06) An air standard Otto cyce has a compression ratio of 7.5. The maximum and minimum cycle temperatures are 1600" and 300 K, and the minimum pressure is 100 kPa. Determine (a) the cycle efficiency (b) the change of entropy during heat addition; (c) the charge of availability per unit mess during the expension process
06) An air standard Otto cyce has a compression ratio of 7.5. The maximum and minimum cycle temperatures are 1600" and 300 K, and the minimum pressure is 100 kPa. Determine (a) the cycle efficiency (b) the change of entropy during heat addition; (c) the charge of availability per unit mess during the expension process
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
Related questions
Question
Given this solution, how does the 0.001404 was calculated? (The value of V)
![Show the complete cancellation of units and finalamaiera:
Trage
06) An air standard Otto cyde has a compression ratio of 7.5.
The maximum and minimum cycle temperatures are 1600" and
300°K, and the minimum pressure is 100 kPa. Determine (e)
hange availability
Step 1
the cycle efficiency (b) the change of entropy during heet
addition; (c) the change of availability per unit mass during the
expension process
Expert Solution
Stap 2
ainakamalanai Dilm uyulai hau a compression resia afT.2. The maximum and minimum ayala tampanašurau an
m pressure is 400/liMa. Deiarmina (a) the uyala añalera, jel ška ahanga af arknapy aluring haai adaikan; (althe
Given:
(compression) = 7.5
ratio
Solution:
tz-tmex=1600°k, tmin = 300° K = t₁
Pminh loa khai l
=
P.V diagram.
T-S diagram,
54
:: 1-2 (isentropic Compression process)
(2) -
=
t₂ =
B2 = Pintong
→ 2 =
·0.4+1
7
300 E 300
{ = 1/₂ = }
(*)** (₁) **
672-65k)
SHARE
100 (7-5)4+1
P2 = 1679.13 kPq](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3df9f2c-7786-4cb0-b8de-6b22c57cabf1%2F07702535-93b1-46c9-aa9d-6ebf599766a3%2Fdg2opmg_processed.png&w=3840&q=75)
Transcribed Image Text:Show the complete cancellation of units and finalamaiera:
Trage
06) An air standard Otto cyde has a compression ratio of 7.5.
The maximum and minimum cycle temperatures are 1600" and
300°K, and the minimum pressure is 100 kPa. Determine (e)
hange availability
Step 1
the cycle efficiency (b) the change of entropy during heet
addition; (c) the change of availability per unit mass during the
expension process
Expert Solution
Stap 2
ainakamalanai Dilm uyulai hau a compression resia afT.2. The maximum and minimum ayala tampanašurau an
m pressure is 400/liMa. Deiarmina (a) the uyala añalera, jel ška ahanga af arknapy aluring haai adaikan; (althe
Given:
(compression) = 7.5
ratio
Solution:
tz-tmex=1600°k, tmin = 300° K = t₁
Pminh loa khai l
=
P.V diagram.
T-S diagram,
54
:: 1-2 (isentropic Compression process)
(2) -
=
t₂ =
B2 = Pintong
→ 2 =
·0.4+1
7
300 E 300
{ = 1/₂ = }
(*)** (₁) **
672-65k)
SHARE
100 (7-5)4+1
P2 = 1679.13 kPq
![2-3 Process (constant volume)
4-1
>> P₂ = P₂.1²
+2
Stap 4
bi
C;
3-4 (isentropic expension)
4-/
>>>
13=3994.06 KPQ
1600 (75)
|tu = 714.65K
Py = (204) * = (2²) "
Py = (3994.06 ) ( 23 )
· ។
Py= 237.86 MPa
:: n
=
9: 9in (v (tz-t₂): 0-718 (-672-65)
1600
665.84 K= /kg
=
2 in
Lout = (y (fy- (₁)
ty
·1600
· 1679-13 x [1622-65]
W
Wmax =
=
Wnet 9gn-yout= 665-84-292-72
368-12 к3 /ку
DQ
Whet
Lin
0.4
=
(714-65-300) (0-718)
gout = 297.72 KJ /kg
368-12
665.84
Y-1
10.553 55.3%
As = (vin (2₂) +Rin (1₂)
= 0-718 in ( 16059265) + 0-287 in (17
In
45 = 0.871kJ/kg
(v (ty-t₁) + To As
=0.718 (714-65-300) + 293 (0-871)
Wmax = 552-9 553 kJ/kg
9₂ (availability) = (P4-1 min)
3494.06
1679.13
$
5330-193
= 552.807 kJ/kg
= 0·00/404 (237.86-100)
92 = 0.193 kJ/kg
Dr](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3df9f2c-7786-4cb0-b8de-6b22c57cabf1%2F07702535-93b1-46c9-aa9d-6ebf599766a3%2Ftmdrvux_processed.png&w=3840&q=75)
Transcribed Image Text:2-3 Process (constant volume)
4-1
>> P₂ = P₂.1²
+2
Stap 4
bi
C;
3-4 (isentropic expension)
4-/
>>>
13=3994.06 KPQ
1600 (75)
|tu = 714.65K
Py = (204) * = (2²) "
Py = (3994.06 ) ( 23 )
· ។
Py= 237.86 MPa
:: n
=
9: 9in (v (tz-t₂): 0-718 (-672-65)
1600
665.84 K= /kg
=
2 in
Lout = (y (fy- (₁)
ty
·1600
· 1679-13 x [1622-65]
W
Wmax =
=
Wnet 9gn-yout= 665-84-292-72
368-12 к3 /ку
DQ
Whet
Lin
0.4
=
(714-65-300) (0-718)
gout = 297.72 KJ /kg
368-12
665.84
Y-1
10.553 55.3%
As = (vin (2₂) +Rin (1₂)
= 0-718 in ( 16059265) + 0-287 in (17
In
45 = 0.871kJ/kg
(v (ty-t₁) + To As
=0.718 (714-65-300) + 293 (0-871)
Wmax = 552-9 553 kJ/kg
9₂ (availability) = (P4-1 min)
3494.06
1679.13
$
5330-193
= 552.807 kJ/kg
= 0·00/404 (237.86-100)
92 = 0.193 kJ/kg
Dr
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