Question

.0.5 kg/s of refrigerant R-134a undergoes a series of steady-state operations as indicated in the figure below: 12: Isobaric heating process. 23: Reversible and adiabatic compression process 34: Isobaric cooling. 45: Expansion process using throttling va Ive. 1) Determine the amount of heat needed during the evaporation process (12) 2) Determine the ch ange of entropy during the compression process and the temperature of the refrigerant at state 3. 3) Calculate the power input for the compression process. 4) Determine the amount of heat extracted during the conden sat ion process (3-4) 5) Calculate the va por quality at state 5. For more data, use R-134 a properties table or P-h diagram Heat Exchanger (Condenser) Sat P3-5 bar Liquid 3 PS-2.5 bar 5 Throttling (Kinetic and potential energy changes are assumed neglected in all processes) compressor valve Heat Exchanger (Evaporater) R-134a Sat 2 Sat. Liquid P1-2bar Vapor

Answer 1

The saturated thermodynamic property of R 134a available in literature is shown below (Table 1)

TABLE 1

D m/kg Ugm/kg h, k]/kg | hg, kJ/kg | $4, kJ/(kg-K) | sp, kJ/(kg-K) Pressure, bar Temp, K Cafs kJ/(kg-K)H, 10* Pa-sk W/(m-K) Pr 0.0039 0.5 0.6 0.8 1.0 169.85 232.69 236.22 242.04 246.80 0.0006285 0.0007062 0.0007113 0.0007199 0.0007272 35.263 .3692 0.3015 .2375 0.1924 -76.68 -0.57 3.85 11.15 17.14 186.50 225.27 227.52 -0.3830 -0.0025 0.0161 0.0467 0.0713 1.1665 0.9669 0.9636 0.9560 0.9507 1147 1.242 1.248 1.258 1.267 2187 0.1121 5.61 480 5.42 0.1078 0.1056 231.19 438 234.15 408 4.90 0.0007276 0.0007421 0.0007543 0.0007648 0.0007743 247.03 256.03 263.09 268.88 273.82 1.013 1.5 2.0 2,5 3.0 0.1902 0.1312 0.0999 0.0806 0.0677 17.50 28.96 38.13 45.75 52.33 234.33 239.86 244.14 247.60 250.50 0.0728 0.1181 0.1533 0.1819 0.2059 0.9503 0.9419 0.9364 0.9326 0.9297 1.268 1.288 1.306 1.322 1.337 406 358.7 326.6 303.2 285.1 0.1054 0.1013 0.0980 0.0954 0.0931 4.89 4.56 4.35 4.20 4.09 4.0 282.08 288.89 294.72 304.47 312.53 0.0007912 0.0008063 0.0008203 0.0008460 0.0008703 0.0512 63.50 72.87 255.22 258.99 262.09 267.01 270.74 0.2458 0.2784 0.3062 0.3522 0.3901 0.9256 0.9232 0.9208 0.9171 0.9144 1.363 1.387 1.410 1.454 1.497 257.7 237.5 221.6 197.6 179.5 0.0893 0.0861 0.0835 0.0790 0.0753 3.93 3.83 3.74 3.64 3.57 5 0.04116 6 8 0.03434 0.02565 0.02035 81.04 95.00 106.86 10 0.01675 0.01414 0.01247 0.01059 0.00931 12 14 16 18 20 0.0008938 0.0009170 0.0009362 0.0009555 0.0009894 319.47 325.57 330.11 336.04 340.63 117.34 126.80 134.00 143.68 151.39 273.65 275.92 277.40 279.01 279.95 0.4227 0.4515 0.4729 0.5013 0.5236 1.541 1.589 .631 698 1.764 0.9120 0.9095 0.9073 0.9041 0.9010 165.1 153.0 144.3 133.2 124.8 0.0721 0.0693 0.0672 0.0645 0.0623 3.53 3.51 3.50 3.51 3.53 0.0010585 0.001144 0.001270 0.001606 0.001948 25 30 35 40 0.00695 0.00528 0.00399 0.00255 0.00195 280.64 278.32 273.52 257.12 241.22 350.73 359.37 366.89 169.30 185.05 203.19 229.24 241.22 0.5738 0.6212 0.6657 0.7292 0.7620 0.8913 0.8807 0.8574 0.8038 0.7620 1.987 2.418 106.6 90.4 0.0577 0.0538 3.67 4.06 373.50 374.18 40.56 t = triple point, c= critical point. h, S= 0 at -40°C 233.15 K.

Flow of refirgerant is 0.5 kg/s

At position 1

Refrigerant is a saturated liquid at 2 bar a

The specific enthalpy (of saturated liquid) of the refrigerant is 38.13 kJ/kg (from the table 1)

Total Enthalpy at 1 = specific enthalpy * flow rate = 38.13 * 0.5 =19.065 kJ/s

(AH1

At position 2

The refrigerant evaporates isobarically, hence the pressure at position 2 is 2 bar a

Refrigerant is a saturated vapor at 2 bar a (given)

The specific enthalpy (of saturated vapor) of the refrigerant is 244.14 kJ/kg (from the Table 1)

The specific entropy (of saturated vapor) of the refrigerant is 0.9364 kJ/kg (from the table 1)

Total Enthalpy at 2 = specific enthalpy * flow rate = 244.14 * 0.5 =122.07 kJ/s

(AH2

The amount of heat required to in evaporation process =
(ДН2) - (ДН1) (АH)
= 122.07-19.065 = 103.005 kJ/s

At position 3

The refrigerant gets compressed to 5 bar (given) in a reversible and adiabatical manner (which is termed as isentropic process)

In a isentropic process, the entopy of the process remains the same.

We will have to use a superheated table for R 134a where the pressure is 5 bar and the entropy is same as position 2 i.e 0.9364 kJ/kg from the following table:

TABLE 2

P=0.50 MPa (15.7°C) |Temp volumeenthalpy entropy °C V(m 3/kg) h(kJ/kg) s(kJ/kg.K) 259.3 263.5 273.0 282.5 292.0 301.5 0.924 0.938 0.970 Sat. 20 0.0411 0.0421 30 0.0443 401 0.0465 1001 50 0.0485 1.031 0.0505 60 1.060 1.088 1.116 1.144 70 0.0524 311.1 320.8 330.6 340.5 350.6 360.7 371.0 381.5 80 0.0543 90 100 110 120 130 140 0.0562 0.0583 1.171 0.0600 1.197 1.223 1.249 1.275 0.0617 0.0635 0.0653

The specific enthalpy of the refrigerant is 263.5 kJ/kg (from the table 2)

Total Enthalpy at 3 = specific enthalpy * flow rate = 263.5 * 0.5 =131.75 kJ/s

(AH3

The power input to the compressor =
[ΔΗ)- ΔΗ
= 131.75 - 122.07 = 9.68 kJ/s

At position 4

The refrigerant losses heat in the condenser isobarically to form saturated liquid at 5 bar a

The specific enthalpy (of saturated liquid) of the refrigerant is 72.87 kJ/kg (from the table 1)

Total Enthalpy at 4 = specific enthalpy * flow rate = 72.87 * 0.5 =36.435 kJ/s

ΔΗ 4

The amount of heat extracted in condenser =
[ΔΗ-ΔΗ)
= 36.435 - 131.75 = -95.315 kJ/s

Note: Negative sign means heat to be removed

At position 5

In a throttling process, the high pressre liquid is converted to low pressure liquid without any change in the total enthalpy (same as position 4). The temperature of refrigerant will be the saturation temperature at that pressure (from table 2). Temperature = 266.88 K

Total Enthalpy at 5 =36.435 kJ/s

(AH5

The specific enthalpy of saturated liquid of the refrigerant is 45.75 kJ/kg (from the table 1)

The specific enthalpy of saturated vapor of the refrigerant is 247.60 kJ/kg (from the table 1)

Let x be the flow of vapor present at positon 5

Therefore flow of liquid present at position 5 = total refrigerant flow - vapor flow = 0.5 - x

Total enthalpy at 5 = Total enthalpy of vapor at 5 + Total enthalpy of liquid at 5

(AH5

36.435 = x * 247.60 + (0.5-x) * 45.75

x = 0.0672 kg/s

Vapor quality = Vapor flow / Total flow = 0.0672 / 0.5 *100 = 13.44 %

Vapor quality =13.44 %