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Solucion P3

The document outlines a thermodynamic analysis involving R134a refrigerant in a mechanical and energy engineering context. It includes calculations for various parameters such as enthalpy, entropy, and efficiency for a refrigeration cycle, along with specific heat capacities and flow rates. The final results provide values for cooling flow rates, net work output, and other relevant thermodynamic properties.

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Manuel Lucas Miralles
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19 views2 pages

Solucion P3

The document outlines a thermodynamic analysis involving R134a refrigerant in a mechanical and energy engineering context. It includes calculations for various parameters such as enthalpy, entropy, and efficiency for a refrigeration cycle, along with specific heat capacities and flow rates. The final results provide values for cooling flow rates, net work output, and other relevant thermodynamic properties.

Uploaded by

Manuel Lucas Miralles
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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File:P3TASept2021.

EES 16/09/2021 13:07:05 Page 1


EES Ver. 10.836: #1645: For use only by INGENIERÍA MECÁNICA Y ENERGÍA UNIVERSIDAD MIGUEL HERNÁNDEZ DE ELCHE

DNI = 5

6
P 1 = 1,4 x 10

T 1 = 180 + 273,15

h 1 = h R134a ; T = T 1 ; P = P 1

s 1 = s R134a ; T = T 1 ; P = P 1

P 2 = 900000

s 2s = s 1

h 2s = h R134a ; s = s 2s ; P = P 2

 TURB = 0,9

h2 – h1
 TURB =
h 2s – h 1

s 2 = s R134a ; h = h 2 ; P = P 2

T 2 = T R134a ; h = h 2 ; P = P 2

P 3 = 900000

x3 = 0

h 3 = h R134a ; x = x 3 ; P = P 3

v 3 = v R134a ; x = x 3 ; P = P 3

T 3 = T R134a ; x = x 3 ; P = P 3

h 4s = h3 + v3 ꞏ P4 – P3

s 3 = s R134a ; h = h 3 ; P = P 3

P4 = P1

 BOMBA = 0,9

h 4s – h 3
 BOMBA =
h4 – h3

T 4 = T R134a ; h = h 4 ; P = P 4

s 4 = s R134a ; h = h 4 ; P = P 4

h1 – h2 – h4 – h3
 Termico =
h1 – h4

3
Wneta = DNI ꞏ 10

Wneta = m R134a ꞏ h1 – h2 – h4 – h3

Caudal de refrigeración
Qsal = m R134a ꞏ h2 – h3

h A = h water ; T = T A ; P = P A

3
P A = 100 ꞏ 10

T A = 20 + 273,15

h B = h water ; T = T B ; P = P B

3
P B = 100 ꞏ 10

T B = 30 + 273,15

Qsal = m agua ꞏ hB – hA

Caudal aceite

Qent = m R134a ꞏ h1 – h4

Qent = m AC ꞏ Cp AC ꞏ T AC1 – T AC2

3
Cp AC = 2700 ꞏ 10

T AC1 = 250

T AC2 = 200

Ärea de captación

Area ꞏ  capt ꞏ Irr = Qent

 capt = 0,4

Irr = 700

SOLUTION
Unit Settings: SI K Pa J mass deg
Area = 429,3 [m2] CpAC = 2,700E+06 [J/kg-K]
DNI = 5 BOMBA = 0,9
capt = 0,4 Termico = 0,04159
TURB = 0,9 h2s = 403597 [J/kg]
h4s = 102034 [J/kg] hA = 84006 [J/kg]
hB = 125821 [J/kg] Irr = 700 [W/m2]
mAC = 0,0008905 [kg/s] magua = 2,755 [kg/s]
mR134a = 0,3796 [kg/s] PA = 100000 [Pa]
PB = 100000 [Pa] Qent = 120216 [W]
Qsal = 115216 [W] s2s = 1281 [J/kg-K]
TA = 293,2 [K] TAC1 = 250 [K]
TAC2 = 200 [K] TB = 303,2 [K]
Wneta = 5000 [W]

1 potential unit problem was detected.

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