Cooling capacity flooded evaporator

[matiobbi]

Hi Everyone!!
This is my first post...
I'm trying to calculate the refrigerating capacity of a flooded evaporator. I was in trouble once I had to decide which enthalpy to take for make the difference with the enthalpy at the suction pressure an temperature.
Should I take, at the evaporating pressure, tha enthalpy of the liquid satured or the one after the expansion valve or...

Any clue???

Thanks

[TXiceman]

What are you trying to calculate? Capacity or mass/volume flow?

Just go back to the simple Rankin cycle for you flow calculations.

If you are after capacity, it is much easier to get the flow and dT on the water side and use that.

Ken

[matiobbi]

What I'm trying to calculate is the capacity as the mass flow x the enthalpy difference.
I know the mass flow and the enthalpy at the compresssor suction.

Rankine????

Cheers

[TXiceman]

You can do it one of two ways. The easy way is to use the NRE or net refrigerating effect. NRE = h(2) -h(1)

h(2) = enthalpy of saturated vapor at evaporator outlet.
h(1) = enthalpy of liquid fed to expansion device and flashed.

The other way is to use the latent heat at you evaporator and then calculate the percent flash gas generated at the expansion device and add this to the mass flow calculated with the latent heat.

capacity = mass x (h(2)-h(1))

I nearly always the first method as it is easier and only one step.

Ken

[Peter_1]

TX Iceman, how do you know h2, in other words, how can you know how many vapor and how many liquid is leaving the evaporator in an overfed evaporator coil?

[John Hunter]

In a flooded evaporator for calculation purposes consider the refrigerant to be at the temperature that relates to the prevailing evaporation pressure. The heat exchange is then a function of what is called the Log Mean Temperature Difference., LMTD. There is a little bit of maths to do this but in basic terms it considers that the refrigerant is at constant temperature and the mean temperature is the difference between (fluid on temperature less the Refrigerant temp) (T1) - ( Fluid off less the refrigerant temp) (T2), but because this is not linear you require to divide that result by the Loge of( T1/T2). The LMTD x by the circulated fluid mass and its specific heat will result in the heat being absorbed by the evaporator in what ever units you are working in . If you know the operating parameters of the refrigeration plant you will know the Refrigeration Effect , i.e. enthalpy per unit mass. This mass multiplied by the specific volume of the refrigerant with some allowance for superheat will give the swept volume requirements of the compressor or the mass which ever you want to work with. Q= Mass. x SF x LMTD . Where Q = Heat value. Mass + Mass flow per unit of time i.e.. Per second. SF = Specific Heat , kj.kg.K or btu.lb.F . LMTD log mean temperature difference.
If the calculation is in kg/sec , the specific heat in kJ.kg K. and the temperature in C , the result will be in kJ/second which is Kilowatts /Second .This calculation is valid for all heat exchangers of the evaporative type it is also valid for steam heating provided the correct specific parameters and units are used . Hope this doesn't confuse you.

[matiobbi]

In a flooded evaporator for calculation purposes consider the refrigerant to be at the temperature that relates to the prevailing evaporation pressure. The heat exchange is then a function of what is called the Log Mean Temperature Difference., LMTD. There is a little bit of maths to do this but in basic terms it considers that the refrigerant is at constant temperature and the mean temperature is the difference between (fluid on temperature less the Refrigerant temp) (T1) - ( Fluid off less the refrigerant temp) (T2), but because this is not linear you require to divide that result by the Loge of( T1/T2). The LMTD x by the circulated fluid mass and its specific heat will result in the heat being absorbed by the evaporator in what ever units you are working in . If you know the operating parameters of the refrigeration plant you will know the Refrigeration Effect , i.e. enthalpy per unit mass. This mass multiplied by the specific volume of the refrigerant with some allowance for superheat will give the swept volume requirements of the compressor or the mass which ever you want to work with. Q= Mass. x SF x LMTD . Where Q = Heat value. Mass + Mass flow per unit of time i.e.. Per second. SF = Specific Heat , kj.kg.K or btu.lb.F . LMTD log mean temperature difference.
If the calculation is in kg/sec , the specific heat in kJ.kg K. and the temperature in C , the result will be in kJ/second which is Kilowatts /Second .This calculation is valid for all heat exchangers of the evaporative type it is also valid for steam heating provided the correct specific parameters and units are used . Hope this doesn't confuse you.

Hi!!!

Your answer is great, but I've the problem calculating it on the refrigerant side in therms on enthalpies times mass flow, considering known the enthalpy at the evaporator outlet and the mass flow...

Cheers

[TXiceman]

Sounds like you need to get a good thermodynamics book or a good refrig book. Look for the book by W. Stoecker.

Ken

I was trying to add a word doc with a sketchof the P-H diagram, but the systems says I do not have acces to this item....what the ?????

[narkom]

2 mattiobi
TXiceman gave you the correct advice.
capacity = mass x (h(2)-h(1))
2 Peter_1
h(2) = enthalpy of saturated vapor at evaporator outlet.
So you need not to know
how many vapor and how many liquid is leaving the evaporator in an overfed evaporator coil.

[Peter_1]

2 mattiobi
TXiceman gave you the correct advice.
capacity = mass x (h(2)-h(1))
2 Peter_1
h(2) = enthalpy of saturated vapor at evaporator outlet.
So you need not to know
how many vapor and how many liquid is leaving the evaporator in an overfed evaporator coil.

Narkom, I know whta h2 is but that was my question, how do you know the gas /liquid ratio leaving the coil?

[TXiceman]

I am getting the feeling that we are going in circles/////

Ken

[matiobbi]

Sounds like you need to get a good thermodynamics book or a good refrig book. Look for the book by W. Stoecker.

Ken

I was trying to add a word doc with a sketchof the P-H diagram, but the systems says I do not have acces to this item....what the ?????

Hi Iceman!!

I do have many books and I dont' have so many problems calculating the refrigerating capacity in a standar cycle.
My problem is to understand how the evaporator is fed in the case of fooded.
If I were in the dry case, I would get the enthalpy I'm looking for, at the condenser outlet before the expansion device considering no exergy loss across it.....
What really drives me crazy is the title of the expanded gas entering the evaporator???

Cheers

[US Iceman]

The expanded gas entering the evaporator is the flash gas. The flash gas is the energy cost of cooling the remaining liquid down to the evaporating temperature "to do useful cooling".

You have two components to this. One is the enthalpy difference due to the boiling liquid. This is the latent heat (hf-hg = hfg) and occurs at a constant pressure (the evaporating pressure). Let's call this case 1.

The other component of this occurs at a pressure reduction or expansion cycle (100% liquid to a mixture of liquid and vapor). In this case the liquid exists at a higher pressure and by undergoing an expansion process some of the liquid is cooled down to a lower pressure. Here you still have hf (at liquid feed temperature) -hg (at evaporating temperature) = delta h. Let's call this case 2.

Now, if we subtract case 1 from case 2 (case 2 - case 1) we have the flash gas amount.

I think what you are getting confused with are the different factors that contribute to the total mass flow.

What you have is: Total mass flow = flash gas mass flow + mass flow due to heat absorption (latent heat).