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01-05-2004, 09:28 AM #1
Re: A question about the pressure drop in coil .
Hi
I would tend to think in the terms of saturated temperature loss say 1k for a condenser. 3 psig is a good round number, but relates to a different temperature drop for each refrigerant, better to talk about a drop in saturation (R12 would have a much greater temperature drop for 3 psig than say R744)
Kind Regards. Andy
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01-05-2004, 01:42 PM #2
Re: A question about the pressure drop in coil .
Maintaining a sufficient two-phase refrigerant velocity would be the primary concern for someone designing an evaporator or condenser coil. This is to allow proper wetting of the inside surfaces of the tubes which provides the desired heat transfer.
Pressure drop, however, is a function of refrigerant velocity. So one can think of a "desired" pressure drop for a coil if it is being related to velocity. But this Prof finds this to be an awkward way of thinking. In his humble opinion, one should be more concerned about a maximum desired pressure drop for the coil.
Specifically, pressure drop is roughy proportional to the length of the circuit and the square of the velocity.Prof Sporlan
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01-05-2004, 02:50 PM #3
Re: A question about the pressure drop in coil .
[QUOTE=Prof Sporlan]
Pressure drop, however, is a function of refrigerant velocity. So one can think of a "desired" pressure drop for a coil if it is being related to velocity. But this Prof finds this to be an awkward way of thinking. In his humble opinion, one should be more concerned about a maximum desired pressure drop for the coil.
QUOTE]
So you are saying pressure drop is unimportant
I have never had the need to design a coil (yet) so please elaborate
I have been taught to work to an agreed temperature drop, but in doing this keep a velocity in mind. In hot gas lines this would be 15m/s I am assuming a little less would be appropiate in a condenser coil.
Please comment as I am learning here
Kind Regards. Andy
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02-05-2004, 12:32 AM #4
Re: A question about the pressure drop in coil .
We in the colonies like to think in terms of pressure drops when we consider heat transfer in closed system heat xchangers. I have never considered t in terms of temperature, although thats probably just as good if one thinks in terms of a particuar application.
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02-05-2004, 04:29 PM #5
Re: A question about the pressure drop in coil .
Pressure drop is a fine metric to use in evaluating coil sizing, particularly with respect to making sure pressure drop is not excessive. But the Prof has a bit of a problem with the statement that a condenser coil must have x-amount of pressure drop to be at its most efficient when we don't know the circuit length, tube OD, refrigerant being used, tube material (steel, aluminum, copper).
It is safer to state that the two phase refrigerant velocity must be within a certain range to obtain optimal performance.
For example, all other things being equal, the condenser coil having longer circuits, or a smaller tube OD, will require greater pressure drop for optimal performance.
Designing coils is a matter of selecting the tube OD, circuit length, number of circuits, and doing the proper circuiting so that refrigerant velocity falls within an acceptable range yet results in minimal pressure drop.Prof Sporlan
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02-05-2004, 04:47 PM #6
Re: A question about the pressure drop in coil .
Hi Professor
Originally Posted by Prof Sporlan
Any chance of a worked example to enlighten thick heads like me
Kind Regards. Andy
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04-05-2004, 03:56 AM #7
Re: A question about the pressure drop in coil .
When modeling pressure drop due to fluid flow thru a pipe, we invariably start with the Darcy-Weisbach equation:
hf = f * Le * v² / (D * 2 * g)
where:
hf = head loss, ft
f = friction factor, dimensionless
Le = equivalent length, ft
v = velocity, ft/sec
D = pipe diameter, ft
g = standard acceleration due to gravity, 32.174 ft/sec²
The effect of velocity and equivalent length is apparent from this equation. The problematic variable here is the friction factor. Friction factor is a function of the effective roughness of the pipe, its diameter, and the Reynold's number. For commercial copper tubing, an effective roughness of 5.0e-6 ft is used.
The Moody diagram can be used to look up the friction factor. The Colebrook equation can also be used to calculate friction factor:
1 / f^0.5 = -2 * log10[ (e / D) / 3.7 + 2.51 / (R * f^0.5) ]
where:
f = friction factor, dimensionless
e = effective roughness, ft, 5.0e-6 for copper tubing
D = pipe diameter, ft
R = Reynold's number, dimensionless
As one will notice in short order, an iterative procedure is necessary to calculate friction factor with the Colebrook equation.
To calculate Reynold's number:
R = v * D * rho / (u * gc)
where:
v = velocity, ft/sec
D = pipe diameter, ft
rho = density, lbm/ft³
u = absolute viscosity, ft²/sec
gc = gravitational conversion factor, 32.174 lbm-ft/(lbf-sec²)
And that's simply if we're looking at a single phase fluid!
The problem we have with refrigerant flow in a condenser or evaporator is we're dealing with two phase flow. With two phase flow, there is no obvious way to determine viscosity, or for that matter, if the Moody diagram or Colebrook equation is even valid.
But as one might guess, past researchers have modeled two-phase flow with the above equations in mind, and have developed useful correlations. The Lockhart-Martinelli correlation was perhaps the first correlation done. Others have followed. Bottom line here if one has a handle on the above equations, one has a good understanding on how pressure drop, line length, and refrigerant velocity are related.Prof Sporlan
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