elektricni strojevi zadaci
TRANSCRIPT
U=230V Pel=1.2kWf=50Hz I=5.4A M=3.5Nm n=2600 o/min
Pel=U ∙ I ∙ cosφ⇒cosφ=Pel
U ∙ I= 1200
230 ∙5.4=0.966
Pmeh=M ∙n9.55
=3.5 ∙26009.55
=952.88W
η=PmehPel
=952.881200
=0.794
S=U ∙ I=230 ∙5.4=1242VA
Q=√S2−P el2 =√12422−12002=320.26 VAr
U=230V f=50Hz R=800ΩL=3H
X L=ω∙ L=2πf ∙ L=2π ∙50∙3=942.5Ω
Z=√R2+ X L2=√8002+9422=1236.25 Ω
I=UZ
= 2301236.25
=0.186 A
P=I 2 ∙R=0.1862 ∙800=27.68W
S=U ∙ I=230 ∙0.186=42.78VA
cosφ= PS
= PU ∙ I
=27.6842.78
=0.647
U=230V η=75% f=50Hz nm=0.5∙nnI=1.5A ηm=0.7∙η nn=1380 o/mincosφ=0.7
Pn=Pel∙ η=U ∙ I ∙ cosφ ∙η=230 ∙1.5 ∙0.7 ∙0.75=181.12W
MM n
=(ωm
ωn)
2
=( nm
nn)
2
⇒
M=M n ∙( nm
nn)
2
=9.55 ∙Pn
nn
∙( 0.5 ∙nn
nn)
2
=9.55 ∙181.121380
∙( 0.5 ∙13801380 )
2
=0.313Nm
PPn
=M ∙ωm
M n ∙ωn
=M ∙nm
M n ∙ nn
=M ∙0.5M n
=0.125
P=0.125∙181.12=22.64 W
I= PU ∙cosφ ∙ηm
= 22.64230 ∙0.7 ∙0.7 ∙0.75
=0.27 A
a)
nn=60 ∙ fp=60 ∙502=1500o/min
M n=9.55 ∙Pn
nn
=9.55 ∙160001440
=106.11 Nm
n=nS ∙(1−M t
nS
nS−n∙M n )=1500 ∙(1− M t
15001500−1440
∙106.11)=1500−0.565 M t
15.5 M t−50=1500−0.565M t⇒
M t=1500
16.065=93.37 Nm
nt=15.5 M t−50=15.5 ∙93.37−50=1397.23o /min
b)
M pokr=1.2 ∙M n=1.2∙106.11=127.32Nm
n=15.5 ∙M t , pokr−50=0⇒M t , pokr=50
15.5=3.23 Nm
M t , pokr=3.23 Nm<M pokr=127.32Nm DA
C)
M e=√∑ ( M i2∙ ti )
∑ ti
=√ 1002∙2+1202 ∙4+1402∙2+802 ∙412
=108.93 Nm
M e=108.93 Nm>M n=106.11 Nm NE
Poredna karakteristika
M e=√∑ ( M i2 ∙ ti )
∑ ti
=√ 902 ∙0.1+1102 ∙0.1+1202 ∙0.6+1002 ∙0.1+602∙0.1+402 ∙0.21.2
=101.41Nm
Pe=M e ∙ n
9.55=101.41 ∙1000
9.55=10618W =10.62kW
Mehanička karakteristika
M e=∑ ( M i ∙ ti )∑ ti
=90 ∙0.1+110 ∙0.1+120 ∙0.6+100∙0.1+60 ∙0.1+40 ∙0.21.2
=96.7 Nm
Pe=M e ∙ n
9.55=96.7 ∙1000
9.55=10125.65 W =10.13kW
M e=√∑ ( M i2 ∙ ti )
∑ ti
=√ 5492 ∙5+982 ∙30+1572 ∙255+30+25+20
=173.6Nm
M n=9.55 ∙Pn
nn
=9.55 ∙16000730
=209.3 Nm
M e=173.6 Nm<M n=209.3 Nm ISPUNJEN PRVI UVJET ! !
M pokr
M n
=2.2⇒M pokr=2.2∙ Mn=2.2∙209.3=460.46 Nm
M pokr=460.46 Nm<Mm=549 Nm ISPUNJEN DRUGI UVJET !!!
v=50km /h⇒ 503.6
=13.9ms
n=60 ∙ v2 π ∙r
= 60 ∙13.92π ∙0.26
=510.52o
min
MV=wV ∙ r=50 ∙0.26=13Nm
J=m∙r2=300 ∙0.262=20.28W S3
M n=Mm−Mt=Mm−MV =Jdωdt
⇒
Mm=MV +Jdωdt
=MV +Jdndt
∙2π60
=13+20.28 ∙2 π60
∙510.52
15=13+72.28=85.28 Nm
P=Mm ∙n
9.55=85.28 ∙510.52
9.55=4558.86W =4.56kW
a)
v=72kmh
⇒ 723.6
=20ms
v=r ∙ω=r ∙2π ∙60∙ n⇒
np=vr
∙30π
= 200.318
∙30π
=600.6o
min
J=m ∙r2
2=100 ∙0.3182
2=5.06W S
3
b)
M p ,U=J ∙dωdt
=J ∙dndt
∙2π60
=5.06 ∙600.6
5∙
π30
=63.65Nm
M p ,max=M p ,U+M tr=63.65+7.1=70.75 Nm
c)
nm=i ∙ np=4 ∙600.6=2402.4o
min
Jm=J ∙( np
nm)
2
=J ∙1i2=5 ∙
142 =0.31kg m2
Mm
M p ,max
=ωP
ωm
∙1η= 1
i ∙ η⇒
Mm=M p , max ∙1
i ∙ η= 70.75
4 ∙0.95=18.62 Nm
d)
Pm, n=Mtr ∙ np
9.55 ∙ η=7.1∙600.6
9.55∙0.95=470.02W
Pm, max=Mm ∙nm
9.55=18.62 ∙2402.4
9.55=4684.05W =4.68 kW
e)
Mm, e=√∑ (M i2 ∙t i )
∑ t i
=√ 160
∙[( M p ,U+M tr
i ∙ η )2
∙5+( Mtr
i ∙ η )2
∙50+( M p ,U−M tr
i ∙ η )2
∙5+0 ∙100 ]=4.34 Nm Pm,e=Mm, e ∙nm
9.55=4.34 ∙2402.4
9.55=1091.77W =1.09kW
f)
ε SN=tZ+t P+ tK
tZ+tP+tK+ tKmY
=5+50+5160
=0.375=37.5 % ;εNORM=40 %
M tr , v=Ptr ,v
0.1047 ∙ nZ
=9.55 ∙20001000
=19.2 Nm
Zamašnamasa
m ∙D2=M tr , v ∙4 ∙30π
∙st
sn
=19.1 ∙4 ∙30π
∙40
1000=29.2kg m2
P=11kWn=2880 o/minm∙D2=0.2kgm2Mk=1.3Mntz=2s
M n=9.55 ∙Pn=9.55∙
1100002880
=36.5 Nm
M k=1.3 ∙ M n=1.3 ∙36.5=47.4 Nm
Iznos zamašnemase
m ∙Dm2 =
38.2 ∙ t ∙M n
n2−n1
=38.2∙2 ∙47.42880
=1.26kg m2
(toliko smije iznositi zamašnamasa radnogmehanizma)
P=30kWn=720 o/minm∙D2=25kgm2v=2m/s s=5mmd=15000kpmt=10000kp
M n=9.55 ∙Pn
nn
=9.55∙30000
720=397.9 Nm
M t=0.8 ∙M n=0.8∙397.9=318.3Nm
Odnos klizanja
s t
sn
=M t
M n
⇒ st=sn ∙M t
M n
=750−720750
∙318.3
3997.9=0.032
nt=750 ∙ (1−0.032 )=726o
min=sn
s=a2
∙ t2 ;v=a ∙t⇒ t=2 sv
=2∙52
=5 s=Δt
m ∙D2=m∙ Dm2 ∙m∙ Dt
2=m∙Dm2 +365 ∙ (md+mt ) ∙( v
mt )2
=25+365∙ (10000+15000 ) ∙27262=94.25kg m2
M u=m∙ D2
38.2∙n t
s=94.25
38.2∙726
5=358.25Nm
Mm=M u+M t=358.25+318.3=676.55Nm; Im=I n ∙Mm
M n
=I n∙676.55397.9
=I n ∙1.7
Nominalanmoment
M n=9.55 ∙Pn
nn
=9.55∙150001450
=98.8 Nm
Moment tereta
M t=9.55 ∙F ∙ vη1 ∙η2
∙1n=9.55 ∙
510 ∙9.81 ∙20.76 ∙0.92
∙1
1450=94.25 Nm
Moment ubrzanja
M u=Mm−M t=1.3 ∙M n−M t=1.3 ∙98.8−94.25=34.2 Nm
Vrijeme zaleta
tZ=∑m∙ D2 ∙n
38.2∙ M n
=0.3+0.005∙ nm+0.4 ∙ nd ∙
nd
nm
∙1η1
+365 ∙510 ∙v2
nm
∙1
η1 ∙ η2
38.2∙ M u
=1.04 s
n=1450 o/minmcu=12kgm∙Dt2=50kgm2Mt=0
a)
Arot=J ∙ωS2 ∙ ( s1−s2)=m∙ D2
365∙ ns
2 ∙∫S2
S1 Mm
Mm−M t
∙ ds=m∙ D2
365∙ ns
2 ∙ (s1−s2 )⇒ s1=1.5; s2=1
b)
ACU 2=J ∙
ωS2
2∙ (s1
2−s22)2
=m∙ D2
365∙ ns
2 ∙∫S 1
S 2 Mm
Mm−M t
∙ s ∙ds=m∙ D2
365∙ns
2
2∙(s1
2−s22 )=192.6 kWs
c)
ACU 2=m∙ D2
365∙ns
2
2∙ (s1
2−s22 )⇒s1=1.5 ; s2=0.5
ACU 2=308.2kWs
Zagrijavanje rotorskog kruga
s ∙ v2=ACU2
mCU ∙ c= 308.2
12∙0.39=65.85 ° C
a)
M n=9.55 ∙Pn
nn
=9.55∙500001500
=318.3 NmMm=1.5 ∙M n=1.5 ∙318.3=477.5 Nm
MU=Mm−M t=477.5−177=300.5 Nm
ACU 2=
m∙ Dm2 +m∙ Dt
2
365∙Mm
M u
∙ [n0 ∙ (n2−n1 )−12
∙ (n22−n1
2 )]=40+10365
∙477.5300.5
∙((1600 ∙1400 )−12
∙ (14002))=274293.18Ws=274.3kWs
b)
Arot=m∙ D2
365∙
Mm
Mm−Mt
∙ n0 ∙n= 50365
∙477.5
477.5−177∙1600 ∙1400=487632.32kWs=487.63Ws
tZ=m∙ D2
38.2∙
nMU
= 5038.2
∙1400300.5
=6.1 s
Srednjavrijednost gubitakau motoru :Pg=2000W
Ag=Pg∙ tZ=2000 ∙6.1=12200Ws=12.2kWs
Ukupna energijaiz mreže potrebna za zalet dobrizne1400o
min
Am=A rot+ Ag=487.63+12.2=499.83kWs