Heat Loss from an Uninsulated Pipe 2

[Timewrap]

Dear All,

Need your help..

I've a got situation where I need to condense and cool a vapour stream from 260 deg C to about 200 deg C using natural convection or by leaving it bare. I need to know the amount of uninsulated length needed to do that. The process line is a 1/2" tubing.

Thanks in advance,

Timewrap

 

[SNORGY]

The heat transfer coefficients for tube to still air were discussed in this thread:

thread391-257339

I think it related to finned tube largely, buit armed with a suitable "U"-value you should have what you need.

Regards,

SNORGY.

 

[Timewrap]

Dear Snorgy,

Can you tell me how this equation works??

Nusset number to be calculated based on the calculated Prantl and Rayliegh number (Both based on air properties. After I get Nu, I can calculate H-T coefficient.

The unit of Q is Btu/hr, I need to calculate heat loss Btu/linear foot. Since I need to know the length of the pipe to be left uninsulated to get to the desired temperature.

Timewrap..

 

[25362]

I assume heat is lost not just by natural convection but also by radiation. A combined heat transfer coefficient for the given data on horizontal pipes taken from very old (Petroleum Refiner-1959) nomographs would be 3.5 BTU/(hr.ft².^oF).

For [Δ]T I'd take the difference between the wall temperature, T_wall, and the surrounding air T_[∞].

The results would be within [±] 30% of the true value.

 

[eadwine]

Try the following formula for heat conduction:

Q=(kA(T2-T1))/l


Where:
Q: heat transferred
k: thermal coductivity
A: heat transfer surface area
T2: hot side temp.
T1: cold side temp.
l: thk. of barrier

 

[Timewrap]

Dear 25362,

Thanks for the e-mail.

I was using a Heat transfer coefficient of 5 btu/hr-deg-ft2, but 3.5 is a more conservative value.

The heat area is the cross section of the pipe, it does not take into account the length of the piping.

Once I solve the heat transfer equation it will give me total heat loss in btu/hr. I need to calculate the heat loss per foot of piping. Can I assume that the total heat loss I get from the equation is the heat loss per foot of piping.

Any comments would be appreciated

Timewrap..

 

[itdepends]

Don't forget you'll get radically different heat transfer depending on wind velocity, humidity and rain. If you go below 200degC is it a problem?

As a chem eng/metallurgist the first part of any answer I give starts with "It Depends"

 

[ione]

For a rough estimate

Q = M*cp*deltaT (1)

Q = thermal power to reduce the stream temperature from 260 °C to 200 °C [W]
M = vapour mass flow rate [kg/s]
Cp = vapour specific heat [J/(kg °C)]
deltaT = temperature decrease [°C]

Cooling process doesn’t take into account change of phase

Q = h*S*deltaT (2)

h = heat transfer coefficient [W/m2 °C]
S = pi*D*L lateral pipe area [m2]. Where D = pipe outer diameter and L = pipe length (your unknown value)

Equal the expression (1) and (2) and solve for L.

Note: As noticed by 25362 you should take into account both natural convection and radiative heat transfer.

 

[ione]

Just to add another thought to my previous post.
It is IMO very important to notice that as cooling proceeds the driving force (deltaT) of the process diminishes. Moreover, as the vapour temperature drops, so does the wall pipe temperature and the properties of air needed to calculate the heat transfer coefficient vary as temperature varies.

 

[25362]

The heat loss is related to the external pipe surface not its cross-section. Therefore, knowing the external surface per unit length will give you the loss per unit length.

 

[ione]

To 25362,

Hope your post was addressed to the OP as I have obviously considered the external surface area (so involving the length of the pipe which is the unknown value)