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## Homework Statement

Would really appreciate if you could lend me a hand with this tube and shell exercise.[/B]

m=0.24kg/s

##D=2.5*10

^{-2}m##

ρ=850 kg/m^3

cp=2000 J/kg*K

μ=2*10

^{-4}kg/m*s

λ=0.3W/m*k

## Homework Equations

Dittus-Boelter:Nu=0.023*Re

^{0.8}*Pr

^{n}

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)}##

## The Attempt at a Solution

i)hi=?

Apply Dittus-Boelter eq Nu=0.023*Re

^{0.8}*Pr

^{0.3}

=>##\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=?

##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

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