Is there someone confirming that 100 $/kW is a good figure to roughly evaluate the price of air-cooled chillers for air-conditiong application in the range 600-1200 kW of refrigeration (170-340 RT)?
Thanks in advance.
Roaling
Thanks Guidoo.
I know the relation between dew point and partial water vapor pressure. What is unknow to me is the relation with the H20-CO2 system that beeing an acid is a condensing agent.
Good day, Roaling
The question is simple and clear.
How does CO2 content effect the dew point of flue gas exhausted from natural gas fuelled gas turbine towards a HRSG?
Thank a lot for your help.
To IRstuff.
Frankly,I know something about plain fins but nothing on serrated. Reference books just to start with the principles and then more detailed info will be very much useful.
I should appreciate to have information, formulas and procedures on how calculate the heat transfer coefficient of tubes with serrated fins.
Thanks for your help.
You have two ways: to make calculation referring to feet or to psi. If you chose psi then you have to convert to feet. Anyway, your assumption to divide atmospheric pressure in feet by S.G. is correct.
Dear Morten,
May be I was not clear. The correction factor to apply is on the NPSHR not NPSHA. In case of gasoline, I think that no correction is needed.
Information from other guys are correct. I suggest you to refer to pipe size or discharge nozzle size assuming respectively a velocity of 2-2.5 m/s or 5-6 m/s. In this way it is very easy to roughly estimate the flow.
For. ex. : 2.5*(3.1416*0.05^2/4)*3600 = 17.7 m3/h (checking the result with...
Yes, MGlycol is right. A demister is more than what required. An adequately sized cyclone separator could be enough. Anyway, go ahead with the demister.
PlantEng,
in your experience have you never seen vessel jacket of such dimensions to be inspected inside ? Do exist regulations requiring to inspect inside the jacket ?
Thanks.
Sorry j I insist.
Assuming 10 mm SS wall: 0.01/15 = 0.00067
8 mm air gap: 0.008/0.025 = 0.32
Assuming 10 mm CS wall: 0.01/40 = 0.00025
1/K = 0.00067 + 0.32 + 0.0025 = 0.32092
K = 3.12 W/sqm°C
This figure differs from that calculated by you (17).
Am I wrong ?
j,
I think you are wrong.
If gap thickness is only 8 mm, I should consider only air conductivity, so the coefficient is much lower. Unfortunately, I have Perry 6th and cannot check your reference.