Condenser Heat Load 7

[dkm0038]

I am designing an air cooled heat exchanger for a steam condenser. The specs that i have recieved for steam is a flow rate of 99,800 lb/hr, 969.89 BTU/lb, and at 4.6 inHg. The heat load given is 86,958,734.

The heat load was calculated by the flow rate times the difference between the enthalpy of vaporization (969.89)and the enthalpy of the condensate (98.56).

My question is why was the heat load calculated as shown above. It seems that the necessary heat load would be the flow rate times the enthalpy of vaporization, which i thought was the energy transfer needed to condense the steam.

Thank You for any help anyone may have.

 

[Montemayor]

The heat load is calculated using 969.89 Btu/lb as the latent heat of condensation --- and that is the correct figure to use. Actually, I get 969.74 Btu/lb using the online NIST database - but that nit-picking.

You obviously have your enthalpies mixed up. The enthalpy of the saturated condensate at 4.6 in Hg = 181.96 Btu/lb
The enthalpy of the saturated steam at 4.6 in Hg = 1,151.70 Btu/lb. The difference is the latent heat of vaporization (or the latent heat of condensation, however you want to describe it). Check it out.

 

[dkm0038]

Thanks for your quick reply.

So the latent heat of vaporization equals the enthalpy of vaporization minus the enthalpy of saturated liquid?

and also is it correct to get these values at the operating pressure (4.6 inHg), or would there ever be a reason to get it at atmospheric pressure? Thats what I'm thinking he got the enthalpy of saturated liquid from.

Thanks Again

 

[chicopee]

As montamayor point out Q= m*(hv-hs) where m is mass flow rate

 

[Montemayor]

dkm0038:

No, the latent heat of vaporization DOES NOT equal the enthalpy of vaporization minus the enthalpy of saturated liquid. Your terminology is wrong and that’s probably what’s giving you the wrong idea and/or concept of what takes place inside the condenser and on a Mollier (P-H) diagram. There is no such thing as “enthalpy of vaporization”. Enthalpy is the property of a substance and not of an action or process. Allow me to explain in detail in order to make sure we both agree on what is happening.

I assume that the steam condenses in the condenser at a constant pressure of 4.6 inches of Hg (gauge). Actually, this is not really true because there has to be a pressure drop within the condenser (otherwise, there would be no flow), but it is what is done in practice and this assumption yields a conservative answer. I also assume that the inlet steam is SATURATED (as opposed to superheated). I’m also assuming that the formed condensate is also saturated (as opposed to supercooled). This means that you must evacuate the condensate as fast as it is formed. Under these conditions, the thermodynamic process is a horizontal line on the Mollier diagram that starts at the saturated vapor curve line on the right hand side of the diagram and extends horizontally to the left portion of the curve that represents the saturated liquid line. This horizontal line should be directly on top of the pressure value of 4.6 inches of Hg (gauge pressure – don’t forget to add atmospheric to convert it to absolute pressure) which can be read on the Ordinate axis of the Diagram.

Note that the horizontal line defines what is happening in the condenser: you are taking saturated steam and condensing it at constant pressure. The point at the saturated vapor curve defines the condenser inlet and its enthalpy can be read directly below, on the Abscissa axis. The point on the saturated liquid curve defines the product condensate and its enthalpy can also be read below, on the abscissa. The definition of the load on the condenser is the heat removed from the steam in order to convert it to condensate and this equates to the enthalpy of the vapor minus the enthalpy of the condensate – as represented by the length of the horizontal line. If you use the Mollier Diagram, you have the enthalpies of both streams. You can also use the NIST free database which you can find at:

http://webbook.nist.gov/chemistry/fluid/.

Either way, you should find that the difference between the enthalpies is the 969.74 Btu/lb.

Additionally, I believe you either have a typo or someone made an error in the calculations’ results you were given. If you multiply the steam mass flow rate by the latent heat of vaporization you were given (& which I confirmed as correct), then you obtain 96,795,022 Btu/hr and NOT 86,958,734 as the heat load you report. Somewhere, something is amiss and I recommend you check it out.

I hope this helps to bolster your confidence in the correct calculated load.

 

[25362]

BTW, as an aside, the accuracy of the estimated heat duty doesn't increase by expressing it in more significant figures than those used for the enthalpies or the flow rates.

 

[speco]

dkm0038,

If this is a real-world application, you need to consider two additional factors. First, there is a pressure drop through the condenser which will lower the total pressure at the outlet. Secondly, there is always a certain amount of air leakage into the system. At the outlet end, this affects the partial pressure of the steam. Usually this is taken as about 80% (per HEI standards) of the total pressure at that point, further reducing the condensing temperature at the outlet. The remaining mixture of steam and air is usually taken to a jet-ejector sytem to maintain vacuum in the condenser and reduce back pressure in the steam turbine. I'm assuming this is a steam turbine condenser based on the conditions you are starting with.

Regards,

Speco

 

[chicopee]

dkm0038,
Based on the original values you gave us, your answer is correct. The change in enthalpy between the inlet and outlet flow times the mass flow rate is your original answer. That's how much heat (btu/hr) was lost by the steam. Any thermo, heat transfer and engineering handbook will bear that answer.

 

[25362]

To dkm0038, if this is exhaust steam from a turbine, what was the quality assumed? The heat duty given to you may have been the result of assuming a certain steam quality.

See please, thread666-143727.

 

[sailoday28]

dkm0038 (Mechanical) did not state that the exhaust is from a turbine. If the only exhaust is from a turbine, the correct heat balance would include the KE of steam. If the condensed exhaust is the only condensate, the condensate can be subcooled and its enthalpy should be considered. (KE of the condensate will be small compared to KE of steam.

Regards

 

[dkm0038]

Thanks again for everyones help.

Let me see if I can summerize all of this i think my trouble is in the different names the same physical quantity.

This is for a 9.50 MW steam turbine generator operating at4.6 inHg(A) with a mass flow rate of 99,800 lbm. I was always under the impression that for pure condensing (no desuper heating or subcooling)the latent heat load is just the mass flow rate times the latent heat of vaporization.

is it true to say that the latent heat of vaporization is the enthalpy of vaporization (hfg) which is further the difference between enthalpy of saturated vapor (hg) and the enthalpy of saturated liquid (hf)?

 

[25362]

To sailoday28. It seems that the kinetic energy of steam entering a condenser at 200-250 f/s should be a very small amount of the steam enthalpy at 4.6 in Hg (A). Would you confirm ?

To dkm0038. This site may give you some ideas about air-cooled turbines exhaust steam condensers:

http://www.hudsonproducts.com/products/stacflo/may22_78.pdf

 

[sailoday28]

Check with turbine mfgrs (if still in business) such as GE. Velocities can be 1000 fps on large nuclear power plant turbine exchaust hoods. My experience however, was on power plant water cooled condensers.

 

[stgrme]

Typically, a turbine vendor will state the exhaust enthalpy as the Used Energy End Point (UEEP) or Turbine End Point (TEP). This enthalpy accounts for the total energy in the steam. Also, the exhaust pressure, when given in "inch Hg", is typically an absolute pressure.

Therefore, for the conditions you state above, I would consider the UEEP equal to 969.89 Btu/lb at an absolute pressure of 4.6 inch Hg. The corresponding liquid enthalpy is 98.56 Btu/lb and the heat load is the mass flow times the difference in these two enthalpies.

Best of luck!

 

[dkm0038]

Stgrme,

thanks for the responce but,

if we do what you say and the heat load would be (99800lb/hr)*(969.89Btu/lb - 98.56Btu/lb) = 86,958,734 Btu/hr

yet the enthalpy of vaporization (or condensation in our case)for steam at 4.6 inHg is aprx 1019.5 Btu/lb. and if this is the energy per pound to condense steam then the heat load to condense would be (99800lb/hr)*(1019.5Btu/lb) = 101,746,100 Btu/hr?

Is this correct. I understand that what stgrme was saying may be standard practice but does this mean that there isn't total condensing going on?

thank you

 

[Montemayor]

To assist everyone on this thread, I'm attaching the calculations for the condenser load at two pressure: 4.6 inches of mercury absolute and gauge.

Note the simple answer.

The original cited load is still wrong - if the steam is saturated. That may be part of the basic data that hasn't been given.

 

[stgrme]

Obviously, the exhaust steam from the turbine is not saturated. In actuality, the exhaust steam is already "wet" (a mixture of steam and water).

If the steam were saturated, the exhaust enthalpy would be 1118.05 Btu/lb. The heat of vaporization is the difference between the enthalpies of saturated steam and saturated water at the same pressure. For the present case, that difference is approximately 1019.5 Btu/lb.

Since the exhaust enthalpy from the turbine (969.89 Btu/lb) is less than the saturation enthalpy (11180.5 Btu/lb), less heat needs to be removed in order to condense the exhaust steam.

I hope this helps!

 

[quark]

969.89 btu/lb steam enthalpy corresponding to 4.6 inHg (a) pressure means 85.5% dryness fraction. Is this common in turbines?

 

[stgrme]

Yes, quark, a dryness factor of 85.5% (moisture content of 14.5%) is very common in a non-reheat, condensing turbine, such as the one discussed in this thread.

 

[rmw]

And so is all the damage; eroded metal parts, pitted and eroded tubing, eroded shell walls, eroded last stage (last several stages) blades/buckets etc that you would expect with that much moisture moving at the velocities commonly present in the latter stages of condensing turbines and condensers.

rmw