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Working out the mass flow rate

 
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novice
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PostPosted: Tue, 13 Apr 2004 11:15:27 UTC    Post subject: Working out the mass flow rate Reply with quote

How can I work out the mass flow rate of air in kg/sec if I know that the volume flow rate is 800 m^3/min. At the inlet the pressure is 100kPa and 27 degrees Celcius, and at the exit the pressure is 95 kPa and temp is 60 degrees Celcius. (this is for a heat exchanger question)

Thanks!
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Leigh Janes
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PostPosted: Wed, 14 Apr 2004 04:20:41 UTC    Post subject: Reply with quote

You need to know the density of air at the given pressures and temperatures. Is the volume flow rate you quoted for inlet or outlet? I expect the mass flow rate to be the same at the outlet as at the inlet, but the volumes could differ with the different conditions.
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PostPosted: Wed, 14 Apr 2004 12:02:42 UTC    Post subject: Reply with quote

Oh I figured I need to use the ideal gas law for this ...

PV = mRT

P V(dot) = m(dot) RT

m(dot) = P V(dot) / (RT)

Except I _think_ V(dot) = 800 m^3/min (quoted for inlet) is the same at the inlet and outlet, cos rate of mass going in = rate of mass going out... But then... I get two different m dots...Which one am I suppose to use?
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Leigh Janes
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PostPosted: Wed, 14 Apr 2004 18:21:17 UTC    Post subject: Reply with quote

The m-dot you use depends on where the v-dot is applicable: inlet or outlet. If the statement of the problem does not tell you, you need to determine it from some other data. I don't have enough to help.
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PostPosted: Thu, 15 Apr 2004 06:17:16 UTC    Post subject: Reply with quote

well the volume flow rate is given for the inlet, and actually when i find the m dot using that together with the temp and pressure at the inlet I got the correct answer.

How come mass flow rate going in = mass flow rate going out but
volume flow rate going in doesn't = volume flow rate going out?
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Leigh Janes
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PostPosted: Thu, 15 Apr 2004 17:26:04 UTC    Post subject: Reply with quote

Mass is conserved, volume is not.
Volume depends on the mass, temperature and pressure; temperature and pressure are different at the outlet than at the inlet.
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