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Samenvatting Oefeningen warmte- en stromingsleer

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Alle uitgewerkte werkcollege oefeningen van warmte- en stromingsleer. Ook een paar oefeningen van op de slides

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, W2-gesloten systeem
- ideaal
gasmodel

Vraag
Zuiger-cilinder
PV = MRT - M
= =
130 .
0,07 = 0. 078
Leg
07m2 Polytroop PC 100 kPa 0 2968 393
V 0
.

=
=
. .

-
Pr = 130 kPa T =
100 ·


C
37
Tr =
120 C PV = MRT - V =
RT
070 0 2968 ,
·




,
.




100


P = 0, 086m

& PV" = PaVan
Pr --
1




=
----
I


M
. 3
1 = 1 ,
229

S
v
V-V
log (1 3) , =
n
log (1 229) ,




n =
1 ,
27

0006-130 0 07
= W Pak-per 10.0 , .


ly
.




=
- = 1, 05
1 -
M
1 27
1 -

,




Vaage 0 2070 285
MRT
V-T2 4 3m3
.




isothermpa
-
·




lucht
,
PV .
1

leg
=
=
2 4 ,
,



= 600 ke 150
P = 150 kPa -




In
T 12 C
.




In
=




zuiger
cilinder W MRT 2, 4 0 , 287 285 .
.
.

= .




A


Pe -
-of
-
= 272
,
14 f
I




Pr-t - -




g
I W =
MRT .


In ()
3 -
1
, 4 =
m()
V Vy
14
= =
E + V = 0, 32m

Vraag3 Pr = -12000 0, 42 + 600
.




6 d
V
S Plus =
Jav +
0,
42m
600
=


P2 -1200 0 12 +
M
-
=


b(v -M)
.

12m3 44
V 0
a
= -
.
=
+
L
P(V) = aV + 6 2 &




11/19
a -12000 KPa/m
8
8lf
=

-
=

,
6 = 600kPa
-




vraag 4
zuig-cilinder
m = 2
ky M
- OV
1

=
, 4
c
W :
R I Tot
Te 360k
P-100kPa-
=




9




0
T 300k
2961360-300
=

0
2
2
.



= ,


s
=
MRT = 1 77,
1 -


0 ,
4

I
= -88 8
b
·
189 29 ,
=




,
-




-



V = . 12
1 m

,Vraag-
5

&
zuiger-cilinder 13
·
lucht
m = 0, 15
kg sothermisch lytro isbaar
P = 2 MPa -
P2-Jookpa
naM Pa
-
Ve


T =
350 : o
S

1 2

V
,




V = RT
= 0 ,
01 Pal =
Py .

- V = 0, 01669m3

V 0 05 Was -PaVe
lf
= =

-34 4
, :
=
,




1 -
Wil =
MRT In .




1) = 37, 69
by
Ws P(V
by
=
-


Vs) = -7, 30
Wat 37 69- 34 4-7 38 4 03
lf
=
, , , = ,




6
Vraag
3m (constant)
=
V
Pr MRT-m 0
33
=
=
1 ,


Hz
P = 250 ka

T
Prese
= 550k
-




Tz = 350k


159 , 09 Pa
= Pr =




Qua = m .
G (TC -
T1)


= 0, 33 .
10 , 83 (350 .

550)

=
= 672 078 ,
f

Kraay7
yx5x7m3
V=
Ain-Quit Win-Wuit All 100 4x717 4
= e
+ = .



m ,


= 10000 ku . 287
0 .

283
(Ain-Quit + Win) At = m .
C (Te-Tu)
Wi = 100W

#10000 -
5000 + 100) At = 172 4 . 0, 718(20-10)
Quit 5000
by In
,
=




At = 830, 76 D

↑1 L ..

Pong = 100 kPa




w = 100 % = 0,
1 kla
& by R
= 10000
by /M = 2 , 78

5000 ty /M
Q 1 , 39
by /n
=
=




:
vraag
-




zuig-cilinder Qin-Quit + Win-Wuit = MCpAT
m = 15
leg -
60 + Win = 15 .
1 005
,
.
(77-25)
Tr =
25
Te = 77c Win = 045 .
9
by
Pr = Pc 300 kPa
59
=




=
Win 0, 235 h
Quit
=
=

Gol
Win =?

, Vags
Thermisch geisoleerde zuig-cilinde
V 1
= 0. 3 m2 CO2

P = 200 kPa
W = m .

Cp At

Tr 27 C
= o
=

m = 10

u = 110

At = 10 min = 600 s Win = 1 , 059 .
0, 846 . 300 =
311
b
Isoboar W v 1 At - 1 =
4 GA
=
= .
. ,




Vz = 0, 6 m3



Vraag
10
-




zuig-cilinder =
mRT-1 1915 ,




lucht Va m A
m = 3
kg = 2
, 583
2 S
P = 200 kPa
* &


le
Ain-Wint All
T = 27C
= =
m(Mz Mr) -
=
2674
P2 =
YookPa
1
=!
·

Rin Wint
Va =
2V1
= P(V -4) = 516
b >
W =
? = 10




1300k by 1 by
S [A17]
214 07
tabellen lucht
=
,




Magron = 933 , 33
l I leg
Yaag
11


lucht
m =
0
, 004 leg
1-2 : adiabatische compressie V =

RT = 0,
0034
P 100kPa


30057
=




=
U


T =
27c Te
Pr =
1MPa
2.3 : isbare warmte toevoer
Qin =
m .
(pAT = 0,
004
Qin 12
.
2 76
=
,
by
3-1 2, 76 0, 004
1, 005 .


(T3-579 2)
Warmteafvoer
= .


: .




P = G . V + C
G Tz = 1265 ,
77k
·


1

W =
Pe
(v -
v) =
PPMRT) = 237ly o >




powerpoint

u = 0 .
5m

yookpa
= 2
P =

M =

T =
27C

1 2A
·


=




At 5min
Win 1 1 At A2 by
=
=
= .
.




u = 120V
isaboar
AT
Aut =

2000
-

Quit + Win = M .

Cp .




Tz = AT
-



28 + 72 = 2 25 1 039
.
.


,
.




29 , 60 = Ja T1 -




Ta =
329 ,
6 k
↳ :

56 , 6 C

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Uploaded on
August 27, 2024
Number of pages
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Written in
2023/2024
Type
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