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Thermodynamics problem -- lunar-based power facility

  1. Aug 5, 2015 #1

    1.Question

    I need to check if i solved it right can someone please check it , thank you. Its worth alot of points
    http://postimg.org/image/vbictz4mt/
    3. The attempt at a solution
    GIven

    P1 = 50 kpa

    T1 = -20 C = 253 K

    1 to 2 is isentropic process

    Therefore we can use

    PT^k/(1-k) = constant

    k = 1.66

    Given

    P2/P1 = 6

    Therefore

    P2 = 6*50 kPa

    P2 = 300 kPa

    P2 = P3

    P3 = 300 kPa

    The

    T3 relation is given in the figure given

    P4 = P1 = 50 kPa

    Let's divide the given graph into two sections

    From 0-1 and 1-2 gas flow rate

    From 0-1

    Here we can write the linear equation

    T3 = a*m+b

    Here a and b are two constant

    (0.200) and (1,150) are points on the line

    Therefore

    200 = b

    150 = a*1+ 200

    a = -50

    The relation for 0-1 gas flow rate is

    T3 = -50*m + 200

    The relation for 1-2 gas flow rate is also linear equation and let it be

    T3 = c*m + d

    Here c and d are constants

    (1,150) and (2,0) are points on the line

    0 = c*2 + d

    2c + d = 0 ---(1)

    150 = c + d -- (2)

    Solving (1) and (2)

    c = -150

    d = 300

    The relation for 1-2 gas flow rate is

    T3 = -150*m + 300

    Since 3 to 4 is isentropic Expansion

    We can use

    P3T3^k/(1-k) = P4T4^k/(1-k)

    T4 = T3*(P3/P4)^(1-k)/k

    T4 = T3*(6)^-0.398

    T4 = 0.49*T3

    The Power developed by the

    P = mCp*(T3-T4)

    Cp = 0.520 kJ/kg/K

    For 0-1 gas flow rate

    P = m*0.520*(T3-0.49*T3)

    P = 0.520*m*0.51*T3

    P = 0.265*m*T3

    P = 0.265*m*(-50m+200)

    P = -13.26m^2 + 53.04*m

    For 1-2 gas flow rate

    P = 0.265*m*T3

    P = 0.265*m*(-150*m + 300)

    P = -39.78*m^2 + 79.56*m

    Optimum at mass flow rate of 1 kg/s

    http://file:///C:/Users/Saad%20khan/AppData/Local/Packages/oice_15_974fa576_32c1d314_2427/AC/Temp/msohtmlclip1/01/clip_image001.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Aug 10, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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