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 - 0.00528t² - 0.0011tetc - 0.000306te²tc + 0.000567tetc² + 3.1x10²tc² 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 - 0.00528t² - 0.0011tetc - 0.000306te²tc + 0.000567tetc² + 3.1x10²tc² 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.
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter28: Special Refrigeration Applications
Section: Chapter Questions
Problem 15RQ: Why is two-stage compression popular for extra-low-temperature refrigeration systems?
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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 technique 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.
![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
UAC = 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 - 0.00528tc²-
0.0011tetc - 0.000306te²tc + 0.000567tetc² + 3.1x10-6t²tc²
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0e69105-edad-4e40-a986-7d2af8f1b9bd%2F15013a3b-a81c-43e4-b31e-1fc183313753%2Faybmw4j_processed.jpeg&w=3840&q=75)
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
UAC = 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 - 0.00528tc²-
0.0011tetc - 0.000306te²tc + 0.000567tetc² + 3.1x10-6t²tc²
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.
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