Post on 22-Feb-2016
description
transcript
Aula Prática 6&7
Balanços integrais
83
85
423
2
41
23
22
3
0
00
03
420
0
3
0
0
UQ
UUQ
yyUdy
yy
UQ
UQ
QQQ
up
S
S
E
upSE
00
000
0
2 QVFFVVQ
VVVV
E
ES
Resolution
dVolgxpdAnunuudVu
t iisurface
iivc
i
..
dAnuubFsurface
i.
Exit Surface
LUbL
yLyyUb
dyLyU
LyUUbdy
LyUUbdAnuub
L
S
LL
surfaceiS
37
22
322
222
22
222.
2
0
322
2
0
22222
0
Momentum flux at the entrance• The fluid leaving through the area of lower momentum entered through an
surface shorter than 2L. Using the mass conservation one gets:
LL
LUbULb
LUbLUbL
yyUb
dyLyUUb
dAu
E
E
L
m
L
m
surfaceim
23
23
23
23
22
222
222
0
2
0
• The Momentum entering is:
LUbE 232
And the force and Cd
LUbLUbLUbLUbF 2222
31
23
67
23
37
22
313
1
2
2
2 LUb
LUb
LUbFCD
• Let us consider a large number of plates, as in a real turbine.
Resolution
• Maximum power is obtained when the derivative in relation to the angular velocity is null:
4
2DVQ
RVV
RVVVV
j
jrS
jrE
rErS
• In this case there is movement of the plate. The linear speed is:RVc
• The relative velocity has the same modulus and opposite sign. The discharge is the jet discharge, if we assume many plates.
RVQF
FRVRVQ
jj
jjj
2
RRVQPot jj 2
• In this case
RV
RRV
RQRRRVQd
dPot
RRVQPot
j
j
jjj
jj
2
022
2
0
RRVRVabsolute
RRVV
rSS
jrS
• Whole jet the kinetic energy would be extracted. Efficiency would be 100%.
• Mass balance• Momentum Balance• Energy Balance
Resolution
21
222
221
221
2
2
1
2
112
21
121
21
21
21
AAUUUP
UPUP
FPAQUQUQUAUA
Resolution
• Bernoulli
1212
22
22
21
2
2112
2121
hUhUhhgF
ghP
UUhQUFPAPAQUQU
1212
22 2
121 ghUghU
• Mass conservation
1112 hUhU
2
1
2
212
1
2
1
2222
22
1
2
21
21
hh
hhgU
ghhhUghU
Resolution considering
2
1
2
21
1
22
22
21
2
1
21
222
22
22
21
1212
22
22
21
1
2121
21
21
hh
hhghhhhhgF
hhhUhUhhgF
hUhUhhgF
21 UU
22122
21 2
21 hhhghhgF
This equation states that the force must balance the hydrostatic pressure plus the momentum exiting from under the sluice gate. The force tends to zero when h2 goes to zero. If the momentum entering into the system was relevant one had to deduce it from the force:
Description of the problem
• Vj is the jet velocity relative to the boat and Vin is the inlet velocity also relative to the boat. Qj is the jet discharge (equal to the inlet discharge).
• The boat “sees” the water entering into the control volume at its own velocity and leaving it at the relative velocity (Vj).
VVjVinV
k
VQkQQV
VQVQkV
kVVVQF
jjjj
jjj
jj
2
4
02
2
2
The boat velocity depends on the jet velocity, on the discharge and on friction. If there was no friction there would be no force and the maximum boat speed would be the jet speed.
• If we had considered a control volume without the green area, we should consider Vin instead of V, but in that case we should have computed the pressure force in front of the boat, computed as:
VVjVin
ininj FVVQ
• This is the force applied over the fluid inside the green volume. The pressure reduction at the interface between both is the symmetrical of that force.
• This force must appear in the application to the blue control volume:
VVQVVVVQF
VVQFVVQ
FFVVQ
jjininjj
injinjj
ininjj
)(
• Which is the same result as before.
Resolution
0
21
21
21
1
222
211
22
222
211
a
a
pghpDVDV
VpVpVp
2
22
21
1
2
22
212
121
2
22
212
122
222
211
22
222
21
1
2
21
21
21
21
21
DD
ghV
DDVghV
DDVVDVDV
VpVpVghp aa
Resolution
apppVVV
pgHVp
pgLVp
12
32
1222
1233
2121
gHV
gHV
2
021
2
22
LHggLgHp
gLVp
pgLVp
22121
021
3
233
1233