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[SeanK]

Hi

I have to design a Rectangular shape Hot box using LP Steam as heating media

Design the hot box to maintain inside the hot box the temp. at 100 deg C, LP steam available at 140 deg C and 4 bar g.

To calculate heat load by using Q = m cp Delta T, i will take the cp of ambient air and ambient air temp in as 25deg C and inside box as 100 deg C, but what about the "m". do i need to take the mass flow rate of steam, if so how can i calculate the steam mass flow rate

on what basis i can fix the hot box dimensions and steam tube diameter

Please help on this

Many thanks for your help

Regards

Sean

[djack77494]

Design the hot box to maintain inside the hot box the temp. at 100 deg C

Sean,
Above you say that you are designing the hot box to maintain the 100C temperature. The you say you are using
Q = m*Cp*dT
to calculate the heat load. This equation would apply if you wanted to calculate the unsteady state heating of the cool air that is initially in your hot box. It does not seem to apply to your situation. I suggest that you repost your query while supplying the readers with additional information as to just what exactly are you attempting to do. Thoroughly define your starting and desired ending points. Only then can we attempt an intelligent response.

[SeanK]

djack77494

Many thanks for your reply

I have to design a hot box to maintain inside the hot box temp. as 100 deg C by using LP steam which is available at 140 deg C

Ambient temp of air inside the hot box is 20 deg C, now i want to increase the temp upto 100 deg C by using steam sending through a tube inside the hot box.

Any help on how to design the hot box, i have to calculate the heat load by using Q= m cp delta T, then by using latent heat of steam i can cal. the amount of steam req. to maintain the temp at 100 deg C

But on what basis i can fix the volume of the box and how to calculate the steam tube length?

please advice me on this

Regards

Sean

[djack77494]

Ambient temp of air inside the hot box is 20 deg C, now i want to increase the temp upto 100 deg C by using steam sending through a tube inside the hot box.

Any help on how to design the hot box, i have to calculate the heat load by using Q= m cp delta T, then by using latent heat of steam i can cal. the amount of steam req. to maintain the temp at 100 deg C
Sean

Sean,
What I tried telling you is that you are not correctly approaching this problem. Please note that the "ambient air" inside the box will be at 100C; the amount of heat needed to raise the air from 20 to 100C is going to be quite small, and it only happens at the initial warmup (assuming yours is a continuous process). What you really need to be thinking about are the heat losses; i.e. from the hot box to ambient. Again with the assumptions, if the hot box is sealed pretty well, than the heat you need to supply after initial warmup is best thought of as being:
Q = U * A * dT
which is the equation describing heat transfer out of the hot box. If this is a batch process involving operations such as heating or drying, then you must consider that. If you need additional halp, please fully define your system.

[SeanK]

djack77494

Many thanks for your reply and valuable information

Sealed Hotbox volume = 0.9 m3, density of air 1.2 kg/m3, so mass of Amb. air inside the hotbox=1.08kg

I have to design a hot box to maintain inside the hot box temp. as 100 deg C by using LP steam which is available at 140 deg C ad 5 bar a

I have taken the enthalpies of air at 20 and 100deg C

I want to know hot to calculate the heat load and the heat of dissipation and length of the steam tube required

if i use Q= U*A* Delta T, where A = 2* Pi*r*l, i can get the length but how to get the Q

i think i cannot use Q= m*cp*Delta T, since m is the mass flow rate and in my case air is not flowing to anywhere but stays inside the hot box, please correct me if i am wrong

please advice me on this

Regards

Sean

[djack77494]

If you need additional help, please fully define your system.

Sean,
It is necessary that you respond to this. I really must know more about your system and problem in order to formulate a better response.

[SeanK]

Djack,

Please find the attached sketch and guide me to design the hot box

Regards

Sean

Attached Files

•  Hot_box.pdf   59.66KB   69 downloads

[djack77494]

OK Sean, the sketch greatly clarifies your situation. You have a (steam) heated enclosure through which you are conveying your antioxidant stream and you wish to avoid cooling of that stream. So yours is a problem of heat maintenance. You needn't worry about anything going on inside the box. It doesn't matter if the box is filled with hydrogen or air or water. The volume, density, and mass inside the box are not important. The only thing you do need to worry about are heat losses, and they are heat losses from the surfaces of the box to ambient. So you must calculate (estimate) the heat losses that occur from the box to the surroundings as step one. For step two you must design a steam coil that will go inside the box and will replace the heat lost. It may be difficult to estimate heat losses from the box, and this parameter is very dependent on the construction of the box, and especially how "tight" the box is made. You may find suggestions in heat transfer books and/or online that will provide you estimates of the value of "U", your average overall heat transfer coefficient. You should be able to readily calculate the box's surface area "A" as well as the temperature difference from the inside of the box to ambient. Based on previous entries, dT is 100 - 20 = 80 degrees C. After this, you will have Q. As part two of this problem, you again apply the heat transfer equation:
Q = U * A * dT
only this time you have Q, you estimate or lookup U, you have dT = 140 - 100 = 40 degrees C, and you calculate A. This new A is not the surface of the box, but is the surface of your coil. As you've previously shown that you can get from here to the length of the coil, I'll drop further description. The important point to note here is that you calculate the heat losses first, then you set the heat loss = needed heat gain, and then you calculate what is needed to obtain the heat gain.

[SeanK]

Djack,

Thank you very much for your valuable suggestion , your suggestion really gave me an idea to design the hot box and pumping my confident levels to high, thanks indeed.

Firstly I have to calculate the heat losses
Secondly I have to calculate the Heat load then calculate the A of the coil

Coming to the second step,
I will calculate Q = U*A*dT, where U i will take for the air to the mild steel, from internet i have found U as 8 w/m2k, A is the surface area of the box, dT is the 100-20=80 degC
Now using the above equation, Q as the calculated one from the above, U will be for the steam to the mild steel, from internet i have found U as 14.2 w/m2k, and dT as 140-100, then cal. the A of the coil, then cal. the length of the coil. but how can my steam out temp. as 100 degC, please clarify me
To calculate the amount of steam required shall i use Q=m*cP*dT, where Q as above cP of steam140 degC and dT as 140-100, please clarify me

Coming to the first step i am working on how to calculate the heat losses, i am very happy if you have any references on the heat losses calculation

Once again many thanks for sparing your valuable time and help
Regards
Sean

[djack77494]

You may find the heat transfer equation in any book dealing with heat transfer or transport operations. Bird, Stewart, and Lightfoot was the Transport Phenomena textbook I used in school, and that would be a good reference.

I think you've mostly got the concept, but let me review once more. You have two separate heat transfer problems to solve. For the first, you have:

Q1 = U1 * A1 * dT1

where Q1 = heat losses from the box to ambient,
U1 = Heat transfer coefficient for same,
A1 = Area of hot box, and
dT1 = Box to ambient temperature difference
= 100 - 20 = 80C

That solves your ambient heat losses. Now you must calculate the heat gain; i.e. from the steam coil to the hot box. So, your second equation to solve is:

Q2 = U2 * A2 * dT2

where Q2 = heat given up by the steam to maintain the hot box temperature,
U2 = Heat transfer coefficient for same,
A2 = Area of the steam coil, and
dT2 = Steam to Box contents temperature difference
= 140 - 100 = 40C

Your third step is to note that, by heat balance, Q1 = Q2. Now, I understand, you will need a fourth step as you want to calculate your steam requirement. Steam, of course, is what supplies all the heat; the heat supplied by the steam is Q (=Q1=Q2). For a process driven by latent heat, you solve for m as follows:

Q = m * dH(lh)

where Q = previously calculated,
m = mass flowrate fo steam required, and
(dH)lh = latent heat of steam at condensing conditions.

You will note that we still do not use the equation Q = m*Cp*dT. It simply does not apply to this type of situation.

[SeanK]

Djack,

Thank you very much for your reply

Firstly I have calculated the heat losses by using the following equation

Q1 = U1 * A1 * dT1

where Q1 = heat losses from the box to ambient,
U1 = Heat transfer coefficient for same,
A1 = Surface Area of hot box, and
dT1 = Box to ambient temperature difference = 100 - 20 = 80C

Now coming to the heat gain calculations i.e. from the steam coil to the hot box. I have to use the following equation

Q2 = U2 * A2 * dT2

where Q2 = heat given up by the steam to maintain the hot box temperature,
U2 = Heat transfer coefficient for same (known value)
A2 = Area of the steam coil, and
dT2 = Steam to Box contents temperature difference= 140 - 100 = 40C(known value)

but here i have two unknown values, one is Q2 and second is A2

I am again struck up here, please guide at this step

coming to the amount of steam required i have to use to following equation

Q = m * dH(lh)

where Q =Q2, heat given up by the steam to maintain the hot box temperature

(dH)lh = latent heat of steam at condensing conditions,

So if i calculate the Q2 then i can calculate the amount of steam required, please guide me on how to calculate the Q2 since i have two unknown values in the second equation

Thanks indeed for sparing your precious time

Regards

Sean

[djack77494]

but here i have two unknown values, one is Q2 and second is A2

Sean,
For the first equation, you know all the variables except Q1, so you calculate it. For the second equation, you do not initially know Q2 or A2, as you have noted. But, you missed a key step; perhaps I wasn't totally clear. What you also know is that:

Q2 = Q1

There is only one Q which is the heat lost from the hot box, but is also the heat gained by the hot box or lost by the steam coil. So set Q2 equal to Q1, and then A2 is the only unknown. Solve for A2, and then calculate the amount of the steam you need and you're done (I think).

[SeanK]

Dear Djack,

Many thanks for your valuable suggestions on this topic so far

I have calculated the area and length of the steam coil required, now i want to calculate the time required to initially heat the amb. air in the hot box from 20 deg c 100 deg c

For this, i have calculated the heat load required initially and the amount of steam required

then time required = mass of air/ amount of steam required= kg/(kg/hr)= hr

I have got 6 mins. time, my doubt is in reality is it really 6 minutes enough to heat the air from 20 to 100deg C in a closed hot box

Please let me know if you have any ideas on this regard

Thanks

Sean

[djack77494]

I have calculated the area and length of the steam coil required, now i want to calculate the time required to initially heat the amb. air in the hot box from 20 deg c 100 deg c

For this, i have calculated the heat load required initially and the amount of steam required

then time required = mass of air/ amount of steam required= kg/(kg/hr)= hr

I have got 6 mins. time, my doubt is in reality is it really 6 minutes enough to heat the air from 20 to 100deg C in a closed hot box

Please let me know if you have any ideas on this regard

Sean,
Your methodology confuses me. The time required to heat up the air is found by starting the amount of energy needed to raise the air's temperature (i.e. kcal or joules).
Then divide this energy by the amount of energy you will get for each kg of steam to get the amount of steam required. Finally divide the amount of steam you require by the flowrate of steam to get the time required. (This procedure assumes a constant rate of heating which may not be completely true.)
Doug

[SeanK]

Dear Djack,

Many thanks for your reply

actually i have completed the calculations, but i am bit confused with this type of system heat calculations, thus i have doubt about my time calculation

I have attached my calculations, could you please check my calculation and correct me if i am wrong.

I request the other members also to guide me on this hot box design

I have tried to upload my excel file but it was failed, that is why sending the word format file

Many many thanks in advance

Regards

Sean

Attached Files

•  Hot_Box_Calculation.pdf   131.54KB   37 downloads

[djack77494]

I did not see any problems with this in a quick look-over. I did not check any of the math however.

[SeanK]

Djack,

Thank you very much for your comments and kind help on this issue

Regards
Sean

[narendrasony]

Dear Sean,
The calculation for heat load during start up is not correct....It should be as mentioned by Djack.

It will be Qs= M * Cpa* dT
Where M = Weight of the air in the box
Cpa= Avg. specific heat of air
dT = Air temperarur temperature rise, 20 Deg C to 120 Deg C.

For Steam flow (m) requirement , Q is taken as starting heat load. It should be heat load during normal operation as you calculated in 2.1 of your calculation sheet.
For time required during start up as mentioned by Djack ts = Qs/m however this can be manipulated by changing steam flow (m) during start-up.

Also, for steam side heat transfer during normal operation (Page-3), air inside the hot-box which is enclosed, is at constant temperature of 120 DegC (and ambient air outside hot-box is 20 DegC). Please correct me if I'm wrong. So, in this case there is no need to calculate mean air temperature. Accordingly Delta T = Steam temp (160) - 120 = 40 DegC.

Hope It helps

Regards
Narendra Kumar

[SeanK]

Dear Narendra Kumar,

Thanks indeed for corrected me

I have calculated Qs = mass of air* avg cP of Air* temp diff. of air = comes in KJ
i have cp of steam in KJ/KG
thus energy required during start up/energy per kg of steam = Qs/cp of steam = KJ/KJ/KG=KG
is the amount of steam required
now for time required = amount of steam/steam flow rate= kg/kg/hr= hr

So this is my time required during start up isn't it
please correct me if i am wrongs

Thanks for your valubale suggestion
Regards
Sean