Pressure drop in condensers.

[PaulLag]

Hi there,

I am looking for a rule of thumb concerning pressure drop in condensers depending on the feeding numnber.

Following is the topic.
Let's assume I have n feedings, and a pressure drop DP1.
in this case pressure drop in proportional to 1/2*v^2

DP1 proportional to 1/2*v^2


Let's assume now that I double the number of feedings, therefore 2n
Since I double the number of feedings, and assuming the same capacity, velocity will decrease to v/2

said this, the new pressure drop will be

DP2 proportional to 1/2*(v/2)^2

said this, I would expect that

Dp1/Dp2 proportional to 4

that is, pressure drop in second case - with the double of circuits - will be something around 4 times less

Please, do you think my argumentation makes sense ?

Thanks

 

[PaulLag]

Sorry I just realized.

DP1 proportional to 1/2*v^2 and l circuit
DP2 proportional to 1/2*(v/2)^2 and l/2 circuit

This means Dp1/dp2 is something proportional to 8.

Please,
does this make sense to you ?
does anybody now a specific rule ?

Thanks !

 

[mk3223]

It isn't clear what are the Dp1 and Dp2 represented for? For a better understanding, a sketch of the system diagram will be helpful.

 

[EdStainless]

Paul, are you talking about single pass verses double pass?
I just pulled an old design up at random.
In double pass it is 7.4 ft/sec velocity and 17.9ft of head loss.
In single pass it is 3.7 ft/sec velocity and 2.66ft of head loss.
It does not work out to a factor of 8 because of inlet and outlet losses.

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P.E. Metallurgy, Plymouth Tube

 

[PaulLag]

Hi

many thanks for your answers.

@mk223
Dp1 is the pressure drop in case 1
DP2 is the pressure drop in case 2 (double of circuits, half of velocity)

@
EdStainless

Yes, more or less.
ok.
so maybe it is an 8^0.9 ?

I don't need an exact number, just a rule of thumb

 

[EdStainless]

Well, maybe.
So for this one condenser you get
14.8 ft/sec 120.4ft
7.4 ft/sec 17.9ft
3.7 ft/sec 2.66ft
For another one
14.0 ft/sec 107.7ft
7.0 ft/sec 16.0ft
3.5 ft/sec 2.14ft

For any real condenser you need to know the tube alloy to select a velocity.
With Cu alloys you stay low, ~6ft/sec for brass and ~7ft/sec for CuNi.
In SS (any stainless) and Ti the modern trend is to run about 9.5-11.5ft/sec.
At these velocities you get better heat transfer and the tubes stay cleaner (which also improves heat transfer).
No one would ever design at <5ft/sec (they would foul) or >15ft/sec (pumping losses would be huge).

You can calculate these, I have found that the results of the HEI method are within 5% of reality. (both deltaP and back pressure)

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P.E. Metallurgy, Plymouth Tube

 

[Afox88]

Paul,

If I am following you correctly, which I think I am, roughly 8 times is correct as a rule of thumb.

For the record, I am assuming that the double circuit design is two units in parallel, and each unit is half the length but same tube count as the full circuit (1 unit).

 

[Afox88]

To be more clear, I assumed this flow diagram. Pardon my handwriting and shorthand.

 

[EdStainless]

So your lower example has double the tube count at half the length? This would have the same surface area.
Why?
You can gain more by a slight change in tube diameter.
How are you modeling the overall heat transfer?
Nearly all real steam condensers are 2 pass, though 4 pass is common also.
The only single pass that I know of uses very long tubes (114').
What is your goal?

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P.E. Metallurgy, Plymouth Tube

 

[Afox88]

I am not the original poster and the drawing is just used to make sure I understand what the OP was talking about. I simplified my drawing by making it 1 pass, but it would work with multiple passes as well. I think the goal of the OP is just to get a rule of thumb or understanding of the options.

The question is with regards to pressure drop and not overall heat transfer. It was not considered in my example. Obviously more tubes = lower velocity = lower heat transfer coefficient.

 

[rotw]

I am not sure if you do mean 1/2 (as "1" for one) or l/2 (as "l" for length) as for the circuits?
I think you should be basically comparing Dp1 with approx. 2 x Dp2 (for 2 x 1/2 circuit). so the factor you are looking at is dp1 / (2 x dp2) and that factor then equates ~2 (instead of 8). Thus your rule of thumb factor is roughly 2.
means same tube diameters, same length, same total capacity, doubling the feeding would reduce the total losses by a factor 2.
Or am I missing something ??

 

[PaulLag]

Hi All

thanks for your answers.
@EdStainless
I am looking for a smart and quick way to evaluate the impact of changing the number of tube feedings in the SAME condenser.

@Afox88
yes, the OP was to get a rule of thumb saying:
what will happen to pressure drop when doubling the number of feedings in the same condenser ?

@rotw
DP1 proportional to 0.5*v^2 and circuit_length
DP2 proportional to 0.5*(v/2)^2 and circuit_length*0.5

thanks

 

[Afox88]

You will get about 1/4 the pressure drop. You had the correct methodology in your original post.

1/2 the flowrate will lead to 1/2 the velocity which will lead to about 1/4 the pressure drop, since it is a function of v^2.

Hope this helps.

 

[PaulLag]

Hi Afox88

please do not forget that the length of circuit will be half of the original.
therefore there is also the circuit_length*0.5

 

[Afox88]

I assumed by the same condenser, you mean that the length of the condenser is the same. If the length of the condenser is half, then the pressure drop is an additional 1/2, which makes 1/8.

The words circuit, feeds, and same condenser, is not typical lingo for what you are asking.

 

[EdStainless]

Though the impact of changing the tube diameter should be considered also.
Many modern condensers with 10-12f/s flows are using 1.125" or 1.250" tubes to reduce pressure drop.
But with higher velocity you still get good heat transfer.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube