2 adiabatic expansion y-1 (1) ..... ..... 3 heating with fixing volume T3 P3 (2) T2 P2 3. T1_P3 . (T3=T1) T2 P2 By substituting P3 %3D P2 (3) P3 In P2 Y-1, P1 P2 %3D y(InP3 – InP2) = (y – 1)(InP1 - InP2) (InP3 – InP2) Y-1_, (InP1-InP2) -1- 1 The table needs to be solved using [(InP3-InP2)] 1- [(InP1-InP2) the written laws 1 (InP1-InP2)-(InP3-InP2) (InP1-InP2)-(InP3+lnP2) %3D InP1-InP2 InP1-InR7 117

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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adiabatic expansion
Y-1
(1)
3 heating with fixing volume
T3 P3
(2)
T2 P2
3
T1 P3
(Т3-TI)
T2 P2
By substituting
(3)
P3 Y-1
In-
P1
In
P2
Y(InP3 – InP2) = (y - 1)(InP1 - InP2)
(InP3 – InP2)
y - 1
1
= 1 -
(InP1 – InP2)
The table needs to be solved using
[(InP3-InP2)
= 1-
1
(InP1- InP2)
the written laws
(InP1-InP2)-(InP3-InP2)
(InP1-InP2)-(InP3+InP2)
InP1-InP2
InP1-InP2
No R
1
(InP1-InP3)
12.3
-3=
2.4
InP1 - InP2
22.4
1.6
1.3
1.
1.6
(InP1-InP2)
11
InP1-InP3
Transcribed Image Text:adiabatic expansion Y-1 (1) 3 heating with fixing volume T3 P3 (2) T2 P2 3 T1 P3 (Т3-TI) T2 P2 By substituting (3) P3 Y-1 In- P1 In P2 Y(InP3 – InP2) = (y - 1)(InP1 - InP2) (InP3 – InP2) y - 1 1 = 1 - (InP1 – InP2) The table needs to be solved using [(InP3-InP2) = 1- 1 (InP1- InP2) the written laws (InP1-InP2)-(InP3-InP2) (InP1-InP2)-(InP3+InP2) InP1-InP2 InP1-InP2 No R 1 (InP1-InP3) 12.3 -3= 2.4 InP1 - InP2 22.4 1.6 1.3 1. 1.6 (InP1-InP2) 11 InP1-InP3
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