Calculate the quality of ammonia at state 2 using the relation,v, = (v,), +x (Va),0.04 = 0.00168 +x, (0.10881)0.35217The pressure remains constant during the process 1-2'.temperatureObtaintheR= P, =1600kPaandatV, = v, = 0.04m / kg from saturated ammonia tables.T = 41°CThe heat transfer takes place between the ammonia and steel is dueto the change in internal energy and change in specific volumeduring the process.Theprocessfrom 1-2' is constant pressure and 2'-2 is the constantvolume process.Draw the process on p-v diagram as shown.PCalculate the work done in the process 1 to 2.W = W2 +Wy-2= P(V, -V,)+0=1600(0.02 – 0.05634)%3DW, =-58.144kJNegative sign indicates work done on the system.

Answered: Calculate the quality of ammonia at…

Calculate the quality of ammonia at state 2 using the relation, v; = (v; ), +x (va), 0.04 = 0.00168 +x, (0.10881) X, = 0.35217

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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Concept explainers

Entropy

In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its molecules are in random motion, vibrating and colliding with each other. During this scenario, the probability of finding a molecule at a particular place twice, is low. Under this circumstance, the system can be said to be in a high entropy.

Question

Calculate the quality of ammonia at state 2 using the relation,
v, = (v,), +x (Va),
0.04 = 0.00168 +x, (0.10881)
0.35217
The pressure remains constant during the process 1-2'.
temperature
Obtain
the
R= P, =1600kPaand
at
V, = v, = 0.04m / kg from saturated ammonia tables.
T = 41°C
The heat transfer takes place between the ammonia and steel is due
to the change in internal energy and change in specific volume
during the process.
The
process
from 1-2' is constant pressure and 2'-2 is the constant
volume process.
Draw the process on p-v diagram as shown.
P
Calculate the work done in the process 1 to 2.
W = W2 +Wy-2
= P(V, -V,)+0
=1600(0.02 – 0.05634)
%3D
W, =-58.144kJ
Negative sign indicates work done on the system.

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