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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

### Refrigeration Plant as a Water Chiller A refrigeration plant operating on a specified cycle functions as a water chiller. Below is the data on individual components: - **Evaporator:** - \( UA_e = 30.6 \, \text{kW/K} \) - Water flow rate \( m_e = 6.8 \, \text{kg/s} \) - **Condenser:** - \( UA_c = 26.5 \, \text{kW/K} \) - Water flow rate \( m_c = 7.6 \, \text{kg/s} \) #### Refrigeration Capacity of the Compressor The capacity is determined by the evaporating (\( t_e \)) and condensing (\( t_c \)) temperatures of the refrigerant. The capacity is expressed by the equation: \[ q_e \, [\text{kW}] = 239.5 + 10.073t_e - 0.109t_e^2 - 3.41t_c - 0.00250t_c^2 - 0.2030t_e t_c + 0.00820t_c^2 + 0.0013t_e t_c^2 - 8.0005 \times 10^{-5} t_e^2 t_c^2 \] #### Power Consumption of the Compressor The power consumption also depends on \( t_e \) and \( t_c \), represented as: \[ P \, [\text{kW}] = -2.634 - 0.3081t_e - 0.00301t_e^2 + 1.066t_c - 0.00528t_c^2 - 0.0011t_e t_c - 0.000306t_e t_c^2 + 0.000567t_e^2 t_c + 3.1 \times 10^{-6} t_e^2 t_c^2 \] #### Energy Balance According to the 1st Law of Thermodynamics, the condenser must reject the energy added in both the evaporator and the compressor. It is noted that constant water properties can be assumed (i.e., no temperature dependence). --- ### Task 1. **Newton-Raphson Simulation Setup:** - Set up equations for a Newton-Raphson simulation. - Use initial
Transcribed Image Text:### Refrigeration Plant as a Water Chiller A refrigeration plant operating on a specified cycle functions as a water chiller. Below is the data on individual components: - **Evaporator:** - \( UA_e = 30.6 \, \text{kW/K} \) - Water flow rate \( m_e = 6.8 \, \text{kg/s} \) - **Condenser:** - \( UA_c = 26.5 \, \text{kW/K} \) - Water flow rate \( m_c = 7.6 \, \text{kg/s} \) #### Refrigeration Capacity of the Compressor The capacity is determined by the evaporating (\( t_e \)) and condensing (\( t_c \)) temperatures of the refrigerant. The capacity is expressed by the equation: \[ q_e \, [\text{kW}] = 239.5 + 10.073t_e - 0.109t_e^2 - 3.41t_c - 0.00250t_c^2 - 0.2030t_e t_c + 0.00820t_c^2 + 0.0013t_e t_c^2 - 8.0005 \times 10^{-5} t_e^2 t_c^2 \] #### Power Consumption of the Compressor The power consumption also depends on \( t_e \) and \( t_c \), represented as: \[ P \, [\text{kW}] = -2.634 - 0.3081t_e - 0.00301t_e^2 + 1.066t_c - 0.00528t_c^2 - 0.0011t_e t_c - 0.000306t_e t_c^2 + 0.000567t_e^2 t_c + 3.1 \times 10^{-6} t_e^2 t_c^2 \] #### Energy Balance According to the 1st Law of Thermodynamics, the condenser must reject the energy added in both the evaporator and the compressor. It is noted that constant water properties can be assumed (i.e., no temperature dependence). --- ### Task 1. **Newton-Raphson Simulation Setup:** - Set up equations for a Newton-Raphson simulation. - Use initial
### Initial Guesses: - **te** (temperature of refrigerant in evaporator): 1 °C - **tc** (temperature of refrigerant in condenser): 40 °C - **qe** (heat transfer in evaporator): 100 kW - **qc** (heat transfer in condenser): 150 kW - **P** (power used by compressor): 50 kW - **tbe** (temperature of water exiting condenser): 35 °C - **tae** (temperature of water exiting evaporator, i.e., the temperature of the chilled water supplied to the end use): 2 °C ### Tasks: #### b) Solve the Problem: Complete the problem using Matlab or other software to solve the system of non-linear equations. You may modify your Newton-Raphson technique to converge to a solution. Calculate the Coefficient of Performance (COP) = qe/P of the system. Submit any code and solutions used. #### c) Simulation and COP Plot: 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) Discussion: Discuss the results obtained in part c.
Transcribed Image Text:### Initial Guesses: - **te** (temperature of refrigerant in evaporator): 1 °C - **tc** (temperature of refrigerant in condenser): 40 °C - **qe** (heat transfer in evaporator): 100 kW - **qc** (heat transfer in condenser): 150 kW - **P** (power used by compressor): 50 kW - **tbe** (temperature of water exiting condenser): 35 °C - **tae** (temperature of water exiting evaporator, i.e., the temperature of the chilled water supplied to the end use): 2 °C ### Tasks: #### b) Solve the Problem: Complete the problem using Matlab or other software to solve the system of non-linear equations. You may modify your Newton-Raphson technique to converge to a solution. Calculate the Coefficient of Performance (COP) = qe/P of the system. Submit any code and solutions used. #### c) Simulation and COP Plot: 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) Discussion: Discuss the results obtained in part c.
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