Temperature Rise in Steel Pipe 2

Hi SpecialA102

Have a look at this site:-


desertfox

The pdf linked to specifically says that it is not applicable to non-steady state applications.

hi SpecialA102

If the outside temperature of the pipe is constant and the fluid temp is constant as it flows along I thought you could do a simple heat transfer by conduction through the pipe wall, then by convection from pipe to atmosphere but if your fluid temp is changing as it flows along the length of the pipe then I agree you can't use those equations.

desertfox

Yeah...why is it not steady state? It sounds like you're saying you have a fluid temperature of 210 deg F and an ambient of 70 deg F. Are you trying to determine the outer surface temperature of the pipe after there is no longer any air flow in the pipe and it's been given 5 to 10 minutes to cool down?

I get about 116°F, assuming no change to gas temperature, so just heat flow across the steel, and passive convection and radiation on the outside.



TTFN

FAQ731-376

Hi SpecialA102

If its unsteady look at this link:-


desertfox

I assumed steady state equations would not apply since 5-10 minutes isn't a very long time and the pipe would most likely still be heating up from the air flow. Initially, the pipe will have no flow, so it will be at ambient temperature. Then, at t0, flow begins at 210ºF. I want to know what the pipe temperature would be at t=5-10min, not what the temperature would be at t=infinity (or a sufficiently long time), which would be steady state.

IRstuff, I'm having a little trouble following where you got 119F from. I mean, I understand the equations you were using, but I don't see a time factor in there, so I don't know how that would tell me what the temperature of the tube is at 5 or 10 minutes. But if that is the temperature at steady state, that would give me a good upper bound, at the very least.

Desertfox, I'll take a look at Newton's Law of Cooling. I had seen it previously before, but wasn't sure if I could apply it to my situation. Thanks.

Hi SpecialA102

I guess what IRstuff as posted is heat loss through the pipe wall as he has stated assuming gas temp constant.

On the link I gave you look for this paragraph it as other links from it too.:-

Lumped system analysis
A common approximation in transient conduction, which may be used whenever heat conduction within an object is much faster than heat conduction across the boundary of the object, is lumped system analysis. This is a method of approximation that suitably reduces one aspect of the transient conduction system (that within the object) to an equivalent steady state system (that is, it is assumed that the temperature within the object is completely uniform, although its value may be changing in time). In this method, a term known as the Biot number is calculated, which is defined as the ratio of resistance to heat transfer across the object's boundary with a uniform bath of different temperature, to the conductive heat resistance within the object. When the thermal resistance to heat transferred into the object is less than the resistance to heat being diffused completely within the object, the Biot number is small, and the approximation of spatially uniform temperature within the object can be used. As this is a mode of approximation, the Biot number must be less than 0.1 for accurate approximation and heat transfer analysis. The mathematical solution to the lumped system approximation gives Newton's law of cooling, discussed below.

This mode of analysis has been applied to forensic sciences to analyse the time of death of humans. Also it can be applied to HVAC (heating, ventilating and air-conditioning, or building climate control), to ensure more nearly instantaneous effects of a change in comfort level setting.[1]

desertfox

What about just taking into account the energy balance?

M(steel)*Cp(steel)* (Te – Ta) = V(air) * Cp (air)* (Ti-Te)

M(steel) = pipe mass [kg] You need this value.
Cp(steel) = 490 [J/(kg*°C)] specific heat of carbon steel
Te = equilibrium temperature [°C] to be computed
Ta = 21.1 °C = 70 °F (ambient temperature)
V(air) = 12460 m^3 = 440000 cf (air volume delivered in 10 mins)
Cp(air) = 0.9544 [J/(m^3*°C)] specific heat of air
Ti = 98.9 °C = 210 °F (initial air temperature)

My analysis was for steady state. The 116°F is the solution to the energy balance, assuming constant inner surface temperature.

A more expanded steady-state analysis drops the temperature to about 105°C, using 75W/m^2-K convection coefficient on the inside.

It takes 2311/m^2 to heat the pipe up 35°F in 5 minutes. There's a maximum heat flux of 5833W/m^2 at the start, and 860W/m^2 in steady state, so it looks like the pipe will probably reach steady state temperature in probably around 5 minutes.



TTFN

FAQ731-376

Now you've all got me confused as to whether this is steady stead or some sort of transient heat flow problem! At least I understand better what SpecialA102 is trying to determine. I take it the real intent of this whole exercise is to determine whether or not insulation should be put on this pipe or not?

Without doing any calc's myself, the numbers like 116 deg F and 105 deg F seem kind of light though. I mean if we do reach steady state in 5 minutes and there's 210 deg fluid in this steel pipe (granted it's air and not water), I would instinctively think the outer surface temperature would still be pretty hot to the touch.

Depending on the length of the pipe, 5-10 minutes may be enough time to establish steady state , so unless I have missed something, what is the length of piping? A good answer should nor be too hard to determine.
For the transient,you write 2 equations involving, temperature and time one for gas and one for the pipe, the solution of which is available in closed form.Or it can be done numerically.The solution will yield
Tpipe=Tpipe(x,t); Tgas=Tgas(x,t)
temperature as a function of distance from source and time.
The steady state solution is easier to formulate and can be done more readily, yielding the temperature of the pipe as a function of distance from the source, or
Tpipe=Tpipe(x), also Tgas-Tgas(x) if needed.

"Depending on the length of the pipe, 5-10 minutes may be enough time to establish steady state , so unless I have missed something, what is the length of piping? A good answer should nor be too hard to determine."

Not in closed form.Involves hypergeometric equation.

Will attempt this numerically when I get some time.

CountOlaf:
"Without doing any calc's myself, the numbers like 116 deg F and 105 deg F seem kind of light though. I mean if we do reach steady state in 5 minutes and there's 210 deg fluid in this steel pipe (granted it's air and not water), I would instinctively think the outer surface temperature would still be pretty hot to the touch."

Just going thru the posts...

You are right, the IRstuff steady state solution (not the one sought) has an obvious error in it since, given his
K/thk=4199 and
hair=5
it is obvious that the tube to air resistance is a few orders of magnitude greater than the resistance thru the pipe, so the surface temp should be close to the inside wall temperature.

Finally got around to this one, which is generally a formidable problem in heat transfer. I could not develop a solution in closed form so I did it numerically using Excel and found that owing to the huge flow rate, 44000CFM, thru a 20" pipe, yielding an internal convection coefficient of h=1000 BTU/HR/Ft^2/degF this problem can be analyzed as a one dimensional one.

From my numerical analysis done for 10 feet of piping it looks like the exiting temperature of the gas reaches values of near the entrance temperature in about a second and after about 6 seconds, the time constant of the one dimensional system,the wall temperature reaches 60%*(210-70)+70= 150 F
The one dimensional heat eq is, excluding the negligible effect of outside temperature:
rho*w*c*T'=h*(Tg-T) whose solution is:
T=Ti+(Tg-Ti)*(1-e^(-t/tau)=70+140*e^(-t/5.8)
where
t= time in seconds
T,Tg,Ti transient temperature of pipe, gas temprature, initial temperature of pipe
rho= 490 lb/ft^3 steel
w 3/8" thick
c=0.12
h= 1000 BTU/hr/ft^2/deg F
tau= time constant=rho*w*c/h=5.8 seconds(after conversions)

With a little more effort, we could provide a more accurate prediction, but in general:
for a steel pipe of thickness s and thermal diffusivity a, the time constant is proportional to s^2/a.

In this case, s=3/8"=0.0313 ft, a=(1/7926 ) ft2/s, so the time constant is about 7.7 seconds. A 1 minute period would be 7.8 time constants==> basically steady state after 1 minute.

Point well taken. My "lumped" solution of 5.8 seconds is slightly inaccurate compared with your 7.7 seconds. However, the steady state in this particular case, i.e. where the surface reaches Tg is only 3 time constants, since the parameter h*thickness/K =2.5.
This is found from Schneider curves for this case.

So, the time for the surface to reach .6*(210-70)+70=150 F
is .8*7.7 sec= 6.16 sec, again, the .8 is taken from Schneider curves for h* thickness/K=2.5
The fact that it comes closer to the lumped value is due to the very high h in this case.