Mechanical Engineering Exam > Mechanical Engineering Questions > A simple saturated refrigeration cycle has th...
Start Learning for Free
A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:
- a)5
- b)3.5
- c)6
- d)Not possible to find with this data
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A simple saturated refrigeration cycle has the following state points....
Most Upvoted Answer
A simple saturated refrigeration cycle has the following state points....
Given Data:
Enthalpy after compression = 425 kJ/kg
Enthalpy after throttling = 125 kJ/kg
Enthalpy before compression = 375 kJ/kg
Solution:
To find the coefficient of performance (COP) of the refrigeration cycle, we need to determine the heat absorbed (Qin) and the work done (W).
Step 1: Determine the heat absorbed (Qin):
The heat absorbed by the refrigerant during the refrigeration cycle can be calculated using the enthalpy values.
Qin = Enthalpy after throttling - Enthalpy before compression
= 125 kJ/kg - 375 kJ/kg
= -250 kJ/kg
Step 2: Determine the work done (W):
The work done in the refrigeration cycle can be calculated using the enthalpy values.
W = Enthalpy after compression - Enthalpy before compression
= 425 kJ/kg - 375 kJ/kg
= 50 kJ/kg
Step 3: Calculate the COP:
The coefficient of performance (COP) of the refrigeration cycle is given by the ratio of the heat absorbed to the work done.
COP = Qin / W
= -250 kJ/kg / 50 kJ/kg
= -5
Since the COP should be a positive value, we take the absolute value of the calculated COP.
COP = | -5 |
= 5
Therefore, the correct answer is option 'A' - 5.
Enthalpy after compression = 425 kJ/kg
Enthalpy after throttling = 125 kJ/kg
Enthalpy before compression = 375 kJ/kg
Solution:
To find the coefficient of performance (COP) of the refrigeration cycle, we need to determine the heat absorbed (Qin) and the work done (W).
Step 1: Determine the heat absorbed (Qin):
The heat absorbed by the refrigerant during the refrigeration cycle can be calculated using the enthalpy values.
Qin = Enthalpy after throttling - Enthalpy before compression
= 125 kJ/kg - 375 kJ/kg
= -250 kJ/kg
Step 2: Determine the work done (W):
The work done in the refrigeration cycle can be calculated using the enthalpy values.
W = Enthalpy after compression - Enthalpy before compression
= 425 kJ/kg - 375 kJ/kg
= 50 kJ/kg
Step 3: Calculate the COP:
The coefficient of performance (COP) of the refrigeration cycle is given by the ratio of the heat absorbed to the work done.
COP = Qin / W
= -250 kJ/kg / 50 kJ/kg
= -5
Since the COP should be a positive value, we take the absolute value of the calculated COP.
COP = | -5 |
= 5
Therefore, the correct answer is option 'A' - 5.
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
|
Explore Courses for Mechanical Engineering exam
|
|
Similar Mechanical Engineering Doubts
Top Courses for Mechanical EngineeringView all
A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer?
Question Description
A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer?.
A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A simple saturated refrigeration cycle has the following state points. Enthalpy after compression = 425 kJ/kg; enthaly after throttling = 125 kJ/kg; enthalpy before compression = 375 kJ/kg. The COP of refrigeration is:a)5b)3.5c)6d)Not possible to find with this dataCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup