3 replies to this topic

[ColorMan]

say i have a heat exchanger with a capacity of 100,000 BTU/hr.

and that steam will be the heating fluid, and oil will be the process fluid. i understand how to calculate for all my needed info for the process fluid. but i am not sure how to calculate the required info for the heating fluid (steam).

i want to be able to figure out how to make sure my heat exchanger is working at its full capacity, but i need to know how much steam i need to get into the system.

anybody have the needed equations for this?

[Art Montemayor]

Color:

In any heat exchanger operation, in order to determine the required flow rate of one fluid, one must make an energy balance:

Q = W LHv

Where,

Q = the heat transfer taking place between both fluids, 100,000 Btu/hr;
W = the mass flow rate of the condensing stream (steam in your case), lb/hr;
LHv = the Latent Heat of Condensation (or vaporization) of the stream in question, Btu/lb.

W = Q / LHv

What year of university studies are you currently in?

[ColorMan]

thanx for the help Art!!

i graduated from West Point in 02, but post Army, i switched from Telecommunications to Project Managerment, so I am trying to get the basics of thermo down.

[djack77494]

ColorMan,
When you delve into the workings of a heat exchanger, there are three simultaneous equations that govern your system. There is an equation describing the heat gained/lost by the shellside fluid. If steam is on the shellside of your exchanger, than that is the equation that Art presented you. There is a similar equation for the tubeside fluid. For your liquid fluid (oil), that would be:

Q = W * Cp * (T2 - T1)

where,

Q = the heat transfer taking place between both fluids, again 100,000 Btu/hr;
W = the mass flow rate of the process fluid (oil in your case), lb/hr;
Cp= the heat capacity of the process fluid, Btu/lb/degreeF
T1,T2= Inlet and outlet oil temperatures, degreeF

Note how the earlier equation presented by Art relates to energy change due to latent heat (condensation of the steam in this case), while the above equation relates to sensible heat effects. It's possible to have both within a single heat exchanger. By conservation of energy, the heat lost by the steam = the heat gained by the oil, or

Ws * LHvs = Wo * Cpo * (T2o - T1o)

(Note that I've added "s" and "o" to the variables to more clearly indicate what is steam and what is oil.)


OK, now I said there were three governing equations and so far we've seen two. The third is the equation governing heat transfer:

Q = U * A * Ft * DTlm

where,

Q = the heat transfer taking place, again 100,000 Btu/hr;
U = the overall heat transfer coefficient for your exchanger, Btu/hr/ft^2/degreeF;
Ft= Correction factor (refer to TEMA Standards), dimensionless;
DTlm= Log Mean Temperature Difference, degreesF.

Refer to any heat transfer or process design book for DTlm calculation.

Your exchanger will always be in heat balance, but its ability to transfer heat may degrade over time. This is due to fouling. Thus, if you monitor the exchanger's performance, starting with a clean exchanger, you will see a slow but steady decrease in Q (or an increase in U). Typically the degradation is initially rapid, but then slows.

Hope that helps,
Doug