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mixer [17]
3 years ago
14

One of the basic requirements of a servomotor is that it must produce high torque at all: a)Frequencies b)-Voltages c)-Loads d)-

Speeds
Engineering
1 answer:
Schach [20]3 years ago
5 0

Answer: D) Speeds

Explanation:

Servomotor is an electromechanical device that contain feedback mechanism.

In servomotor, torque curve basically depicts that a motor can produce the amount of torque accordingly to the operating speed.

Servomotor are able to hold the heavier loads that is why it has ability to produced high torque in all range of speeds.

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Each of the following activities are commonly performed during the implementation of the Database Life Cycle (DBLC). Fill in the
kicyunya [14]
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3 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
3 years ago
CODE
zalisa [80]

Answer:

Explanation:

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6 0
3 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
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Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

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