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A refrigeration plant that operates on the cycle shown below serves as a water chiller. Data on individual components are as follows: Evaporator: UA = 30.6 kW/K Condenser: Water flow rate = me = 6.8 kg/s UA = 26.5 kW/K Water flow rate = mc = 7.6 kg/s The refrigeration capacity of the compressor is a function of the evaporating and condensing temperatures of the refrigerant, te and tc, respectively. That capacity is given as: qe [kW]= 239.5 +10.073te - 0.109te²3.41tc-0.00250tc²- 0.2030tetc + 0.00820te²tc + 0.0013tetc²-8.0005x10-5 te²tc² The power consumed by the compressor is also a function of te and tc and is given as: P [kW] = -2.634 -0.3081te -0.00301te²+ 1.066tc - 0.00528t.² - 0.0011tetc - 0.000306t²tc +0.000567tete²+ 3.1x10t²tc² Appreciate that, due to the 1st Law of Thermo, the condenser must reject the energy added in both the evaporator and the compressor. Also, you can assume constant properties for the water (i.e., no temperature dependence). a) Set up the equations that would be used for a Newton-Raphson simulation. Given the initial guesses provided below, use the N-R method to obtain the guesses for iteration #2. Provide the matrix used to solve each iteration along with the guesses for iteration #2. For this part, the inlet water temperature to the evaporator, ta, is 10 °C, and the inlet water temperature to the condenser, tb, is 25 °C.

A refrigeration plant that operates on the cycle shown below serves as a water chiller. Data on individual components are as follows: Evaporator: UA = 30.6 kW/K Condenser: Water flow rate = me = 6.8 kg/s UA = 26.5 kW/K Water flow rate = mc = 7.6 kg/s The refrigeration capacity of the compressor is a function of the evaporating and condensing temperatures of the refrigerant, te and tc, respectively. That capacity is given as: qe [kW]= 239.5 +10.073te - 0.109te²3.41tc-0.00250tc²- 0.2030tetc + 0.00820te²tc + 0.0013tetc²-8.0005x10-5 te²tc² The power consumed by the compressor is also a function of te and tc and is given as: P [kW] = -2.634 -0.3081te -0.00301te²+ 1.066tc - 0.00528t.² - 0.0011tetc - 0.000306t²tc +0.000567tete²+ 3.1x10t²tc² Appreciate that, due to the 1st Law of Thermo, the condenser must reject the energy added in both the evaporator and the compressor. Also, you can assume constant properties for the water (i.e., no temperature dependence). a) Set up the equations that would be used for a Newton-Raphson simulation. Given the initial guesses provided below, use the N-R method to obtain the guesses for iteration #2. Provide the matrix used to solve each iteration along with the guesses for iteration #2. For this part, the inlet water temperature to the evaporator, ta, is 10 °C, and the inlet water temperature to the condenser, tb, is 25 °C.

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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Part c & d

A refrigeration plant that operates on the cycle shown below serves as a water chiller. Data on individual components are as follows: Evaporator: UA = 30.6 kW/K Condenser: Water flow rate = me = 6.8 kg/s UA = 26.5 kW/K Water flow rate = mc = 7.6 kg/s The refrigeration capacity of the compressor is a function of the evaporating and condensing temperatures of the refrigerant, te and tc, respectively. That capacity is given as: - qe [kW]= 239.5 + 10.073tė – 0.109te² – 3.41tc - 0.00250tc²-0.2030tetc + 0.00820t²tc + 0.0013tetc²-8.0005x10-5 te²tc² The power consumed by the compressor is also a function of te and tc and is given as: P [kW] = -2.634 - 0.3081te -0.00301te²+ 1.066tc - 0.00528tc²- 0.0011tetc - 0.000306te²tc + 0.000567tetc² + 3.1x10t²tc² Appreciate that, due to the 1st Law of Thermo, the condenser must reject the energy added in both the evaporator and the compressor. Also, you can assume constant properties for the water (i.e., no temperature dependence). a) Set up the equations that would be used for a Newton-Raphson simulation. Given the initial guesses provided below, use the N-R method to obtain the guesses for iteration #2. Provide the matrix used to solve each iteration along with the guesses for iteration #2. For this part, the inlet water temperature to the evaporator, ta, is 10 °C, and the inlet water temperature to the condenser, tb, is 25 °C.
Transcribed Image Text:A refrigeration plant that operates on the cycle shown below serves as a water chiller. Data on individual components are as follows: Evaporator: UA = 30.6 kW/K Condenser: Water flow rate = me = 6.8 kg/s UA = 26.5 kW/K Water flow rate = mc = 7.6 kg/s The refrigeration capacity of the compressor is a function of the evaporating and condensing temperatures of the refrigerant, te and tc, respectively. That capacity is given as: - qe [kW]= 239.5 + 10.073tė – 0.109te² – 3.41tc - 0.00250tc²-0.2030tetc + 0.00820t²tc + 0.0013tetc²-8.0005x10-5 te²tc² The power consumed by the compressor is also a function of te and tc and is given as: P [kW] = -2.634 - 0.3081te -0.00301te²+ 1.066tc - 0.00528tc²- 0.0011tetc - 0.000306te²tc + 0.000567tetc² + 3.1x10t²tc² Appreciate that, due to the 1st Law of Thermo, the condenser must reject the energy added in both the evaporator and the compressor. Also, you can assume constant properties for the water (i.e., no temperature dependence). a) Set up the equations that would be used for a Newton-Raphson simulation. Given the initial guesses provided below, use the N-R method to obtain the guesses for iteration #2. Provide the matrix used to solve each iteration along with the guesses for iteration #2. For this part, the inlet water temperature to the evaporator, ta, is 10 °C, and the inlet water temperature to the condenser, tb, is 25 °C.
Use these initial guesses: te (temperature of refrigerant in evaporator) tc (temperature of refrigerant in condenser) qe (heat transfer in evaporator) qc (heat transfer in condenser) P (power used by compressor) tbe (temperature of water exiting condenser) tae (temperature of water exiting evaporator, i.e., the temperature of the chilled water supplied to the end use) 1 °C 40 °C 100 kW 150 kW 50 kW 35 °C 2 °C b) Solve the problem to completion. (Feel free to use Matlab or other software to solve the system of non-linear equations, or modify your Newton-Raphson technic developed in part (a) to converge to a solution.) Provide the Coefficient of Performance (COP) = qe/P of the system. Please turn in any code that you use and your solutions. C) Use your simulation to plot the COP as a function of condenser inlet water temperature for evaporator inlet temperatures of 10 °C, 15 °C, and 20 °C. Examine a range of condenser inlet water temperatures from 20 °C to 40 °C. d) Discuss the results obtained in part C
Transcribed Image Text:Use these initial guesses: te (temperature of refrigerant in evaporator) tc (temperature of refrigerant in condenser) qe (heat transfer in evaporator) qc (heat transfer in condenser) P (power used by compressor) tbe (temperature of water exiting condenser) tae (temperature of water exiting evaporator, i.e., the temperature of the chilled water supplied to the end use) 1 °C 40 °C 100 kW 150 kW 50 kW 35 °C 2 °C b) Solve the problem to completion. (Feel free to use Matlab or other software to solve the system of non-linear equations, or modify your Newton-Raphson technic developed in part (a) to converge to a solution.) Provide the Coefficient of Performance (COP) = qe/P of the system. Please turn in any code that you use and your solutions. C) Use your simulation to plot the COP as a function of condenser inlet water temperature for evaporator inlet temperatures of 10 °C, 15 °C, and 20 °C. Examine a range of condenser inlet water temperatures from 20 °C to 40 °C. d) Discuss the results obtained in part C
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