A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic
1 answer:
Answer:
The answer is below
Explanation:
A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.
Given that:
number of poles (p) = 4, frequency (f) = 60 Hz
1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:
2) The slip (s) = 0.05
The speed of the motor (n) is the speed of the rotor, it is given as:
3) s = 0.04
The rotor frequency is the product of the supply frequency and slip it is given as:
4) At standstill, the motor speed is zero hence the slip = 1:
The rotor frequency is the product of the supply frequency and slip it is given as:
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Answer:
A) See Attachment B) See Attachment C) 186.153
Explanation:
C) Per phase Voltage= 220/∛3
= 6.037
S=V²/Z
= (6.037)²/(0.01+i0.05)
= 62051.282-i310256.41
Active power losses per phase= 62.051kW
total Active power losses= 62.051×3
=186.153kW
Answer:
V_t=6 = 32 m/s
Explanation:
The origin is at point 0 with the positive motion to the right
The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:
a=dv/dt
From here we can express dv as:
dv = a dt
Replace a by 2t — 1
dv = (2t — 1) dt
Integrate both sides of equation
v=t
a=t_0
putting these value in integral
<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>
We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values
v — 2 = (t^2 — t) — (0^2 — 0)
From here we can write the expression for v as:
v_t=6=6^2-6+2 (1)
So the velocity at t = 6 s is:
v_t=6 = 32 m/s
V_t=6 = 32 m/s
In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero
To do that, we'll just replace v by 0 in expression (1)
0 = t^2 — t + 2
The roots of the quadratic equation are:
t_1/2=1± √(1^2-4*2*1)/2
Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right
Now that we have all the details, we can correctly draw the path of the particle
We can see from the sketch that the total distance traveled is:
s^T=Δs_0-1
s^T=| s_1 - s_0 |
Replace s_0 by its value
s^T=| s_1 - 1 | (2)
In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time
Since we have already wrote expression for v as function of time (step 2), we'll use expression to get the expression for s
v= ds/dt
Multiply both sides of equation by dt
v dt = ds
Replace v by expression (1)
(t^2 — t + 2) dt = ds
Integrate both sides of equation
t=s
b=(s=0)
x=(t^2 — t + 2)
dx=ds
putting these value in integral
(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0
Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6
(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1
