Carters Widget Revealed!

[mad fridgie]

OK, keeping it simple, if we had NO superheat at the exit of the evap, would the flash gas in the evap have any effect on the net cooling of the room. (energy only)

[desA]

OK, keeping it simple, if we had NO superheat at the exit of the evap, would the flash gas in the evap have any effect on the net cooling of the room. (energy only)

Yes.

Simply put, as flash gas is reduced, 'x' will reduce (ie. less vapour in incoming 2-phase fluid) - this will increase Q'evap (refrigeration effect). (A few simplifying assumptions in this, so far)

[mad fridgie]

Yes.

Simply put, as flash gas is reduced, 'x' will reduce (ie. less vapour in incoming 2-phase fluid) - this will increase Q'evap (refrigeration effect). (A few simplifying assumptions in this, so far)

Are you sure?

I will re word. Liquid is at fixed pressure and temp (saturated). It enters the expansion valve, expands (reduction in pressure) So just before it enters the evaporator we a % of vapour by mass (weight) and a % of liquid by mass.
If there in no pressure drop in the evap and no superheat at the exit of the evap, 100% by mass is all vapour at saturation leaving the evap.
Does the original flash gas (that entering the evap) have any benefit to the net cooling.?

We are looking at energy only, not effect on heat transfer co-efficients

[desA]

Are you sure?

I will re word. Liquid is at fixed pressure and temp (saturated). It enters the expansion valve, expands (reduction in pressure) So just before it enters the evaporator we a % of vapour by mass (weight) and a % of liquid by mass.
If there in no pressure drop in the evap and no superheat at the exit of the evap, 100% by mass is all vapour at saturation leaving the evap.
Does the original flash gas (that entering the evap) have any benefit to the net cooling.?

We are looking at energy only, not effect on heat transfer co-efficients

If you split the flows as follows:
1. Vapour only (your 'flash gas') at m'fg;
2. Liquid only at m'l...
3. m'r = m'l + m'fg (total flow before split)

... then, if we know the pressure of each stream, we can locate the individual state points as follows:
1. Vapour - assume saturated vapour - located on right of bell;
2. Liquid - assume saturated liquid - located on left of bell.

If the fluids are mixed, then the mixture properties are constructed based on 'x', the vapour percentage in the mixture.

If the vapour is transported away, then only m'l goes towards the evaporator. The net refrigeration effect (assuming fully saturated liquid) is:

Q'evap = m'l*hfg

Where :
Q'evap = evaporator capacity [W]
m'l = liquid mass flowrate [kg/s]
hfg = latent heat of vapouraziation [J/kg]

So, the real amount of duty [W] managed by the evaporator will depend on how much liquid is actually going to it.

[Tayters]

OK, keeping it simple, if we had NO superheat at the exit of the evap, would the flash gas in the evap have any effect on the net cooling of the room. (energy only)

Right time to jump in with both feet.

The flash gas is caused where the energy contained in the liquid refrigerant entering the TEV is used to lower the temperature of itself to get to the boiling point in the evaporator.

i.e. Te is -30*C, Tc 35*C subcooling 5K means as ref. enters the TEV at 30*C it has to drop temperature by 60K to boil. Ideally the temperature of liquid feed to the TEV would be as low as possible as this shifts the expansion line on a PE chart over to the left into the higher % region of liquid/vapor mix. If it was 100% that means the liquid at the TEV was already at -30*C. If it was you wouldn't really need a refrigerant circuit as you somehow had the means to get refrigerant at the same temperature as Te so you'd use that means instead.

So in answer the flash gas has no net effect in cooling the room. It only cools the refrigerant to the evaporators boiling point.

How does that sound?

Cheers,
Andy.