LPG fueled Burner 2

[lucino]

Good day.
As part of an improvement team created by my company, I am presently making a study of minimizing LPG usage to an oven using Shellane (LPG with heating value of 49.95 MJ/Kg) as fuel to preheat a food product conveyed inside a chamber with a control temperature of 180oC.
The oven is a Japanese made unit with two burners with specifications as below:
Zone 1=105,000 kCal/Hr
Zone 2=80,000 kCal/hr.
My observation revealed that the product temperature before the oven is 70oC however to my amazement the temperature after the oven is 63oC.

How can I compute for the consumption basing on the above data (by the way no dedicated LPG metering is installed on this oven)
Can I relate product temperature to the LPG usage?
Thank you for anybody's help.

Lucino

 

[quark]

Strange. Food seems to be heating the heating chamber.

The theoretical consumption shoud be (105000+80000)x 4.18/49950 Kgs/Hr.(should be about 15.5kg/hr)
(1 kCal = 4.18 kJ)

This link gives you specific heats for various food items.

http://www.engr.usask.ca/classes/ABE/323/notes/specificheat1.pdf

Use Q = mCp(T1-T2)/t with correct temperature readings.
m = mass of food product
Cp = its specific heat
T1 = inlet temperature in proper units
T2 = outlet temperature
t = time for heating the food product.

Compare this with heat input. Let us know what happened further.

Regrds,

 

[lucino]

Quark,
Thank you for your comments as immediate as i need.
I have yet to explore the internet site you recommend.
Anyway, as an added info:
Product speed: 1.1 meter per minute
Heating chamber length: 10 meters.
Incoming Product rate : 450 kg/hr.
With the following heating control:
Zone 1 : 105oC
Zone 2 : 79oC
As I understand, the burner will heat the air supplied by a fan which in turn will heat the product.
As of this moment I still don't know how to correlate the LPG consumption with these data.
I still have to analyze and use the formula you have illustrated.
If you have further suggestion please write.
Thanks.
Regards

Lucino

 

[quark]

Lucino!

Ok. I am ready to do some homework. But you should give me these details.

1. What food you are processing?
2. Initial temperature and final temperature required for
the stuff.
3. Ambient condition at the exit of the oven.
4. If possible, the flow rate of fan.

The second stage seems to be a stabilizing zone for already heated food in the first zone. If you still observe low product temperature, one reason may be evaporation of moisture from the product. This may reduce the temperature of product. Is it possible to check the weight of product before and after heating?

Regards,

 

[quark]

Lucino!

In the absence of the critical data required, I assumed the values of specific heat and moisture content of your product. (at 2 kJ/Kg ⁰C and 50% wb) These are the average values taken from the list I provided above. Considering moisture evaporation above 100⁰C, the first chamber seems to be designed well but I think you can save some energy at the second chamber.

Regards,

 

[lucino]

Quark,
Thank you for your never ending assistance to those of us that needed an expertise such as yours.

As of the moment I am facilitating the analysis of the moisture content before and after the Oven (Pre Dryer)
so that I can have a basis for comparison.

The ambient temperature at the exit of the oven is 32oC while the temperature of the product before entering the oven is 70oC.

Fan flow rate is 80m3/min for zone 1 and 45m3/min for zone 2.

The product is breakfast cereal with yellow corn grits as base material.

I hope posted data is now complete that can merit engineering calculations possible to compute for LPG savings, if any.

Actually we are producing a variety of cereal products that as of now I am continously observing the oven and Temperature / moisture behaviour.

As new information will pile up I will make a post for your updates.

Thanks once again.

Lucino

 

[quark]

Lucino!

For such low air flow rates, the heat capacity seems to be very much higher. Infact it is difficult to assess the system without having a physical look. Anyhow I will give you some leads.

1. Check DB and WB temperature of hot air and the exit air. To check WB, just put a wet cotton wick on the bulb. Always ensure that the wick is wet. Now you can get enthalpy of air from a psychrometric chart.Download high temperature psychrometric charts from here.

http://www.handsdownsoftware.com/Downloads.htm

Now heat gain by the air = 4.5 x cfm x (H1-H2) Btu/Hr
(heat lost to the product)

1 kCal/Hr = 3.968 Btu/Hr and 1 M³ = 35.31 Ft³

As your process is already going on it should be easy for you to check these values. Ideally this should be the heat input by LPG.

2. As the air flowrate is low, you can use desiccant dehumidifiers (Honey comb type or with activated alumina) to remove moisture from the air so that you can recirculate back the air. This will eliminate exhausting hot air.But if your process involves steam cooking prior to dryer, you should be careful about case hardening of the grits.

3. Contacting manufacturer will be a better idea because there may be new modifications with the machine.

Note: Nothing needs to be told much, after all it is all in a days work.

Regards,

 

[lucino]

Quark,

Now I have the idea how to do with the savings initiative.
Your suggestion of recirculating the hot air through moisture removal by desiccant dehumidifier is well taken.
I am now downloading the chart you recommend and will be using it after I can gather DB&WB values.
Regards.

Lucino