78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
Answer:
There are 6 types of pressure control valves and their function is to regulate the pressure below a threshold level within safe limits and to maintain and control pressure of a particular circuit.
Explanation:
The six type of Pressure valve with their functions are given below:
a. Unloading Valve:
These type of pressure valve are used to pour fluid into the container at very low or no pressure.
b. Safety valve:
These are used when the pressure within the vessel is in excess as inside temperature is near about preset [point point then these valves are open to release the extra pressure and are closed once normal conditions are regained.
c. Pressure Reducing Valve:
These are basically used for the control of the pressure in downstream not exceeding the design limits.
d. Pressure Relief Valves:
These are basically used to limit and regulate the pressure of any system.
e. Counter Balance Valve:
These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.
f. Sequence Valve:
These are used to maintain sequence or order in the operations of two parts or branches.
Answer:
The answer is below
Explanation:
1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:
2) The speed of the rotor is the motor speed. The slip is given by:
3) The frequency of the rotor is given as:
4) At standstill, the speed of the motor is 0, therefore the slip is 1.
The frequency of the rotor is given as:
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
