Capacity calculation

[Peter_1]

How do I calculate the needed capacity to cool 5000 kg/h vegetables in a cooling tunnel from 20°C to 1°C with a time in the tunnel of 6 minutes?
5000 kg/hx 0.8 cal/kg°C x 19Kbut what about the latent heat?

[frank]

5000kg/h = 500kg per 6 minutes so your calc should be for 500kg (the actual load in the tunnel at any one time.

The latent process can be ignored as you are not going below freezing (phase change)

So, your calc should be
Q = m x C x dT all divided by the time is seconds and chilling factor.
What vegetables are they Peter?

[Andy]

How do I calculate the needed capacity to cool 5000 kg/h vegetables in a cooling tunnel from 20°C to 1°C with a time in the tunnel of 6 minutes?
5000 kg/hx 0.8 cal/kg°C x 19Kbut what about the latent heat?

Peter I make it 104kW including 10% fan load. thats taking the product specific heat capacity of 3.7 kJ/kG

No allowance for air change or infiltration in that.

Probably a 125kW tunnel and plant, giving a good safety factor

Kind Regards Andy

[Peter_1]

I explained/translated it the wrong way I see now.
It should have been respiration heat, due to the fact that vegetables are a living metabolism and produce heat, other then meat.

[frank]

I explained/translated it the wrong way I see now.
It should have been respiration heat, due to the fact that vegetables are a living metabolism and produce heat, other then meat.

I don't think you did Peter. Specific heat Capacity before freezing is just the same for meat or vegetables.

Have a revision look in Dossat.

[frank]

Using a chilling factor of 0.85 I get 114.8KW. Add to that a safety factor and I agree with Andy's suggestion of 125kw.

[Peter_1]

See once this link http://www.agr.gc.ca/cal/epub/1532e/1532-0026_e.html

[Josip]

Hi, Peter_1

I explained/translated it the wrong way I see now.
It should have been respiration heat, due to the fact that vegetables are a living metabolism and produce heat, other then meat.

See attachment,

Best regards, Josip

[Peter_1]

Thanks all,

I need to use figures which are correct.
It's for the case with the potatoes cooling tunnel.

I had a small problem with the HD in my head

I had a table which gave BTU/lb and thought there was something wrong with it and I see now that should have been BTU/lb/hour.

Is following calculation then correct if I count the load per hour
5.000 kg/hour (11.023 lb/hour) entering at 20°C (68°F) and leaving at 1°C (33.8°F), specific heat of 0.8 and a respiration heat of 0.028 kcal/kg°C (or 0.028 BTU/lb/hour)
gives (5000 x (20-19) x 0.8) + (5000 x 0.028 x 19) = 76.000 kcal + 2.660 kcal = 78.660 kcal or 91 kW + eventually additional load like Andy said for the fans but the motor of the fans are outside the tunnel and the fanblade itselves is connected to the inside via a long shaft/rod on the motor.

Or (11.023 x (68-33.8) x 0.8 ) + (5000 x (68-33.8) x 0.028 = 301589 + 10555 = 312144 BTU/hour)

[Josip]

Hi, Peter_1

Are you sure you can cool 5000 kg of potato from 20C down to 1C in one hour

Maybe I miss something

http://www.uidaho.edu/ag/plantdisease/pstore.htm

http://www.uidaho.edu/ag/plantdiseas...re.htm#anchor4

Best regards, Josip

[Peter_1]

Corrected givest his:

5.000 kg/hour (11.023 lb/hour) entering at 20°C (68°F) and leaving at 1°C (33.8°F), specific heat of 0.8 and a respiration heat of 0.028 kcal/kg°C (or 0.028 BTU/lb/hour)
gives (5000 x (20-19) x 0.8) + (5000 x 0.028) = 76.000 kcal + 140 kcal = 76.140 or 88.5 kW + eventually additional load like Andy said for the fans but the motor of the fans are outside the tunnel and the fanblade itselves is connected to the inside via a long shaft/rod on the motor.

Or (11.023 x (68-33.8) x 0.8 ) + (11.023 x 0.028) = 301589 + 309 = 301897 BTU/hour

Im' busy with the report this Sunday. Already 35 pages.