to 8 if the cooling is only at 8)

After comments which show that people many times comment without knowing what it should be said may I give you my opinion.

I made following computations based on the assumptions:

- The ice you add is to compensate the amount of heat given by the carcasses and maintain temperature constant at 8°C

You add 250 kg/h = 4.167 kg/minute. The latent melting heat of ice at 0°C is 3.33E4 J/kg, the cp of water is 1180 J/(kg*K°) this means that per minute you compensate

Q= 4.17*(3.33E4+1180*8) =1.782E5 J

Carcass is at about 90% water = 1.26 Kg. The amount of heat will be 1.26*1180*(32-19)=1.933E4 J.

If my assumptions are correct output is 1.782E5/1.933E4=9.2 carcass/min = 553/h.

Of course those values are to be taken only as order of magnitude.

Now to the cooling possibilities, carcass walls are not very thick has contact with cooling water from both sides and water heat diffusibility is not very low so that in a 1^st approximation we can consider the temperature as same through the "wall".

Then the equation will be where

M= carcass mass [kg]

Cp= specific heat of carcass (assumed equal to water) [J/(kg*K°)]

t= time

α= convection coefficient [W/m²*K°]

S= convection surface [m²]

The equation can be simplified by referring all values to θe→ u= θ-θe→

Taking into consideration that M*Cp/(S*α) has the dimension of a "time" we set it to be the time constant of the cooling process "τ".

The solution will then be u= C*exp (-t/τ) with the initial condition at t=0 u= u(0)= θo -θe

θ-θe = (θo - θe)*exp(-t/τ) or θ = θe + (θo - θe)*exp(-t/τ) Based on your data the time constant is τ≈10'.

The figure shows how carcass temperature decreases versus time. It is 32°C at entry, 19°C after 8' and goes further down depending on the time it spends in the cool water (considered constant at 8°C). It is up to you which decision you take. It is now clear that, as was written, you cannot reach 8°C if the water is at this temperature. Water has a quite high convection coefficient so that a flow increase will not bring more. But as you see you can go down to ≈12°C in about 16'. This can be done either by increasing length or with 2 lines in parallel at a reduced advance speed (half of actual one) in order to maintain output. In any case more ice should be added. You can also make a economical optimization and define for instance the length (and thus the time) as function of cost per unit of length and corresponding decrease in temperature. As length increases it becomes less economical to go further.

I hope this will help more than the previous comments.