User Name Remember Me? Password

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum Feb 10th 2018, 07:52 PM #1 Junior Member   Join Date: Feb 2018 Posts: 3 Thermodynamics problem An industrial boiler is fitted with a sight‐glass device to indicate its level of liquid and the boiler contains saturated steam at 200°C. During the routine "blowdown" procedure, liquid water is purged from the bottom (mud-drum) at a slow rate, and the temperature in the boiler remains constant during this process. If the cross‐ sectional area of the boiler is 2.5 m^2 (assumed constant), and the level of the liquid drops by 150 mm, determine the mass of liquid withdrawn. Not sure how to attempt this question. I cant use isothermal process equation as I have no Q values. I was thinking maybe i could use ideal gas law. Initial pressure of steam = initial pressure of whole container(?) Use table to find pressure of sat steam at 200 degrees PV=mRT. I think i have to form an equation of initial stage = final stage. As T is the same so T= PV/mR (P1V1)/(m1) = (P2V2)/(m2) but i dont have P2 and im guessing i need to get a (v1-v2) so that i can substitute the 150mm x 2.5m^2 in too.   Feb 10th 2018, 09:11 PM #2 Senior Member   Join Date: Apr 2017 Posts: 498 the volume equivalent to the level drop is 2.5 x 0.15 = 0.375 cu meters this volume is now filled with steam at 200C If you find out mass of this extra steam ... then subtract this from the mass of water at 200C occupying 0.375 cu meters to get the answer.   Feb 11th 2018, 03:16 AM #3 Junior Member   Join Date: Feb 2018 Posts: 3 Okay, I interpolated steam at 200C to have vg of 0.1273 v=Xvg. Since its steam so dryness factor is 1. v=vg=0.1273 v=V/m so m=V/v = 0.375/0.1273=2.946kg. 0.375cu meters of water at 200 degrees is 375kg as v=0.001. 375-2.946 = 372.05kg But the answer key says that its 321.15kg Last edited by RustyPotato; Feb 11th 2018 at 03:22 AM.   Feb 11th 2018, 05:04 AM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Well I get Specific vol of water at 200 = 0.001157 cu m/kg Specific vol of steam at 200 = 0.127360 cu m/kg So if V = 0.375 cu m then Wt = (0.375/0.001157) - (0.375/0.12736) = (375/1.157) - (375/127.36) = 324.11 - 2.94 = 321.17 kg If you are going to do many of these sorts of questions, I recommend getting a copy of Thermodynamics, an Engineering Approach by Cengel and Boles The book has lots of useful tables in the back (in both metric and imperial measures), including the water/steam tables I extracted the 200C values from. Last edited by studiot; Feb 11th 2018 at 06:06 AM.   Feb 11th 2018, 06:34 AM   #5
Junior Member

Join Date: Feb 2018
Posts: 3
 Originally Posted by studiot Well I get Specific vol of water at 200 = 0.001157 cu m/kg Specific vol of steam at 200 = 0.127360 cu m/kg So if V = 0.375 cu m then Wt = (0.375/0.001157) - (0.375/0.12736) = (375/1.157) - (375/127.36) = 324.11 - 2.94 = 321.17 kg If you are going to do many of these sorts of questions, I recommend getting a copy of Thermodynamics, an Engineering Approach by Cengel and Boles The book has lots of useful tables in the back (in both metric and imperial measures), including the water/steam tables I extracted the 200C values from.
Oh I get it now, the water that is in the question isnt compressed water but the water in the 2 phase mixture (area under pv curve), no wonder i cant take v=0.001!

Thanks so much, oz93666 and studiot!  Tags problem, thermodynamics Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post engrmansoor2534 Advanced Thermodynamics 6 Dec 16th 2017 02:23 PM HelpNeeded Advanced Thermodynamics 0 Dec 19th 2016 06:38 PM HelpNeeded Advanced Thermodynamics 0 Dec 19th 2016 06:34 PM alicegranda Advanced Thermodynamics 3 Nov 24th 2015 02:45 PM baghaloo Thermodynamics and Fluid Mechanics 0 Nov 4th 2014 03:36 AM