# Tube and Shell exercise

## 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*10-2m##
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*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)}##

## 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=?

##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=?
Need more hints here

Last edited:

Chestermiller
Mentor
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?

williamcarter
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?

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?

Chestermiller
Mentor
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?
$$Q=UA(\Delta T)_{lm}$$

williamcarter
$$Q=UA(\Delta T)_{lm}$$
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

Chestermiller
Mentor
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.

williamcarter
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.

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

Chestermiller
Mentor
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
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)?

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)?

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

Chestermiller
Mentor
m'=m/5=0.048

williamcarter
m'=m/5=0.048
Thank you.
I assume that's because the pipes are in parallel and all identical
hence m'=m/5

Chestermiller
Mentor
Thank you.
I assume that's because the pipes are in parallel and all identical
hence m'=m/5
Sure.

williamcarter
m'=m/5=0.048
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{Q} {U*ΔTlm}##

##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:
Chestermiller
Mentor
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?
No. The heat transfer area is ##5\pi DL##

williamcarter
No. The heat transfer area is ##5\pi DL##
##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

Chestermiller
Mentor
I get Re = 10186, Pr = 1.6, Nu = 42.6, and ##h_{in}=511##

I get Re = 10186, Pr = 1.6, Nu = 42.6, and ##h_{in}=511##

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

Chestermiller
Mentor
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
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.

williamcarter
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.
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:
Chestermiller
Mentor
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.
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.

williamcarter