Cooling Water Flow 1

I also know that I can use the effectiveness of the heat exchanger but my "problem" is when there is condensation, the deltaT is minimum or even zero. Is this method the correct one? with the effectiveness?

As long as the enthalpy loads (including latent heats) on each side balance out, i.e., the heat lost by the warm fluid equals the heat gained by the cold fluid, barring heat losses to the surroundings, the estimated cooling water mass flowrate is OK.

Effectiveness of heat exchangers is another issue altogether. I suggest you visit thread391-58702 or thread24-60807.

25362 is correct. If the heat left the process (gas), it had to go somewhere which is the water. If there were environmental heat losses from the process, then the calculated water flow would be more than the acutal, not less as you first stated.

I first stated that I should have more CW flow rate and not less.

An example: 87483 kg/h of Raw Gas, cp 0.598 kcal/kgºC enters the heat exchanger at 71.6ºC and leaves at 59ºC. CW is used to cool down the gas with cp=1 kcal/kgºC which enters the tube side at 19.9ºC and leaves at 28.3ºC

The effectiveness is Eff = (71.6-59)/(71.6-19.9)=0.239

My real question is:

CW = 87483*0.598*(71.6-59)/(1*(28.3-19.9)) = 78472 kg/h or

CW = (87483*0.598*(71.6-59)/(1*(28.3-19.9)))/Eff = 328336

My confusion must be the concept of Effectiveness. With 24% of effectiveness, many heat is lost to the surroundings or the performance of the HE is very low?

Thanks
AndreChE

You can't use temperatures this way. I've never used 'effectiveness' in looking at heat exchangers. Approach temperatures yes (which is 39.1C in your case, quite high so maximum cooling was not the design target of your heat exchanger).

The amount of heat transferred by the gas is the mass flow rate * its heat capacity * the temperature change plus any latent heat effects if condensation occurs.

For the cooling water side, it's the mass flow rate of cooling water * its heat capacity * the temperature change.

Calculate the Q for the gas, it is the same as the Q gained by the water except for any temperature losses to the ambient surroundings which for a shell and tube exchanger should be minor.

to AndreChE, I'll try to concentrate on the subject of efficiency which seems to bother you.

The effectiveness &quot;e&quot; of a heat exchanger is expressed, as you did, on the fluid whose &quot;m.c&quot; is the lowest of both, because it represents the fluid than could in theory attain the maximum delta t. In this example, it is the gas side, since 87483*0.59<78472*1.

&quot;e&quot; values are affected by, and are a function of the type of HE, and depend quite strongly on two dimensionless numbers: R=(mc)_min/(mc)_max, and on NTU_max (number of transfer units)=UA/(mc)_min.
U is the overall HTC, and A is the area.

The mathematical relation of &quot;e&quot; with R and NTU is complicated and exponential. However, some general conclusions can be formulated:

For low NTU values (<1) as in the example you gave, any increase in water flow, no matter how large, will not increase markedly the &quot;e&quot; values resulting from additional cooling of the gas, as it may be seen at the table given at the end of this message.

However, by reducing the water flow rate by, say, 75% one may decrease the heat load by 17% and increase the &quot;e&quot; to above 0.53! I'll try to explain:

When one reduces the water mass flow rate this stream becomes the one with (mc)_min, and &quot;e&quot;=delta t_w/(71.6-19.9).

For the submitted example UA is taken as constant at about 17000 kcal/(h*^oC). By dropping the water flow rate to, say, 20,000 kg/h, NTU=17000/(20000*1)=0.85.
R=20000*1/(87483*0.598)=0.38. From charts &quot;e&quot;=0.53.

Thus the cooling water temperature will rise by 0.53*(71.6-19.9)=27.4^oC.
The duty will drop to 27.4*20,000=548,000 kcal/h, i.e., by (659,165-548,000)/659,165=16.8%

The gas will cool only by 548,000/(87483*0.598)=10.5^oC, down to 71.6-10.5=61.1^oC instead of the original 59^oC.

And the unit will become more &quot;effective&quot; at lower water flow rates (surprised ?). As you see, the influence of varying water flow rates in a given exchanger can be analysed, with outlet temperatures and heat duties changing with changes in effectiveness.

Reducing water flow rates may bring about undesirable results on the process side, thus it should be contemplated in a wider context.

Increasing A in order to improve NTU, to obtain a better &quot;e&quot; is a good approach. However, this means spending money in installations, operations, maintenance, etc., so it is all a matter of economics.

There are books with charts, and I pressume computer programs, to estimate and optimize &quot;e&quot;, especially when the outlet temperatures are not known. Although the effectiveness method is not new at all, it offers advantages for analysing problems in the selection of the HE type best suited to accomplish a particular HE objective.
It may help to decide on whether to split the duty into two units in parallel or in series, it helps in analysing what happens when flows are changed in one given exchanger, and so forth.

Here is a short table comparing effectiveness values for two type of heat exchangers:

NTU R crossflow counterflow
0.25 0 0.23 0.23
0.25 1 0.22 0.19
0.50 0 0.40 0.42
0.50 1 0.32 0.31
1 0 0.62 0.63
1 0.5 0.50 0.55
1 1 0.48 0.49
2 0 0.87 0.87
2 0.5 0.72 0.77
2 1 0.61 0.67
4 0 0.98 0.99
4 0.5 0.87 0.92
4 1 0.72 0.81

Good luck in your HE analyses.

to 25362

Thanks for your explanation. I've been reading a lot about this subject and all your information is correct and precious.

What was bothering me was try to explain the effectiveness of the heat exchanger with the heat lost to the surroundings, which is not correct. I already understood that.

Again, thanks a lot.

AndreChE

I think that there is condensation and probally you are no considering it. Also, it is a common industry practice to report gas cpmpositions in dry basis, so you nedd to consider the potential presence of condensing water. So you need to determine the transfered heat as: sensible heat + latent heat, which results (approx) in mCpDT(gas)+mw.DH.
m=total gas flow,mw=condensing water, DH Laten heat for water.
If you need to determine how much water in in the gas (assuming it as saturated)you need to considered the water partial pressure as vapor pressure of water at inlet temperature. Then asume that the gas will be also saturated at outlet conditions. So determine the partial pressure of water at oulet and consequently the condensed water