Please state assumptions and tables. ### Refrigeration Cycle Problem#### Problem Statement:In a refrigerator, Refrigerant-134a (R-134a) enters the evaporator (the cold heat exchanger) at \(-17^\circ C\), with a quality \(x = 25\%\) and leaves at \(-17^\circ C\), with a quality \(x = 1\). The Coefficient of Performance (COP) of the refrigerator is 2.3 and the mass flow rate is 0.006 kg/s. Find the net work input \(\dot{W}\) to the cycle and the total heat rejection rate \(\dot{Q}_H\) to the surroundings. (You may need to do interpolation between the temperatures \(-16^\circ C\) and \(-18^\circ C\)).#### Explanation:- **State 1 (Entering Evaporator):** - Temperature (\(T_1\)): \(-17^\circ C\) - Quality (\(x_1\)): 0.25 (25%) - **State 2 (Leaving Evaporator):** - Temperature (\(T_2\)): \(-17^\circ C\) - Quality (\(x_2\)): 1 (100%)#### Given Data:- Coefficient of Performance (COP): 2.3- Mass Flow Rate (\(\dot{m}\)): 0.006 kg/s#### Objectives:1. Net Work Input (\(\dot{W}\))2. Total Heat Rejection Rate (\(\dot{Q}_H\))#### Methodology:To solve this problem, you would typically:1. Use property tables or software to find the enthalpy values for R-134a at the given temperatures and qualities.2. Apply the definition of COP for the refrigerator (\(COP_{refrigerator} = \frac{\dot{Q}_L}{\dot{W}}\)).3. Determine the work input and heat transfer rates using energy balance equations.(Here, interpolation might be required to find precise values for the enthalpies at \(-17^\circ C\)).Note: As detailed numerical interpolation and property values are beyond the scope of this transcription, you should consult R-134a refrigerant tables for further calculation steps.### Detailed Explanations and DiagramsFor comprehensive understanding and step-by-step calculations, students are encouraged to refer to their

Given data, Assumption:- The operation of the refrigeration cycle is in a steady-state condition.…

In a refrigerator, R-134a enters the evaporator (the cold heat exchanger) at –17°c , x= 25% and leaves at –170c, x = 1. The COP of the refrigerator is 2.3 and the mass flow rate is 0.006 kg/s. Find the net work input W to the cycle and the total heat rejection rate Q# to the surroundings? (You may need to do interpolation between the temperatures –16°C and – 18°C) |

In a refrigerator, R-134a enters the evaporator (the cold heat exchanger) at –17°c , x= 25% and leaves at –170c, x = 1. The COP of the refrigerator is 2.3 and the mass flow rate is 0.006 kg/s. Find the net work input W to the cycle and the total heat rejection rate Q# to the surroundings? (You may need to do interpolation between the temperatures –16°C and – 18°C) |

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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Question

Please state assumptions and tables.

### Refrigeration Cycle Problem

#### Problem Statement:
In a refrigerator, Refrigerant-134a (R-134a) enters the evaporator (the cold heat exchanger) at \(-17^\circ C\), with a quality \(x = 25\%\) and leaves at \(-17^\circ C\), with a quality \(x = 1\). The Coefficient of Performance (COP) of the refrigerator is 2.3 and the mass flow rate is 0.006 kg/s. Find the net work input \(\dot{W}\) to the cycle and the total heat rejection rate \(\dot{Q}_H\) to the surroundings. (You may need to do interpolation between the temperatures \(-16^\circ C\) and \(-18^\circ C\)).

#### Explanation:
- **State 1 (Entering Evaporator):**
- Temperature (\(T_1\)): \(-17^\circ C\)
- Quality (\(x_1\)): 0.25 (25%)

- **State 2 (Leaving Evaporator):**
- Temperature (\(T_2\)): \(-17^\circ C\)
- Quality (\(x_2\)): 1 (100%)

#### Given Data:
- Coefficient of Performance (COP): 2.3
- Mass Flow Rate (\(\dot{m}\)): 0.006 kg/s

#### Objectives:
1. Net Work Input (\(\dot{W}\))
2. Total Heat Rejection Rate (\(\dot{Q}_H\))

#### Methodology:
To solve this problem, you would typically:
1. Use property tables or software to find the enthalpy values for R-134a at the given temperatures and qualities.
2. Apply the definition of COP for the refrigerator (\(COP_{refrigerator} = \frac{\dot{Q}_L}{\dot{W}}\)).
3. Determine the work input and heat transfer rates using energy balance equations.

(Here, interpolation might be required to find precise values for the enthalpies at \(-17^\circ C\)).

Note: As detailed numerical interpolation and property values are beyond the scope of this transcription, you should consult R-134a refrigerant tables for further calculation steps.

### Detailed Explanations and Diagrams
For comprehensive understanding and step-by-step calculations, students are encouraged to refer to their

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