lbs/hr of steam 1

[joecontrol]

Gents,

I am looking to control Rh in 5000 cu/ft of air with air replacement changes every 4 hours(low air flow).
The temperature range is 85 F(98%Rh) to 120 F(20%Rh)

The High Rh(95%-90%) at lower temps 85 F to 100 F and lower Rh(75% -20%) at higher temps 100F to 120F.
I am using dry heat (fin tubes)to heat and was going to use steam sprays to add Rh. I need to know what to expect in lbs/hr use of steam to size steam boiler for jets.

I figure there is a formula for volume of steam to increase Rh by 1% per cu/ft of air at various temps.

Is there such a beast??

With the volume of air I think the heat added by the steam will be small enough not to affect the temp wildly.Also if I place many small jets along the lenght of the space the Rh will be somewhat uniform.

space is 7'X 7'X 98'==4802 cu/ft(5000)

Air will enter from floor many small holes (1/2" on 4" centers) and exit out 4 evenly spaced draw fans(5-20 cfm/fan w vfd)
Spray misters above floor plate at floor level spaced at 10'?? along center line. Two temp/Rh sensors (rototronic) at 30' spacing(to avg). Fin coil under floor plate with holes 1/2" plate(to help with uniformity)and walk on. Hot water heated 150F to 200F.

1 small supply air fan 20 cfm to 100 cfm w/vfd control

Space pressure control by slightly faster supply fan then exhaust fan with dp xmitter and controller.
space pressure +.05 to -.05 " H20

Thanks,Joe

 

[cme]

It would help knowing the application

I am looking to control Rh in 5000 cu/ft of air with air replacement changes every 4 hours(low air flow).
The temperature range is 85 F(98%Rh) to 120 F(20%Rh)
............... You have to know your mixed air conditions, how much outside air?

The High Rh(95%-90%) at lower temps 85 F to 100 F and lower Rh(75% -20%) at higher temps 100F to 120F.
I am using dry heat (fin tubes)to heat and was going to use steam sprays to add Rh. I need to know what to expect in lbs/hr use of steam to size steam boiler for jets.

.......... this doesn't seem like an occupiable space

I figure there is a formula for volume of steam to increase Rh by 1% per cu/ft of air at various temps.

Is there such a beast??

With the volume of air I think the heat added by the steam will be small enough not to affect the temp wildly.Also if I place many small jets along the lenght of the space the Rh will be somewhat uniform.

space is 7'X 7'X 98'==4802 cu/ft(5000)

Air will enter from floor many small holes (1/2" on 4" centers) and exit out 4 evenly spaced draw fans(5-20 cfm/fan w vfd)
......... I don't know where you are getting a vfd for that low of cfm


Spray misters above floor plate at floor level spaced at 10'?? along center line. Two temp/Rh sensors (rototronic) at 30' spacing(to avg). Fin coil under floor plate with holes 1/2" plate(to help with uniformity)and walk on. Hot water heated 150F to 200F.
................. suggest some kind of ducted system

1 small supply air fan 20 cfm to 100 cfm w/vfd control .............. again, I question the vfd

Space pressure control by slightly faster supply fan then exhaust fan with dp xmitter and controller.
space pressure +.05 to -.05 " H20

what is the heat source? electric, steam boiler?

 

[joecontrol]

i will be drying a clay product. It is not a space to be occupied.We will not be recycling any air.

the exhaust fans are for chimneys and come in that small size. I had a vendor quote them to me.

We need mostly radiant heat. At High Rh amounts and we can move the air more as the water is driven out of the product.

It is a 4 day cycle!! pretty slow! we are going to try to make it go faster but we know what we can get away with right now.
we will be using fresh air(90F 76% Rh in summer and 30f and 20 % in winter) outside air in northeast USA (Erie,Pa)

Thanks for your interest,Joe

 

[joecontrol]

OOps,

Forgot to add:

420,000 btu/hr boiler for main heat source(150F to 200F hot water)

Need to know expected steam needed to size for Rh control. I would figure on a small electric steam generator.Could be gas if size is larger.

Thanks Again,Joe

 

[walkes]

You can download humidification software from Nortec at

http://www.humidity.com/

 

[joecontrol]

I got the software it calculates a number of 65 Lbs/hr but i have to increase the make up air(higher then i want) to get it to calculate for me.

i would like min air movement inside the space(20 cfm).

I guess i am looking to calculate how much Rh i will lose while i am heating 20 cfm of fresh air moving through 5000 cf of air at 90F & 98% Rh.
Would it just be the difference between the make up air and the 5000 cf of air inside my 7' X 7' X 98?

Of course in the summer it will be a lower difference then in the winter so i will need good turndown.On/off with a steam jet would be full turndown in my mind.

Would i just calculate to highest Rh load(biggest difference)?
Which would be at start up in the winter of 10F/20%Rh(outside) up to 88F/95% Rh (inside space).



Joe

 

[walkes]

The air in the space at 95-98% RH has a certain amount of moisture in it at a certain temperature. Grains of moisture/lb of air. The 20 CFM of air has a certain amount of moisture in it depending on the temperature and RH. You will need to humidify the space based on the difference in grains per lb of air introduced.

The document at the following link should explain the process.

http://axairnortec.humidity.com/data/docs/15054/03-124B.pdf

 

[joecontrol]

Thanks i think that will get me in the ballpark and i can get some idea what to expect.

Thanks,Joe

I can come up with a baseline with several different conditions and work fron there.

 

[tombmech]

With respect to a lot of these posts, this entire subject was made much more complicated than needed. Some of that came from confusion in defining the problem, though. It seems clear that you want outside air to be humidifed to a final condition after heating. Flow is important, outside temperature and humidity is not. Below freezing, the humidity numbers are so small it makes little difference. We need to determine your final requirement, instead. There are different numbers quoted all over the place, but it looks like you want something close to 95%RH at 90F.

Humidity Ratio, 90F, 95%RH = 0.0294 lbs water/lbs dry air
Humidity Ratio, 10F, 20-80%RH =~ 0.001-0.002 lbs water/lbs dry air (air below freezing has very little moisture, period)

Steam needed: 0.0294 - 0.001 = 0.0284 lbs water/lbs dry air

Air at 20 cfm =~ 20 cu.ft/min / 12 cu.ft./lb =~ 1.67 lbs/min dry air

Steam needed = 0.0284 lbs water/lbs dry air * 1.67 lbs/min * 60 min/hr = 2.84 lbs/hr.

Or, you could multiply the humidity ratio needed times the specific volume:
(0.0284 lbs-w /lbs-a * 60 min/hr) / (12 cu.ft./lbs-a) = 0.142 lbs steam per cfm

 

[ROCHESTER]

I've used the formulas in this publication to calculate the Psychrometrics:

http://web.umr.edu/~tien/218/Chap2.pdf

 

[joecontrol]

tombmech,
Well the process goes from 90 F 90% Rh up to 120F 10% Rh. I thought the hardest part would be at the point where temp and Rh are close together 90F 90% Rh . I guess i was mistaken. But the caculation you have worked up is much better then all the others i have found. I still have one question for you. Where do you think the highest demad for Rh would be? I would think it would be winter.

I expect to be drawing the Rh% douwn by heating and exhausting the moisture out of the space while not adding steam to the incoming air.

I was going about this as 2 seperate items(heating coils(heat), steam jets(Rh)).

Am I looking at this the wrong way.
I understand the best way to get Rh out of air is cooling but the process i am preforming(drying terra cotta) does not allow for low temp Rh reduction the terra cotta will crack. Also large air movement will cause the items to dry unevenly.
So my goal is to dry with radiant heat(hot water coils)running at 140 F water glycol mix with low air flow 20 cfm.
As the heat temp rises increasing the air flow up to 150 to 200 cfm to help extraxt the moisture.
And using steam injection to add moisture to incoming air to control Rh.
I am going to use a Rototronic temp/Rh sensor which i have had good luck with.

Thanks for you time and info,Joe

 

[quark]

Calculating steam consumption is the trivial part of this problem, but as I echo tombmech, the objective is not yet defined. What is the exact process and what you want to achieve? Is it heating the product or drying it or both? What is the product and what is the mass flowrate of product? What is the maximum temperature the product can be exposed to?

If I want to shoot in the dark, I widen the target

 

[tombmech]

Well, no offense - but quark has made an understatement. I hesitated to get specific with your "ranging," but you just did it again. So far, this is what you've thrown out:

85 deg.F. @ 98% RH
120 deg.F. @ 20% RH
85 deg.F. @ 95% RH
100 deg.F. @ 90% RH
100 deg.F. @ 75% RH
120 deg.F. @ 20% RH
90 deg.F. @ 98% RH
88 deg.F. @ 95% RH

Not to mention you also threw in multiple outside values to confuse the issue more:

90 deg.F. @ 76% RH
30 deg.F. @ 20% RH
10 deg.F. @ 20% RH
88 deg.F. @ 95% RH

Now you've just thrown out:
90 deg.F. @ 90% RH, and
120 deg.F. @ 10% RH.

I was prepared to offer you some insight to the physical process at hand, and how to identify your worst case and control it. Unfortunately, it's almost impossible with your effort so far. One could say, "Make up your mind!" but that's not an adequate response to what you've been doing.

Please - define your problem - period. Organize your thoughts - for once and for all - and ask your questions consistently. There are people waiting to help you.

 

[joecontrol]

Well,
Here goes!!
I need to dry a clay product without cracking it. To accomplish this i need to contol the temperature and Rh and air flow.
For heating :Hot water heating fin coil 90' long 5'wide(180f to 200f water adj)Under product racks.

Humidity:Steam generator ?? size

fresh air :Inlet fan 20 cfm up to 250 cfm vfd controlled

remove moist air: Exhaust fan 20 cfm to 250 cfm w/vfd

Recirculating fan 20 to 250 cfm w/vfd (try to save energy)

The drying program starts with the box closed 25,000 lbs of product on steel open hole racks in two lines of equal length 95'long 3'wide 6'6"'tall 2"of space between the slabs of clay product starting 4" from the floor up to 4" from ceiling in box.

The temp/Rh cycle starts at 98F 90%Rh and holds for 48 hours
then increases 5F per hr to 180F while % Rh reduces in 4.5% Rh increments per hour down to 10% Rh at 180F

The space will be slightly positive during the 48 Hr hold +.005"H20 then neutral for completion of cycle

The customer has found that large variations in air currents will make the product dry unevenly but, they also know that not moving the air causes uneven drying. That is why i would like to use vfd for incoming,exhausting and recirculating air.

I know we will always be introducing 20 cfm min.They have found that at the higher temps the air can move more and not cause uneven drying up to a point that is why 250 cfm is max incoming and outgoing air.From history we know 250 cfm is rare usually 200 cfm is max throughput for fresh air and exhaust at higher temps 120f and above.

I am sorry for all of the confusion I am finding out that the customer changes his "idea" of how it should work almost daily. So i am trying to come up with the best way to use the space and technology available at an affordable price to accomplish the task of drying the product.

I am used to working on Kilns and Large furnaces. But i feel if i get the right information i can help my customer find a solution to his problem.

We have figured we need 700,000 btu/hr to heat the product with allowance for the racks and fresh air load.

The Steam amount needed is the problem I have.

thanks everyone for your help and interest,Joe

 

[joecontrol]

Sorry,
This is in Erie,Pa. I think you might need this for avg temp and rh numbers.

Thanks again,Joe

 

[tombmech]

joecontrol,

Thank you for not being insulted at the complaints of varying parameters. It is easy to relate if you have customers that are hard to pin down. That's what we get paid for, and I will try to help.

I picked up this phrase as key:
"The temp/Rh cycle starts at 98F 90%Rh and holds for 48 hours
then increases 5F per hr to 180F while % Rh reduces in 4.5% Rh increments per hour down to 10% Rh at 180F"
Does this define the desired process?

Looking at it, your starting point should be the essential RH (steam) requirement. It should be possible to control the desired incremental RH reduction after that just by varying the fan, and as a natural consequence of increasing temperature. (I guess that's the point!) I don't have the Psych calcs or a chart in front of me, but I would plot RH vs. Temperature in those increments above 98F - without additional steam - and compare them with moisture removal rates over a range of outside conditions. I will try to do it for you Monday if no one comes up with a better response over the weekend.

All parameters should be reduced to the Humidity Ratio, which is the true measure of the amount of water vapor in the air - lbs of water vs. lbs of dry air are the most convenient units, since that will directly result in the needed steam. The humidity ratio is constant, regardless of temperature. It can only be altered in your scenario by removal of the exhaust air or injecting steam. As a matter of fact, the whole process should be modeled on a mass flow basis for the water vapor/steam into and out of the oven.

Since Humidity Ratio is the true measure of moisture content, that will be the true measure of drying your product. I have run into this scenario with paint drying processes (on a grand scale), and it is difficult for the customer to understand. Heating lowers the RH, but it is a facade for removing moisture unless there is flow to replace the heated air with air of a lower moisture content. It seems like a strange contradiction, but moisture removal does not occur without cooling. Drying your product will depend only on the humidity ratio of the cooler outside air, and the rate at which it replaces the oven air. For instance, if your outside air is already at the same moisture content (Humidity Ratio) of 98F and 95%RH (approx. 0.03 lbsw/lbsa), your goose is cooked.

You are also correct in your earlier guess that winter presents the worst case conditions. Developing the starting basis of 90% RH at the elevated temperature of 98F., at winter temperatures, will be the worst case. As I stated earlier, the differences in relative humidity at freezing temps and less matter little - it's all very low, and it increases exponentially as temperature increases. So, the end point effectively defines the amount of steam. There is very little difference for a wide range of less-than-freezing RH's. Unlike cooling, all heating processes are "sensible" as termed in the HVAC industry, meaning that no moist air processes are involved - the process is essentially "dry." So any heating will necessarily result in a drop in Relative Humidity. (Humidity Ratio - actual moisture content - remains unchanged.)

All that being said, I'm afraid you will also have a quality-control precision issue at that starting point. 90% humidity is very close to 100% and total saturation. The level of normal HVAC RH control is on the order of +-5%. which should be sufficient. However, you should be careful in your selection of humidity sensors and steam valves. Redundant failsafes are always recommended. Over-controlled condensing humidity is almost as undesired as freezing. Therefore, failsafe high-limits are often employed analogous to freezestats on water coils.

Good luck, and I will try to look at some numbers and a graph on Monday (and give you a less verbose post).

 

[joecontrol]

Tombmech,

Yes, the defined process is :The temp/Rh cycle starts at 98F 90%Rh and holds for 48 hours then increases 5F per hr to 180F while % Rh reduces in 4.5% Rh increments per hour down to 10% Rh at 180F:

We are putting condsiderations in for water all over the place.

The inner walls will be either plastic/Galvanized sheet. and the lower half will be concrete with drains in the floor.

From what i have seen operating in the customers plant the product brings plenty of moisture with it to the process. So the primary need for steam is to deal with the fresh air lower Rh and the heating/exhausting loses of Rh.

I think that recirculating some of the heated and humidified air will reduce the need for large additions of steam and heat. I was thinking to recycle anywhere from 20 cfm to 250 cfm depending on what the clay will let me get away with.

Thanks,Again,Joe

I never get upset with engineers,they are sometime so focused they don't see that everyone is not as smart as they expect.But with a little explanation others can understand.

 

[quark]

Joe,

The humidity ratio of air at 98F and 90% RH is about 0.03622lb/lb. The maximum quantity of moisture it can hold corresponds to 100% RH condition and is about 0.04051lb/lb. The specific volume of air at 98F and 90% RH is 14.871cu.ft/lb(20 cfm is 20/14.871 = 1.35lb/min). If I consider a maximum moisture content of 20%, this comes out to be 0.2 x 25000 = 5000 lbs.

1.35 lb/min of air can carry (0.04051-0.03622 =)0.00429lbs of moisture. So you should should provide 5000 x 1.35/0.00429 = 1573426 cf in one minute for total drying or 26223 cfm(1573426/60) if you want to dry out in 1 hour or 546 cfm(26223/48) if you want to dry the product in 48 hours. If the inlet air is dry, you can reduce the flowrate and no. of working hours further.

Note that humidification process is isenthalpic(except for steam injection)and this will reduce the air temperature. By humidification, you are reducing the moisture carrying capacity of air thus increasing the drying time substantially and also losing lot of heat.

My idea of doing this will be first to check the allowable shrinkage rate of the pottery to avoid the cracking. Check the actual moisture content of the product by taking wet and dry product weights. Devise your control scheme so that the drying rate will not exceed the allowable shrinkage rate. I will approximate maximum total shrinkage as the volume of water that was evaporated from the product.

Instead of putting moist air, I try to control the drying rate. Download PsyChart from the link

http://www.heatcraft.com/resources/software.asp

You can perform some high temperature calculations with this tool. Once you exactly check the moisture content, calculate the stabilization temperature and cfm for moisture carryover, the way I did earlier. Further drying, in the absence of shrinkage data,can be designed by trial and error as you already have variable speed fans.

I am eager to see what tombmech comes up with.

Regards,

 

[tombmech]

OK, a good Engineer is lazy, so I altered your start point to 100 F. That turns your increments of temperature and humidity into nice, round numbers (5 deg. F and 5% RH). Converting for Erie,PA altitude (670 ft) and iterating numerous times, I developed this table and graph for you. Essentially, it shows what I tried to describe up there: Relative Humidity is meaningless in a drying process. RH is always used, because that's the only quantity that most people understand. As you can see though, it leads to undesireable effects in a drying scenario.

To maintain this proposed process, your worst case steam load actually occurs at 155F and 35%RH. That means you will be adding the most moisture in the middle of the drying "turn down." This is plainly obvious in the graph. So, if you are ultimately held to this process, your fan and steam injector will be doing a tap dance. However, I don't think your customer actually wants the amount of moisture in the brick to bounce up and down.

Ignoring the 48 hr sink time, the humidity ratio of outside air will not vary much over the 16 hr period represented by this process. Unless this 16 hr period encompasses morning dew or rain, the amount of moisture will not appreciably change. Therefore, it should be possible to control moisture removal with a much more linear control on fan flow. It would be much better to explain this phenomena to your customers. You should suggest to them that they want to control the drying process by actually controlling the moisture content in the air (vs. product).

[tt]
No Humid Add Desired
Temp RH H. Ratio RH H. Ratio
100 90.00% 0.0384 90.00% 0.0384
105 75.20% 0.0384 85.00% 0.0423
110 62.95% 0.0384 80.00% 0.0495
115 52.79% 0.0384 75.00% 0.0504
120 44.35% 0.0384 70.00% 0.0545
125 37.32% 0.0384 65.00% 0.0584
130 31.45% 0.0384 60.00% 0.0620
135 26.53% 0.0384 55.00% 0.0652
140 22.41% 0.0384 50.00% 0.0678
145 18.94% 0.0384 45.00% 0.0694
150 16.03% 0.0384 40.00% 0.0670
155 13.56% 0.0384 35.00% 0.0691
160 11.48% 0.0384 30.00% 0.0666
165 9.72% 0.0384 25.00% 0.0620
170 8.23% 0.0384 20.00% 0.0552
175 6.96% 0.0384 15.00% 0.0457
180 5.88% 0.0384 10.00% 0.0335
[/tt]

In the graph below, %RH and Humidty Ratio times 1000 is the Y axis, and degrees F. is the X axis. %RH is graphed in black. The black, smooth, dotted arc represents the physical change in %RH if the temperature is increased with no moisture addition (no change in Humidity Ratio). It resembles a natural logarithmic decay and is a measure of the increasing ability of the heated air to hold moisture. The solid black line represents the %RH "desired" for the process.

The dotted red line represents the Humidity Ratio, in lbs-water/lbs-mass x 1000 at the start conditions (100F, 90%RH) and at every other position if no moisture is added. As referenced in the previous paragraph, this causes the %RH to form the logarithmic curve.

The solid red line represents the Humidity Ratio that must be achieved if the straight line process for %RH is desired. As you can see, it is quite a "tap dance". Again, the Humidity Ratio is the actual measure of moisture in the air - not the %RH - and is the real quantity that must be removed if drying is to occur. Quark and myself provided adequate reference for how to calculate the total amounts using the fan cfm and the oven volume (use mass flow rates to calculate the change per time).

Finally, please note that references were given for the equations governing the physical relationships. However, almost all of these calculations depend on the empirically obtained thermodynamic properties of water. Calculating the partial pressure of water at various temperatures is therefore a choice among several, higher-ordered curve-fitted equations. (In other words, this was a lot of work.)

A full size version may be found here:
Drying Process

 

[tombmech]

Oops! Please note that the Y-scale for the Humidity Ratio (red lines) goes from 30 to 80 (divided by 1000). I cut it off in the interests of making the graph large enough, yet still within the 400 pixel width limit.