Turn on thread page Beta

Anyone help - thermodynamics (urgentish) watch

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    Hi folks,

    Hmm, 13 months since my last visit and my postcount is still 4 posts per day. That pretty much summarises my history here. True to form I'm posting at 2am...

    I'm studying mechanical engineering, but badly, and have remedial coursework to complete by Friday. The work is to write a lab report about the performance of a refrigerator. I have this pressure-enthalpy graph of R-12 refrigerant:
    /!\CAREFUL, IT'S MASSIVE IMAGE and I can't conveniently resize it.
    Spoiler:
    Show


    I have plotted the refrigerant's cycle on the graph, but I need to determine the specific volume of the refrigerant. I don't have a clue how. I see those lines marked "log10 v = -2" but it's not at all obvious how to use them to find the specific volume...

    EDIT: Ahh shoot. It might appear that my plan to 'post graph on TSR, then sit on hands until someone explanes those log(v)=c lines' might have to be replaced with 'search Google again' or even the dreaded 'email tutor for help'...
    Offline

    8
    ReputationRep:
    Oh dear...
    The lines already on the graph are isochoric. That is each line has a volume associated with it. Since each one is so far in volume from the others they are stated in their log v, which I assume means log (v/1m3kg-1). In short...just read off the graph...

    THAT MEANS its OBVIOUS that you want to manipulate "log10 v/1m3kg-1 = -2", to find v.
    which means
    10^[log10(v/1m3kg-1)] = 10^-2
    v/1m3kg-1 = 0.01
    v = 0.01m3kg-1

    Please learn to use logs and powers in future.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Mehh)
    Oh dear...
    The lines already on the graph are isochoric. That is each line has a volume associated with it. Since each one is so far in volume from the others they are stated in their log v, which I assume means log (v/1m3kg-1). In short...just read off the graph...

    THAT MEANS its OBVIOUS that you want to manipulate "log10 v/1m3kg-1 = -2", to find v.
    which means
    10^[log10(v/1m3kg-1)] = 10^-2
    v/1m3kg-1 = 0.01
    v = 0.01m3kg-1

    Please learn to use logs and powers in future.
    What, the curved lines marked 'Log10 (v) = -n' are isochoric? Yes, I know what 'isochoric' means, and I can calculate logs, all I need to know is...

    ...dammit, I've been stupid...

    ...I needed to know where exactly I needed to find the specific volume. By scribbling on the graph in MSPaint, the answer is clearly Log10(v)=-1.96. Hmm, the last 18 hours have been somewhat wasted. Thanks for the reply.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: August 16, 2006

University open days

  • University of Lincoln
    Brayford Campus Undergraduate
    Wed, 12 Dec '18
  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 12 Dec '18
  • Buckinghamshire New University
    All undergraduate Undergraduate
    Wed, 12 Dec '18
Poll
Do you like exams?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.