Quote Originally Posted by US Iceman View Post
Double risers are only required if the system/compressor has capacity control. If the system load decreases you need to have a method for unloading the compressors (speed or displacement). When this is done the swept volume through the suction line decreases and reduces the piping velocity. That is why you need double risers (suction or discharge) for oil return back to the compressor. Otherwise, the oil will simply lay there until the gas velocity increases to the point where the oil slug moves.


If you have vapor condensing in a gas pipe you of course need to provide a method for minimizing the liquid slug once it begins to move in the system. This happens a lot in steam system and is called drip draining or something similar to that. This is something different from an oil slug from a P-trap.


You could say an oil separator at the bottom of the riser would work, if you can return the oil automatically to the compressor. There a lot of ways to manage oil in a refrigeration.


The best way is to do it such that no gadgets are used or a method that requires human intervention to make it work.

Thank you iceman.


Marc covered double risers in detail in other powerpoint slides. He talks a lot more than he writes about each point so we have a lot of writing to do during lectures. He said you also need to use double risers if you have unloading evaporators even if the compressor does not unload which is rare these days.


I received an email reply from him today. I asked Marc about the Trane newsletter which is attached.


Here is Marc O'Briens reply


Double risers are certainly not a conventional practice no longer needed. They are still as important today as they ever were. But I can imagine an explanation for why the Trane document says so. They say the scrolls they use unload. The scroll compressor then is probably the Copeland digital which disengages the scrolls every 10 seconds for a period inversely proportional to the load proportion. This means the suction riser gas velocities do, for a moment every 10 seconds, rise to design levels.


Required suction riser velocity can be calculated...
V = 0.723(gD((Rl/Rv)-1))^(1/2)
g = gravity
Rl = density of oil and refrigerant solution
Rv = density of refrigerant vapour


The lower the evaporating temperature and the higher the superheat or pipe diameter the higher the velocity required.


It is a nonsense that a trap atomises a pool of oil better than an elbow or bend does. It is nonsense too anyway that atomising the oil pool even helps. Oil is simply swept up the pipe wall.


It is certainly a nonsense that during normal operation, unloaded, a slug can form in a trap ready to be suddenly transported on the instance of completing its formation as US Iceman claims.


In physics the principle by which an oil slug is shot up the riser is called an impulse and an impulse is a near instantaneous change in momentum, which can only happen at a trap on start up or upon a substantial compressor capacity transient. It can also occur with destructive forces on pumped liquid systems after hotgas defrost when the suction stop valve is suddenly opened. You will recall this from the liquid hammer videos given during my lecture on pumped liquid systems.


Imagine there was no continuous flow of gas behind an impulse driven oil slug. The height it would reach then, in a frictionless system, is simply proportional to the square of its initial velocity. The height reached is a conversion from kinetic energy to potential energy KE=PE so 0.5mv² = mgh and in terms of h is 0.5v²/g or if we needed to hit 10m the impulse velocity needed would be (2gh)^(1/2) = (2x9.81x10)^0.5 = 14m/s. If we assume the density of the oil is 800kg/m³ we can estimate that in a ½” pipe trap there might be 18 grams of oil or 0.018kg. Now impulse is the product of force and change in time Fdt. If we assume a time of impulse of 10 milliseconds then the force to accelerate the 18grams of oil instantaneously to 14m/s would be:


Impulse = m(Vf – Vo)/dt = 0.018(14-0)/0.01 = 25N


The inside area of a ½” outside diameter pipe would be about 10.9mm squared times Pi. Dividing that into 25N gives 66733Pa. That means that an instantaneous or over 10milliseconds at least, pressure of 0.66Bar is required to shoot the slug up 10m. In reality the time over which the force is applied is much longer which compensates for the more gradual, though sometimes not much more gradual, development of pressure differential. Then the gas flow up the pipe continues to also help by the sweeping effect.


The larger the pipe the greater the impulse force pressure required but we don't see the height between traps being recommended to be less with increasing pipe diameters. Because traps are intended only to limit the size of the slugs potentially arriving at the compressor on start-up.


So, the numbers above help illustrate the principle of the use of traps only as a means to limit the size of slug arriving at the compressor rather than as a means of assisting oil return during continuous operation.


You see, if we admit that the size of traps have to be limited to limit the size of slugs arriving at the compressor then we are forced to also admit that oil traps cause oil return, the one is an implied statement of the other. But then their assistance in oil return would only be relevant to systems stopping and starting frequently which is not how we anyway design systems. We instead design systems for as much continuous operation as possible.


With regard to continuous operation, a trap will not atomise an oil pool better than an elbow will and anyway the oil still has to be swept up the wall atomised or not. And since during normal operation no significant impulse forces occur on account of any rapid or instantaneous pressure transients a trap cannot cause a slug to be transported in any near-whole form, if one could even be formed, we are left then with only the principle of suction line velocities for sweeping oil return. A properly designed system does not allow slugs of oil to collect in any trap other than on the larger trap of a double riser during low loads.



The maximum pressure that could build up behind a slug is mgh where m is 800kg/m^3 x volume and since a slug would be liquid it would be caused by such a tiny pressure forming to be instantly spread across the pipe wall area otherwise just sit there while the gas bubbles its way through.



One of my favourite quotes and very relevant here:


Wittgenstein once asked a friend, "Tell me, why do people always say it was natural for man to assume that the Sun went round the Earth rather than that the Earth was rotating?" His friend replied, "Well, obviously, because it just looks as though the Sun is going round the Earth." To which Wittgenstein responded, "Well, what would it have looked like if it had looked as though the Earth was rotating?"


You are welcome to copy and paste this to the discussion board.


I used to police the thinking on that discussion board so it is no longer as sharp as it used to be.