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