Transient Heat Transfer 2

[dsengine]

I have a bit of a tricky heat transfer problem and was wondering if someone could point me in the right direction if possible.

There is an above ground stainless steel pipeline containing crude oil @ 10degC and 130 m3/hr. The pipe has OD 508mm and thickness 6.35mm. A heating band is then placed around the pipe diameter and heated until the outside of the pipe reached 250degC. The heating band is then removed - how long will it take the outside of the pipe to reach ambient conditions? I'm only considering radial heat transfer and assuming the pipeline is infinitely long.

In my mind this is a forced convection/conduction/natural convection problem. Forced convection from the moving oil inside the pipe, conduction through the pipe wall and finally natural convection with the ambient air. The tricky part is where the point of interest is - on the external pipe wall.

I could probably work out the steady-state heat transfer between the external pipe wall and air and between external pipe wall and oil (assuming I can find a relevant convection heat transfer coefficients) but obviously this is a transient problem - how would that help calculate the cooling time.

I'm unsure whether this could be treated as a lumped system (vague recollections about Biot number). Are there any assumptions which could be made to come to an analytical solution?

Appreciate any help or direction you can give. Many thanks in advance.

 

[gwolf2]

The strongest heat transfer route is axially by conduction, this is a 2D problem not 1D. FE is your best route.

 

[zekeman]

Unless I am mistaken, you have laminar flow inside the pipe.
If you can get the inside film coefficient, the problem can be done analytically.

 

[corus]

At 250C you can't neglect radiation from the outer surface, which makes the problem more non-linear. I would use numeric techniques for this, and preferably in 2D, as gwolf says.

Tata

 

[athomas236]

Is the pipe insulated and if so what is the insulation and thickness.

Regards,

athomas236

 

[athomas236]

Sorry forgot to ask, how long is the pipe?

Regards,

athomas236

 

[dsengine]

Thanks guys. I'm thinking FE might be the best option but is there any way of simplifying the problem. Assume no insulation. I don't know how long the heating band will be so assuming 1m at the moment (i.e. pipe length = 1m).

For steady state (or time =0) I've calculated the following (mainly using the relations provided in Fundamentals of Thermo Fluid Sciences (Cengel & Turner):

At time 0
Internal pipe convection coefficient - forced convection

Nu = 1.86* (Re Pr D / L)^1/3 * (dynamic visc bulk / dynamic visc surface)^0.14 dynamic surface visc evaluated at 250degC

Nu = 364.45 for 1m length of pipe >> than fully developed value of 3.66

h = k / D * Nu
h int = 105 W/m2K for 1 metre length of pipe
Area = 1.596 m2

Heat loss Q 40119.3 W / m Convection dominates - ignore radiation

External pipe convection coefficient - natural convection

Properties of air at film temperature Tf = (250+ Tamb) /2 Tf = 132.5 degC 405.5 K


From Table A-18 p960 Fundamentals of Thermo Fluid Sciences @ Pamb = 1atm (1 bar)

thermal conductivity k air 0.033 W/mK
Pr air 0.703
kinematic viscosity air 2.664E-05 m2/s
Beta (1 / Tf) 0.002466091 1/K


Raleigh = (g * beta * (Ts - Tamb) D^3 Pr) / (kinematic viscosity air)^2

= 7.380E+08 Use equation 18-16 p 789 Thermo Fluid Sciences

Nu = (0.6 + (0.387Ra^1/6 / (1+ (0.559/Pr)^9/16)^8/27))^2

= 105.07

h ext 6.919 W/m2K
Area 1.596 m2 (assume 1m pipe length)

Heat loss Q 2594.9 W / m

External pipe radiation
rad = emissivity * A * stefan boltzmann * (Ts^4 - Tamb^4)

= 6147.7 W / m Assuming pipe is black body

Obviously the above only considers steady state heat loss. As soon as the pipe cools these values will alter due to the decreasing temperature difference.

Any way of calculating this analytically or in time steps? I've looked at Newton's logarithmic law of cooling but unsure how to apply this when the heat transfer is to the oil and to the air.

Again appreciate everyone's continuing help. Thanks.

 

[zekeman]

One of my concerns is that the viscosity of crude drops dramatically with temperature, doubling for every 10 deg drop , so by the time it approaches 10 deg c, you have laminar flow.
This complicates an analytic solution, but does not rule out a decent solution as long as the Reynolds no>2100.
A one dimensional equation can be written as

(1) rho*w*C*dT/dt=-hi(T-10) -ho(T-15)

rho= density steel
w wall thickness
c specific heat of steel
hi,ho= inside and outside film coefficients
Assuming the doubling formula
mu=muref^2(250-T)/10)
muref= viscosity at 250 C
And from your correlation of Nu, Re and Pr ( I can not confirmh this from my information)it looks like
h varies inversely as mu^0.14, but I wouldn't bet the farm on this.
So, if you can get hi as a function of T, you can solve the nonlinear eq (1) numerically or by integrating between T=250 to T
-1/[(hi(T)*(T-10)+ho*(T-15)].So
rho*w*C*int(-1/[(hi(T)*(T-10)+ho*(T-15)]=t+constant
get constant from initial condition
T=250 when t=0

 

[dsengine]

Thanks Zekeman but I'm assuming the oil does not change temperature but stays at 10degC. I believe this is a reasonable assumption for a short section of pipe or discrete time step.

I think I have a numerical method of solving it through timesteps using a similar equation as your eqn 1. I'm using the stored energy in the pipe due to heating as my starting point then subtracting the heat loss via each of the mechanisms for each step. Then recalculating the coefficients for subsequent steps. I hope this will deliver an asymptotic temperature decay as I would expect.

 

[zekeman]

That is also my assumption.
but my point is that your evaluation of hi depends on the viscosity of the average temperature of wall and oil, so in the course of cooling down, the wall will approach 10 degrees and the Reynolds number will go below the transition number making your Nu,Re, Pr correlation invalid.

My other concern is the correlation, itself. Please check that since my literature shows more dependence on viscosity.

 

[Dave41A]

For the internal convection coefficient, I suggest using the
fully-developed Nusselt (3.66) instead of the expression you are proposing if the flow is indeed laminar. The expression you are proposing is only for low Prandtl numbers (less than 5), but oil has one of the highest possible. Further, that correlation is for "entrance region" problems (flow not fully developed), which is unlikely here since the pipe is iso long. If the flow is turbulent, I suggest Dittus-Bolter, Seider-Tate or Gneilinski correlations (the last is best for transitional values of Re, down to 3000).

Since the pipe is stainless (fairly high thermal conductivity, i.e. k approx 15 W/mK at 300K), I would dis-regard radial heat transfer and concentrate on axial conduction (along the pipe) and heat loss to the
atmosphere and to the oil. You'll likely find that the "delta T" across the pipe wall is pretty small, but this can be verified through calculation.

I suggest developing a control volume for a short length of pipe wall, and then modeling the heat loss to the oil, the air, and to the rest of the pipe. External convection will be hardest to model, due to the variations in the environment (I assume this pipe is outdoors).
Radiation is possibly significant, although the blackbody assumption is too high. I suggest setting the emissivity at about 0.8 or so (what kind of surface does the pipe have? Painted? Shiny?)

The easiest way to treat this would be as a 1-D conduction problem (heat transfer down the pipe wall), with convection to the oil and the environment.

A finite-difference approximation is pretty easy to implement for this type of problem, and can be done even in a spreadsheet program. Your energy balance for each unit length of pipe (choose 1 meter or so) will then be:

rho_pipe*pi*D*t*c*L*(dT/dt)=q_conv+q_rad+q_cond

Where:

rho_pipe=denisty of pipe material
pi=3.14159...
D=diameter of pipe
t=thickness of pipe
c=specific heat of pipe
dT/dt=rate of change of pipe wall temp
T=temp of pipe wall
L=unit length of pipe

q_conv=h_internal*pi*D*L*(T-T_oil)+h_external*pi*D*L*(T-T_air)

q_rad=epsilon*sigma*pi*D*L*(T^4-T_air^4)

q_cond=heat lost by conduction=2*pi*D*L*t*k_wall*(T-T_adj)

where T_adj is the temp of the adjacent section of the pipe (which must also be found by finite difference). The "2" is due to symmetry (the control volume has two ends).

I suggest cylinder in cross flow correlations for the external convective coefficient, but ground effects are likely to disrupt the air flow, so any computed result will only be a rough guess. You should be able to use 10 meters of pipe in either direction for a good "chunk" of pipe before you get far enough away where the temp is un-affected.

You should run your simulation "before" and "after" you apply the heating element to fully capture the steady-state and the cool-off from that state.

Good luck.

Dave

 

[athomas236]

I would still be intrested to know the pipe length.

Regards,

athomas236

 

[corus]

Dave41A appears to recommend a quasi 2D approach where a 1D model down the length of the pipe is assumed with heat loss to the oil etc as a heat sink in the orthogonal direction, ie. an equation fo the form pCp.dT/dt=div(k.div(T)+H(T-Ta) where H would have units of W/m^3 K. This apporach is ok but assumes that the temperature gradient through the wall thickness is negligible. In this case I doubt it is (with an outer temperature of 250C and the inner surface cooled by a fluid at 10C) and as such a better approximation would be to assume radial flow, and that the heating element is sufficiently long to ignore axial conduction. Another approximation would be to assume that the pipe has a large radius such that you can use cartesian co-ordinates to give you the equation pCp.dT/dt=kd^2T/dx^2 for constant conductivity, instead of the axisymmetric form of the equation.
Use a htc to the outside amobient of about 10 W/m^2 K and your htc to the bulk oil temperature at 10C, and you should be able to get a reasonable answer to the problem, though you'll still have to use finite differences unless there's an analytical solution out there.

Tata

 

[Dave41A]

Corus:

You do in fact correctly interpret my recommendation..a quasi- 2D approach.

However, the significance of the temp drop through the steel depends greatly upon the flow. Assuming laminar flow inside the pipe, and assuming a pipe diameter (D) of 1 meter, and a wall thickness (t) of 3 cm (of stainless steel), with an outside temperature of 250C, a quick analysis of the steady-state situation returns the following:

Nu=3.66
h=Nu*(k_oil)/D = 1.52 W/m^2K (based on k_oil=0.147 W/mK)
k_ss (stainless steel) = 15 W/mK

Assuming a unit length (L) of 1 meter...

Thermal resistance (oil to pipe wall) = 1/(h*A) = 1/(h*pi*D*L)=0.209 K/W

Thermal Resistance (through pipe wall) = t/(k_ss*A) = 0.03/(15*pi*D*L)=0.0006366 K/W

The ratio of the thermal resistances is 328 to 1. Therefore, the temperature drop across the pipe wall will be only a tiny fraction of the total temperature drop (in this case, less than one degree C), and the approach I propose will likely work very well.

Of course, this all changes if the flow is turbulent. In these cases, Nu will be many, many times higher (in the 100's, even the 1000's) and then the temp drop inside the steel will become significant.

This might be a good place to mention (for everyone) that most radial systems can be modeled as flat walls with very little error. The ratio of outer diameter to inner diameter has to approach 2:1 before the "flaw wall approximation" acquires significant error (more than 5%). At 1.5:1 it is down to about 1%, and the error is likely trivial for this case (large diameter oil pipeline).

Hence, Corus' suggestion to consider a Cartesian system is a good one.

Dave

 

[corus]

Good point Dave. Thinking about it, an assumed 1m pipe length is really too short to just use a radial temperature temperature distribution.
Given the doubt over the Nusselt number I'd opt for a full 2D analysis but if I was looking for a worst case scenario, ie. trying to calculate the longest time it would take to cool, then a 1D radial solution (assuming no heat loss axially) would be the preferred and easiest solution should the OP have no access to software to solve the problem. For that case I'd use the proper terms for radiation and natural convection to the ambient rather than a constant value I suggested as a 'quick' method.
Whether the OP does a quasi 2D axial solution or a radial solution, some assumptions must be made. It depends on knowing which assumption gives a resonaable error in the direction you know.

Tata

 

[zekeman]

Dsengine,

I finally caught up with your correlation,
Nu = 1.86* (Re Pr D / L)^1/3 * (dynamic visc bulk / dynamic visc surface)^0.14 dynamic surface visc evaluated at 250degC

I hadn't realized we are dealing with laminar flow and can now confirm your equation above.It is indeed for laminar flow which is your case.
The problem is fairly straightforward, except as I noted the Nu is proportional to mu^-.14 and since mu doubles for every 10 deg C
I get,
mu=mu(250)*2^-0.14(250-T)/10
Therefore
Nu=Nu(250)*2^-0.14(250-T)/10
and
h=h(250)*2^-0.14(250-T)/10
I believe the problem is almost 1 dimensional and the wall thickness is very small vs diameter, so we can use cartesian coordinates and assume the temperature uniform thru the wall; writing the energy equation for a unit surface area of the cylinder we have

rho*w*c*dT/dt=-2*sigma*T^4-hi(250)*2^-(0.14*(250-T)/10*(T-10)
-h0*(T-To)

where emissivity is conservatively assumed =1 and the factor 2 on the radiation term is for for both sides of the walland where

T absolute wall temperature deg K ( I changed it to deg R in what follows)
rho= density of steel=494lb/ft^3
w = wall thickness=.02ft
c= specific heat=0.13 BTU/lb-degF
Using lb- BTU -F hr units and your values of h1 and h0 converted(conversion factor 0.188), and changing T to Rankine, I get
1.35*dT/dt=-2*.173*10^-8*T^4-19*2^-[(942-T)/18]*(T-510)
-1.3*(T-519)
You can solve this numerically with a spreadsheet using delta t or direct integration. I usually chose integration and do it by solving for dt from the form
dT/f(T)=dt
integrating both sides
which has t the independent variable
Thus
integral 1/f(T)(limits 942 to 519) = t
f(T)=-2*.173/1.35*10^(-8)*T^4-19/1.35*2^-[(942-T)/18]*(T-510)
-1.3*(T-519)/1.35

I will try this on my TI-83 calculator and post results later.

 

[Dave41A]

I want to correct what I wrote earlier regarding the appropriate Nusselt number for the internal convection coefficient.

The value Nu=3.66 assumes fully developed hydrodyanic and thermal boundary layers. Since the heater is applied locally, the thermal boundary layer will not have fully developed.

However, the "entrance region" correlation attributed to Seider and Tate, namely:

Nu=1.86 8{[Re*Pr/(L/D)]^(1/3)}*(mu/Mu-s)^0.14

is normally not used here since it only applies to cases where Pr<5. Since Oil's Pr is so high, I suggest a correlation from Hausen:

Nu=3.66+0.0668*(D/L)*Re*Pr/{1+0.04*[(D/L)*Re*Pr]^(2/3)}

The above are from Incropera, et al "Fundamentals of heat and Mass Transfer" (Wiley, 6th Ed, 2007).

However, the difference between the Hausen and Seider and Tate correlations is not that great within the entrance region, so you could probably use either in that region. The difference begins to grow as you move into the fully developed region, with the Hausen becoming more accurate (it asymptotically approaches the fully develped value of 3.66 as L --> infinity, while the Seider and Tate correlation does not).

Good luck,

Dave

 

[zekeman]

Dave,

"Nu=1.86 8{[Re*Pr/(L/D)]^(1/3)}*(mu/Mu-s)^0.14

is normally not used here since it only applies to cases where Pr<5. Since Oil's Pr is so high, I "

I.m puzzled over this statement, since the the bulk viscosity is eliminated in the correlation. The Seider-Tate correlation
can be written as

Nu*(Mu-s/Mu)^0.14=1.86*(4*w'C/(PI*K*L))^.33
w'=lb/hr of flow.
No dependence on Prandtl number.

 

[dsengine]

Thanks chaps - appreciate the superb advice. I'll get back onto this next week and let you know how I get on.

 

[zekeman]

Results:

First off, the expression I posted,
"dT/f(T)=dt
integrating both sides
which has t the independent variable
Thus
integral 1/f(T)(limits 942 to 519) = t
f(T)=-2*.173/1.35*10^(-8)*T^4-19/1.35*2^-[(942-T)/18]*(T-510)
-1.3*(T-519)/1.35"

I corrected it to include the radiation from the oil and ambient which I missed So it now reads

f(T)=-2*-19/1.35*2^-[(942-T)/18]*(T-510)
-1.3*(T-519)/1.35 +.173/1.35*10^(-8)*510^4+.173/1.35*10^(-8)*519^4

I performed the integration on my graphing calculator with the following results
TEMPERATURE-C TIME-minutes
250 0
200 1.1
150 3.4
100 8.5
50 21
15 40