1 - PROJEKTNI
June 13, 2018 | Author: Dušan Marković | Category: N/A
Short Description
Projektni zadatak iz termodinamike sa termotehnikom...
Description
UNIVERZITET U NIŠU FAKULT FAKULTET ET ZAŠTITE ZA ŠTITE NA RADU
PROJEK PROJEKTNI TNI ZADA ZADATAK Predmet: TERMODINAMIKA SA TERMOTEHNIKOM
Mentor: Prof. Miomir Rao
Student: Du!an Mar"o#i$ %% %%%&' %&' U Ni!u( )*%+
Za ,#rto -ori#o ede$e- /ro0entuano- ata#a:
C &&(+1
H
O
N
S
A
W
2(+1
+('
*(&*
3(*&
2%(2*
3()*
ra,unati tra4ene /arametre na#edene u 4adat"u. H d
1) Donj Donja a top topo otn tna a mo! mo!
"#ra$%nata &DI 'orm%om #a (r*to +or"o dato+
*a*taa:
H d =340 ∙ C + 1200 ( H −
(
O 8
)+ 105 ∙ S −25 ∙W
H d =340 ∙ 44,58+ 1200 3,58 −
)+
5,9 8
105 ∙ 7,04 −25 ∙ 7,2=15719,24 [ kJ / kg ]
Lmin
,) M"n"m M"n"man ana a -o" -o"("n ("na a a#d a#d%.a %.a
3
"*tra/ena %
m / kg potre0na #a
*a+oreanje dato+ +or"a N234 5 " O2,1 5):
,616 Teori5"a Teori5"a 4a/remina "ieoni"a:
O2 min=
O2 min=
1 100
[
1 100
[
(
1,865 ∙ C + 5,6 H −
(
1,865 ∙ 44,58 + 5,6 3,58 −
5,90 8
O 8
)+ ]
]
)
3 + 0,7 ∙ 7,04 =1,03987 [ m N O2 / kg ]
,6,6 Teori5"a Teori5"a 4a/remina #a4du6a: #a 4du6a:
L
min=
O2 min
1,03987 1,03987 6
0,21
0,21
=
3
= 4,95179 [ m N O2 /kg ]
,676 St#arna "oi0ina #a4du6a:
7 ¿ 1,0 …1,5 … 2,0 L= λ ∙ Lmin 3
λ =1,0 L =1,0 ∙ Lmin =1,0 ∙ 4,95179 = 4,95179 [ m N L / kg ] 3
λ =1,1 L=1,1 ∙ Lmin =1,1 ∙ 4,95179 =5,447 [m N L / kg ]
)
0,7 ∙ S
3
λ =1,2 L=1,2 ∙ Lmin =1,2 ∙ 4,95179 =5,942 [ m N L / kg ] 3
λ =1,3 L=1,3 ∙ Lmin =1,3 ∙ 4,95179 =6,437 [ m N L / kg ] 3
λ =1, 4 L=1,4 ∙ Lmin =1,4 ∙ 4,95179 = 6,933 [ m N L / kg ] 3
λ =1,5 L=1,5 ∙ Lmin =1,5 ∙ 4,95179 =7,428 [ m N L / kg ] 3
λ =1,6 L =1,6 ∙ Lmin =1,6 ∙ 4,95179 =7,923 [ m N L / kg ] 3
λ =1,7 L=1,7 ∙ Lmin =1,7 ∙ 4,95179 = 8,418 [ m N L / kg ] 3
λ =1,8 L= 1,8 ∙ Lmin =1,8 ∙ 4,95179 =8,913 [ m N L / kg ] 3
λ =1,9 L =1,9 ∙ Lmin =1,9 ∙ 4,95179 =9,408 [ m N L / kg ] 3
λ =2,0 L =2,0 ∙ Lmin =1,8 ∙ 4,95179 = 9,904 [ m N L / kg ]
7) Zaprem"na a/n". " *%". prod%-ata *a+oreanja8 " -on*tr%-$"ja d"ja+rama V x − λ 1,0 … 2,0 #a 9 ¿ *a -ora-om od 1816
7616 Teori5"a 4a/remina /rodu"ata a-ore#an5a:
CO2
a)Troatomn" +a*o"
V RO =V CO + V SO = 2
V RO = 2
2
1 100
0) A#ot
V RO = 2
V RO = 2
2
100 1 100
100
SO2
)
(1,865 ∙ C + 0,7 ∙ S )
3 ( 1,865 ∙ 44,58 + 0,7 ∙ 7,04 )=0,880697 [m N / kg ]
N 2
1
1
"
)
( 0,8 ∙ N + 79 ∙ Lmin)
( 0,8 ∙ 0,4 +79 ∙ 4,95179 )=3,9151141 [ m N 3 / kg ]
$) &oda
2
V RO = 2
V RO = 2
1 100 1 100
( 11,2 ∙ H + 1,24 ∙W )
( 11,2 ∙ 3,58 +1,24 ∙ 7,2 )=0,49024
3 [ m N / kg ]
d) &a#d%.
7 ¿ 1,0 …1,5 … 2,0 V L= ( λ−1 ) ∙ L min
[
λ =1,0 V L =( 1−1 ) ∙ 4,95179 =0 m N / kg 3
]
[
λ =1,1 V L =( 1,1−1 ) ∙ 4,95179 =0,495 m N / kg 3
[
λ =1,2 V L =( 1,2−1 ) ∙ 4,95179 =0,990 m N / kg 3
[
λ =1,3 V L =( 1,3−1 ) ∙ 4,95179 =1,486 m N / kg 3
[ λ =1,5 V =( 1,5−1 ) ∙ 4,95179 =2,476 [ m λ =1,6 V =( 1,6 −1 ) ∙ 4,95179 =2,971 [ m
]
] ]
] / kg ] / kg ]
λ =1,4 V L =( 1,4 −1 ) ∙ 4,95179 =1,981 m N / kg 3
3
L
N
L
N
3
[ λ =1,8 V =( 1,8−1 ) ∙ 4,95179 =3,961 [ m
] / kg ]
λ =1,7 V L =( 1,7 −1 ) ∙ 4,95179 =3,466 m N / kg L
3
3
N
[ λ =2,0 V =( 2,0− 1 ) ∙ 4,95179 =4,952 [ m
] / kg ]
λ =1,9 V L =( 1,9−1 ) ∙ 4,95179 = 4,457 m N / kg L
3
3
N
76,6 U"u/na 4a/remina #a8ni6 /rodu"ata a-ore#an5a V W =V RO + V N + V H + V L 2
2
2
O
&
[
3
C W =V RO +V N + V H = 0,880697 + 3,9151141+ 0,49024 =5,2860511 m N / kg 2
2
2
O
[
V W ( λ=1,0 )=C W + V L ( = )=5,286 m N / kg λ 1,0
3
] [
V W ( λ=1,1 )=C W + V L ( = ) =5,28605 + 0,495=5,781 m N / kg λ 1,1
3
[
V W ( λ=1,2 )=C W + V L ( = ) =5,28605 + 0,990 = 6,276 m N / kg λ 1,2
3
[
V W ( λ=1,3 )=C W + V L ( = )=5,28605 + 1,486 =6,772 m N / kg λ 1,3
3
[
V W ( λ=1,4 )=C W + V L( = )= 5,28605 + 1,981 =7,267 m N / kg λ 1,4
3
[
V W ( λ=1,5 )=C W + V L ( = )=5,28605 + 2,476 =7,762 m N / kg λ 1,5
3
] ] ] ] ]
V W ( λ=1,6 )=C W + V L( = )=5,28605 + 2,971 =8,257 m N / kg 3
λ 1,6
[
V W ( λ=1,7 )=C W + V L( = )=5,28605 + 3,466 =8,752 m N / kg λ 1,7
3
[
V W ( λ=1,8 )=C W + V L ( = )=5,28605 + 3,961 = 9,247 m N / kg λ 1,8
3
[
V W ( λ=1,9 )=C W + V L ( = )=5,28605 + 4,457 =9,743 m N / kg λ 1,9
3
[
] ] ]
V W ( λ= 2,0 )=C W + V L( = )=5,28605 + 4,952=10,238 m N / kg λ 2,0
3
]
7676 U"u/na 4a/remina u#i6 /rodu"ata a-ore#an5a: V S =V W −V H 2
V S ( λ = 1,0 )=V W
O
−V H =5,286 −0,49024 =4,796 [ m N / kg ] 3
λ = (1,0)
2
O
[
V S ( λ =1,1 )=V W =( )− V H =5,781 −0,49024 =5,291 m N / kg λ
1,1
2
O
3
]
3 − V H =6,276 −0,49024 =5,786 [ m N / kg ]
V S ( λ =1,2 )= V W
λ =( 1,2)
2
O
V S ( λ =1,3 )=V W =( ) −V H =6,772 − 0,49024 = 6,281 m N / kg 3
λ
1,3
V S ( λ =1,4 ) = V W
2
O
−V H =7,267−0,49024 =6,777 [ m N / kg ] 3
λ = (1,4 )
2
O
[
V S ( λ =1,5 )=V W =( ) −V H =7,762 − 0,49024 =7,272 m N / kg λ
V S ( λ =1,6 )= V W
1,5
λ = (1,6 )
2
O
3
]
3 −V H = 8,257−0,49024 =7,767 [ m N / kg ] 2
O
+
]
V S ( λ =1,7 )= V W
λ = (1,7
3 − = − = / kg ] V m 8,752 0,49024 8,262 [ H N ) O
2
[
V S ( λ =1,8 )=V W = ( ) −V H =9,247 −0,49024 =8,757 m N / kg λ
V S ( λ = 1,9 )=V W
1,8
2
O
3
]
−V H =9,743 −0,49024 =9,252 [ m N / kg ] ) 3
λ =( 1,9
2
O
[
V S ( λ =2,0 )=V W = ( ) −V H =10,238 −0,49024 =9,748 m N / kg λ
2,0
2
O
3
]
Do9i5eni ra4utati /ri"a4u5u e ta9earno i -rafi,"i:
Ta0ea16 λ
&
&;
&*
1
0,00
4,95
5,29
4,80
1,1
0,50
5,45
5,78
5,29
1,2
0,99
5,94
6,28
5,79
1,3
1,49
6,44
6,77
6,28
1,4
1,98
6,93
7,27
6,78
1,5
2,48
7,43
7,76
7,27
1,6
2,97
7,92
8,26
7,77
1,7
3,47
8,42
8,75
8,26
1,8
3,96
8,91
9,25
8,76
1,9
4,46
9,41
9,74
9,25
2
4,95
9,90
10,24
9,75
View more...
Comments