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 105°F condensing and a 20°F 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>20°F</sub> = 43.1 psig

h<sub>g 20°F</sub> = 106.2 Btu/lb<sub>m</sub> (sat vapor enthalpy)

40°F vapor entering compressor (20°F 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>105°F</sub> = 210.8 psig

<font face="SYMBOL">r</font><sub>f 105°F</sub> = 70.3 lb<sub>m</sub>/ft<sup>3</sup> (liquid density)

h<sub>f 105°F</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>= 165°F (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>

<html><p>The Prof submits the above R-22 P-H (pressure-enthalpy) diagram for review, showing a refrigeration cycle with a 105°F condensing and a 20°F 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>20°F</sub> = 43.1 psig

h<sub>g 20°F</sub> = 106.2 Btu/lb<sub>m</sub> (sat vapor enthalpy)

40°F vapor entering compressor (20°F 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>105°F</sub> = 210.8 psig

<font face="SYMBOL">r</font><sub>f 105°F</sub> = 70.3 lb<sub>m</sub>/ft<sup>3</sup> (liquid density)

h<sub>f 105°F</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>= 165°F (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>