Homework Help is Here – Start Your Trial Now!
Engineering
Mechanical Engineering
air-conditioning system is shown in the figure below in which air flows over tubes carrying Refrigerant 134a. Air enters with a volumetric flow rate of (AV)1 = 130 m3/min at 32°C, 1 bar, and exits at 22°C, 0.95 bar. Refrigerant enters the tubes at 5 bar with a quality of 20% and exits at 5 bar, 20°C.   Ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at stead

air-conditioning system is shown in the figure below in which air flows over tubes carrying Refrigerant 134a. Air enters with a volumetric flow rate of (AV)1 = 130 m3/min at 32°C, 1 bar, and exits at 22°C, 0.95 bar. Refrigerant enters the tubes at 5 bar with a quality of 20% and exits at 5 bar, 20°C.   Ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at stead

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
See similar textbooks
icon
Related questions
Question
An air-conditioning system is shown in the figure below in which air flows over tubes carrying Refrigerant 134a. Air enters with a volumetric flow rate of (AV)1 = 130 m3/min at 32°C, 1 bar, and exits at 22°C, 0.95 bar. Refrigerant enters the tubes at 5 bar with a quality of 20% and exits at 5 bar, 20°C.
 
Ignoring heat transfer at the outer surface of the air conditioner, and neglecting kinetic and potential energy effects, determine at steady state:

(a) the mass flow rate of the refrigerant, in kg/min.

(b) the rate of heat transfer, in kJ/min, to the air from the refrigerant.
Air P1 = 1 bar T = 32°C = 305 K (AV), 3 Refrigerant 134a R-134a P3 = 5 bar Xz = 0.20 4 + R-134a P4 = 5 bar T = 20°C Air 2+P2 = 0.95 bar T = 22°C = 295 K
Transcribed Image Text:Air P1 = 1 bar T = 32°C = 305 K (AV), 3 Refrigerant 134a R-134a P3 = 5 bar Xz = 0.20 4 + R-134a P4 = 5 bar T = 20°C Air 2+P2 = 0.95 bar T = 22°C = 295 K
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Refrigeration and Air Conditioning
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY
SEE MORE TEXTBOOKS