A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor ent
ers the compressor at 2 bar, and saturated liquid exits the condenser at 8 bar. The isentropic compressor efficiency is 80%. The mass flow rate of refrigerant is 7 kg/min.
1-Write the assumptions no less than three
2-Draw clear Schematic
(3) the compressor power, in kW.
(4) the refrigeration capacity, in tons.
(5) the coefficient of performance.
1-Write the assumptions no less than three
2-Draw clear Schematic
(3) the compressor power, in kW.
(4) the refrigeration capacity, in tons.
(5) the coefficient of performance.
1 answer:
You might be interested in
Answer:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
Explanation:
Engineering Stress is defined as Load applied to the original cross-sectional area which we have taken in the start.
True stress is defined as the load divided by area of cross-section of specimen at that instant.
Engineering stress and true stress can be expressed by relation
Where
The tension in the cable and the <em>net</em> acceleration of the block B are 291.814 newtons and 0.112 meters per square second, respectively.
<h3>How to determine the tension and the acceleration of block B</h3>
Based of the representation seen in the image, an <em>external</em> force moves the block A <em>downwards</em> and the block B moves <em>upwards</em> and with an net acceleration of same magnitude due to the cable. By Newton's laws we construct the free body diagrams for each mass and that can be seen in the image seen below.
The <em>motion</em> equations are described below:
Mass A
250 N - 0.4 · Na - T + (30 kg) · (9.807 m / s²) · sin 30° = (30 kg) · a (1)
Na - (30 kg) · (9.807 m / s²) · cos 30° = 0 (2)
Mass B
- T + 0.4 · Na + 0.3 · Nb + (15 kg) · (9.807 m / s²) · sin 30° = - (15 kg) · a (3)
Nb - Na - (15 kg) · (9.807 m / s²) · cos 30° = 0 (4)
By (2):
Na = (30 kg) · (9.807 m / s²) · cos 30°
Na = 254.793 N
By (4):
Nb = 254.793 N + (15 kg) · (9.807 m / s²) · cos 30°
Nb = 382.190 N
Then, we have the following system of <em>linear</em> equations:
T + (30 kg) · a = 250 N - 0.4 · Na + (30 kg) · (9.807 m / s²) · sin 30°
T + 30 · a = 295.188
T - (15 kg) · a = 0.4 · Na + 0.3 · Nb + (15 kg) · (9.807 m / s²) · sin 30°
T - 15 · a = 290.127
The solution of the <em>linear</em> system is T = 291.814 N and a = 0.112 m / s².
To learn more on Newton's laws: brainly.com/question/27573481
#SPJ1
