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Tube and Shell exercise

  1. Aug 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Would really appreciate if you could lend me a hand with this tube and shell exercise.

    tubeandshell.JPG
    m=0.24kg/s
    ##D=2.5*10-2m##
    ρ=850 kg/m^3
    cp=2000 J/kg*K
    μ=2*10-4 kg/m*s
    λ=0.3W/m*k
    2. Relevant equations
    Dittus-Boelter:Nu=0.023*Re0.8*Prn
    where n=0.4 heating and n=0.3 for cooling, so in our case n=0.3
    Log mean temp##ΔTlm=\frac {ΔT1-ΔT2} {ln(ΔT1/ΔT2)}##
    3. The attempt at a solution
    i)hi=?

    Apply Dittus-Boelter eq Nu=0.023*Re0.8*Pr0.3
    =>##\frac {h*D} {λ}##= ##0.023*(\frac {4*m} {\pi*D*μ})^{0.8}## ##* (\frac{cp*μ} {λ})^{0.3}##
    We have all data from above
    ##h=λ*0.023*(\frac {4*0.24} {\pi*2.5*10^-2*10^-4})^{0.8}## ##* (\frac{2000*2*10^-4} {0.3})^{0.3}##
    hi=50.72 W/m^2*K

    ii)ΔTlm=?(log mean temp)
    ##ΔTlm=\frac {ΔT1-ΔT2} {ln(ΔT1/ΔT2)}##
    Where ΔT1=left hand side temp diff, and ΔT2 is right hand side temp difference.
    Oil is on hot side, water on cold side.
    Oil:Tin=420K; Tout=320K
    Water: tin=290K ,tout=?
    up.png

    ##tcout=tcin+\frac{Qhot} {mcold*cpcold}##
    ##Qhot=m*cp*ΔT=0.24*2000*(420-320)##
    ##Qhot=48000J##
    =>##tcout=290K+\frac{48000} {0.24*4180}##
    =>tcout=337.84K
    =>ΔT1=320-290k
    ΔT1=30K

    ΔT2=420-337K
    ΔT2=83K
    =>##ΔTlm=\frac{30-83} {ln(30/83}##
    =>ΔTlm=52K

    iii)ho=4.5 KW/m^2*K
    L=?
    I think we need to apply again Dittus-Boelter but not sure about this?
    Need more hints here
     
    Last edited: Aug 17, 2016
  2. jcsd
  3. Aug 17, 2016 #2
    You don't need to apply Dittus-Boelter again. You know that the the inside heat transfer coefficient is 50.72 and the outside heat transfer coefficient is 4500. What is the overall heat transfer coefficient U?
     
  4. Aug 18, 2016 #3
    Are i) and ii) correct?

    U=##\frac{1} {1/hi+1/ho}##

    U=##\frac{1} {1/50.72+1/4500}##

    U=50.15

    But how to get the length?
     
  5. Aug 18, 2016 #4
    $$Q=UA(\Delta T)_{lm}$$
     
  6. Aug 18, 2016 #5
    Right.

    ##A=\pi*D*L##
    ##L=\frac{A} {\pi*D}##
    ##A=\frac{Q} {U*ΔTlm}##
    A=48000/(50.15*52.08)
    A=18.37m2
    ##L=\frac{18.37} {\pi*2.5*10^{-2}}##

    L=233.99m
    L~234 meters
     
  7. Aug 18, 2016 #6
    Go back and recheck your analysis. There are 5 pipes, not one. This would affect your answers in parts i and iii. The total rate of heat flow should be in W, not J.
     
  8. Aug 18, 2016 #7
    Q=48000 W
    hi=5*50.72 => ##hin=253.6 W/m^2*K ##
    hout=4500 W/m^2*K
    ##U=\frac{1} {1/hi+1/ho}##
    ##U=\frac{1} {1/253.6+1/4500}##

    =>U=240.07
    ##A=\pi*D*L##
    ##L=\frac{A} {\pi*D}##
    ##A=\frac{Q} {U*ΔTlm}##
    A=48000/(240.07*52.08)
    A=3.83m2
    ##L=\frac{3.83} {\pi*2.5*10^{-2}}##

    L=48.88 m
     
  9. Aug 18, 2016 #8
    This is definitely not correct. You don't just multiply the inside heat transfer coefficient by 5. Please go back and redo part (i) correctly. What should m really be in part (i)?
     
  10. Aug 18, 2016 #9
    Q=48000 W
    m'=5*m
    where m=0.24kg/s (mass flowrate for 1 pipe)
    =>m'=1.2 kg/s mass flowrate
    hout=4500 W/m^2*K

    ##hin=0.3*0.023*(\frac{4*1.2} {\pi*2.5*10^{-2}*10^{-4}})^{0.8}##*##(\frac{2000*2*10^{-4}} {0.3})^{0.3}##
    =>##hin=320.07 W/m^2*K##

    ##U=\frac{1} {1/hi+1/ho}##
    ##U=\frac{1} {1/320.07+1/4500}##

    =>U=298.81

    ##A=\pi*D*L##
    ##L=\frac{A} {\pi*D}##
    ##A=\frac{Q} {U*ΔTlm}##
    A=48000/(298.81*52.08)
    A=3.08m2
    ##L=\frac{3.08} {\pi*2.5*10^{-2}}##

    L=39.27 m
     
  11. Aug 18, 2016 #10
    m'=m/5=0.048
     
  12. Aug 18, 2016 #11
    Thank you.
    I assume that's because the pipes are in parallel and all identical
    hence m'=m/5
     
  13. Aug 18, 2016 #12
    Sure.
     
  14. Aug 18, 2016 #13
    so m'=0.048

    =>##hin=0.3*0.023*(\frac {4*0.048} {\pi*2.5*10^{-2}*10^{-4}})^{0.8}## *##(\frac {2000*2*10^{-4}} {0.3})^{0.3}##
    hin=24.38 W/m2*K
    hout=4500 W/m2*K

    U=1/(1/hi+1/ho)=>U=1/(1(24.38)+1/(4500))
    U=24.23

    for Area
    ##A=\frac{48000} {24.23*52.08}##

    A=38.03 m2
    ##L=\frac{A} {\pi*D}##

    ##L=\frac{38.03} {\pi*2.5*10^{-2}}##

    =>L=484.31 m

    Is this correct?
     
    Last edited: Aug 18, 2016
  15. Aug 18, 2016 #14
    No. The heat transfer area is ##5\pi DL##
     
  16. Aug 18, 2016 #15
    Thank you for your answer
    ##A=\frac{48000} {24.23*52.08}##

    A=38.03 m2

    ##A=5*\pi*D*L##

    =>##L=\frac {A} {5*\pi*D}##

    ##L=\frac {38.03} {5*\pi*25*10^{-2}}##

    L=96.86 m
     
  17. Aug 18, 2016 #16
    I get Re = 10186, Pr = 1.6, Nu = 42.6, and ##h_{in}=511##
     
  18. Aug 18, 2016 #17
    I did ##Re=\frac {4*m'} {\pi*D*μ}##

    ##Re=\frac {4*0.24/5} {\pi*2.5*10^{-2}*2*10^{-4}}##

    this gave me Re=12223

    for Prandlt I did
    ##Pr=\frac {cp*μ} {λ}##
    ##Pr=\frac {2000*2*10^{-4}} {0.3}##
    this gave me Pr=1.3
     
  19. Aug 18, 2016 #18
    OK. I mistakenly used ##\mu=2.4\times 10^{-4}##.
    What about hin. It looks like you forgot to divide by the diameter. I get 560 now.
     
  20. Aug 21, 2016 #19
    Thank you, yes indeed hi=560. W/m2*K

    for ii) I did like this
    Qh=mhot*cphot*ΔT=0.24/5*2000*(420-320)
    Qh=9600W
    ##Tcout=Tcin+\frac{Qh} {mcold*cpcold}##
    ##Tcout=290+\frac{9600} {0.24/5*4180}##
    =>Tcout=337.84K

    ##ΔTlm=\frac{ΔT1-ΔT2} {ln(ΔT1/ΔT2)}##
    ##ΔTlm=\frac{30-83} {ln(30/83)}##
    =>ΔTlm=52.08K

    for iii)
    ho=4500W/m2*K
    hi=560 W/m2*K from i)

    ##U=\frac{1} {1/hi+1/ho}##
    ##U=\frac{1} {1/560+1/4500}##

    U=497

    ##Q=U*A*ΔTlm##

    =>##A=\frac {Q} {U*ΔTlm}##

    ##A=\frac {9600} {497*52.08}##

    =>A=0.370m m2

    but ##A=\pi*D*L##

    so ##L=\frac {A} {\pi*D}##

    ##L=\frac {0.370} {\pi*10^{-1}}##

    =>L=1.18m


    I would really appreciate it if you could tell me if what I did is correct.

    Thank you very much in advance.
     
    Last edited: Aug 21, 2016
  21. Aug 21, 2016 #20
    In part (ii), I would use the total flow rate for both streams. And, in part (iii), I would use that total heat load 9600 x 5 and the total area 5 pi D L. I know that answer comes out the same both ways, but that's how I would do it.
     
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