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[1] Repeat Example 9.1 (pages 330-331) but with the following exceptions: refrigerated space is maintained at -15 °C, cooling water is at 19 °C; and the compressor efficiency is n = 0.75. On the PH diagram for R134a, add the five points, and sketch out the four solid and one dashed curves such that it looks similar to Figure 9.2.

[1] Repeat Example 9.1 (pages 330-331) but with the following exceptions: refrigerated space is maintained at -15 °C, cooling water is at 19 °C; and the compressor efficiency is n = 0.75. On the PH diagram for R134a, add the five points, and sketch out the four solid and one dashed curves such that it looks similar to Figure 9.2.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Figure 9.2: Vapor-compression refrigeration cycle on a PH diagram. In P 2 @= H Example 9.1 A refrigerated space is maintained at -20°C, and cooling water is available at 21°C. Refrigeration capacity is 120,000 kJ-h-1. The evaporator and condenser are of suffi- cient size that a 5°C minimum-temperature difference for heat transfer can be realized in each. The refrigerant is 1,1,1,2-tetrafluoroethane (HFC-134a), for which data are given in Table 9.1 and Fig. F.2 (App. F). (a) What is the value of w for a Carnot refrigerator? (b) Calculate w and m for a vapor-compression cycle (Fig. 9.2) if the compressor efficiency is 0.80. 248.15 Solution 9.1 (a) Allowing 5°C temperature differences, the evaporator temperature is -25°C = 248.15 K, and the condenser temperature is 26°C = 299.15 K. Thus, by Eq. (9.3) for a Carnot refrigerator, Const S = 4.87 299.15 248.15 (b) With HFC-134a as the refrigerant, enthalpies for states 2 and 4 of Fig. 9.2 are read directly from Table 9.1. The entry at -25°C indicates that HFC-134a vaporizes in the evaporator at a pressure of 1.064 bar. Its properties as a saturated vapor at these conditions are: H₂ = 383.45 kJ-kg-1 S2=1.746 kJ-kg-1.K-1 The entry at 26°C in Table 9.1 shows that HFC-134a condenses at 6.854 bar; its enthalpy as a saturated liquid at these conditions is: H4 = 235.97 kJ .kg-1 If the compression step is reversible and adiabatic (isentropic) from saturated vapor at state 2 to superheated vapor at state 3', S3 S₂ = 1.746 kJ.kg-¹.K-1 This entropy value and the condenser pressure of 6.854 bar are sufficient to spec- ify the thermodynamic state at point 3'. One could find the other properties at this state using Fig. F.2, following a curve of constant entropy from the saturation curve to the condenser pressure. However, more precise results can be obtained using an electronic resource such as the NIST WebBook. Varying the temperature at a fixed pressure of 6.854 bar shows that the entropy is 1.746 kJ.kg-¹.K-¹ at T= 308.1 K. The corresponding enthalpy is: H₂ = 421.97 kJ-kg-1 and the enthalpy change is: (AH)s=H₂-H₂=421.97-383.45 = 38.52 kJ kg-1 By Eq. (7.17) for a compressor efficiency of 0.80, the actual enthalpy change for step 2 → 3 is: H3-H₂= (ΔΗ), 38.52 0.80 n Because the throttling process of step 1 →4 is isenthalpic, H₁ = H4. The coefficient of performance as given by Eq. (9.4) therefore becomes: @= 48.15 kJ-kg-1 H₂-H4 383.45-235.97 H3 - H₂ 48.15 = and the HFC-134a circulation rate as given by Eq. (9.5) is: ġc 120,000 H₂-H4 383.45-235.97 m = = 3.06 = 814 kg.h-1
Transcribed Image Text:Figure 9.2: Vapor-compression refrigeration cycle on a PH diagram. In P 2 @= H Example 9.1 A refrigerated space is maintained at -20°C, and cooling water is available at 21°C. Refrigeration capacity is 120,000 kJ-h-1. The evaporator and condenser are of suffi- cient size that a 5°C minimum-temperature difference for heat transfer can be realized in each. The refrigerant is 1,1,1,2-tetrafluoroethane (HFC-134a), for which data are given in Table 9.1 and Fig. F.2 (App. F). (a) What is the value of w for a Carnot refrigerator? (b) Calculate w and m for a vapor-compression cycle (Fig. 9.2) if the compressor efficiency is 0.80. 248.15 Solution 9.1 (a) Allowing 5°C temperature differences, the evaporator temperature is -25°C = 248.15 K, and the condenser temperature is 26°C = 299.15 K. Thus, by Eq. (9.3) for a Carnot refrigerator, Const S = 4.87 299.15 248.15 (b) With HFC-134a as the refrigerant, enthalpies for states 2 and 4 of Fig. 9.2 are read directly from Table 9.1. The entry at -25°C indicates that HFC-134a vaporizes in the evaporator at a pressure of 1.064 bar. Its properties as a saturated vapor at these conditions are: H₂ = 383.45 kJ-kg-1 S2=1.746 kJ-kg-1.K-1 The entry at 26°C in Table 9.1 shows that HFC-134a condenses at 6.854 bar; its enthalpy as a saturated liquid at these conditions is: H4 = 235.97 kJ .kg-1 If the compression step is reversible and adiabatic (isentropic) from saturated vapor at state 2 to superheated vapor at state 3', S3 S₂ = 1.746 kJ.kg-¹.K-1 This entropy value and the condenser pressure of 6.854 bar are sufficient to spec- ify the thermodynamic state at point 3'. One could find the other properties at this state using Fig. F.2, following a curve of constant entropy from the saturation curve to the condenser pressure. However, more precise results can be obtained using an electronic resource such as the NIST WebBook. Varying the temperature at a fixed pressure of 6.854 bar shows that the entropy is 1.746 kJ.kg-¹.K-¹ at T= 308.1 K. The corresponding enthalpy is: H₂ = 421.97 kJ-kg-1 and the enthalpy change is: (AH)s=H₂-H₂=421.97-383.45 = 38.52 kJ kg-1 By Eq. (7.17) for a compressor efficiency of 0.80, the actual enthalpy change for step 2 → 3 is: H3-H₂= (ΔΗ), 38.52 0.80 n Because the throttling process of step 1 →4 is isenthalpic, H₁ = H4. The coefficient of performance as given by Eq. (9.4) therefore becomes: @= 48.15 kJ-kg-1 H₂-H4 383.45-235.97 H3 - H₂ 48.15 = and the HFC-134a circulation rate as given by Eq. (9.5) is: ġc 120,000 H₂-H4 383.45-235.97 m = = 3.06 = 814 kg.h-1
[1] Repeat Example 9.1 (pages 330-331) but with the following exceptions: refrigerated space is maintained at -15 °C, cooling water is at 19 °C; and the compressor efficiency is n = 0.75. On the PH diagram for R134a, add the five points, and sketch out the four solid and one dashed curves such that it looks similar to Figure 9.2.
Transcribed Image Text:[1] Repeat Example 9.1 (pages 330-331) but with the following exceptions: refrigerated space is maintained at -15 °C, cooling water is at 19 °C; and the compressor efficiency is n = 0.75. On the PH diagram for R134a, add the five points, and sketch out the four solid and one dashed curves such that it looks similar to Figure 9.2.
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