Where's the energy going?? Heating Air with Steam 1

[tbaygen]

Hey Folks,
I'm interested in an open discussion for a phenomenon I'm seeing. Basically I am heating an incoming air stream on a drier with steam, I calculated the Q requirement to heat my air stream be on the order of 5MM BTU/hr, however the flow controller and steam pressure says im using 10MM BTU/hr of steam to get there.

I'm confused, where is the other 5MM BTU/hr of steam going??? I'm just guessing that a lot of the steam is not condensing and its got to be going out with the condensate before giving up its energy...I imagine if I had a 5MM BTU/hr leak this is something that would be pretty apparent. What is a typical range of efficiency of a fin/tube heat exchanger with air on the fins and steam on the tubes? Wheres it going?

Thanks

 

[zdas04]

You've got a given mass flow rate of steam at a fixed pressure, temperature, and enthalpy coming in. The continuity equation would say that you have the same mass flow rate of fluid (some mix of steam and condensate) at a different pressure, temperature, and enthalpy going out. These numbers can be very precisely determined without a huge amount of uncertainty. If that calculation works out to 10 MMBTU and you've gotten the steam quality right in your outlet enthalpy (not an easy determination, but let's say you got it right), then the missing energy either went into the air stream or leaked out into some heat sink that you are not measuring. That is what "efficiency" means in a heat exchanger--the amount of heat lost from the heating stream divided into the amount of heat added to the heated stream.

At 100% efficiency (i.e., zero heat leakage), you are actually adding 10 MMBTU to the air. Sources of error on the air side are huge. We don't very often have air-flow measurement much better than +/-20%. We measure wall temperature and assume that it is representative (it rarely is). We don't measure the pressure on the air stream at the heater (makes a significant difference). We rarely account for the humidity of the air properly. Does all that add up to 50%? Very possibly.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

 

[tbaygen]

Interesting thoughts. One adder to this is that I do see a resin (plastics) carryover into this heater box causing fouling on the heater coil. This could be part of the heat sink effect. Second I used the static pressure and fan curve to estimate the quantity CFM of air based on the fan curves.

I wonder if what your saying about the air flow measurement being +/- 20% is playing a big role here. This is something that could be verified with a pitot tube, yes?

 

[zdas04]

Are the fan curves adjusted for your elevation? Humidity? Inlet temperature? Some are, some are not. Are your fan blades in pristine condition? Do the belts ever slip? Do you have exactly the right sized sheaves? Are you assuming an rpm or reading one (line voltage has an acceptable range, the power factor on the motor has an acceptable range, if both are at the end of their range that maximizes rpm then you may be 10% more rpm than you expect). A properly calibrated and properly installed pitot tube would give you much better flow rates than your fan curve.

Fouling of the heater core does not affect the calculation that I would be doing for your problem. The fouling will simply require more mass flow for the same dS. You have calculated a dS and that calculation does not include any insulating effects within the heat exchanger, it is simply inlet enthalpy minus outlet enthalpy times mass flow rate.

David Simpson, PE
MuleShoe Engineering

"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The plural of anecdote is not "data"

 

[LittleInch]

Tbaygen, its been many years since I looked at stream tables, but as zdas04 says, the net stream heat input is enthalpy in minus enthalpy out. I don't know what you're using to measure your energy in, but a flow controller and pressure would seem to give you energy in, but no temperature pressure or steam and condensate flow out. That will add up to a lot of energy, but how are you measuring it?

Post a simple diagram with steam flow, pressure and temp in and out and then someone more familiar with stream can tell you how much energy is going in and still coming out.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[tbaygen]

Thanks for the replies. I did have a realization though, and likely had I provided more information it would make more sense, but this is a closed loop drier, and likely some water vapor or even liquid is en-trained in the air stream, so not only am I heating the air the large heat sink could be the moisture.

Zdas, the fan curve was calibrated for a specific temperature, density and elevation. I did not assume an RPM, I used the static pressure correlation on the fan curve to determine the CFM of air moving, its a fixed drive fan no belts. I can use the AMP readings to determine power.

I think it's backing up to the moisture in the system taking the extra energy.

 

[rotw]

I can imagine that if after including the efficiency of the heat exchanger, part is still missing in your energy balance then the energy might be absorbed as latent heat:
- Steam might be condensing at exit and small droplet are carried with the flow.
- Air is not completely saturated, possibly the relative humidity is increased between air inlet and outlet.

Can it be that your system is combination of heat exchanger and condenser?
Maybe you should monitor the steam temperature and pressure at outlet and relative humidity of the air between inlet and outlet.

Question: where the steam goes when it exit the system?

 

[TBP]

To heat air with steam, all you need to know is the CFM of air to be heated, and the inlet & outlet air temps across the coil.

There will be about 1,000 BTU of useable heat in each lb of steam.

(CFM X 1.08 X delta T)/1,000 = #/hr steam required.

This formula is from "Hook-Ups - Design Of Fluid Systems" by Spirax Sarco. I've used this thin, soft covered book for applications like this for 35 years, and it's never let me down.

If you really are using twice the steam that you figured, then you're moving twice the CFM, heating the air to twice the delta-T, or some combination of the two.

 

[tbaygen]

Thanks TBP,
What I ended up doing when I originally calculated this was I looked at our flow tag that trends live on the DCS, assuming its accurate it says we flow 10,000+ lbs/hr of steam. I took this times the hfg of 150 psi steam (857 BTU/lb) to get the 10MM BTU/hr figure.

When I looked at the air side, I used the inlet and outlet temperatures for dT, and I used the static pressure on the calibrated fan curve to determine the mass rate (CFM of air). I used an average specific heat for Cp of the inlet/outlet temps. This came out to the 5MM BTU/hr figure.

I think what is occuring is alot of the steam energy is going into the specific humidity of the air stream, or water droplets in the air stream (its a closed loop ring dryer air is pulled off the top of the scrubber). This is causing much more energy to be put in to heat the water back up.

Rotaryw,
Steam goes to a condensate reciever.


Tom

 

[LittleInch]

Am I missing something here? Steam comes in with a certain amount of energy at your 10MMBTU, but also exits from the HX with quite a large amount of energy assuming the conservation of mass. Even if all your steam turned to condensate, which it doesn't sound like, you would have about 2MM BTU going out of your HX. If you have more steam than zero lbs/hr exiting this figure will rise quite considerably.

I think a lot of your "lost" energy is flying out the other side of your HX as steam / hot water.

however maybe I've got the wrong end of the stick...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way