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

[frank]

Thats alright but you are talking about 6 MINUTES so I think you can ignore it

the following was taken from your link

he longer the time required to cool the produce, the greater the total respiration heat.

[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]

It are in fact fresh French Fries, at a rate of 5.000 kg/u. Tehy have to be cooled down to 1°C before packing.
It's a coninue line from raw potatoes till packed FF.

[Josip]

Hi, Peter_1

It are in fact fresh French Fries, at a rate of 5.000 kg/u. Tehy have to be cooled down to 1°C before packing.
It's a coninue line from raw potatoes till packed FF.

It is clear now

Best regards, Josip

[US Iceman]

Peter,

You were asking for some comments on the load calculations. I see you are still working on the french fry cooling tunnel problem.

...for the fans but the motor of the fans are outside the tunnel and the fan blade itselves is connected to the inside via a long shaft/rod on the motor.

It does not matter if the fan motors are external to the tunnel. The motors are still adding energy to the air (and load) inside the tunnel.

The fan motors will have some inefficiency, this is the heat that is produced by the motor. Since this is external to the tunnel this heat does not become part of the refrigeration load. This is normally about 15-20% for small motors.

The remaining energy is used to turn the fan blades. Since the fan blades are inside of the cooling tunnel, the majority of the energy used by the motors is absorbed by the air.

Does this cooling tunnel also have an exhaust fan to remove the CO2 or nitrogen ( I forgot which one you were using) from the tunnel. The volume of air this fan moves could also be considered part of the cooling load (at least as far as the cyrogen is considered).

You will also have some air cooling load that enters the tunnel. This might be part of the exhaust fan volume flow, I'm not sure what you have.

Also, the respiration heat is based on Energy/Mass-Time. So in IP units, you are looking at Btu/pound-time (the time rate could be minutes, hours, or 24 hours).

In your calculation you show:

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

The 5000 shown in red should be 11.023 (lb/hour). However, the respiration heat should not include the dT for the product.

If the respiration heat is 0.028 Btu/lb-hr, then you should be multiplying 11.023 pounds X 0.028 Btu/pound-hr to get 308.6 Btu/hr. The pounds(mass) cancel out each other, so you are left Btu/hr.

If you multiple 11.023 pounds per hour X 0.028 Btu/pound-hr X dT the units get scrambled up into something that does not makes sense. (Btu-F/hr)

Does that help?

[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.

[US Iceman]

Hi Peter,

The last post you made I agree with. You should also add any additional motor power input to the load calc.'s, i.e., conveyor, etc.

Good luck with the report.