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A refrigerator operates on the ideal vapor-compression refrigeration cycle and use R-134a as the working fluid. The condenser operates 1.4 MPa and the evaporator at 0.140 MPa. (a) Show the cycle on a T-s diagram, indicating the enthalpies at the end of each process, with respect to saturation lines. For a mass flow rate of 0.100 kg/s, determine (b) the net power input to the compressor, in kW, (c) the heat rejection to the environment, in kW, (d) the heat rejection from the cold refrigerated space to the evaporator, in kW, and (e) the coefficient of performance.

A refrigerator operates on the ideal vapor-compression refrigeration cycle and use R-134a as the working fluid. The condenser operates 1.4 MPa and the evaporator at 0.140 MPa. (a) Show the cycle on a T-s diagram, indicating the enthalpies at the end of each process, with respect to saturation lines. For a mass flow rate of 0.100 kg/s, determine (b) the net power input to the compressor, in kW, (c) the heat rejection to the environment, in kW, (d) the heat rejection from the cold refrigerated space to the evaporator, in kW, and (e) the coefficient of performance.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Refrigeration Cycle: A refrigerator operates on the ideal vapor-compression refrigeration cycle and use R-134a as the working fluid. The condenser operates 1.4 MPa and the evaporator at 0.140 MPa. (a) Show the cycle on a T-s diagram, indicating the enthalpies at the end of each process, with respect to saturation lines. For a mass flow rate of 0.100 kg/s, determine (b) the net power input to the compressor, in kW, (c) the heat rejection to the environment, in kW, (d) the heat rejection from the cold refrigerated space to the evaporator, in kW, and (e) the coefficient of performance.

 

use the formula in the picture

Formula and Property Tables: QL = m(h1 – h4) Win = m(h2 – h) QH = m(h2 – h3) h1 - h4 COP = h2 – h СОР in Saturated R134a - Pressure Table P, kPa Tat °C Specific Enthalpy Specific Entropy he, kJ/kg h kJ/kg h kJ/kg Sp. kJ/kg K Stg. kJ/kg.K $. kJ/kg K 120 - 22.32 22.49 214.48 236.97 0.09275 0.85503 0.94779 140 - 18.77 27.08 212.08 239.16 0.11087 0.83368 0.94456 160 - 15.60 31.21 209.90 241.11 0.12693 081496 0.94190 1200 46.29 117.77 156.10 273.87 0.42441 0.48863 0.91303 1400 52.40 127.77 148.90 276.12 0.45315 0.45734 0.91050 1600 57.88 135.93 141.93 277.86 0.47911 0.42873 0.90784 Superheated R134a P= 0.140 MPa, (Tat=-18.77°C) Superheated RI34a P= 1.40 MPa, (Tat = 57.88°C) T, u h T, u h °C n/kg kJ/kg kJ/kg kJ/kg.K °C m/kg kJ/kg kJ/kg kJ/kg.K Sat 0.14014 215.54 239.16 0.9446 Sat 0.01212 258.47 277.86 0.9078 - 20 60 0.01237 260.89 280.69 0.9163 - 10 0.14605 225.91 246.36 0.9724 70 0.01343 271.76 293.25 0.9535 0. 0.15263 233.23 254.60 1.0031 80 0.01436 282.09 305.07 0.9875
Transcribed Image Text:Formula and Property Tables: QL = m(h1 – h4) Win = m(h2 – h) QH = m(h2 – h3) h1 - h4 COP = h2 – h СОР in Saturated R134a - Pressure Table P, kPa Tat °C Specific Enthalpy Specific Entropy he, kJ/kg h kJ/kg h kJ/kg Sp. kJ/kg K Stg. kJ/kg.K $. kJ/kg K 120 - 22.32 22.49 214.48 236.97 0.09275 0.85503 0.94779 140 - 18.77 27.08 212.08 239.16 0.11087 0.83368 0.94456 160 - 15.60 31.21 209.90 241.11 0.12693 081496 0.94190 1200 46.29 117.77 156.10 273.87 0.42441 0.48863 0.91303 1400 52.40 127.77 148.90 276.12 0.45315 0.45734 0.91050 1600 57.88 135.93 141.93 277.86 0.47911 0.42873 0.90784 Superheated R134a P= 0.140 MPa, (Tat=-18.77°C) Superheated RI34a P= 1.40 MPa, (Tat = 57.88°C) T, u h T, u h °C n/kg kJ/kg kJ/kg kJ/kg.K °C m/kg kJ/kg kJ/kg kJ/kg.K Sat 0.14014 215.54 239.16 0.9446 Sat 0.01212 258.47 277.86 0.9078 - 20 60 0.01237 260.89 280.69 0.9163 - 10 0.14605 225.91 246.36 0.9724 70 0.01343 271.76 293.25 0.9535 0. 0.15263 233.23 254.60 1.0031 80 0.01436 282.09 305.07 0.9875
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