Temperature Increase in Pipeline 4

[gator1991]

I have a 16 inches (outside diameter)chilled water pipeline (cast iron) that is 80 feet long. Water is flowing through it with an initial temp of 45 degF. The line is exposed to an air temperature of 55 degF and I would like to know the temperature at the end of the pipeline. The flow rate is 1000 gpm.
I'm trying to generate some kind of pipeline heat gain table using the exposed pipe vs the insulated pipe.
Can anybody helps?

 

[25362]

Here are somme comments for your consideration:

Is the (black) pipe exposed to solar radiation ? If it is, the heat gain during light hours may be markedly larger than by the convection-wise heat transfer from the surrounding air.

Convection HT coefficients for still -indoors- air (natural convection) differ dramatically from those of forced convection (winds).

An estimated value of 0.35 Btu/(h.ft^2oF) for still air can increase dramatically for windy conditions.

Besides, what about air humidity ? Could it be that some moisture condenses on the uninsulated pipe, adding some resistance to heat transfer ?

The location of the pipe, whether near ground level or elevated, with or without vertical sections, may also affect heat transfer.

Your comments are appreciated. Thanks.

 

[chicopee]

and dont forget the heat gain from the frictional effect although for a 16" dia. pipe w/ a flow of 1000gpm would not be significant and can even be disregarded.

 

[denniskb]

Gator,

I have calculated this case and find that the heat transfer values are as follows;

Bare pipe, in shade, no wind
Heat Gain 92.3 W/m of pipe
Heat Gain 72.3 W/m2 of pipe surface
Temp Gain 0.15°C per 100m or 0.08°F/100ft
Temp Gain 0.06°F

Very little heat transfer should be happening under these conditions.


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[25362]

To denniskb,

My sources indicate a HTC for free convection of 2.2-2.5 W/(m²*^oC) for a horizontal cylinder in air.

Even taking the higher value of 2.5 and a total [Δ]T of 10/1.8=5.6⁰C, the results show a heat gain of
13.9 W/m², much lower than the 72.3 W/m² of your findings.

Where is my error ? Thanks

 

[corus]

The temperature increase of a fluid moving within a pipe of radius R, length L, at a speed of V m/sec, given a pipe wall temperature of Tp, a heat transfer coefficient to the water of h W/m^2 C, and an initial water temperature of Tw, at a density p, and specific heat C, is given by

Delta T = 2*L*h*(Tp-Tw)/(p*C*V*R)

by my calculation. The heat transfer coefficient to the water can be found from such references as MAdams, Heat Transfer.

I'd be interested if anyone has a reference to confirm my calculation.

corus

 

[sailoday28]

corus (Mechanical)
Your formula on the right hand side assumes the water temp to remain constant over the length of pipe even if the assumption of pipe wall temp is also constant. The water temp should at least vary.
Start with a differential equation, such as in deriving temp rise in cooling water thru a condenser tube and see a better approximation.
I'm glad to see someone referencing McAdams. How about references to Jakob?

 

[gator1991]

Guys...thanks a lot for all the input...it really helps.
Denniskb...what analysis did you use to get the answers...did you use any of the spreadsheets that you have on your website...if you did..please tell me which one.
Have a good day.

 

[denniskb]

I re-checked my calc and found I had used an incorrect algorithm for convection so pls consider the following;

Bare pipe, in shade, no wind
Heat Gain 104.4 W/m of pipe
- composed of Convection 12.15 W/m, Radiation 92.25 W/m
- Convective Coeff (internal) 1037 W/m2.K, (external) 1.73 W/m2.K
- Radiative Coeff (external) 13.15 W/m2.K
Heat Gain 81.8 W/m2 of pipe surface
Temp Gain 0.165°C per 100m or 0.091°F/100ft
Temp Gain 0.073°F

25362,
My previuos calc had an error which resulted in 0 W/m2.°K so everything was due to radiation. The algorithms used for both external and internal convection are from Incropera and De Witt and the external one is also used by the 3E Plus software.

corus,
The problem with your formula is that gator1991 does not know the pipewall temperature Tp. My calc puts the outside wall temperature at 7.31°C [45.16 °F].

sailoday28,
Since the temperature change along the pipe length is so small there is no need for a differential equation in this case.

Sorry that most of my data is in SI units but the coversions are a pain.


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[corus]

sailoday,
You misread my formula. The water doesn't stay constant as the idea was to evaluate the increase in water temperature, ie. the delta T. The value on the right hand side Tw is the initial water temperature as it enters the pipe. The delta T is that increase in water temperature over the pipe length, ie. the outlet temperature will be Tw + delta T.

denniskb, you're correct in that you would need to know the pipe wall temperature for my expression. Presumably with your value inserted into the formula then the outlet water temperature could be found.

corus

 

[Tremolo]

gator1991,

I performed a heat transfer calculation using 2.5 W/m2/C as suggested by 25362. My result: a heat flux of 13.8 W/m2 (nearly identical to 25362), or 16.8 W/m (17.5 Btu/hr/ft; 1048 Btu/hr over the 60 foot pipe length).

So, over the length of the 60 ft pipe, there would very little temperature increase in the water (.002 F).

Even with denniskb's calculation with higher heat transfer, the temperature increase is still imperceptibly small.

TREMOLO

 

[25362]

To Corus, your formula would be right except for the term
(T_p-T_w), which should be LMTD or, at least, the arithmetic average of the temperature differences at both ends of the pipe.

Even in the later case, only when [Δ]T_w is very small [≈] 0, meaning that that the fluid's temperature almost doesn't change, one could use your formula as a good approximation.

Your formula is the result of equating:

V.[π] R².[ρ].C_p.[Δ]T_w = h. 2[π] R.L.(LTD)​

resulting in:

[Δ]T_w = 2.h.L.(LTD)/(V.R.[ρ].C_p)​

Do you agree ?

 

[corus]

25362: You are correct in my derivation although I have no idea what the acronyms mean and, as a cry to the world, would wish that people in general would stop using them, ASAP (joke).

If I read it correctly you are saying that the LTD is the temperature difference between the pipe wall and the mean water temperature in the pipe, rather than the inlet water temperature, as I have used. I agree with this but this puts the unknown outlet water temperature into the right hand side of my equation.

With some manipulation and using the '25362' correction factor, my expression would then become

?Tw = 2.h.L.(Tp-Tw)/(V.R.?.Cp + L.h)

with all values on the right side of the equation being known.

Thanks for the correction.


corus

 

[corus]

Note that the question marks in the above relate to the delta symbol and the greek rho.
So much for cutting and pasting.

corus

 

[25362]

Corus, I'm sorry if the acronym disturbs you. But LMTD is a very well known abbreviation for Logarithmic Mean Temperature Difference.

Denniskb, there may again be something wrong with my calculations. The radiation transfer coefficient as well as the radiation heat flux that I found for a pipe in the shade -not exposed to solar radiation- at the given temperatures are much lower than those reported by you, even when using 0.9 for black iron absorptivity. The recommended value for low temperature radiation is only 0.21. Kindly comment.

 

[corus]

My equation would be an approximation as it assumes that the pipe wall temperature remains constant along the length. As a rough calculation I'd simply assume that the pipe was at 50F, half way between the inlet water temperature and air temperature, presuming there was no additional heating from the sun. In the case of an insulating jacket I'd assume that the pipe was at 45F, the inlet water temperature. That should give a quick and reasonable estimate of the increase in water temperature with the formula I posted and give a comparison between the two cases.

corus

 

[denniskb]

gator1991,
The calculation I use is one of my many Excel applications, but no it is not currently available from my website. Sorry but it took way too much time and effort to simply give away.

I am concerned that many of the recent postings are distracting readers from finding a realistic solution to the original problem. The temperature change along the 80 foot length is negligible so simply ignore it. This heat transfer case is, like most real world HT calcs, complicated enough without introducing an additional dimension that is not needed for a solution.

Using typical HT coefficients from texts is useful but often leads to incorrect assumptions about the real world. Look back at my numbers and see that 88% of the heat transfer is via radiation. There is little to be gained from chasing down details of convection coefficients and LMTDs.

This case involves several layers of HT all acting in series and parallel at the same time;
> forced convection in the water flowing in the pipe
> conduction through an inside film layer
> conduction through the pipe wall
> conduction through an external film layer
> natural convection in the air outside the pipe
> radiation transfer between the outer pipe wall and the surroundings

The two keys to solving the HT are 1) determining the HT properties of the various fluids and materials involved and 2) finding the inside and outside surfaces of the pipe. The second of these is the most important and the most difficult.

My application includes a library of environment, materials and fluid properties (temperature sensitive in many cases) as well as a library of convection formulas for both the inner surface and outer surface of the pipe (which provide the Nusselt No values). What the application does is to set up a series of simultaneous equations for the various heat transfer modes and then solves them iteratively by adjusting the surface temperatures until all of the energy flows balance.

The heart of the calculation is the general formula for calculating the thermal resistance of the various layers involved in a pipe.
R = 1/U = 1/(2.Pi.Ri.(hci + hri))+ln(Ro/Ri)/(2.Pi.kio)+1/(2.Pi.Ro.(hco + hro))
U = Overall HT Coefficient
Ri = inside radius
hci = inside convection ceofficient
hri = internal radiation coefficient (= 0)
kio = thermal conductivity of pipe wall

With the application running I can set up and solve simple cases like this one in about 20 minutes and then spend some time checking the sensitivty of the result to various changes in conditions (e.g. wind velocity, surface coefficients, solar radiation levels). For cases where the fluid in pipe temperature changes significantly I simply re-run the calculation for the conditions further along the pipe.

If I had the time I would write out each of the equations used but that may need to wait until retirement?

The point is that this HT case can be reliably solved with an appropriate effort while the simple approach of using typical textbook values can lead to significant errors and a misunderstanding of what is really happening. While many textbooks address the theory I find Incropera and De Witt the most useful.

Sorry to be so long winded but as you can tell I had some real world cases, including personnel safety, to find solutions to and had to go the hard yards on this subject.


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[25362]

Corus is right. For this particular case

[Δ]t_w [≈] 0.1 MTD​

where MTD is the mean temperature difference at both ends.

A few iterations assuming, "a grosso modo", water outlet temperatures for logical pipe temperatures, would show the heat-up of water is practically negligible.

 

[corus]

I'm not sure what objections denniskb has to the advice offered. Although in this case the answer may be negligible it is important for other cases that people may have, like myself. Which solution is incorrect and which is correct and have any measurements been made to validate the calculations?

From all the answers involving heat gain I'm a little surprised that none require the velocity of the water, for instance. Surely for infinite water velocity the increase in water temperature along the length of pipe is zero?

corus

 

[denniskb]

corus,

A few temperatures from my simulation may help.

Water 7.22°C 45.00°F
Inside Film 7.25°C 45.05°F
Inside Wall 7.28°C 45.10°F
Outside Wall 7.31°C 45.16°F
Outside Film 10.05°C 50.09°F
Air 12.77°C 50.00°F

The two decimal places does not represent the accuracy but are need to show the small temperature differences.

To use a pipewall temperature of 50°F will produce a wrong result.

The radiative properties I used for the pipe outer wall are;

Background Temperature 12.77°C 50.00°C
Surface Absorptivity 0.80
Surface Emmisivity 0.66
- these figures are for carbon steel with light rust
Background Absorptivity 0.66
Background Emmisivity 1.00

The calculated radiative loads are;

Absorption from background 303.3 W/m2
Emmission from pipe surface -231.0 W/m2
Radiative heat balance 72.3 W/m2
Radiative Coefficient 13.15 W/m2.°K 2.31 BTU/hr.ft2.°F

I'm not sure where your 0.21 value comes from or what the units are.

Please note how the heat transfer is hugely dominated by the radiative component and how sensitive the result will be to changes in the absorptivity and emmisivity properties.

Of course the convective load will rise quickly if there is wind movement of the air. My calculation indicates that with a wind of 5 m/sec the convective heat load (128.25 W/m) will be 10 times the still air case yet the overall heat load (0.347°C/100 m) will only have doubled. Still a very small change in the water temperature.


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb