Calculating Heat Loss from a Heat Exchanger

[brobertson89]

Heat Exchanger!

I have steam flowing past two pipes which have cold water flowing in them. I want to find out how much heat is lost from the cold water and thus the final temperature of the cold water when it leaves the pipe.

I am doing this as an experiment and before I start I want to know what calculations or equations I will need to use and how to use them. I can find out things like the temperature of the steam the pipe size and all that but I'm not sure which values I will need.

I have read multiple articles on heat exchangers however I am still confused with this problem and need help desperately.

Thank you for the help!

 

[timthereaper]

Here are a few questions you might need to answer first:

Is your heat exchanger counter-, cross-, or parallel-flow?
Are the fluids both mixed, both unmixed, or mixed-unmixed?
Are you going to take into account the steam condensation?
If it's a shell-tube heat exchanger, how many passes does the tube make?
How big are the fouling factors (estimates) on your exchanger?

All of these questions will help you to determine bulk exit temperatures for the cold water.

 

[brobertson89]

Here are a few questions you might need to answer first:

Is your heat exchanger counter-, cross-, or parallel-flow?
Are the fluids both mixed, both unmixed, or mixed-unmixed?
Are you going to take into account the steam condensation?
If it's a shell-tube heat exchanger, how many passes does the tube make?
How big are the fouling factors (estimates) on your exchanger?

All of these questions will help you to determine bulk exit temperatures for the cold water.

Thank you for the help so far, i didn't even think of mentioning these thinks, so here goes:

The flow is perpendicular (so cross-flow).

The fluids are both unmixed.

I would like to take the steam condensation into account however it is not necessary.

It is not a shell-tube hx its just steam flowing over 2 pipes with cold water flowing inside.

I would assume a fouling factor of 0.00009 (m^2K/W) for both the steam and water.

 

[timthereaper]

Is the steam acting in forced or free convection?

 

[brobertson89]

free convection

 

[brobertson89]

i know that the temperature of the water at the outlet will be found using:

Tc0 = Tci + q/(m. Cp)

I know the mass flow rate and Cp and the inlet temperature but i don't know q

 

[firavia]

you should use the NTU method in order to solve this , visit this link for help :
http://www.firavia.com/NTU.html

 

[timthereaper]

I was going to post something similar to firavia's link. I'd give that a good read-through for sure. However, I think you might have something slightly different than what is described there. Since you have steam, you'll have to account for the change in latent heat somehow as it cools down. If it wasn't steam but a saturated liquid, you could use the NTU method for sure. I'm not exactly sure on how to account for the steam, but the book that the link is referencing might have something about that.

 

[edgepflow]

You can do an back-of-the-envelope (may take a few envelopes or set up on the computer) calculation as follows:

q = UA (LMTD).

q = heat transfer rate
U = overall heat transfer coefficient (defined below)
A = surface area of cold water pipes in contact with steam
LMTD = log mean temperare difference (defined below)

U = 1 / [ 1/hi + (ro ln (ro/ri) /2k) + 1/ho ]

where,

hi = convection heat transfer coefficient on inside of water pipe (use "Dittus Boelter" correlation).
ro and ri = pipe outside and inside radius
k = thermal conductivity of pipe
ho = convection heat transfer coefficient on outside of water pipe. This will be a "condensation heat transfer coefficient." You can use a correlation by Chato. Note that for condensation, ho >> hi most of the time so you can neglect the hi term to get started.

You can estimate the LMTD as follows:

LMTD = DT1 - DT2 / ln (DT1/DT2)

where,
DT1 = Tsteam - Tinlet_water
DT1 = Tsteam - Toutlet_water

Assuming the steam is not superheated, it will condense at constant temperature.

Once you figure q, the rate you will condense steam is:

mstean = q / hfg

where hfg is the latent heat of vaporization at the steam pressure.

 

[firavia]

The usear is asking how much heat is lost from the steam to the water and how much the final temprature of the water is , so the temprature of the water at outlet is unknown and you cannot use the LMTD method , only the NTU method can be used. cause you only need the inlet tempratures of the steam and the water in the NTU method.

 

[edgepflow]

The usear is asking how much heat is lost from the steam to the water and how much the final temprature of the water is , so the temprature of the water at outlet is unknown and you cannot use the LMTD method , only the NTU method can be used. cause you only need the inlet tempratures of the steam and the water in the NTU method.

The LMTD can be applied in this case although many may prefer the NTU.

To apply the LMTD method, simply assume a value of liquid outlet temperature and compute:

QHX1 = mdot_LIQ * cp_LIQ * (Tout_LIQ - Tin_LIQ)

and compare this to the value from UA (LMDT). Adjust the liquid outlet temperature until the heat rates match.

I took some documents I had set up before in MathCAD for the LMTD method and incorporated this procedure and found the solution in a few minutes.

 

[firavia]

yes it works you are totaly right it just need some time

 

[appannusa]

when 50% cut baffles are used?

 

[docfreezzzz]

If you can measure the inlet and outlet temperature of the steam you can predict the total heat transfer to the cold water side by looking at the total states on a T-h diagram for steam. I'm assuming that you are using saturated steam (as is common in industry). This is as simple as finding the two temperature values on the saturation line and then computing the enthalpy difference. Any enthalpy change will either heat the pipe in the heat exchanger or at steady-state, be transferred to the other working fluid.