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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This question is incomplete, the complete question is;
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.
Use the cold air standard assumptions.
Answer:
a) The compression ratio is 18.48
b) The maximum temperature of the cycle is 1893.4 K
c) The cutoff ratio, v₃/v₂ is 1.946
Explanation:
Given the data in the question;
Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K
Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K
Net work per cycle = 590.1 kJ/kg
Heat transfer input per cycle Qs = 925 kJ/kg
a) compression ratio;
As illustrated in the diagram below, 1 - 2 is adiabatic compression;
so,
Tγ = constant { For Air, γ = 1.4 }
hence;
⇒ V₁ / V₂ = T₂ / T₁
so we substitute
⇒ V₁ / V₂ = 973 K / 303 K
= 3.21122
= 18.4788 ≈ 18.48
Therefore, The compression ratio is 18.48
b) maximum temperature of the cycle
We know that for Air, Cp = 1.005 kJ/kgK
Now,
Heat transfer input per cycle Qs = Cp( T₃ - T₂ )
we substitute
925 = 1.005( T₃ - 700 )
( T₃ - 700 ) = 925 / 1.005
( T₃ - 700 ) = 920.398
T₃ = 920.398 + 700
T₃ = 1620.398 °C
T₃ = ( 1620.398 + 273 ) K
T₃ = 1893.396 K ≈ 1893.4 K
Therefore, The maximum temperature of the cycle is 1893.4 K
c) the cutoff ratio, v₃/v₂;
Since pressure is constant, V ∝ T
So,
cutoff ratio S = v₃ / v₂ = T₃ / T₂
we substitute
cutoff ratio S = 1893.396 K / 973 K
cutoff ratio S = 1.9459 ≈ 1.946
Therefore, the cutoff ratio, v₃/v₂ is 1.946
Answer:
D. N= 11. 22 rad/s (CW)
Explanation:
Given that
Form factor R = 8
Speed of sun gear = 5 rad/s (CW)
Speed of ring gear = 12 rad/s (CW)
Lets take speed of carrier gear is N
From Algebraic method ,the relationship between speed and form factor given as follows
here negative sign means that ring and sun gear rotates in opposite direction
Lets take CW as positive and ACW as negative.
Now by putting the values
N= 11. 22 rad/s (CW)
So the speed of carrier gear is 11.22 rad/s clockwise.