The photo below is the refrigeration cycle of Ammonia, Im just confused on where h2 and Tsup=105 C came from? what kind of table usedand how it is used? So the enthalpy at state 1T1 = -16°CPL = 227.04 kPahi = h, = 1442. 60 kJ/kgFor state 2,S2 = S1At superheated stateP = Psat @T = 32°CРн 3D 1239.6 КРaTsup = 105°Ch2 = 1690 kJ/kgstate 3saturated liquid at T = 32°Chz = 351.47 kJ/kgstate 4hz = h4 = 351.47h-h4СОР —h2-h1442.6–351.471690–1442.6СОР — 4.4

To Show the concept how cardinal points are calculated in ammonia refrigeration cycle.

Answered: e photo below is the refrigeration…

e photo below is the refrigeration cycle of Ammonia, Im just confused on where h2 and Tsup=105 C came from? what kind of table usedand how it is used?

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

Work and Heat Transfer

It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. There are various mechanisms by which, transfer of heat takes place i.e. thermal conduction, convection, radiation or by change of phase. All these mechanisms have their own specific features which sometimes occur simultaneously within the same system.

Question

The photo below is the refrigeration cycle of Ammonia, Im just confused on where h2 and Tsup=105 C came from? what kind of table usedand how it is used?

So the enthalpy at state 1
T1 = -16°C
PL = 227.04 kPa
hi = h, = 1442. 60 kJ/kg
For state 2,
S2 = S1
At superheated state
P = Psat @T = 32°C
Рн 3D 1239.6 КРa
Tsup = 105°C
h2 = 1690 kJ/kg
state 3
saturated liquid at T = 32°C
hz = 351.47 kJ/kg
state 4
hz = h4 = 351.47
h-h4
СОР —
h2-h
1442.6–351.47
1690–1442.6
СОР — 4.4

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