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Old Feb 10th 2018, 07:52 PM   #1
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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 200C. 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of the boiler is 2.5 m^2 (assumed constant), and the level of the liquid drops by 150 mm, determine the mass of liquid withdrawn.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.
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Old Feb 10th 2018, 09:11 PM   #2
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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.
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Old Feb 11th 2018, 03:16 AM   #3
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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.
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Old Feb 11th 2018, 05:04 AM   #4
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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.
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Old Feb 11th 2018, 06:34 AM   #5
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Originally Posted by studiot View Post
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!
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