Abot Condensing Water Vapor

What you want to know is how to solve the problem, not the answer....right?

I had no formula. There is no heat exchanger attechment possible. Direct surface condensation should be done. It is precticle application

If you asked me, I'd say no.

But you didn't.

Why?

Are you using a heat exchanger for cooling or some other method?

Is this a real application, or theoretical exercise?

CR4 doesn't do cost.

it would be much cheaper to use vater as the coolant for this vapour.

You have no idea!

It's homework or an interview question. OP was given all the information necessary (or can look it up) except the knowledge of how to do the heat balance. Hint- you already know the mass flow rates and delta T, now do the math; the nitrogen price you'll have to Google yourself.

I dont know power plant engineering. The power plant engineers have eqations which can tell me how much ln2 required for cooling of such a hot mixture of air and water vapor?

www.steamtablesonline.com

Come back with any questions.

Correct, Google on "Heat Exchanger Calculations".

Have you got a set of Steam Tables, Mildred?

Maybe this one is Auntie, rather than Mildred.

Feel free to share that with anyone who cares.

Isnt that the same thing?

There seems a lot of confusion in postings and some incoherent data.

You mention in a subsequent post that the fluid to be cooled is water / air but are you sure that is the case. You have a hot pressurised fluid and I cannot think of a case in my experience (which 20+ yrs in oil and gas industries) where you would have this mixture air in steam systems is a complete nightmare so please check. Also if you want to condense the entire stream then there cannot be any air present or do you want to condense the water out of a steam / air mix?

I suspect it is more likely that you have a steam / water vapour exhaust that is pure water vapour.

I have assumed that this cooling is to be achieved in a heat exchanger of some sort rather than by directly mixing the liquid nitrogen and water vapour.

Your base data seems confused.

Is the pressure 7bara or 7barg (ie absolute pressure or guage pressure)?

Both are possible but neither correlates with your temperature. From steam stables I get

7bara 215C (488K) or 7barg 290C (563 K) to give 3.13kg/m3 both much cooler than your 839K seems very high - even at 120barg (typical HP Steam pressure) this would be 250degrees of superheat but at 7 bar its around 400 degrees of superheat.

The water vapour / steam velocity seems ok but you have not provided a pipe diameter so we have no way of determining the heat load (Q).

Q (kW) = pipe area (m2) * 30.85 (m/s) * 3.13 (kg/m3) * condensing duty in kJ / kg

The condensing duty will be the sum of:

1. energy required to desuperheat the steam from 839K to ~170C

2. latent heat of steam at 7bar (approx 2400kJ/kg)

3 cooling duty for hot water from ~170C to 290K

For the cooling duty again you have given us far too much to guess.

Firstly I want to say that in an industrial use it would be madness to cool this stream with liquid nitrogen (lin). It is a standard principle to use the cheapest cooling to do as much of the duty as possible and lin is very expensive. If this is a lab scale process then I can understand that you dont want several different cooling streams to cool one process stream. In an industrial process I would expect to use air coolers, then a water cooling (either colosed loop cooling water or sea / river water) and then use the liquid nitrogen, although to go down to 290K really only needs an easy domestic refrigerant.

You said that water couldn't be used because it would heat to 280C and become superheated steam. This reveals a number of flaws in either your undertanding of heat transfer problems or how you have explained the problem. The water used for cooling will heat up as it cools the water vapour by:

Q (kW) = m (mass flow kg/s) x Cp (heat capacity kJ/kgK) x DT (Temp change in K)

hence DT = Q / (Cp * m)

Water Cp = 4.2kJ/kgK as a working estimate

The Q is as specified above. Clearly all you need to do is increase the flow of water until you get DT in a sensible range (typically 20K). Also if the cooling water is pressurised to 80barg then even at 283C it will still be liquid.

You have not specified any of the lin conditions and whether you are cooling by evaporating the lin (I assume this is the case but the pressure at which evaporation occurs changes the quantity of lin required).

Mass flow lin (kg/s) = Q (kW) / latent heat lin (kJ/kg)

Latent heat lin = 115kJ/kg for 20 bara lin or 153kJ/kg for 10 bara lin

At these pressures lin is iquid at 100 - 120K which is far too cold a temperature for cooling to 290K.

As an engineer they keep me away from costs so I cannot help you with that part but liquid nitrogen systems are very expensive - pressures, exotic materials, cryogenic equipment, massive insulation issues