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Prof Sporlan
07-04-2001, 04:19 AM
http://walden.mvp.net/~aschoen/r22p-h.bmp

<html><p>The Prof submits the above R-22 P-H (pressure-enthalpy) diagram for review, showing a refrigeration cycle with a 105F condensing and a 20F evaporating temperatures.<br>
Ok, he's not a genius when it comes to producing nice looking images, but he figures this diagram is worth perhaps 250 words, if not 1000.... :)<br>
The thermodynamic properties are defined as follows, since they are no doubt difficult to read:<br>
P<sub>20F</sub> = 43.1 psig
h<sub>g 20F</sub> = 106.2 Btu/lb<sub>m</sub> (sat vapor enthalpy)
40F vapor entering compressor (20F superheat)
<font face="SYMBOL">r</font><sub>g </sub>= 1.01 lb<sub>m</sub>/ft<sup>3</sup> (vapor density entering compressor)
h<sub>g </sub>= 109.6 Btu/lb<sub>m</sub> (vapor enthalpy entering compressor)
P<sub>105F</sub> = 210.8 psig
<font face="SYMBOL">r</font><sub>f 105F</sub> = 70.3 lb<sub>m</sub>/ft<sup>3</sup> (liquid density)
h<sub>f 105F</sub> = 40.8 Btu/lb<sub>m</sub> (sat liquid enthalpy)<br>
For the purpose of discussion, isentropic (ideal) compression will be assumed when determining the state of the vapor leaving
the compressor. Therefore:<br>
<font face="SYMBOL">T</font><sub>g </sub>= 165F (vapor temperature leaving compressor)
<font face="SYMBOL">r</font><sub>g </sub>= 3.42 lb<sub>m</sub>/ft<sup>3</sup> (vapor density leaving compressor)
h<sub>g </sub>= 125.0 Btu/lb<sub>m</sub> (vapor enthalpy leaving compressor)<br>
For the sake of argument, lets say the evaporator is rated for 18,000 Btu/hr. So, with all this information, what do we know about the
system? :)</p></html>

Prof Sporlan
07-04-2001, 03:27 PM
<html><p>Enthalpy is the amount of useful energy in a substance, and it is defined as follows:<br>
h = u + p * <font face="Symbol">n</font><br>
where:
h = enthalpy, Btu/lb<sub>m</sub>
u = internal energy, Btu/lb<sub>m</sub>
p = pressure, lb<sub>f</sub>/ft<sup>2</sup>
<font face="Symbol">n</font> = specific volume, ft<sup>3</sup>/lb<sub>m</sub><br>
Internal energy accounts for the potential and kinetic energies in a substance, and it is a function of the temperature of the substance. It takes energy to increase the temperature of a substance. Also, it takes energy to increase pressure and volume (as in blowing up a ballon). If pressure and volume are decreased, useful energy is given up.<br>
Some of you might notice the p * <font face="Symbol">n</font> term is not consistent with enthalpy in the above equation. That is why the equation is normally written:<br>
h = u + p * <font face="Symbol">n</font> / J<br>
where:
J = 778 ft-lb<sub>f</sub>/Btu, conversion factor
when using english units<br>
When determining the amount of heat absorbed or rejected, you will be comparing the enthalpy of the substance (in our case, the refrigerant) entering and leaving the process.</p></html>

Prof Sporlan
10-04-2001, 02:01 AM
I'm going to program my TI-82 to convert all the imperial expressions used by the Prof, then I can jump on board and go for a ride through the thermophysical wonderland I see the Prof pointing at on the near horizon here The Prof would be amused if he had helped in any small way to return the world back to Imperial units. :)
How many forms of kinetic and potential energies, should I ask how many instances of these two energies are there through the phases we work with. That's through subcooled to saturated liquid to saturated two phase to saturated vapour to superheated vapour to gas? Interestingly, any change in state of the refrigerant results in a change of energy, resulting in work. Work done per unit of time results in power, e.g., Btu/hr or tons of refrigeration. Energy can be broken down to its potential and kinetic components, but this would be rarely needed to understand and model a thermodynamic process. The Prof would liken this to studying the machine code resulting from a Visual Basic program. :)


[Edited by Prof Sporlan on 10-04-2001 at 01:58 PM]

subzero*psia
17-04-2001, 01:54 AM
Looks like you are starving the evaporator... maybe superheat setting, bulb location or the wrong valve. Just a hunch... based on the entropy etc. I would love to work it out but I don't have an R22 PT chart. Some guys can really go to town on these things!

Prof Sporlan
19-04-2001, 11:44 PM
<i>For the sake of argument, lets say the evaporator is rated for 18,000 Btu/hr. So, with all this information, what do we know about the system?</i>

First, we can calculate the necessary refrigerant flow rate. This is done as follows:

NRE = h<sub>g</sub> - h<sub>f</sub> = 106.2 - 40.8 = 65.4 Btu/lb<sub>m</sub>

where NRE is the <i>net refrigerating effect</i>, i.e., the quantity of heat that each pound of refrigerant absorbs while flowing thru the evaporator. To determine the necessary refrigerant flow rate:

W = 18,000 Btu/hr / 65.4 Btu/lb<sub>m</sub> = 275.2 lb<sub>m</sub>/hr = 4.59 lb<sub>m</sub>/min

So we know we must flow 4.59 lb<sub>m</sub>/min of R-22 thru the evaporator to achieve 18,000 Btu/hr of cooling. We can also determine the power requirements for the compressor by determining the enthalpy difference between states [1] and [2]:

<font face="Symbol">D</font>h during compression = 125.0 - 109.6 = 15.4 Btu/lb<sub>m</sub>

Theoretical horsepower req'd = 15.4 Btu/lb<sub>m</sub> * 4.59 lb<sub>m</sub>/min / 42.4 Btu/(min-hp) = 1.67 hp
where the 42.4 Btu/(min-hp) term is necessary to convert heat flow rate to horsepower.

Of course the 1.67 hp figure seems small to handle 18,000 Btu/hr cooling. The reason, of couse, is we are assming a 100 percent efficient compressor and ideal compression. If we assume 80 percent efficiency for the compressor:

horsepower req'd = 1.67 / 0.8 = 2.09 hp.

Allowing entropy to increase during compression, which will be the case, will increase horsepower requirements further.

Prof Sporlan
20-04-2001, 02:11 AM
Calculating heat of rejection becomes a simple matter, as we need only the enthalpy difference between states [2] and [3]:

<font face="SYMBOL">D</font>h = 125.0 - 40.8 = 84.2 Btu/lb<sub>m</sub>

heat of rejection = 84.2 Btu/lb<sub>m</sub> * 275.2 lb<sub>m</sub>/hr = 23,172 Btu/hr

where the 275.2 lb<sub>m</sub>/hr refrigerant flow rate was calculated in the previous post.

Note that the heat of rejection will always be greater than the cooling capacity. In fact, it will be the cooling capacity plus the work done by the compressor plus any heat absorbed thru the suction line.

Is there any additional information that can be learned from the P-H diagram?

Dan
21-04-2001, 03:31 AM
Professor, I think you have a wonderful tutorial going on here. But I cannot see the drawing. It appears as a broken image. Is it just my browser?

Dan

Prof Sporlan
21-04-2001, 03:45 AM
It looks fine here, though the Prof uses a 1024 x 768 resolution on a 19" monitor. The detail is a bit difficult to see, thus the need to specify the values. How about it Dez? Is the image that bad? :(

Steve
21-04-2001, 07:19 AM
I can see the image OK, and am also using a 1024 x 768 resolution on a laptop. It takes a bit of time to load though.

WebRam
21-04-2001, 10:16 AM
I see the immage OK here Prof, detail is not great but if I squint my eye's and keep drinking the wee heavy ...she sure looks good ;)

Dan
21-04-2001, 04:49 PM
Got it on MSIE. It's a netscape thing.

subzero*psia
25-04-2001, 12:20 AM
I have misplaced my Roy Dossat book... and I am ashamed to admit that I don't remember everything I once knew about PH charts.

Just looking at it again and you seem to have alot of entropy occuring like premature flashing may be occuring in the liquid line to subcool the liquid and also on the low side. I know I am reaching here without my book...

Prof Sporlan
25-04-2001, 02:59 AM
Once again referring to the P-H diagram, the Prof finds the process from state [3] to [4] most interesting. This process is, of course, the refrigerant flow thru the expansion device.

Flow thru the expansion device is assumed to be <b><i>isenthalpic</i></b>, i.e., enthalpy of the refrigerant entering and exiting the expansion device will be the same. This will be the case regardless of the expansion device used: thermostatic expansion valve, automatic expansion valve, hand valve, capillary tube, short tube restrictor (plug orifice). An isenthalpic process means, by definition, no change in useful energy (enthalpy) is gained or lost. Folks who amuse themselves with the study of thermodynamics will also refer to this process as a <b><i>throttling</b></i> process.

If the Prof is allowed to elborate a bit, an <b><i>isenthalpic</i></b> process is categorized as an <b><i>adiabatic</b></i> process, a process in which no energy is exchanged. This is also true of an <b><i>isentropic</b></i> process, which we assumed for the process from state [1] to [2], the work done by the compressor. The difference between these two processes is that the <b><i>isentropic</b></i> process is said to be a <b><i>reversible</b></i> process, which is to say in this case the work used to compress the gas can be retrieved by allowing the gas to reexpand. The <b><i>isenthalpic</i></b> process, however, is an <b><i>irreversible</b></i> process. Once the refrigerant has reached state [4], it cannot return to state [3] unless energy is expended. This concept might seem odd at first, for no change in useful energy has occurred with this process. Why should should energy be necessary to return the refrigerant to state [3]? The answer is entropy increases with an <b><i>isenthalpic</i></b> process, which is a measure of energy that is no longer available to do useful work. So useful energy stays the same, unuseful energy increases, which is where we "lose" our energy.

State [3] is the refrigerant leaving the expansion device and entering the evaporator coil, and it will be normally located under the "dome", the two-phase (liquid and vapor) region of the P-H diagram. We can use the P-H diagram to determine the <b><i>quality</i></b> of the refrigerant at this state, i.e., the percentage vapor of the two-phase flow by mass. First, we need to know liquid enthalpy at the evaporating temperature:

h<sub>f 20F</sub> = 16.0 Btu/lb<sub>m</sub>

Quality is then calculated: (h<sub>f 105F</sub> - h<sub>f 20F</sub>) / (h<sub>g 20F</sub> - h<sub>f 20F</sub>) = (40.8 - 16.0) / (106.2 - 16.0) = 0.275

or 27.5 percent of the refrigerant entering the evaporator is vapor by mass. Since the vapor entering the evaporator has essentially no refrigerating effect, we can say that the remaining 72.5 percent of the refrigerant is doing all the work!

Prof Sporlan
25-04-2001, 03:57 PM
Knowing the quality of the refrigerant entering the evaporator, we can determine the <b><i>void fraction</i></b> of the two-phase refrigerant flow. This is the percentage of the volume the vapor portion of the two-phase flow takes up.

<font face="SYMBOL">y</font> = 1 / (1 + (1 - x) * <font face="SYMBOL">r</font><sub>V</sub> / (x * <font face="SYMBOL">r</font><sub>L</sub>))

where:

<font face="SYMBOL">y</font> = void fraction
x = quality
<font face="SYMBOL">r</font><sub>V</sub> = vapor density
<font face="SYMBOL">r</font><sub>L</sub> = liquid density

For this equation, the units of density need only be consistent.

The density of the liquid and vapor:

<font face="SYMBOL">r</font><sub>L 20F</sub> = 81.4 lb<sub>m</sub>/ft<sup>3</sup>
<font face="SYMBOL">r</font><sub>V 20F</sub> = 1.07 lb<sub>m</sub>/ft<sup>3</sup>

and the quality is 0.275 from the previous post.

Plugging in the values into the equation, we find:

<font face="SYMBOL">y</font> = 1 / (1 + (1 - 0.275) * 1.07 / (0.275 * 81.4) = 0.967

Or 96.7 percent of the refrigerant entering the evaporator coil is vapor by volume! Consider what these numbers mean. The vapor portion of the two-phase flow represents 27.5 percent of the weight of the refrigerant entering the coil, but it takes up 96.7 percent of the volume. The liquid portion of the two-phase flow represents 72.5 percent of the weight of the refrigerant entering the coil, but it takes up only 3.3 percent of the volume!

Visualizing this, the TEV is actually feeding a "wet gas" into the evaporator coil, not liquid as often assumed.

Dan
28-04-2001, 05:15 PM
Originally posted by Prof Sporlan
Knowing the quality of the refrigerant entering the evaporator, we can determine the <b><i>void fraction</i></b> of the two-phase refrigerant flow. This is the percentage of the volume the vapor portion of the two-phase flow takes up.

Or 96.7 percent of the refrigerant entering the evaporator coil is vapor by volume! Consider what these numbers mean. The vapor portion of the two-phase flow represents 27.5 percent of the weight of the refrigerant entering the coil, but it takes up 96.7 percent of the volume. The liquid portion of the two-phase flow represents 72.5 percent of the weight of the refrigerant entering the coil, but it takes up only 3.3 percent of the volume!

Visualizing this, the TEV is actually feeding a "wet gas" into the evaporator coil, not liquid as often assumed.


Void fraction. Odd term, but fascinating view of what the TEV and evaporator are doing. I have never thought that way about it. I don't know what to do with it, but it is a facinating turn of thought for me.:)

Dan
07-05-2001, 05:37 AM
The diagrams... the PE or PT or psychromeric diagrams both fascinate and elude me. I would be interested in who the artists/authors are and how these visual aids developed.

I think that whomever made up these devices were kind people. They did the hard work with all the curvy lines so we could apply straight lines to predict properties and state. Who invented the Pressure enthalpy chart?

Dan

[Edited by Dan on 07-05-2001 at 06:05 AM]

subzero*psia
07-05-2001, 11:50 AM
I might be wrong about this but I think that is a Carnot chart? It would be interesting to know the story behind it. I would love to find a book about the Bernouli brothers too.

Prof Sporlan
07-05-2001, 07:03 PM
The Carnot cycle is your ideal heat engine. Like the ideal refrigeration cycle, it consists of four processes, in this case: isentropic compression and expansion of the working fluid, and a heat transfer to (boiler/evaporator) and from (condenser) the working fluid. Not only is it thermodynamically efficient as possible, it is also a reversible process. And when one reverses the Carnot cycle, you get the ideal refrigerator, thus the reason it is of interest to refrigeration folks who like studying the implications of the second law of thermodynamics.

But you might ask, isn't the refrigeration cycle you posted also "ideal"? It is, but it's not as thermodynamically efficient as the reverse Carnot cycle. The problem, the Prof has to admit, it with the expansion device. As previously posted, flow thru the expansion device is not "reversible". With the reverse Carnot cycle, however, the "expansion device" simply the same piston/cylinder arrangement used to compress the fluid.

<i>I might be wrong about this but I think that is a Carnot chart?</i>

P-H (pressure-enthalpy) chart would be the correct term for the chart posted, which is the commonly used chart for refrigeration cycle analysis. Interestingly, the Carnot cycle is normally drawn on a T-S (temperature-entropy) chart in thermodynamic texts. The Carnot cycle displays simply as a square box on a T-S chart, whereas it would show as a curvy box on a P-H chart. One might refer to a T-S chart as a Carnot chart, though.

<i>Who invented the Pressure enthalpy chart?</i>

Good question! The principles necessary to generate the P-H and T-S charts we known early in the 19th century. It would be interesting to know when useful charts were able to be generated.

Prof Sporlan
07-05-2001, 10:53 PM
Not that the Prof wants to solely blame the expansion device for not allowing the ideal refrigeration cycle to be as efficient as the reverse Carnot cycle. There is one other source of "irreversibility" in the ideal refrigeration cycle: the process of desuperheating refrigerant vapor in the condenser coil. Why is this a source of inefficiency, you may ask?

Let it suffice to say at the moment that the reverse Carnot cycle requires heat transfer thru the evaporator and condenser to be isothermal (constant temperature) processes. For the ideal refrigeration cycle, heat transfer thru the evaporator is essentially an isothermal process. Ok, using a TEV requires we run the evaporator at some amount of superheat at the evaporator outlet, so once again we look disapprovingly at the expansion device for not allowing Carnot cycle efficiencies. But also keep in mind that compressor manufacturers do not care much to have their compressors operating with saturated refrigerant vapor entering their compressors. :) Heat transfer thru the condenser, however, is clearly not an isothermal process for the ideal refrigeration cycle. A significant portion of the condenser must simply desuperheat the refrigerant vapor before condensation occurs.

The Prof recalls that HySave had promoted the use of their refrigerant pumps to desuperheat refrigerant vapor entering the condenser. Any thoughts on how desuperheating vapor before it enters the condenser might affect system efficiency?

Prof Sporlan
11-05-2001, 02:10 AM
<i>Then probably our biggest selling point for liquid injection is the reduced scaling, maintanance and treatment required for water cooled condensers</i>

Which suggests designing in some loss in thermodynamic efficiency may well prove beneficial in overall system design...:)