THERMODYNAMICS question...pls help me :(
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I've got this question which i do not really know how to do. Can anyone be kind enough to teach me the method of doing it?
Q.) A rigid tank contains water vapour at 300 degrees celceius and an unknown pressure. When the tank is cooled to 180 degree celcieus, the vapour starts condensing. Estimate the initial pressure in the tank. -
volume = constant
therefore can ignore the V part..
P1/T1 = P2/T2
at 180 degrees, it's 1 atmosphere as the vapour condenses...
go from there.. and remember to change the temp to Kelvin
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P1 V1 T1(kelvin)=P2 V2 T2(kelvin)Originally posted by Mariner:I've got this question which i do not really know how to do. Can anyone be kind enough to teach me the method of doing it?
Q.) A rigid tank contains water vapour at 300 degrees celceius and an unknown pressure. When the tank is cooled to 180 degree celcieus, the vapour starts condensing. Estimate the initial pressure in the tank.
Since volume = constant
Therefore
P1 T1=P2 T2
At 180 degrees celcius pressure = atm
so basicly you just put no. get answer lah . . . -
Originally posted by the Bear:volume = constant
therefore can ignore the V part..
P1/T1 = P2/T2
at 180 degrees, it's 1 atmosphere as the vapour condenses...
go from there.. and remember to change the temp to Kelvin -
from superheated (300) to saturated (180) waterOriginally posted by Mariner:I've got this question which i do not really know how to do. Can anyone be kind enough to teach me the method of doing it?
Q.) A rigid tank contains water vapour at 300 degrees celceius and an unknown pressure. When the tank is cooled to 180 degree celcieus, the vapour starts condensing. Estimate the initial pressure in the tank.
find at 180 saturated water the Pressure is wat (which is 1.0021 Kpa) then use
P1/P2 = T2/T1 cos rigid tank V dun change
therefore P1=0.6xP2 = 0.601 Kpa = 601 Pa -
need to use steam table isit?? few months nvr study, rusty liao
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uhh.. wrong lah..
how can heated can have a pressure less than the cooler can?Originally posted by chunyong:from superheated (300) to saturated (180) water
find at 180 saturated water the Pressure is wat then use
P1/P2 = T2/T1 cos rigid tank V dun change
therefore P1=0.6xP2 -
the answer i got is...o.126MPa....but the answer given by the teacher is 1.325MPa
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true also then i dunno liao, so long ago liao....Originally posted by the Bear:uhh.. wrong lah..
how can heated can have a pressure less than the cooler can? -
alamak, need interpolation
i got it too
first u find saturated water at 180, find the specific volume of sat vapor, since rigid tank, sat vapor's specific vol remains the same, then u go to superheated table...then interpolate between 1.4 and 1.2Mpa...
sat vap's specific vol is 0.19405
for P=1.2Mpa, v=0.2138
for P=1.4Mpa, v=0.18228
interpolate to get answer, viola -
Hi thanksOriginally posted by chunyong:alamak, need interpolation
i got it too
first u find saturated water at 180, find the specific volume of sat vapor, since rigid tank, sat vapor's specific vol remains the same, then u go to superheated table...then interpolate between 1.4 and 1.2Mpa...
sat vap's specific vol is 0.19405
for P=1.2Mpa, v=0.2138
for P=1.4Mpa, v=0.18228
interpolate to get answer, viola
got the answer.... -
carefu leh, alot of people fail thermo just last sem....but heng i took it in sem 1 which was an easy paper
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u in marine engineering course at SP ? lol
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if u mean me, i ntu mae
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no..da marinerOriginally posted by chunyong:if u mean me, i ntu mae