Elet 360		homework #6		William Crenshaw
ELET 360 HW 6 – Due date: Friday February 1, 2019, 5:00 PM

Make sure to include a copy of the question on your HW solution
and make SURE to staple your HW solution.

1.	A 3-phase 460V, 40 HP, IM is supplied with 60A
at pf of .83 lagging. Its stator and rotor copper losses are
4000 W and 1000 W respectively. Its core losses are 3000W
while it’s Friction and Windage losses are
750W and stray losses are 300W. Determine

	a. The air gap power of the machine
	b. The mechanical power developed
	c. The shaft output power
	d. Efficiency of the motor
	e. Also draw the diagram depicting the energy flow
		in the machine

a)	Pag=Pin-Pstcu-Pcore	Pin=sqrt(3)*Vl*Il*cos(theda)
Pin=sqrt(3)*460*60*0.83 = 39677.8199 W
Pag=39677.8199-4000-3000 = 32677.8199 W

b)	Pmech=Pag-Prot.
32677.8199-1000 = 31677.8199 W

c)	Pout=Pmech-Pfrict.+Pwind.+Pstray
31677.8199-(750+300) = 30627.8199 W

d)
(30627.8199/39677.8199)*100 = 77.19 %

e) on paper


2. A 3-phase, 4 pole 50 Hz induction motor provides 20 HP to
a load at a speed of 1480 RPM. If the
mechanical losses of the motor are assumed to be 0, calculate
	a. Slip at given speed
	b. The developed (or Mechanical) Torque
	c. The shaft speed of the motor if the Torque
		is increased by 125%
	d. The output power at the Torque calculated in part c

a)	ns=(120f)/p	S=(ns-n)/ns
(120*50)/4=1500	rpm	(1500-1480)/1500 = 1/75

b)	Tem=pmech/(omega)m	Pmech=Pout
(20*746)/(2*3.1415*(1480/60)) = 96.2701 Nm

c)	recall T ~= S at low slip
	Snew=2.25*S	Tem_new=Tem*2.25
Tem_new=96.2701*2.25 =			216.607725 Nm
2.25*(1/75)=0.03
0.03=(1500-n)/1500 -> (0.03)*1500=(1500-n)
 -> -(((0.03)*1500)-1500)=n =		1455 rpm

d)	Tout=Tem if no losses
	Tem=Pem/(omega)m -> Pout=Tout*(omega)m
216.6077*2*3.1415*(1455/60) = 33002.9448 W


3.	A 3-phase, 2 pole 35 HP, 480 V 60Hz, Y connected
motor has the following equivalent circuit impedances:
R1 = .3 Ω, R2′ = .2 Ω, X1 = .6 Ω, X2′ = .5Ω, Xm = 12 Ω.
The rotational losses are 1800 W and assumed to be constant.
For a rotor slip of 2.5%, at rated voltage

	a. Draw the single phase equivalent circuit
		of the machine using given parameters
	b. Speed of the motor in rpm and rad/s
	c. The stator current
	d. Power factor
	e. The developed power Pmech and the output power Pout
	f. The developed torque Tem and the output power Tout
	g. The efficiency of the motor
		at the given operating condition.

a)	on paper

b)	ns=120f/p	0.025=(ns-n)/ns
(120*60)/2=3600 rpm	0.025=(3600-n)/3600 ->
0.025*3600=(3600-n) -> -((0.025*3600)-3600)=n
	= 3510 rpm
2*3.1415*(3510/60)
	= 367.5555 rad/s

c)	Istat.=(V/sqrt(3))/Zeq	Zeq=R1+jX1+[jXm||((R2'/S)+jX2')]
Zeq=0.3+j0.6+[j12||((0.2/0.025)+j0.5)=5.5304+j4.4275 ohms
(480/sqrt(3))/5.5304+j4.4275 = 39.1184<-38.6796 A

d)	p.f=cos((theda)v-(theda)i)
cos(0-(-38.6796) = 0.78 lagging

e)	Pin=3*|Vl-n||I|cos((theda)v-(theda)i)
	Pag=Pin-3*(Ist.)^2*R1	Pmech=Pag(1-S)	Pout=Pmech-Prot.
Pin=3*|(480/sqrt(3))||39.1184|cos(0-(-38.6796))=25388.72937 W
Pag=25388.72937-3*(39.1184)^2*0.3=24011.5051 W
Pmech=24011.5051*(1-0.025) =			23411.2175 W
Pout=23411.2175-1800 =				21611.2175 W

f)	Tem=Pmech/(omega)m	Tout=Pout/(omega)m
Tem=23411.2175/(2*3.1415*(3510/60)) =	63.6944 Nm
Tout=21611.2175/(2*3.1415*(3510/60)) =	58.7972 Nm

g)	eff.=Pout/Pin*100%
(21611.2175/25388.72937)*100% = 85.12 %


4.	Consider the machine in the previous question,
and assume the machine is at standstill (start-up).
Calculate:
	a. The slip at start-up
	b. Current at the start-up (starting current)
	c. Power factor at start-up
	d. The developed power Pmech
		and the output power Pout at start-up
	e. The developed torque Tem
		and the output power Tout at start-up

a)	S=ns-n/ns
S=3600-0/3600 = 1

b)	Istat.=(V/sqrt(3))/Zeq	Zeq=R1+jX1+[jXm||((R2'/S)+jX2')]
Zeq=Zeq=0.3+j0.6+[j12||((0.2/1)+j0.5)=0.4842+j1.0829 ohms
Ist.=(480/sqrt(3))/0.4842+j1.0829 = 233.6223<-65.909 A

c)	p.f=cos((theda)v-(theda)i)
p.f.=cos(0-(-65.909)) = 0.41

d)	Pin=3*|Vl-n||I|cos((theda)v-(theda)i)
	Pag=Pin-3*(Ist.)^2*R1	Pmech=Pag(1-S)	Pout=Pmech-Prot.
Pin=3*|(480/sqrt(3))||233.6223|cos(0-(-65.909))=79282.1467 W
Pag=79282.1467-3*(233.6223)^2*0.3=30160.7056 W
Pmech=30160.7056(1-1) =	0 W
Pout=			0 W

e)	Tem=Pag/(omega)s	Tout=Pout/(omega)m
Tout = 0 Nm
Tem=30160.7056/(2*3.1415*(3600/60)) = 80.0061 Nm


5.	A 3-phase, 460 V, 4 pole, 60Hz, wound rotor
induction motor with the following parameters:
R1 = .24 Ω, R2′ = .22Ω, X1 = .45 Ω, X2 = .5 Ω,
Xm = 25Ω, operates at 1760 rpm at full load. The rotational
losses of the machine are 1600 Watts.
Calculate:
	a. The synchronous speed ns in rpm
	b. Starting current
	c. Full load slip
	d. Full load current
	e. Ratio of starting current to full load current

a)	ns=120f/p
ns=(120*60)/4 = 1800 rpm

b)	Istart=(V/sqrt(3))/Zeq	Zeq=R1+jX1+[jXm||((R2'/S)+jX2')]
Zeq=0.24+j0.45+[j25||((0.22/(1))+j0.5)]=0.4514+j0.942 ohms
Istart=(460/sqrt(3))/(0.4514+j0.942) = 254.2405<-64.395 A

c)	slip=(ns-n)/ns
S=(1800-1760)/1800 = 1/45

d)	Istfl=(V/sqrt(3))/Zeq	Zeq=R1+jX1+[jXm||((R2'/S)+jX2')]
Zeq=0.24+j0.45+[j25||((0.22/(1/45))+j0.5)]=8.5092+j4.1506 ohms
Istfl=(460/sqrt(3))/(8.5092+j4.1506) = 28.0519<-26.0021 A

e)	254.2405/28.0519 = 9.0632

6.	Draw the family of T-s curves depicting the rotor
resistance control method of speed control of Induction
machines. How does the change in rotor resistance impact
the breakdown slip, and the maximum torque?
	on paper

7. Draw the family of T-s curves depicting the line
voltage control method of speed control of Induction
machines. How does the change in stator voltage impact
the breakdown slip, and the maximum torque?
	on paper