A 6-pole, 50 Hz squirrel cage induction motor has rotor resistance and standstill reactance referred to stator of 0.2 ohm and 1
ohm per phase, respectively. With rated voltage and rated frequency it runs at full load with 4 % slip. Neglect stator resistance and rotational losses. Determine the operating speed of the motor, when the stator voltage impressed is reduced to 0.707 times the rated voltage, frequency remaining the same, if (a) the load torque remains constant at the rated motor (b) the load is of fan-type having rated torque at rated speed.
1 answer:
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Answer:
material remove in 3 min is 16790.4 mm³/s
Explanation:
given data
length L = 80 cm = 800 mm
width W = 30 cm
height H = 15 cm
make grove length = 80 cm
width = 8 cm
depth = 10 cm
mill toll diameter = 4 mm
axial cutting depth = 20 mm
to find out
How much material removed in 3 minutes
solution
first we find time taken for length of advance that is
time =
here advance is given as 0.001166 mts / sec
so time =
time = 686.106 seconds
now we find material remove rate that is
remove rate = mill toll rate × axial cutting depth × advance
remove rate = 4 × 20×0.001166 ×1000
remove rate = 93.28 mm³/s
so
material remove in 3 minute = 3 × 60 = 180 sec
so material remove in 3 min = 180 × 93.28
material remove in 3 min is 16790.4 mm³/s
Answer:
The efficiency of this fuel cell is 80.69 percent.
Explanation:
From Physics we define the efficiency of the automotive fuel cell (), dimensionless, as:
(Eq. 1)
Where:
- Maximum power possible from hydrogen flow, measured in kilowatts.
- Output power of the automotive fuel cell, measured in kilowatts.
The maximum power possible from hydrogen flow is:
(Eq. 2)
Where:
- Volume flow rate, measured in cubic meters per second.
- Density of hydrogen, measured in kilograms per cubic meter.
- Heating value of hydrogen, measured in kilojoules per kilogram.
If we know that ,
,
and
, then the efficiency of this fuel cell is:
(Eq. 1)
(Eq. 2)
The efficiency of this fuel cell is 80.69 percent.