Determine the minimum force P to prevent the 30 kg uniform rod AB from sliding. The contact surface at B is smooth, whereas the
1 answer:
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Answer:
a) see attached pic
b) No-load speed: 1368 rpm
c) IL at 1% below no-load speed = 22.4 A
Explanation:
a)
(See attached pic)
b) The No- Load Speed:
Since, Ea = KФn [ Ea=Internal Generated Voltage, k=Machine constant,Ф=flux, n=speed]
As the Vt (terminal voltage) and field resistance (Rf) is constant, If (field current is constant. Assuming negligible armature reaction, the following relationship can be written,
Ea2 / Ea1 = K Ф n2 / K Ф n1
Since K and flux is contant,
Ea2 = (n2 / n1) x Ea1
Or, n2 = (Ea2 / Ea1) x n1 -------- (1)
Since, Ea = Vt - IaRa
But at no load, Ia = 0, therefore,
Ea = Vt
At load condition, IL = 29 A
Therefore,
Ia = IL - Vt/Rf
Ia = 29 - 240/120
Ia = 27 A
Therefore, Ea at this load will be,
Ea = Vt - IaRa
Ea = 240 - (27)(0.12)
Ea = 236.76 V
The No-Load speed of motor can be calculated using equation (1),
n2 = (Ea2 /Ea1 ) x n1
1350 = (236.76/240) x n1
n1 = 1368 rpm = No-load Speed
c)
Now, if the speed is 1% below the no-load speed, which is,
n1 = no load speed = 1368 rpm
n2 = 1% less speed = 1354 rpm [ 1368 - (0.01)(1368) ]
Using equation (1),
n2 = (Ea2/Ea1) x n1
1354 = (Ea2/240) x 1368
Ea2 = 237.5 V
Since, Ea = Vt - IaRa
Ia = (Vt - Ea)/ Ra
Ia = (240-237.5)/(0.12)
Ia = 20.4 A
Now,
IL = Ia + If
IL = 20.4 + 240/120
<em>IL = 22.4 A</em>
When, Speed = 1368 rpm, IL = 4.75 A
When , Speed = 1354 rpm, IL = 22.4 A
When, Speed = 1350 rpm., IL = 29 A
<em>(See attached pic) </em>
